Quantum Mechanics

Following John Preskill’s 1998 Quantum Information and Computation notes. Introduction Landauer’s principle: It takes energy to erase information (since erasure always compresses phase space, such processes are irreversible) You can store a bit of information as one molecule in a box. If it’s on the left, it’s on, else it’s off. If you slowly compress the volume in half, you’re gaurenteed to be in the LHS. The change in entropy is k ln 2, which has some associated work that needs to be performed Logic gates used to perform computation are typically irreversible For instance, NAND is irreversible, since one bit of information is lost for each gate NOT is an example of a reversible gate Charles Bennett observed that any computation can in principle be done reversibly You can construct a Toffoli gate: Input is (a,b,c) Output is $(a,b, c \oplus (ab))$ So a and b get mirrored, and the third bit gets flipped if the first two bits are both 1 (otherwise, it mirrors the input) a Toffoli gate is universal (provided you throw away the ancilla bits when interpreting You can in principle do any computation up to the end, print out a copy of the answer (logically reversible process), then step back the computation back to the beginning Maxwell’s Demon The aforementioned ideas allows a resolution of the Maxwell’s demon paradox The original formulation is as follows: You have a partitioned box (split into A and B parts) and some demon which observes the molecules in the box If a fast particle is moving from A to B and it will cross the partition, then the demon allows it through If a fast particle is moving from B to A across the partition, the demon blocks it Over time, you will get faster particles on the right and slower ones on the left with minimal work. This has the effect of having heat flow from a cold place to a hot place at no cost, in violation of the 2nd law of thermodynamics The resolution to this is that the demon needs to have some memory/ keep information on the molecules in the box. If the demon has a finite memory capacity, eventually, information will need to be erased, which results in work being done Quantum Influence The quantum nature of reality changes the definition of information Quantum mechanics is a truly random process, which has no place in deterministic classical dynamics The uncertainty principle implies that the act of acquiring information from a system invariably disturbs the system The no-cloning theorem of quantum mechanics states that quantum information cannot be copied with perfect fidelity In classical computation, you can clearly copy a state with impunity The main fundamental difference of quantum mechanics is Bell’s theorem. This states that quantum mechanics is not a local hidden variable theory. All of the information in a quantum system is encoded in nonlocal correlations that have no classical analog Quantum Complexity Classically, the indivisible unit of information is the bit The quantum analog is the qubit, which is a vector in a 2D complex vector space with an inner product ie. $|\phi > = a|0> + b | 1>$ Performing measurements on $|\phi>$ gives a probabilistic output Generalizing to N qubits, you need $2^{N}$ basis vectors to completely specify the state This can be compactly written as $\Sigma_{x=0}^{2^{N}-1} a_{x} | x>$ where $a_{x}$ are complex numbers Thus, any quantum computation consists of applying unitary transformations onto this N qubit representation. You can then measure the state by projecting onto one of the basis vectors A quantum computation is probabilistic by nature, so multiple runs are not guaranteed to yield the same result All of these operations can be simulated on a classical computer (re: vector representations, matrix multiplications, inner products) The trouble arises if you want to simulate a large number of qubits. Even 100 qubits requires manipulating way more than $10^{30}$ complex numbers as a vector You can’t divide and conquer the complexity of a quantum system due to the nonlocal correlations imparting the vast majority of the information content Standard computers are Turing complete: If given an infinite amount of time and infinite memory, any computation can be done Problems can be classified as “hard” or “easy” depending on how much time and memory they consume Describe how “hard” a problem is should be universal: it should not depend on the hardware you’re running on The standard distinction is between polynomial time algorithms and exponential time algorithms Simulating a quantum computer on a classical computer is not a polynomial time algorithm In light of this physical reality, a classical Turing machine is not an appropriate model for quantum computers Quantum Parallelism (Deutsch Problem) Imagine a function f(x) which takes a long time to compute. We want to know if f(0)==f(1) (constant) or f(0)!=f(1) (balanced). A classical computer would need to calculate both values, then compare Define a 2 qubit system x,y. Define a unitary operator U such that $|x>y> \rightarrow |x>|y \oplus f(x)>$ ie. flip the bit if f(x) acting on x is 1. Otherwise, don’t do anything Imagine that you prepare $x = \frac{1}{\sqrt{2}} (|0> + |1>)$ and $y = \frac{1}{\sqrt{2}} (|0> -|1>)$ We then project the first qubit onto the basis $|\pm > = \frac{1}{\sqrt{2}}(|0>\pm |1>)$ We see that we get $|+>$ if the function is constant and $|->$ if the function is balanced. Hence, we only need to do 1 pass of the machine to get the answer! By feeding in a superposition of states, we can create a final state from which we can extract information (hallmark of quantum parallelism) Can generalize this readily to N bits: Set the input to a known superposition, calculate the function f(x), then choose some funny basis or do additional transformations to teach out global properties of f Errors Information is stored in nonlocal correlations of the system. A large quantum system can couple to its’ environment, which “spreads out” the correlation to the environment. You can’t measure the environment perfectly, so you inevitably lose information Can think of this as decoherence: the enviornment is continually “measuring” the state, which causes it to drift over time A separate problem is that we don’t have perfect quantum gates. Trying to implement perfect unitary transformation ain’t happening. There will always be some error of order $\epsilon$ from the ideal You need some sort of error correcting codes to deal with this reality Classically, there are a ton of error correcting codes (think Hamming codes) The simplest is just repetition: make N copies of your data, then do majority polling to determine the correct output Quantum mechanically, more things can go wrong You can have the standard bit errors like in classical computing You can have phase errors, where, for instance, $|0> \rightarrow - |0>$ This phase shift is continuous, unlike the jump discontinuities of classical bit flips You can’t clone a quantum system with perfect fidelity You can’t measure the system without disturbing it Unsurprisingly, Peter Shor developed the first quantum error-correcting code, which can be thought of as an extension of the N-bit repetition code. For simplicity, suppose that we want to encode one qubit as 3 qubits So the state $a|0> + b |1> = a|000> + b|111>$ We want to be able to detect errors without destroying this superposition If I measure the first qubit and get 0, then all the states collapse to 0 and I lose information about the coefficients a and b What if you instead measure pairs of qubits? Let the 3 qubit state be represented as $|x,y,z>$. Define $x \oplus y$ denote XOR-ing the bits Define the two-bit observable $(y \oplus z, x \oplus z)$ For our 3-qubit system, this observable will dictate which index an error occurred at 0 for no error, and then 1,2,3 (in binary) to denote that qubit 1,2,3 has been flipped (going left to right) What if there is a small deviation in state? Say that the following perturbations occur: $|000> \rightarrow |000> + \epsilon |100>$ and $|111> \rightarrow |111> + \epsilon |011>$ When you project onto the eigenstate of $(y \oplus z, x \oplus z)$, most of the time (with probability $1-|\epsilon|^{2}$), the state will get reprojected back to the original state. The $|\epsilon|^{2}$ outcome just sets the observable to (0,1), indicating a bit flip in position 1 This scheme fails with a probability of $|\epsilon|^{4}$ if multiple bit-flips happen The above scheme allows us to: Make a measurement of the system without damaging information ( you gain information about the error location, but not the exact configuration of your system) Small continuous errors either get corrected immediately, or produce a large discrete error which can be corrected Avoid the no cloning theorem ($a|000>+b|111>$ is not the same as $(a|0>+b|1>)^{3}$) The only source of error left are phase errors Do a similar trick used to address the other problems: Define a set of 9-qubit states $|0> = \frac{1}{2^{\frac{3}{2}}}(|000> + |111>)(|000> + |111>)(|000> + |111>)$ and $|1> = \frac{1}{2^{\frac{3}{2}}}(|000> - |111>)(|000> - |111>)(|000> - |111>)$ Define a “cluster” as state within each parentheses If a bit flip occurs within a cluster, you can correct it with the aforementioned scheme Observe that the phases between each cluster is aligned (ie. they are all + or all -). If a phase flip occurs, then the phases between each cluster won’t be aligned (ie. one is + and one is 1) You can generate a similar index observable, but now it’s a 6-bit observable. If this observable is non-zero, than you can detect which cluster has a different sign compared to the others and correct for it The most general single-qubit unitary transformation can be expanded to order $\epsilon$ in terms of the Pauli matrices: $U = 1+ i\epsilon_{x}\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}+ i \epsilon_{y}\begin{pmatrix} 0 & -i \\ 0 & i \end{pmatrix}+ i\epsilon_{z} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ Each term can be though of as a bit flip, a phase flip, and a combination of the two respectively The key takeaways of the quantum repetition code are: That the errors became digitized: Either you are in a state of no error, or you are in a discrete set of error states that you know how to recover from You can measure the error without measuring the data (re: sampling all of the qubits) The errors are local (re: uncorrelated), and the encoded information is nonlocal, so as long as you keep your measurements to single qubits, then you can’t get information out of the system States and Ensembles Quantum Mechanics Axioms A state is a ray in Hilbert space A Hilbert space is a vector space over the complex numbers You can define an inner product which maps an ordered pair of vectors to C that is strictly positive, linear, and skew symmetric (ie. swapping order of vectors is complex conjugation) An additional constraint is that there is some notion of a norm: $||\phi|| = <\phi| \phi>^{\frac{1}{2}}$ A ray is an equivalence class of vectors which differ by a multiplication by a nonzero complex scalar. We choose the representative of this class to have a unit norm An observable is a property of a physical system which can be measured Must be a linear, self-adjoint operator(ie. $<\psi| A \psi> = < A^{\dag} \psi | \psi >$ ) As a result of the self-adjoint property, we can write $ A= \Sigma_{n} a_{n} P_{n}$ where the projection operator $P_{n}$ satisfies $P_{n}P_{m} = \delta_{n,m} P_{n}$ and $P^{\dag} = P$ We can make a measurement of a state by using the projection operator The dynamics of the system is unitary (re: probability preserving) and is governed by $\frac{d}{dt}|\phi> = -i H |\phi>$ We can describe the time evolution of a state as $|\phi(t)> = U(t) | \phi(0)>$ If H is time independent, then $U = exp(-i t H)$ Schrodinger equation is linear in the Hamiltonian If we want to compose systems A and B, we take the tensor product of their Hilbert spaces. So states $|\phi_{A}>$ and $|\phi_{B}>$ get combined to the state $|\phi_{A}> \otimes |\phi_{B}>$ You can compose a basis of the combined system as a tensor product of the bases of each original state: $|i_{A}> \otimes \mu_{B}>$ You can define the operator $X = A \otimes I$ to only act on the first subsystem Qubits A qubit is a state in a 2D Hilbert space that can take the form $a|0> + b|1>$ Symmetries Any symmetry of a quantum system must leave the probabilities untouched This implies that a symmetry is a automorphism of the Hilbert space which preserves the absolute values of inner products for all members of the space Each symmetry maps onto either a unitary or anti-unitary operator Anti-unitarity doesn’t matter for continuous symmetries Compositions of symmetries should also be a symmetry (up to an overall phase factor). This follows from group theory Symmetries should commute with the dynamics of a system Let R be the symmetry in question. The following must hold: $U(R) exp(-itH) = exp(-itH) U(R)$ Namely, that symmetries should commute with the time evolution operator For a continuous symmetry, we can let R get arbitrarily close to the identity: $R = I + \epsilon T$. This implies that, to first order, $U = 1-i\epsilon Q$ where Q is unitary. This in turn implies that Q commutes with H We call Q the generator Since any finite transformation can be written as a product of infinitesimal ones: $R = (1+\frac{\theta}{N} T)^{N} \rightarrow U = exp(i\theta Q)$, knowing how the infinitesimal symmetry transformations are represented allows you do finite transformations Rotations A finite rotation is given by $R(\hat{n}, \theta) = exp(-i\theta\hat{n}\cdot J)$ $\hat{n}$ is the axis of rotation, $\theta$ is the rotation angle, and $\vec{J}$ is the angular momentum The associated commutation relationship for angular momentum is $[J_{k}, J_{l}] = i \epsilon_{klm} J_{m}$ The simplest non-trivial irrreducible representation of angular momentum is 2D, as given by the Pauli matrices: $J_{k} = \frac{1}{2}\sigma_{k}$ The Pauli matrices satisfy the anti-commutation relationship: $\sigma_{k}\sigma_{l}+\sigma_{l}\sigma_{k} = 2\delta_{lk} I$ We can use the Pauli matrices to write any finite rotation as $U(\hat{n},\theta) = exp(-i\frac{\theta}{2} \hat{n}\cdot \vec{\sigma})$ There is a $2\pi$ ambiguity, which gives rise to spinor representations For rotation, you have the action $U(\hat{n} ,\theta = 2\pi) = -1$ The components of angular momentum transform under rotations like a vector $U(R)J_{k} U(R)^{\dag} = R_{kl} J_{l}$ This implies that if state $|m>$ is an eigenstate of $J_{3}$, then $U(R)|m>$ is an eigenstate of $RJ_{3}$ with the same eigenvalue The above implies that we can construct eigenstates of angular momentum along the axis $\hat{n} = < \sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta >$ by applying a rotation through $\theta$ to the z axis More explicitly: all direction measurements can be performed by first rotating the $\hat{n}$ axis to the z axis, and then measuring along z The axis $\hat{n’} = < -\sin \phi, \cos \phi, 0>$ is the axis around which you perform the counter clockwise rotation by $\theta$ Density Operator In real quantum systems, we don’t have a closed system: there is always some interaction with the environment This can cause some axioms of quantum mechanics to appear to be violated States are not rays Measurements are not orthogonal projections Evolution is not unitary Say that you have a two qubit system A and B, but you can only observe A. How can you characterize the observations made on A alone? We can define the density matrix of the system as $\rho = \Sigma_{i} p_{i} |i><i|$ This of this as a representation of the ensemble of possible quantum states, each with their own probability We can then define expectation values of observable Q as $tr(Q \rho)$ This idea extends to any bipartite system: You can calculate the expectation value of one system by partial tracing over the subsystem In general, we have that $\rho_{A} = tr_{B}(|\phi><\phi|)$ A general density matrix (in diagonal form) can also be written as $\rho_{A} = \Sigma_{a} p_{a} |a>< a|$ The density matrix has the following properties: self-adjoint ($\rho_{A} = \rho_{A}^{\dag}$) $\rho_{A}$ is positive definite $tr(\rho_{A}) = 1$ For so called “pure states”, we have that $\rho^{2} = \rho$ You can think of a pure state as when the subsystem is also a ray (re: no mixing of components) Bloch Sphere We can write any 2x2 self-adjoint matrix in the basis of the Pauli matrices and the identity $\rho(\vec{P}) = \frac{1}{2}(I+\vec{P}\cdot \vec{\sigma})$ where $\vec{P}$ is some 3D vector The $\frac{1}{2}$ arises because $tr(\rho)=1$ and the Pauli matrices are traceless, hence we need to scale down the identity so that $\rho$ is a density operator The eigenvalues for $\rho$ must be non-negative (since they are interpreted as probabilities) Looking at the determinant of $\rho$ we get $det(\rho) = \frac{1}{4}(1-\vec{P}^{2})$ The above constrains $\vec{P}^{2} \leq 1$, which corresponds the the unit ball So Bloch “Sphere” is a bit of a misnomer The actual sphere (ie. $|\vec{P}| = 1$) corresponds to density matrices with a vanishing determinant. Since the trace is 1, the eigenvalues are either 0 or 1. Hence, the boundary of the ball are pure states Accordingly, we can write a pure state as: $\rho(\hat{n}) = \frac{1}{2}(I+\hat{n}\cdot \vec{\sigma})$ We can write this in spherical coordinates as well: $\rho(\theta,\phi) = \frac{1}{2} I + \frac{1}{2} \begin{pmatrix} \cos \theta & \sin \theta exp(-i\phi) \\ \sin \theta exp(i\phi) & -\cos(\theta) \end{pmatrix}$ This is easily derived from the vector $\Phi(\theta,\phi)> = \begin{pmatrix} exp(-i\phi/2) \cos(\theta/2) \\ exp(i\phi/2) \sin(\theta/2)\end{pmatrix}$ We remember that $\hat{n} = (\sin\theta\cos\phi, \sin\theta\sin\phi, \cos\theta)$ You can construct the density matrix of a system by measuring $\hat{p}\cdot\vec{\sigma}$ across the 3 linearly independent axes Schmidt Decomposition The standard orthonormal basis for a bipartite system is $|\phi_{AB}> = \Sigma_{\alpha, \mu} \phi_{\alpha\mu} |a_{A}> \otimes |\mu_{B}>$ An alternative representation is the so called Schmidt decomposition: $|\phi_{AB}> = \Sigma_{i} \sqrt{p_{i}} |i_{A}> \otimes | i’_{B} >$ This is just the SVD of the original operator $\tilde{i_{B}}> = \Sigma_{\mu} \phi_{i\mu} |\mu_{B}>$ These turn out to be orthogonal to each other the transformation from unprimed to primed can be encoded in a unitary transformation $U_{B}$ The $p_{i}$ are the singular values This decomposition holds for any matrix Can derive as follows: Assume some general composite state: $|\phi> = \Sigma_{i,\mu} |i_{A}>\otimes \mu_{B}> = \Sigma_{i} |i> \tilde{i}>$ Choose a basis for which $\rho_{A}$ is diagonal: $\rho_{A} = \Sigma_{i} p_{i} |i><i|$ We know that $\rho_{A} = tr_{B}(|\phi><\phi|)$ Comparing the two results shows you that $<\tilde{i}|\tilde{j}> = p_{i} \delta_{ij}$ So $\tilde{i}>$ are actually orthogonal. We normalize them by rescaling each basis by $\sqrt{p_{i}}$ Calculating the density matrices $\rho_{A}$ and $\rho_{B}$ by tracing on the other subsystem. Applying the Schmidt decomposition to each, we find that $\rho_{A}$ and $\rho_{B}$ share the same non-zero singular values If A and B are different dimensions, the difference is made up with zeros Entanglement Define the Schmidt number as the number of non-zero singular values in a bipartite pure state If the schmidt number is greater than one, then the state is entangled. Otherwise, it’s seperable So if the state can be written as a direct product of the subspaces (re: $|\phi_{AB}> = |\phi_{A}>\otimes |\phi_{B}>$) then it’s seperable A seperable state is not necessarily uncorrelated: If you have $\uparrow_{A}>\uparrow_{B}>$, then the system is seperable but correlated An entangled state is fundamentally different in that there are non-local quantum correlations (re: there must be interactions between the subsystems) Ensemble Interpretation Ambiguity The density operator is self-adjoint, nonnegative and has $tr(\rho) = 1$ From the above, you can construct a density matrix as a convex linear combination between two other density matrices which still satisfies all the properties: $\rho(\lambda) = \lambda \rho_{1} + (1-\lambda) \rho_{2}$ where $0 \leq \lambda \leq 1$ The above implies that the density operators are a convex subset of the real vector space of dxd Hermetian operators Pure states can’t be defined as a complex sum of other states Define the pure state $\rho = |\phi><\phi|$ and define $\phi_{\perp}>$ as some orthogonal vector to $\rho$ Suppose that $\rho$ can be written as a convex linear combination $<\phi_{\perp}|\rho|\phi_{\perp}> = 0 = \lambda <\phi_{\perp}|\rho_{1}|\phi_{\perp} + (1-\lambda) <\phi_{\perp}|\rho_{2}|\phi_{\perp}>$ For the above to hold, both terms must vanish. If $\lambda$ is 0 or 1, then $\rho_{1}= \rho_{2} = \rho$ Otherwise, $\rho_{1}$ and $\rho_{2}$ are orthogonal to $|\phi_{\perp}>$. Since $\phi_{\perp}>$ can be any orthogonal vector, you arrive at the same conclusion These pure states are extremal points of the set You can see this structure on the Bloch sphere (extremal points are on the boundary) d=2 is a special case where the extremal points are all pure states. This does not hold for d>2 The convexity of the set of density matrices has a physical interpretation: Calculating the expectation value of some observable M, we have $<M> = tr(\rho(\lambda) M)$ Suppose that we have two states $\rho_{1}$ and $\rho_{2}$, where the former has a probability of $\lambda$ and the latter has probability of $1-\lambda$ Taking the expectation value of M on this state also gives $tr(M \rho(\lambda))$ So this preperation of $\rho$ gives the same observable, regardless of $\rho_{1}$ and $\rho_{2}$ used in preparation Since pure states can’t be a convex linear combination of other density matrices, there is an unambiguous way of preparing a pure state This ambiguity in mixed states stands in stark constrast to classical systems, where there is a unique preparation method to generate a probability distribution Faster Than Light Communication?! (No…) Suppose that qubit A has the density matrix $\rho_{A} = \frac{1}{2}(|\uparrow_{z}><\uparrow_{z}|+|\downarrow_{z}><\downarrow_{z}|$ This density matrix could arise from an entangled bipartite pure state $|\phi_{AB}>$ with the Schmidt decomposition $\rho_{A} = \frac{1}{2}(|\uparrow_{zA}><\uparrow_{zB}|+|\downarrow_{zA}><\downarrow_{zB}|)$ We can realize the ensemble interpretation of A via measuring qubit B Since $\rho_{A}$ has degenerate eigenvalues, this Schmidt basis is not unique, so any direction works (just apply a unitary tranformation V to A and $V^{*}$ to B to rotate your basis) Possible faster than light information propagation: Prepare many copies of this entangled state. Alice takes all the A qubits to one location, while Bob takes all the other B qubits to another Bob measures along the z axis for all of the prepared states. Alice, concurrently, measures all of her qubits to see which axis was read. For now, ignore the fact that the measurements need to be simultaneous (re: Bob calls Alice and tells her to measure something) The main problem with this is that density matrix $\rho_{A}$ is the same between the two setups, so it’s fundamentally impossible to distinguish between the two states Follows from the convexity of the density matrices set Quantum Erasers The density matrix $\rho_{A} = \frac{1}{2} I$ is an incoherent mixture between $|\uparrow>$ and $\downarrow>$ A coherent superposition would be $|\phi> = \frac{1}{\sqrt{2}}( |\uparrow > \pm \downarrow>)$ The distinction is that the relative phase of a coherent superposition is observable Entanglement causes decoherence Imagine an entangled state like that in previous section Bob makes a measurement along the x axis and sends his measurement result to Alice This forces Alice’s spin to be a pure state along the x axis, which in turn can be interpreted as a coherent superposition of z axis spins So even though Alice initially had an incoherent state, Bob’s measurement caused Alice’s state to become coherent Another thought experiment: Bob uses a Stern-Gerlach experiment to measure his z axis spin. This precludes Alice from having a coherent superposition along the z axis Bob refocuses the two beams of the Stern-Gerlach which then passes through a Stern-Gerlach along the x axis. Alice’s coherence along the z axis is now restored! (re: Alice is in a known x axis configuration, but the z axis position is lost) This situation is called a quantum eraser. Coherence of a state can be restored if an orthogonal measuremnt occurs Information is physical: measuring an system fundamentally changes the physical description of the system HJW Theorem How do you extend the quantum eraser to multiple qubits with a more general density matrix? Consider the general density matrix: $\rho_{A} = \Sigma_{i} p_{i} |\phi_{i}><\phi_{i}|$ where $\Sigma p_{i} = 1$ Don’t assume that $|\phi_{i}>$ are orthogonal, but do assume they are normalized Imagine some bipartite system such that performing a partial trace over the subsystem B yields $\rho_{A}$ $|\phi_{1AB}> = \Sigma_{i} \sqrt{p_{i}} |\phi_{iA}>\otimes |\alpha_{iB}>$ $<\alpha_{i}|\alpha_{j}> = \delta_{ij}$ $tr_{B}(|\phi_{1AB}><\phi_{1AB}|) = \rho_{A}$ The construction of this bipartite state is called purification. It can be though of as representing a mixed state as a pure state in a higher dimensional Hilbert space Performing a measurement on B is projecting onto a $|\alpha_{iB}>$ basis, which in turn forces system A to be in the pure state $|\phi_{i}><\phi_{i}|$ Is there a different ensemble interpretation of $\rho_{A}$ that we can construct by making a different measurement of B? Let $\rho_{A} = \Sigma_{\mu} q_{\mu} |\phi_{\mu}>< \phi_{\mu}|$ so a different ensemble of pure states There is a similar purification of A: $|\phi_{2AB}> = \Sigma_{\mu} \sqrt{q_{\mu}} |\phi_{\mu A}>\otimes |\beta_{\mu B}>$ How are $|\phi_{1}>$ and $\phi_{2}>$ related? Partial tracing over B yields the same density matrix in both cases. They both have Schmidt decompositioons: $\phi_{1}> = \Sigma_{k} \sqrt{\lambda_{k}} |k_{A}> \otimes | k_{1B}>$ $\phi_{2}> = \Sigma_{k} \sqrt{\lambda_{k}} |k_{A}> \otimes | k_{2B}>$ Alternatively: $|\phi_{1AB}> = (I_{A} \otimes U_{B}) | \phi_{2AB}>$ Since $k_{1}>$ and $k_{2}>$ are both orthonormal bases for B, there is a unitary transformation $U_{B}$ between the two Hence, $|\phi_{1}>$ and $|\phi_{2}>$ are the same purification. You just need to change which direction in B you measure along Suppose that we have many ensembles that realize $\rho_{A}$, where the max number of ensembles is d We can then choose a Hilbert space $H_{B}$ of dimension d and a pure state $|\phi_{AB}> \in H_{A} \otimes H_{B}$ such that any of the ensembles can be realized by measuring a suitable observable of B This is the HJW theorem The general density matrix mixes the pure states incoherently (re: can’t detect the relative phases of these states) So you can erase information by making a measurement in B, and restore the coherence by making a different measurement Fidelity Suppose that you have two density operators $\rho$ and $\sigma$. The fidelity is defined as $(tr(\sqrt{\rho^{\frac{1}{2}} \sigma \rho^{\frac{1}{2}}}))^{2}$ Think of this as the distance metric between two mixed states This is well defined since $\rho$ and $\sigma$ are positive definite matrices; hence, you can take the square root via the spectral theorem The fidelity is bound between 0 and 1. 1 occurs if $\rho$ and $\sigma$ are identical If $\rho = |\phi>< \phi|$ (ie. a pure state), the fidelity becomes $F(\rho,\sigma) = <\phi|\sigma|\phi>$ If both density matrices are pure states, the fidelity reduces to the Born interpretation An alterrnative definition of the fidelity is $F(\rho,\sigma) = ||\sigma^{\frac{1}{2}} \rho^{\frac{1}{2}}||_{1}$ $||A||_{1} = tr \sqrt{A^{\dag}A}$ For Hermetian matrices, take the sum of the absolute values of the eigenvalues The symmetry between the arguments of the fidelity is more manifest in this notation, since for any Hermetian matrices A and B, we have that $||AB|| $ $= ||BA||$ This follows from the fact that BAAB and ABBA have the same eigenvalues (hence the same traces) Uhlmann’s Theorem How does the fidelity of two density operators relate to the overlap of their purifications? Define $|\Phi_{AB}>$ as the purifcation of the density operator $\rho_{A}$ where $\rho_{A} = tr_{B}(|\Phi><\Phi|)$ Suppose that $\rho = \Sigma_{i} p_{i} |i><i|$ where $|i_{A}>$ is a orthonormal basis for system A The associated purifcation is then $\Phi_{\rho}> = \Sigma_{i} \sqrt{p_{i}} |i_{A}> \otimes |i_{B}>$ By the HJW theorem, the general purification is $\Phi_{\rho}(V)> = I \otimes V |\Phi_{\rho}>$ where V is unitary This can also be written as $(\rho^{\frac{1}{2}} \otimes V) |\tilde{\Phi}>$ where $\tilde{\Phi}> = \Sigma_{i} |i_{A}> \otimes | i_{B}>$, which is a normalized maximally entangled state Suppose that we have two density operators $\rho$ and $\sigma$ acting on A. What is the inner product of their purifications? $<\Phi_{\sigma}(W)|\Phi_{\rho}(V)> = <\tilde{\Phi}| \sigma^{\frac{1}{2}} \rho^{\frac{1}{2}} \otimes W^{\dag} V |\tilde{\Phi}>$ Using the fact that $U\otimes I |\tilde{\Phi}> = I \otimes U^{T} |\tilde{\Phi}>$, we can write the inner product of the purifications as $<\tilde{\Phi}|\sigma^{\frac{1}{2}}\rho^{\frac{1}{2}}U \otimes I | \tilde{\Phi}> = tr( \sigma^{\frac{1}{2}}\rho^{\frac{1}{2}}U )$ where $U = (W^{\dag}V)^{T}$ We use the polar decomposition: $A = U’ \sqrt{A^{\dag}A}$ This is the “obvious” fact that any square complex matrix can be factorized as $A=UP$, where U is a unitary matrix and P is a positive semi-definite Hermetian matrix so $ tr( \sigma^{\frac{1}{2}}\rho^{\frac{1}{2}}U ) = tr(UU’\sqrt{\rho^{\frac{1}{2}}\sigma \rho^{\frac{1}{2}}})$ This is very close to the square root of the fidelity. If we select $U’ = U^{-1}$, then they are the same This selection can be thought of as maximizing the inner product of the purifications (this can be seen by Schmit decomposing $ \sqrt{\rho^{\frac{1}{2}}\sigma \rho^{\frac{1}{2}}}$ into its’ non-negative eigenvalues $\lambda_{a}$ and eigenvectors $|a>$, and realizing that $<a|UU’|a>$ is maximized when $UU’ = I$ The final result of Uhlmann’s theorem is that $F(\rho,\sigma) = (tr( \sqrt{\rho^{\frac{1}{2}}\sigma \rho^{\frac{1}{2}}}))^{2} = max_{V,W} | <\phi_{\sigma}(W)|\phi_{\rho}(V)>|^{2}$ An immediate corrollary to this is the monotonicity of fidelity: $F(\rho_{AB},\rho_{AB}) \leq F(\rho_{A},\rho_{A})$. In words: tracing out a subsystem cannot decrease the fidelity of two density operators (follows from Uhlmann’s because purification of $\rho_{AB}$ also purfies $\rho_{A}$ Measurement and Evolution Suppose that we have some system S embedded in a larger system. We want to make a measurement on S. This process can be described as a orthonormal projection operator on the whole system. However, an orthonormal projection on the whole system might not be an orthonormal one on S Let’s generalize the notion of measurement: to measure an observable M, you need to modify the Hamiltonian of the world by coupling the observable M to to some other variable which represents our apparatus. This auxilliary system used to facililate the measurement is called “the pointer”, “the meter”, or “the ancilla” We can then prepare an eigenstate of M by observing the pointer More concretely, think of the pointer as a particle of mass m that propagates freely apart from the coupling We want to measure the position of the pointer, so we need the wavefunction to be spatially confined, but not too confined since otherwise the packet widens too quickly via uncertainty The width of the wavepacket will evolve like $\Delta x (t) \approx \Delta x + \frac{\hbar t}{m\Delta x}$ which minimizes when $\Delta x \approx \sqrt{\frac{\hbar t}{m}}$. This is a upper bound on the resolution of the pointer’s position $\Delta x = \sqrt{\frac{\hbar t}{m}}$ is called the standard quantum limit (SQL) For simplicity, assume the pointer mass is heavy enough for the SQL to not matter The Hamiltonian takes the form $H = \mathbf{H_{0}} + \frac{\mathbf{P^{2}}}{2m} + \lambda(t) \mathbf{M} \otimes \mathbf{P}$ where $\mathbf{H_{0}}$ is the unperturbed Hamiltonian, $\frac{\mathbf{P^{2}}}{2m}$ is the kinetic energy, $\lambda$ is the coupling Assume the kinetic energy term is negliglible due to the large mass of the pointer For simplicity, assume that $[H_{0}, M] = 0$, or that the measurement is quickly done so as to not perturb the evolution. The Hamiltonian then becomes $H \approx \lambda(t) M \otimes P$ Assume that $\lambda$ suddenly turns on a t=0 and turns off at t=T. The time evolution operator is then $U(T) \approx exp(i\lambda T \mathbf{M}\otimes \mathbf{P}) = \Sigma_{a} |a> exp(i\lambda T M_{a} \mathbf{P}) <a|$ where we diagonalize $\mathbf{M}$ Recall that $exp(-i x_{0} \mathbf{P}) \phi(x) = \phi(x-x_{0})$ (Follows from $P = -i\frac{d}{dx}$ and Taylor expansion) Suppose that our initial state is in a superposition of M eigenstates, initally unentangled with the position-space wavepacket $\phi(x)>$ (ie. $\Sigma_{a} \alpha_{a} | a> \otimes \phi(x)>$. Acting U(T) on this state gives $\Sigma_{a} \alpha_{a} |a> \otimes |\phi(x-\lambda T M_{a}) > $ The time evolution correlates the position of the pointer with the value of observable M Hence, observing the shift from x, we can figure out the associated eigenstate of M with probability $|\alpha_{a}|^{2}$ A classical example is the Stern-Gerlach experiment: you have some observable $\sigma_{3}$. You pass through a B field of the form $B_{3} = \lambda z$. The perturbed Hamiltonian is then $-\lambda \mu z \sigma_{3}$. This coupling imparts an impulse on the pointer (z is a translator of $P_{z}$) which can be observed as spin up or spin down Imagine that we can couple any observable to a pointer as described above. Observing the pointer will be some orthogonal projection in the Hibert space Let $[E_{a}]$ be some set of orthongonal projectors such that: $E_{a} = E_{a}^{\dag}$ $E_{a}E_{b} = \delta_{ab} E_{a}$ $\Sigma_{a=0}^{N-1} E_{a} = 1$ Define the unitary transformation $U = \Sigma_{a,b} E_{a} \otimes |b+a> <b|$ Addition of b + a is modulo N Act this unitary transformation on the initial state of the system: $|\Phi> = |\phi> \otimes |0>$ to get $U|\Psi> = \Sigma_{a} E_{a} |\phi> \otimes |a>$ The effect of this transformation on the density matrix due to measurement is to transform it like $\rho \rightarrow \Sigma_{a} E_{a} \rho E_{a}$ The probability of observing a then becomes $<\phi| E_{a}|\phi>$ So we can see that by coupling the system to a macroscopic apparatus, and doing appropriate unitary transformations, we can observe the pointer in its’ fiducial basis to perform any orthogonal measurement on the system

The Basics Hibert Space Review In an infinitely dimensional space, we have some abstract vector $\Psi$. We can choose coordinate axes such that all values can be taken by the position x, so that we can describe $\Psi$ as as set of components $\Psi(x)$ Could have also chosen p instead for our coordinate axes $(v,w) \geq 0$, where (a,b) denotes an inner product This is linear in the right argument, and anti-linear in the first argument (ie. take complex conjugates) (v,v) = 0 implies v = 0 $v = \Sigma_{i} e_{i} (e_{i}, v)$ always holds in Hilbert spaces (ie. can always decompose into an eigenbasis, where $(e_{i}, e_{j}) = \delta_{ij}$) This can be extended to non-orthonormal bases (ie. $(e_{i}, e_{j}) = G_{ij}$), which implies that $(v,w) = \Sigma_{ij} (v_{i}e_{i})G_{ij}^{-1} (e_{j}, w)$ You can add Hilbert spaces of different dimensions together The new Hilbert space has the equivalence relationship of $<v_{1}+v_{2},w_{1}+w_{2}> = <v_{1},w_{1}> + <v_{2},w_{2}>$ The total dimensionality of the new Hilbert space is just the sum of the previous two You can also multiply two Hilbert spaces together The equivalence relation which must hold for the product is that $\lambda <v,w> = <\lambda v,w> = <v,\lambda w>$ You can decompose this product Hilbert space as a sum of tensor products of the eigenbases of each prior Hilbert space (ie. $H_{prod} = \Sigma_{ia} \Phi_{ia} e_{i} \otimes e_{a}$) Observables Can think of any operator as any mapping of the Hilbert space on itself (ie. A transforms $\phi$ such that $A\phi$ is in the same Hilbert space) Real and complex numbers commute with any other operators Real observables are postulated to be linear and Hermetian (ie. self adjoint) Linear means that $A(\phi+\phi’) = A\phi+A\phi'$ the adjoint means that $(\phi’, A^{\dag} \phi) = (A\phi’,\phi)$, which means $A^{\dag} =A$ If we have a complete set of orthonormal basis vectors, we can embed the operator as a matrix: $A_{ij} = (\Phi_{i},A\Phi_{j})$ The eigenvalues for Hermetian observables are real The trace of an operator is defined by $Tr(A) = \Sigma_{i} (\phi_{i},A\phi_{i})$ This is basis independent This is only well defined for finite dimensional Hilbert spaces Projection Operator We can define a special operator $P = |\phi_{i}><\phi_{i}|$ called the projection operator $\Sigma_{i} P_{i} = Id$ You can rewrite any Hermetian operator A with eigenvalues $\alpha_{i}$ and a complete set of orthonormal eigenvectors $\phi_{i}$ as $A = \Sigma_{i} \alpha_{i} (\phi_{i}\phi_{i}^{\dag})$ Derived from seeing that the operator $A - \Sigma_{i} \alpha_{i} (\phi_{i}\phi_{i}^{\dag})$ annihilates any $\phi_{i}$ Symmetries Cannonical Transforms We know that canonical transformations are an important part of classical Hamiltonian mechanics. What is the analog in quantum? We want the probability to be conserved under a canonical transformation! (ie. $P(\Phi) = |(\Phi,\Phi)|^{2}$) We also want to preserve linearity (ie. $(\phi_{1}+\phi_{2})’ = \phi_{1}+\phi_{2}$) The only transformations which satisfy these properties are either Unitary or Anti-Unitary (Wigner’s Theorem) Unitary $U(a\phi+b\psi)= aU\phi + bU\psi$ (can applies either before or after linearity) $UU^{\dag} = I$, where the adjoint of an operator is defined as $(\psi O^{\dag} \phi) = (\phi O \psi)$ Needs to be invertible Operator O transforms like $U^{\dag}O U = O'$ transform inner product, do adjoint on the left, compare Suppose that you have some unitary operator U which is arbitrarily close to 1 (ie. $U = 1+i\epsilon T$) U need to maintain unitarity up to order $\epsilon$, which implies $T = T^{\dag}$, or that T is Hermetian T is called the generator of the symmetry Define $\epsilon = \frac{\theta}{N}$ where $\theta$ is some finite N-independent parameter. Imagine applying the symmetry transformation N times, and letting N go to infinity $(1+\frac{i\theta T}{N})^{N} = exp(i\theta T) = U(\theta)$ Under a symmetry transform $\phi \rightarrow U\phi$, any observable A must transform like $A \rightarrow U^{-1} A U$ For an infinitesimal transformation, A transforms like $A \rightarrow A-i\epsilon[T,A]$ Anti-Unitary $U(a\phi+b\psi)= a^{*}U\phi + b^{*}U\psi$ (ie. must maintain anti-linearity) $(U\phi, U\psi) = (\psi, \phi) = (\phi, \psi)^{*}$ Operators transform like $U^{-1} O U$ Parity Defined as $\Pi^{\dag} \vec{x} \Pi = -\vec{x}$ We also define $\Pi* \Pi = 1$, which implies $\Pi = \Pi^{-1}$ and that the eigenstates are $\pm 1$ Suppose that $\phi$ is an eigenstate of $\Pi$. Then $(\phi x \phi) = 0$ (Apply parity definition, transfer parity operator to states, apply eigen equation $\Pi \psi = \epsilon \psi$), and then note that you get $-q =q $, which implies q=0 As a consequence of this, if you Hamiltonian commutes with parity, then you know that you can write you state as a sum of even and odd states Rotations Defined by $\Sigma_{i} R_{ij}R_{ik} = \delta_{jk}$ Alternatively, any real linear transformation that leaves the scalar product $x\cdot y = \Sigma x_{i}y_{i}$ Alternatively any matrix which satisfies $R^{T}R = Id$ and $det(R)=1$ det(R)=-1 are spatial inversions like parity Imagine you have an operator V representing a vector observable (things like the coordinate vector X or the momentum vector P). A unitary rotation must act on this vector operator like $U^{-1}V_{i}U = \Sigma_{j} R_{ij}V_{j}$ For infinitesimal rotations, we know that unitarity must hold. We can write any infinitesimal rotation as $U(1+\omega) = 1+\frac{i}{2\hbar} \Sigma_{ij}\omega_{ij}J_{ij}$ where $\omega_{ij} = -\omega_{ji}$ and J is some set of Hermetian operators Can use composition of rotations in order to extract commutation relationships for J. In 3 dimensions $[J_{i},J_{j}] = i\hbar \Sigma_{k} \epsilon_{ijk} J_{k}$ More generally: $[J_{i},V_{j}] = i\hbar \Sigma_{k} \epsilon_{ijk} V_{k}$, where $V_{k}$ is any 3 dimensional vector operator Can define $S=J-L$, where S obeys the same commutation rules as J and L, but is independent of position and momentum operators. S also commutes with L, X and P Spin et Cetera Eigenvalues of $J^{2}$ and $J_{3}$ To derive the eigenvalues of $J^{2}$ and $J_{3}$, we have the following prescription: Realize that $J^{2}$ and $J_{3}$ commute with each other, which means they can have simultaneous eigenstates Define raising and lowering operators $J_{\pm} = J_{1}\pm iJ_{2}$ Realize that acting $J_{3}$ on $J_{\pm}\phi_{m}$ shows that $J_{\pm}\phi_{m}$ is an eigenvector of $J_{3}$ with eigenvalue $(m\pm 1)\hbar$ Realize that there is some min/max value of eigenvalues of $J_{3}$ since $J_{3}$ is a part of the vector of $J^{2}$ Define the min and max eigenstates, then act the raising and lowering operators on both of them (NOTE: j and j’ currently have no relationship to each other at this point in time) $(J_{1}+iJ_{2})\phi_{j} = 0$ $(J_{1}-iJ_{2})\phi_{j’} = 0$ Since the raising and lowering operators are atomic, we need to apply an integer number of them to transition from $\phi_{j}$ to $\phi_{j’}$, which implies that $j-j’$ must be a whole number Using the commutator relationships, you can show that $J_{\mp}J_{\pm} = J^{2}-J_{3}^{2}\mp \hbar J_{3}$ Can use the above to find the eigenvalues of $\phi_{j}$ and $\phi_{j’}$. Equating the eigenvalues to each other (since they both represent the same spectrum) yields that either $j’=-j$ or that $j’=j+1$ The second is impossible, since j’ is the minimum eigenvalue, and thus can’t be higher than the maximum Because the distance between j and j’ is an integer, j must be either an integer or a half integer To calculate the normalization of acting the raising and lowering operators on $\phi_{j,m}$, use $J_{\pm} \phi_{j,m} = \alpha_{\pm}(j,m)\phi_{j,m\pm 1}$, take the norm of this expression, use the commutators to show that $J_{\pm} \phi_{j,m} = \hbar \sqrt{j(j+1)-m^{2}\mp m} \phi_{j}^{m}$ The eigenstates for J are just the spherical harmonics Addition of Angular Momenta $\phi_{j,j’,j’’}^{m} = \Sigma_{m’m’’} C_{j’j’’}(jm;m’m’’)\phi_{j’,j’’}^{m’,m’’}$ The only non-vanishing Clebsch–Gordan coefficients occur when $m=m’+m’'$ To construct the Clebsch–Gordan coefficients, start at the highest state, imagine applying the lowering operator for each particle with the appropriate normalization, then recursively do this In practice, look this up in a table Wigner Eckart Theorem Suppose that you have a set of 2j+1 operators $O_{j}^{m}$, with m ranging from -j to j in integer increments Also suppose that this set of operators obeys the angular momemtum commutation relationships: $[J_{3},O_{j}^{m}] = m O_{j}^{m}$ $[J_{\pm},O_{j}^{m}] = \sqrt{j(j+1)-m^{2}\mp m} O_{j}^{m}$ We want to show that $(\phi_{j’’}^{m’’}O_{j}^{m}\psi_{j’}^{m’}) = C(j’’m’’;j’m’) (\phi || O|| \psi)$ This is the Wigner Eckart Theorem. What the hell does it mean? $O_{j}$ is a spherical tensor operator Spherical denotes the fact that we are using a spherical basis of coordinates (think spherical harmonics) Tensor implies that the set of $O_{j}$ transforms like a tensor (ie. applying a rotation on the set just yields a linear combination of the original set) The C are the Clebsch-Gordan coefficients between a given set of spherical harmonics The remaining element is called the reduced matrix element. The important part is that this proportionality factor is independent of m, m’, or which $O_{j}$ you use This allows you to calculate a bunch of matrix elements by only calculating one matrix element, then using the proportionality to “rotate” to another matrix element Indistinguishable Particles For simplicity, assume that we have two particles, whose total Hamiltonian is $H_{0}=H_{1}+H_{2}$. Naively, we would write the total wavefunction as $\phi_{0} = \phi_{1}\phi_{2}$, but if we have a boson or a fermion, this isn’t invariant under particle swap So, we need to create a more symmetric state: $\Phi = \frac{1}{\sqrt{2}}(\phi_{1}\phi_{2} \pm \phi_{2}\phi_{1})$ where the plus is for bosons and minus is for fermions Generalizing to N particles yields the idea of the Slater determinant. For bosons, you just turn all of the minus signs in the determinant to plus signs Calculating the probability density yields that $P = \int d^{3}\vec{x_{1}}d^{3}\vec{x_{2}} |\phi_{1}(x_{1})|^{2}\phi_{2}(x_{2})|^{2}+|\phi_{1}(x_{2})|^{2}\phi_{2}(x_{1})|^{2} \pm 2 Re(\phi_{1}(x_{1})\phi_{2}(x_{2})\phi_{1}^{*}(x_{2})\phi_{2}^{*}(x_{1}))$ The last term arises from the bosonic and fermionic interactions To generalize to N particles: Suppose that each particle has it’s own hamiltonian h and wavefunction $\phi$ which satisfy $h e_{\alpha} = E_{\alpha} e_{\alpha}$ and $\phi = \Sigma_{\alpha} \phi_{\alpha} e_{\alpha}$ We can write the total hamiltonian of N particles as the sum of each particle’s hamiltonian, and the wavefunction as $\psi = \Sigma \phi_{\alpha_{i}…\alpha_{n}} e_{alpha_{1}} \otimes … e_{alpha_{n}}$ If everything was distinguishable, then $\phi_{\alpha_{i}…\alpha_{n}} = \Sigma_{i=1}^{N} \phi_{i}(x_{i})$ What is the size of the Hilbert space of we have indistinguishable particles? We go by induction. With 0 particles, we have 1, represented by $\Omega_{\Sigma_{i=1}^{N} n_{i}}$ How to we add particles? we define a creation operator $a^{\dag}_{i}$ such that a new particle is created in the ith state an anihilation operator such that $a_{i}$ such that a particle is removed from the ith slot. If $n_{i}=0$, then the overall resulting state is 0 We can construct the cannonical bosonic commutator relationships $a_{i}a_{j}-a_{j}a_{i} = 0$ $a_{i}^{\dag}a_{j}^{\dag}-a_{j}^{\dag}a_{i}^{\dag} = 0$ $a_{i}a_{j}^{\dag}-a_{j}^{\dag}a_{i} = \delta_{ij}$ We can motivate these by examining $a_{j}a_{i}^{\dag}\Omega = \delta_{ij}$ and $a_{i}^{\dag}a_{j} \Omega =0$ We can also define the normalization from these commutators: $a_{i}^{\dag} \Omega_{…n_{i}…} = \sqrt{n_{i}+1} \Omega_{…(n_{i}+1)…}$ and $a_{i} \Omega_{…n_{i}…} = \sqrt{n_{i}} \Omega_{…(n_{i}-1)…}$ We can construct an analogous set of anticommutator relationships for the fermions $b_{i}b_{j}+b_{j}b_{i} = b_{i}^{\dag}b_{j}^{\dag}+b_{j}^{\dag}b_{i}^{\dag} = 0$ $b_{i}b_{j}^{\dag}+b_{j}^{\dag}b_{i} = \delta_{ij}$ Hydrogen Spectrum Derivation Classically, there is an additional vector conserved in central potential celestial mechanics called the Runge-Lenz vector $R = \frac{-Ze^{2}x}{r}+\frac{1}{2m}(p\times L - L \times p)$ This commutes with the Coulomb Hamiltonian $H = \frac{p^{2}}{2m}-\frac{Ze^{2}}{r}$ You can use a bunch of commutator relationships between x, p, and L in order to show that $R^{2} = Z^{2}e^{4}+\frac{2H}{m}(\vec{L}^{2}+\hbar^{2})$ Since H is in this expression and since R commutes with H, you can calculate the energies if you know the eigenvalues of $R^{2}$ $[R_{i},R_{j}] = \frac{-2i}{m}\hbar \Sigma_{k}\epsilon_{ijk} H L_{k}$ from a bunch of commutator identities Since R is a vector, we also get that $[L_{i}, R_{j}] = i\hbar \Sigma_{k} \epsilon_{ijk} R_{k}$ We can define raising and lowering operators $A_{\pm} = \frac{1}{2}(L\pm \sqrt{\frac{m}{-2H}} R)$ which obeys the following commutation relationships $[A_{\pm i}, A_{\pm j}] i\hbar \Sigma_{k}\epsilon_{ijk} A_{\pm k}$ $[A_{\pm i}, A_{\pm,j}] = 0$ Since we know that $R\cdot L=0$, we can see that $A_{\pm}^{2} = \frac{1}{4}(L^{2}+\frac{m}{-2H}R^{2})$ From the angular momentum eigenvalue derivation, we know that the eigenvalues of $A_{\pm}^{2}$ take the form of $hbar^{2}a(a+1)$. Defining a principle quantum number as $n = 2a+1 = 1,2,3…$, once can show that $E = \frac{-Z^{2}e^{4}m}{2\hbar^{2}n^{2}}$, just like the standard Bohr atom derivation This also gives the degeneracy at each level to be $n^{2}$ Approximations for Energy Eigenvalues 1st Order Non-degenerate Time Independent Perturbation Theory We have some unperturbed Hamiltonian with some orthonormal eigenvectors $H_{0}\phi_{a} = E_{a}\phi_{a}$ $(\phi_{a},\phi_{b}) = \delta_{ab}$ Suppose that we have some small perturbation to the Hamiltonian called $\delta H$ which is proportional to some $\epsilon$ This produces a change in the state vector $\delta_{1} \phi_{a}$ and some small change in energy $\delta_{1} E_{a}$ Make these perturbations to all of the variable, eliminate terms of order $\epsilon^{2}$ and higher, contract the entire expression with $\phi_{a}$ and eliminate like terms to get $\delta_{i} E_{n} = (\phi_{a}, \delta H \phi_{a})$ This is fine and dandy if you have no degenerate eigenstates. This problem can be made lucid by contracting with $\phi_{b}$ instead of $\phi_{a}$ you end up with $(\phi_{b} \delta H \phi_{a}) = (E_{a}-E_{b}) (\phi_{b} \delta \phi_{a})$ If a and b are degenerate,then you have $(\phi_{b} \delta H \phi_{a})=0$, which is no gaurenteed The work around for this solution is to diagonalize the degenerate subspace in some new orthonormal basis In math terms, this means you diagonalize the subspace such that $(\phi_{b}, \delta H \phi_{a}) = 0$ Given the above $(\phi_{b}, \delta_{1}\phi_{a}) = \frac{(\phi_{b},\delta H \phi_{a})}{E_{a}-E_{b}}$ The above needs to be normalized which gives the condition $0 = (\phi_{a} \delta_{i} \phi_{a})$ Hence, the $\delta_{1} \phi_{a} = \Sigma_{b\neq a} \phi_{b} \frac{(\phi_{b}\delta H \phi_{a})}{E_{a}-E_{b}}$ Zeeman Effect The shift of atomic energies in the presence of an external magnetic field is called the Zeeman effect We can define the peturbation as $\delta H = \frac{e}{2m_{e}c} \vec{B} \cdot (\vec{L} + g_{e} \vec{S})$ Comes from classical contribution to energy $\frac{e}{2m_{e}c} \vec{B} \cdot \vec{L}$ as well as quantum spin term with the g factor which is ~ 2 Can use Wignar-Eichart theorem $(\phi_{nlj}^{m’}, (L+g_{e}S) \phi_{nlj}^{m}) = g_{njl} (\phi_{nlj}^{m’}J \phi_{nlj}^{m})$ where $g_{njl}$ is the Lande g-factor J commutes with $J^{2}$, so you write the eigenvector $J\phi_{njl}^{m}$ as a linear combination of $\phi_{nlj}^{m’’}$ with various m'' In math: $\Sigma_{i} (\phi_{nlj},(L+g_{e}S) \phi_{nlj}^{m}) = g_{njl} \Sigma_{i} (\phi_{nlj}^{m’}J_{i}J_{i} \phi_{nlj}^{m})$ Can use identities that $J = L+S$, and eigenspectrum of J,L, and S to show that $g_{njl} = 1+(g_{e}-1)(\frac{j(j+1)-l(l+1)+\frac{3}{4}}{2j(j+1)})$ Note that $g_{njl}$ is independent of n This means that we need to compute the matrix elements $(\phi_{njl}^{m’}, \delta H \phi_{njl}^{m}) = \frac{eg_{l}}{2m_{e}c}(\phi_{njl}^{m’}, \vec{B}\cdot \vec{J} \phi_{njl}^{m})$ Choose a coordinate system such that the z axis aligns with B. This implies that $BJ_{z} \phi_{njl}^{m} = \hbar m B \phi_{njl}^{m}$ This gives that $\delta E_{njlm} = \frac{e\hbar g_{jl} B}{2 m_{e} c} m$ Second Order Perturbation Theory You need 2nd order perturbation theory if the 1st order vanishes, or if you want more precision Expanding to 2nd order, we find that $H_{0} \delta_{2} \phi_{a} + \delta H \delta_{1} \phi_{a} = E_{a} \delta_{2} \phi_{a} + \delta_{1} E_{a} \delta_{1} \phi_{a} + \delta_{2} E_{a} \phi_{a}$ We know what $\delta_{1}$ and $\delta_{1} \phi_{a}$ Using a similar procedure as 1st order perturbation, we see that $\delta_{2} E_{a} = \Sigma_{b\neq a} \frac{|(\phi_{b} \delta H \phi_{a})|^{2}}{E_{a}-E_{b}}$ You diagonalize degenerate subspaces as needed Variational Method Let $\phi$ be some normalizable wave function $(\phi H \phi) \geq E_{ground}$ This sets an upper bound on the ground state energy Born-Oppenheimer Approximation Consider a molecule with N nuclei and n electrons. The Hamiltonian becomes $H = \Sigma_{\alpha=1}^{N}\frac{p_{\alpha}^{2}}{2M_{\alpha}}+V_{NN}(\vec{R})+\Sigma_{i=1}^{n}\frac{p^{2}}{2m}+V_{eN}(\vec{R},\vec{r}) +V_{ee}(\vec{r})$ $\alpha$ variables are associated with the nuclei and i variables are associated with the electrons We consider the nuclei essentially fixed (ie. $\vec{R}$ is fixed). This gives rise to a Schrodinger equation for the electrons: $(-\frac{\hbar^{2}}{2m}\Sigma_{i=1}^{n} \Delta_{\vec{r_{i}}}^{2} + V_{eN}(\vec{R},\vec{r})+V_{ee}(\vec{r}))\phi_{\vec{R}}^{i}(\vec{r})= E_{e}^{i}(\vec{R})\phi_{\vec{R}}^{i}(\vec{r})$ We can try and write the full ansatz as $\Phi(\vec{R},\vec{r}) = \Sigma_{i} \eta(\vec{R})^{i} \phi_{\vec{R}}^{i}(\vec{r})$ This is too complicated to solve, so we make the simpler ansatz $\Phi(R,r) = \eta(R)\phi(R,r)$ Time Dependent Perturbation Theory (TDPT) Suppose that you have some Hamiltonian $H(t) = H_{0}+H’(t)$ where $H’(t)$ is small compared to $H_{0}$ The eigenvector satisfies $i\hbar \frac{d\phi}{dt} = H(t) \phi(t)$, where $\phi_{n}$ is a orthonormal basis of time-independent unperturbed eigenvectors of $H_{0}$ The time evolution is then $\phi(t) = \Sigma_{n} c_{n}(t) exp(-\frac{-iE_{n}t}{\hbar}) \phi_{n}$ The perturbation acting on $\phi_{n}$ can be expanded as a linear combination of the original Hamiltonian eigenstates $H’(t) \phi_{n} = \Sigma_{m} \phi_{m}(\phi_{m},H’(t) \phi_{n}) = \Sigma_{m}H_{mn}’(t) \phi_{m}$ where $H_{mn}’ = (\phi_{m}, H’(t)\phi_{n})$ Plugging in the above equations for $\phi$ and $H’$ acting on $\phi$ into the Schrodinger equation for the total Hamiltonian, you can Cancel out the terms proportional to $E_{n}$, interchange the labels m and n on the RHS, and equate the coefficients of $\phi_{n}$ to eventually get $i\hbar \frac{dc_{n}(t)}{dt} = \Sigma_{m} H_{nm}’ c_{m}(t) exp(i\frac{(E_{n}-E_{m})t}{\hbar})$ To make a perturbative approximation, we see that the rate of change of $c_{n}$ is proportional to the perturbation (ie. $H_{nm}’$). Hence, we can replace $c_{m}(t) \approx c_{m}(0)$ (ie. $c_{m}$ doesn’t change that fast…). This allows ups to solve the differential equation $c_{n}(t) \approx c_n(0)-\frac{i}{\hbar}\Sigma_{m} c_{m}(0) \int_{0}^{t} dt’H_{nm}’(t’) exp(i\frac{(E_{n}-E_{m})t’}{\hbar})$ TDPT Interaction Picture Same starting point: $H = H_{0}+V(t)$ and $i \frac{d}{dt}H = H \phi$ Assuming $H_{0}$ is time-independent, define the time evolution operator $U(t) = exp(itH_{0})$. This obeys $i\frac{d}{dt}U = H_{0}U$ Act the operator differential equation in $\phi$, then multiply by $U^{-1}$ to get $i \frac{d}{dt}\phi = U^{-1} V U \phi = V_{I}\phi$ Hence, we get a differential operator equation for the interaction potential. We can solve this for U: $U_{I}(T)-1 = -i \int_{0}^{T} dt V_{I} U_{I}$ This is equivalent to the Schrodinger picture, but now you assign the evolution to the operators, not the state Monochromatic Perturbations Assume that $H’(t) = U exp(-i\omega t) + U^{\dag} exp(i\omega t)$ where U is some unitary operator Think of this as a sinusoidal perturbation with angular frequency $\omega$ Using time dependent perturbation theory (straightforward integrals), you can write $c_{n}(t) = c_{n}(0)+\Sigma_{m} U_{nm} c_{m}(0) (\frac{exp(i\frac{(E_{n}-E_{m}-\hbar\omega)t}{\hbar})-1}{E_{n}-E_{m}-\hbar\omega})+U_{nm}^{*} c_{m}(0) (\frac{exp(i\frac{(E_{n}-E_{m}+\hbar\omega)t}{\hbar})-1}{E_{n}-E_{m}+\hbar\omega})$ Examine the case where $c_{n}(0)=0$ for all n except n=1. Then $c_{n}(t)$ for $n \neq 1$ is $c_{n}(t)U_{n1}\frac{exp(i\frac{(E_{n}-E_{m}-\hbar \omega)t}{\hbar})}{E_{n}-E_{m}-\hbar \omega}+c_{n}(t)U_{n1}^{*}\frac{exp(i\frac{(E_{n}-E_{m}+\hbar \omega)t}{\hbar})}{E_{n}-E_{m}+\hbar \omega}$ At t=0, both terms vanish Both terms increase from time 0 to time $t = \frac{i(E_{n}-E_{1}-\hbar\omega)t}{\hbar}$, after which they just oscillate Suppose that $E_{n} \approx E_{1}+\hbar\omega$ This corresponds to the absorption of one quanta of energy This means that the ramp up time can be very long This causes the 2nd term to fall off much faster than the 1st term After a long enough time: $|c_{n}(t)|^{2} \approx 4 |U_{n1}|^{2}\frac{\sin^{2}(\frac{(E_{n}-E_{1}-\hbar \omega)t}{2\hbar})}{(E_{n}-E_{1}-\hbar \omega)^{2}}$ For very large times, we can approximate $\frac{2\hbar \sin^{2}(\frac{Wt}{2\hbar})}{\pi t W^{2}} \approx \delta(W)$ where $W = E_{1}+\hbar\omega -E_{n}$ Hence: $|c_{n}(t)|^{2} = 4|U_{n1}|^{2} \frac{\pi t}{2\hbar} \delta(E_{1}+\hbar \omega -E_{n})$ We can define the transition rate $\Gamma = \frac{|c_{n}(t)|^{2}}{t}=\frac{2\pi}{\hbar}|U_{1n}|^{2} \delta(E_{n}+\hbar\omega-E_{1})$ This is Fermi’s Golden Rule We can extend this derivation to a continuous final spectrum of the Hamiltonian Imagine putting the system in a big box with sides of length L with periodic boundary conditions (ie. $\phi(x+c) = \phi(x)$). The solution to this system has eigenstates $\phi(\vec{x}) = \frac{1}{\sqrt{L^{3}}} exp(i\vec{k}\cdot \vec{x})$ where $k_{i} = \frac{2\pi n_x}{L}$ The phase space volume is then $d^{3} k = \frac{(2\pi)^{3}}{L^{3}} d^{3}n$ We occasionally want to write $d^{3}n = \rho(E) dE$, where $\rho(E)$ is the density of states of the system (ie. it’s easier to work with hypersurfaces of E than 3 components of momentum) Since $E = \frac{\hbar^{2}k^{2}}{2m}$, we can write the momentum integral in spherical coordinates we can show that $\rho(E) = (\frac{L}{2\pi})^{2} \frac{m}{\hbar^{2}} k d \Omega$ The final transition rate is just summing over this continuum of final states. Each little volume of momentum space has some density associated with it. Hence $\Gamma_{i\rightarrow f} = \int P_{i\rightarrow f}(t_{0}) \rho(E_{f}) dE_{f}$ Ionization by an EM Wave Consider a hydrogen atom in the ground state placed in a light wave. If $\lambda » a$ (ie. the wavelength of light is much larger than the Bohr radius), then the perturbation Hamiltonian only depends on the E field at the location of the atom (neglecting magnetic field on non-relativistic charged particles b/c those are negligible in comparison) We can write this perturbation as $H’(t) = e\Epsilon\cdot X exp(i\omega t) + e\Epsilon^{*} \cdot X exp(i\omega t)$ $\Epsilon$ is a constant, and X is the electron position operator We can make the association $U = e\Epsilon X_{3}$ if we align $\Epsilon$ with the $X_{3}$ direction We want to ionize the electron (ie. rip it from the atom). Hence, our initial state is the ground state of the hydrogen atom $\phi_{1s}(x) = \frac{\exp(\frac{-r}{a})}{\sqrt{\pi a^{3}}}$ and our final state is a free electron with momentum $\hbar k_{e}$ (ie. $\phi_{e}(x) = \frac{exp(ik_{e}\cdot x)}{(2\pi\hbar)^{\frac{-3}{2}}}$) We are assuming that the final electron energy is much larger than the hydrogen binding energy The matrix element takes the form $U_{e,1s} = \frac{e\Epsilon}{(2\pi\hbar)^{\frac{3}{2}}\sqrt{\pi a^{3}}}\int d^{3}x e^{-i\vec{k_{e}}\cdot \vec{x}} x_{3} exp(\frac{-r}{a})$ This can be solved by examining the integral $\int d^{3}x e^{-i\vec{k_{e}}\cdot \vec{x}} f(r)$ Transform to spherical coordinates, integrate out the spherical components to $4\pi$, take the derivative w.r.t. $k_{3}$, then solve the remaining straightforward (if tedious) integrals The final result is $\frac{-4\pi ie\Epsilon k_{e3}}{k_{e}^{3}(2\pi\hbar)^{\frac{3}{2}}}\sqrt{\pi a^{3}} \frac{8k_{e}a^{5}}{(1+k_{e}^{2}a^{2})^{3}}$ You can make the approximation $k_{e}^{2}a^{2} » 1$ since the final electron energy is large In this limit, we have $U_{e,1s} = \frac{-8 \sqrt{2}i e \Epsilon \cos \theta}{\pi \hbar^{\frac{3}{2}}k_{e}^{5}a^{\frac{5}{2}}}$ where $\cos \theta$ represents the angle between $k_{e}$ and the polarization vector of the EM wave The differential ionization rate then becomes: $d\Gamma(1\rightarrow k_{e}) = \frac{2\pi}{\hbar}|U|^{2} \delta(\hbar c k_{\gamma} -E_{e}) \hbar^{3} k_{e}^{2} dk_{e} d\Omega$ You can integrate out $k_{e}$ dependency via the delta function Fluctuation Perturbations Imagine that we have some perturbation which fluctuates randomly in time We define $\overline{H_{nm}’(t_{1})H_{nm}’^{*}(t_{2}) }= f_{nm}(t_{1}-t_{2})$ In words, we assume that the correlation between two points in time only depends on the differences in times Assuming $c_{n}(0) = \delta_{n1}$, we have that $\overline{|c_{n}(t)|^{2}} = \frac{1}{\hbar}\int_{0}^{t} dt_{1} \int_{0}^{t} dt_{2} f_{n1}(t_{1}-t_{2}) exp(i\frac{(E_{n}-E_{1})(t_{1}-t_{2})}{\hbar})$ We can Fourier transform $f_{nm}$ to $F_{nm}$, use Fubini’s theorem to convert the two time integrals into the modulus squared of a single time integral over $t_{1}$, and then use the approximation as in previous sections to get $\Gamma(1\rightarrow n ) = \frac{\overline{|c_{n}(t)|^{2}}}{t} = \frac{2\pi}{\hbar} F_{n1}(\frac{E_{n}-E_{1}}{\hbar})$ where $F_{nm}(\Omega)$ is the Fourier transform Absorption and Stimulated Emission of Radiation $H_{nm}’(t) = e \Sigma_{N} <x_{N}>_{nm} E(t)$ where N denotes the position of each electron in the atom Good approximation since scale of E field variation is much larger than atom radius $<x_{N}>_{nm} = \int \phi _{n}^{*}(x) x _{N} \phi _{m}(x) \Pi _{m} d^{3}x _{M}$ Assume correlation of E field goes like $\overline{E_{i}(t_{1})E_{j}(t_{2})} = \delta_{ij} \int_{-\infty}^{\infty} d\omega P(\omega) exp(-i\omega (t_{1}-t_{2}))$ You can write the average correlation of the perturbation as a fourier transform, with $F_{nm}(\omega) = e^2|\Sigma_{N} <x_{N}>_{nm}|^{2}P(\omega)$ Using $\delta_{ij}$ assumes that there is no preferred direction for the electric field $P(\omega) = P(-\omega)$ via realness of E field This allows us to define the correlation function of the perturbation, which in turn allows us to define $F_{nm}(\omega) = e^{2} | \Sigma_{N} <x_{N}>_{nm} |^{2} P(\omega)$ As per the previous section, we can write the transition rate from 1 to n as $\Gamma(1\rightarrow n) = \frac{2\pi e^{2}}{\hbar^{2}} | \Sigma_{N} <x_{N}>_{nm} |^{2} P(\omega)$ Adiabatic Approximation Suppose that the Hamiltonian depends on some set of parameters s which are slowly varying functions in time (ie. s(t)) We can find the solution of the time-dependent Schrodinger equation by use of the adiabatic approximation The fundamental results of the adiabatic theorem are: The state that you are currently in does not change, but the eigenstates you are in do change You pick up an overall phase from two source: a dynamical phase and a Berry phase For an adiabatic approximation, we want to show that $H(t) |\phi_{n}(t)> = E_{n}(t)| \phi_{n}(t)>$ where the $|\phi_{n}(t)>$ form an orthonormal basis for all t Can think of $\phi_{n}(t)$ as some instantaneous eigenstates which vary with time (this is the state that you are stuck at) Importantly, $\phi_{n}(t)$ does not solve the Schrodinger equation in general. We want to make it approximately solve it though We are also assuming that there is no degeneracy in the problem We can decompose any solution to the Schrodinger equation in this orthonormal basis: $\Phi(t) = \Sigma_{n} c_{n}(t) \phi_{n}(t)$ Plug this into the Schrodinger equation and act with $<\phi_{k}(t)|$ from the left to yield $i\hbar \dot{c_{k}} = (E_{k}-i\hbar <\phi_{k}|\dot{\phi_{k}}> c_{k}) - i\hbar \Sigma_{n\neq k} <\phi_{k}|\dot{\phi_{n}}> c_{n}$ We can show that $<\phi_{k}(t) \dot{\phi_{n}} = \frac{<\phi_{k}|\dot{H}(t)| \phi_{n}>}{E_{n}(t) E_{k}(t)} = \frac{\dot{H_{kn}}}{E_{n}-E_{k}}$ Take the time derivative of $H(t) |\phi_{n}(t)> = E_{n}(t)| \phi_{n}(t)>$, contract with $<\phi_{k}(t)$ from the left, then rearrange as needed Since we are assuming an adiabatic process, then $\dot{H_{kn}}$ is approximately 0. This leaves $i\hbar \dot{c_{k}} = (E_{k}-i\hbar <\phi_{k}|\dot{\phi_{k}} c_{k})$, which can be solved to yield $c_{k}(t) = c_{k}(0) e^{i\theta_{k}(t)}e^{i\gamma_{k}(t)}$ $\theta_{k}(t) = \frac{-1}{\hbar}\int_{0}^{t} E_{k}(t’) dt'$ $\gamma_{k}(t) = i \int_{0}^{t} <\phi_{k}| \dot{\phi_{k}}> dt'$ Berry Phase $\gamma_{k}(t)$ is geometric in nature. It depends on the path through parameter space of the Hamiltonian from s(0) to s(t), but not on the time-dependence of travel along this path So you could take a nanosecond or a year to travel between two points in configuration space. $\gamma_{k}(t)$ is the same in either case! You can see this by using the chain rule to expand $ |\dot{\phi_{k}(\vec{s}(t))}>$ as $\nabla_{s} \phi \cdot \frac{\vec{s}}{dt}$. When you plug this into $\gamma$, the dt cancel, leaving you with an integral over the path in configuration space (call this $\Gamma$) In math: $\gamma_{n}(\Gamma) = \int_{\Gamma} i <\phi_{n}(\vec{s})| \nabla_{s} | \phi_{n}(\vec{s}) \cdot d \vec{s}$ $i <\phi_{n}(\vec{s})| \nabla_{s} | \phi_{n}(\vec{s}) = \vec{A_{n}}(s)$ is called the Berry connection. This is not unique. if we change our eigenstates by a arbitrary phase $e^{i\beta(\vec{s})}|\phi_{n}>$, then the Berry connection goes like $\vec{A_{n}}(s) = \vec{A_{n}}(s)+ \nabla_{s}(\beta)$ So in this configuration space, A transforms like a vector potential This also implies that $\gamma’ = \gamma+ \beta(R_{f})-\beta(R_{i})$ From this, if $\Gamma$ is closed, then this phase becomes observable (since then this choice in wavefunction phase doesn’t show up). This is called Berry’s phase Since $<\phi_{k}| \dot{\phi_{k}}> $ is imaginary, then no Berry’s phase occurs if $\phi_{k}$ is real If we are in 1D, then there is no Berry’s phase Scattering Scattering as Time-Dependent Perturbation We can imagine our base Hamiltonian as $H_{0}+V(x)$ where $H_{0} = \frac{p^{2}}{2m}$ whose eigenvalues are $E_{k} = \frac{\hbar^{2}k^{2}}{2m}$ and whose eigenvectors are plane waves We can view scattering as a perturbation which is “turned on” when the potential is large enough The transition from an initial plane-wave to a final plane wave can be written as: $<n|U_{t}|i> = \delta_{ni}-\frac{i}{\hbar}\Sigma_{m}<n|V|m> \int_{t_{0}}^{t} exp(i\omega_{nm}t’) <m|U_{t}(t’,t_{0})|i> dt'$ Since our initial and final states are asymptotic (ie. at $\pm \infty$), we need to massage the integral a bit to prevent things from exploding. We define the transition matrix T as $<n|U_{t}|i> = \delta_{ni}-\frac{i}{\hbar} T_{ni} \int_{t_{0}}^{t} exp(i\omega_{ni}t’+\epsilon t’)$ where $\epsilon >0$ and $t« \frac{1}{\epsilon}$ Prevents blow up at both extremes From the above, we can define the scattering matrix as $S = lim_{t \rightarrow \infty} \lim_{\epsilon\rightarrow 0 } <n|U(t,-\infty) | i> = \delta_{ni} - \frac{i}{\hbar} T_{ni} \infty_{-\infty}^{\infty} exp(i\omega_{ni} t’) dt’ = \delta_{ni}-2\pi i \delta(E_{n}-E_{i}) T_{ni}$ Transition Rates and Cross Sections The transition rate is defined as $w(i\rightarrow n) = \frac{d}{dt}|<n|U(t,-\infty)|\i>|^{2}$ Plugging in the above definition and ensuring $i\neq n$, you can take the appropriate limits (first $\epsilon$, and then $t$), you eventually get that $w(i\rightarrow n) = \frac{2\pi}{\hbar}|T_{ni}|^{2} \delta(E_{n}E_{i})$ Like with Fermi’s Golden Rule, we can integrate over the density of states: $\rho(E) = \frac{mk}{\hbar^{2}}(\frac{L}{2\pi})^{3} d\Omega$ which results in $w(i\rightarrow n) = \frac{mkL^{3}}{(2\pi)^{2}\hbar^{3}}|T_{ni}|^{2} d\Omega$ The cross section is then just the transition rate divided by the flux. In this case: $j = \frac{\hbar k}{m L^{3}}$ and $\frac{d\omega}{d\Omega} = (\frac{mL^{3}}{2\pi \hbar^{2}})^{2}|T_{ni}|^{2}$ General Scattering Far from the potential, we can write the solution as a superposition of wavepackets: $\phi_{g}(t) = \int d^{3}k g(\vec{k}) exp(\frac{-i \hbar t |k|^{2}}{2\mu}) \phi_{k}^{in}$ where $\phi_{k}^{in}$ is the particular solution of the eigenvalue $H \phi_{k}^{in} = \frac{\hbar^{2}k^{2}}{2\mu} \phi_{k}^{in}$ In the limit as $t\rightarrow \infty$, we have that $\phi_{k}^{in} = \Phi_{k}$, where $\Phi_{k}$ is the eigenvector of the momentum operator $P \Phi_{k} = \hbar k \Phi_{k}$ In this limit, $\Phi_{k}$ is also an eigenvector of the free Hamiltonian $H_{0}$ We want to satisfy the above limit. We can rewrite the Schrodinger equation as an integral equation: $(E|k|-H_{0})\phi_{k}^{in} = V \phi_{k}^{in}$ The solution to this is $\phi_{k}^{in} = \Phi_{k}+(E(|k|)-H_{0}+i\epsilon)^{-1} V \phi_{k}^{in}$ This is known as the Lippmann-Schwinger equation The $\epsilon$ is a math trick to let you do contour integration Solving Lippmann-Schwinger We consider the incident particle as a linear superposition of plane waves: $V \Psi_{k}^{in} = \hbar^{3} \int d^{3}q \Phi_{q}<\Phi_{q}|V\Psi_{k}^{in}>$ where $|\Phi_{q} = exp(i \vec{q}\cdot \vec{x})$ We define the Green’s function of the Lippmann-Schwinger equation as $G(\vec{x}-\vec{y}) = <\Phi_{x}| (E(|k|)-H_{0}+i\epsilon)^{-1} | \Phi_{y}>$. The full solution is then given by $\phi_{k}(\vec{x}) = (2\pi \hbar)^{-\frac{3}{2}} e^{i\vec{k}\cdot \vec{x}}+ \int d^{3} y G_{k}(x-y) V(y) \phi_{k}(y)$ $G(\vec{x}-\vec{y}) = \int \frac{d^{3}q}{(2\pi\hbar)^{\frac{3}{2}}} \frac{exp(i \vec{q}\cdot(\vec{x}-\vec{y}))}{E(k)-E(q)+i\epsilon} = \frac{-2\mu}{\hbar^{2}} \frac{e^{ik|x-y|}}{4\pi |x-y|}$ For potentials which fall off fast enough: $\phi_{k}(\vec{x}) = (2\pi \hbar)^{-\frac{3}{2}} (exp(i\vec{k}\cdot \vec{x})f_{k}(\hat{x})\frac{e^{ikr}}{r})$ $f_{k}(\hat{x})$ is the scattering amplitude, and is given by $f_{k}(\hat{x}) = \frac{-\mu}{2\pi \hbar^{2}}(2\pi\hbar)^{\frac{3}{2}} \int d^{3}y e^{-ik \hat{x}\cdot \vec{y}} V(y) \phi_{k}(y)$ We can get the differential cross section via $\frac{\sigma(\hat{x},\vec{k_{0}})}{d\Omega} = |f_{k}|^{2}$ Basic Setup for Spherical Symmetry Why do we care? If we have a complicated potential, then one way we can tease out information about it is to launch particles at it and see how they scatter off of the target Some restrictions We study processes of the form $a+b \rightarrow a+b$ We will look at elastic scattering (ie. the internal states of the particles does not change) For now, we ignore spin Non-relativistic regime (no spawning additional particles) Assume potentials only depend on relative distance between particles (so V(r)) Our Hamiltonian becomes $H = \frac{p^{2}}{2m}+V(r)$ We send in a free-particle wavefunction, with energy $E= \frac{\hbar^{2}k^{2}}{2m}$ since at very long distances, the effect of the potential vanishes For V=0, we have solutions $\phi = e^{i\vec{k}\cdot \vec{x}}$ We also have a spherically symmetric solution $\phi = f(\theta,\phi) \frac{e^{ikr}}{r}$ for the resulting outward propagation when you are very far away from the center ($\frac{1}{r}$ for conservation of probability, and $f(\theta,\phi)$ for generality) Hence, we expect the solution to this scattering problem to look like a super position between the incident and outgoing waves: $\Phi(r»a) = e^{ikz}+ f(\theta,\phi) \frac{e^{ikr}}{r}$ We align our coordinate system with the z axis $f(\theta,\phi)$ encodes the cross section of the process $\frac{d\sigma}{d\Omega}$ In words, $\sigma(\theta,\phi)$ is the number of particles scattered per unit time into the solid angle $d\Omega$ at $(\theta,\phi)$ divided by the flux of the incident particles For the above problem, we can write the incident particle fux as $\frac{k}{m} Im(\phi^{*} \nabla \phi) = \frac{\hbar k}{m} \hat{z}$ where $\phi = e^{ikz}$ A similar probability current calculation yields the outgoing flux. Can make a physical argument about what it is though Imagine a little pill box at distance r with thickness $dr$ which spans angles $\phi$ and $\theta$ The number of particles inside the box is $|\frac{f(\theta,\phi)e^{ikr}}{r}|^{2} r^{2} d\Omega dr$ (ie. multiply the probability of the outgoing wavefunction by the differential volume of the box) The particles spend $dt = \frac{dr}{v} = \frac{dr}{\frac{\hbar k}{m}}$ inside the box Taking the ratio of the two yields $\frac{dn}{dt} = \frac{\hbar k }{m}|f(\theta,\phi)|^{2} d\Omega$ Hence, we can write the differential cross section as ${d\sigma} = |f(\theta,\phi)|^{2} {d\Omega}$ Solving the Problem We don’t expect any $\phi$ dependency due to spherical symmetry around the z axis For a free particle in spherical coordinates, the radial portion takes the form $\frac{u_{l}}{r} = A_{l}\rho j_{l}(\rho)+ B_{l} \rho n_{l}(\rho)$ $\rho = kr$ j and n and the spherical bessel functions of the 1st and 2nd kind This allows us to write the incoming plane wave in terms of the spherical free particle solution: $e^{ikz} = e^{ikr\cos\theta} = \sqrt{4\pi} \Sigma_{l=0}^{\infty} \sqrt{2l+1} i^{l} Y_{l,0}(\theta) J_{l}(kr)$ We drop $n_{l}$ since those diverge as r goes to 0 For large values of r, we can expand $j_{l}(x) \approx \frac{1}{x} \sin(x-\frac{l\pi}{2})$ for $x=rk$, we have $\frac{1}{k}(\frac{e^{i(kr-\frac{l\pi}{2})}}{r}-\frac{e^{-i(kr-\frac{l\pi}{2})}}{r})$ Based on all of the above, we make the ansatz that the final wavefunction has the form $\Phi(r) = \sqrt{\frac{4\pi}{k}} \Sigma_{l=0}^{\infty} \sqrt{2l+1} i^{l} Y_{l0} \frac{1}{2i} (\frac{exp(i(kr-\frac{l\pi}{2})+2i\delta_{l})}{r}- \frac{exp(-i(kr- \frac{l\pi}{2}))}{r})$, where each outgoing l mode is shifted by some phase $\delta_{l}$ The physical argument for this is that conservation of probability forces the incoming and outgoing waves to have the same amplitudes, but doesn’t constrain the phase at all We can then write $f_{k}(\theta) = \frac{\sqrt{4\pi}}{k} \Sigma_{l=0}^{\infty} \sqrt{2l+1} Y_{l0}(\theta) e^{i\delta_{l}} \sin \delta_{l}$ Plugging the above into the cross section formula $\sigma = \int |f_{k}(\theta)|^{2} d\Omega$, which you can easily solve using the orthogonality of the spherical harmonics: $\sigma = \frac{4\pi}{k^{2}} \Sigma_{l=0}^{\infty} (2l+1) \sin^{2}(\delta_{l})$ AT very large distances, we can calculate these $\delta_{l}$ as $\tan \delta_{l} = \frac{-B_{l}}{A_{l}}$ Example: Hard Sphere We have the potential $V(r\leq a) = \infty$ and 0 else where The general solution outside the sphere is once again $\phi(r,\theta) = \Sigma_{l} (A_{l}j_{l}(kr)+ B_{l} n_{l} (kr)) P_{l}(\cos \theta)$ This wavefunction needs to vanish at $r=a$ due to the hard sphere This implies $(A_{l}j_{l}(ka)+ B_{l} n_{l} (ka)) = 0$ for all l due to the independence of the $P_{l}$ This implies $\tan \delta_{l} = \frac{j_{l}(ka)}{n_{l}(ka)}$ or alternatively $\sin^{2} \delta_{l} = \frac{j_{l}^{2}}{j_{l}^{2}+n_{l}^{2}}$ Optical Theorem This is a general statement, whose name makes more sense in an EM context For simplicity, you have a sphere, and you shine a light on it (re. EM plane waves). This creates a shadow where no light exists. Think in the particle picture: the shadow represents a region where no photons got scattered to. Hence the dearth of photons in the shadow needs to be compensated by an increase in the scattered photons in other regions Going back to quantum, what is the scattering amplitude along this forward direction? $Y_{l0}(\theta) = \sqrt{\frac{2l+1}{4\pi}} P_{l}(\cos(\theta))$. At $\theta=0$, $P_{l}(1) = 1$, hence $f_{k}(\theta=0) = \frac{1}{k}\Sigma_{l} (2l+1) e^{i\delta} \sin \delta$ If you take the imaginary part of the above, you pick up a $\sin^{2}\delta_{l}$ in the sum, which let’s you relate the cross section to the forward scattering amplitude: $\sigma = \frac{4\pi}{k} Im(f_{k}(\theta=0))$ Born-Approximation This approximation holds for “weak” potentials. More precisely, if relevant matrix elements of the potential V are much less than typical matrix elements of the kinetic energy $H_{0}$, then we can approximate the incoming particle as a free particle wave function $\phi_{in} = \frac{1}{(2\pi\hbar)^{-\frac{3}{2}}}exp(i\vec{k}\cdot \vec{x})$ This implies that $f_{k}(\hat{x}) = -\frac{\mu}{2\pi \hbar^{2}} \int d^{3}y V(y) exp(i(\vec{k}-k\hat{x})\cdot \vec{y})$ Yukawa Potential Example Assume that $V(r) = \frac{Z_{1}Z_{2}e^{2}}{r}exp(-\kappa r)$ is our potential of interest Like a shielded Coulomb potential In the Born approximation, we need to solve: $f_{k}(\theta, \phi) = -\frac{2\mu Z_{1}Z_{2}e^{2}}{q\hbar^{2}}\int_{0}^{\infty} dr exp(-\kappa r)\sin (qr)$ where $q = |\vec{k}-k\hat{x}|$ Eikonal Approximation Fundamental assumption is that V(x) varies very little over some length scale $\lambda$. V doesn’t need to be small like in Born This allows use to represent the wave function as $|\Phi> = Exp(\frac{i S(x)}{\hbar})$ where S satisfies the Hamilton Jacobi equation $\frac{(\nabla S)^{2}}{2m} + V = E = \frac{\hbar^{2}k^{2}}{2m}$ We can calculate S by assuminig that the “trajectory” of the particle is a straight line, which holds for small deflections at high energies Aligning our coordinate system with the z-direction, we have that $\frac{S}{\hbar} = \int_{\infty}^{z} (k^{2}-\frac{2m}{\hbar^{2}}V(\sqrt{b^{2}+z’^{2}})) dz’ +C$ b is the scattering parameter (ie. minimum distance from the origin to the trajectory) C is chosen such that $\frac{S}{\hbar} = kz$ as V approaches 0 so as to recover the plane wave form in the zero potential limit If $E = \frac{\hbar^{2}k^{2}}{2m}$, you can use the high energy limit to expand the square root We can then write the scattering amplitude by calculating $<k’|V|\Phi>$ ie. $f(k’,k) = -\frac{1}{4\pi} \frac{2m}{\hbar^{2}}\int d^{3}x’ exp(-i k’ x) V(\sqrt{b^{2}+z^{2}}) exp(ikx’) exp(-\frac{im}{\hbar^{2}k}\int_{-\infty}^{z’} V(\sqrt{b^{2}+z’’^{2}}) dz’’)$ This integral can be done by going to cylindrical coordinates $d^{3} x’ = b db d\phi_{b} dz'$ making the approximation $(k-k’) \cdot x’ \approx k’ \cdot b$ (small deflections) constraining the scattering to the xz plane Using the identities $\int_{0}^{2\pi} d\phi_{b} exp(-ikb\theta cos\phi_{b}) = 2\pi J_{0}(kb\theta)$ $\int_{-\infty}^{\infty} dz V exp(\frac{-im}{\hbar^{2}k}\int_{-\infty}^{z} V dz’) = \frac{-\hbar^{2}k}{m}exp(\frac{-im}{\hbar^{2}k}\int_{-\infty}^{z} V dz’)|_{-\infty}^{\infty}$ End result is: $f(k,k’) = -ik \int_{0}^{\infty} db b J_{0}(kb\theta) (exp(2i\Delta(b))-1)$ where $\Delta(b) = \frac{-m}{2k\hbar^{2}}\int_{-\infty}^{\infty} V(\sqrt{b^{2}+z^{2}}) dz$ Low-Energy Scattering and Bound States In the low energy regime, only the l=0 spherical wave matters (centrifugal potential severely dampens all of the others) Imagine very low energy ($k \approx 0$). If you are sufficiently far away, then for l=0, we have that $\frac{d^{2}u}{dr^{2}}=0$ A straight line solves this (ie $u = c(r-a)$). This can be imagined as the infinitely long wavelength limit of the standard radial function: $lim_{k\rightarrow 0} \sin (kr+\delta_{0})$ a is the scattering length. This can differ by orders of magnitude from R (ie. the range of the potential) For an attractive potential, it possible for the scattering length to be far greater than the potential’s range In this limit, the wavefunction kind of acts like a decaying exponential (ie. the wavefunction is very flat). This is the wavefunction of a bound state! We equate the logarithmic derivative (z’/z) of the standard bound state with that of this low-energy scattering state $\frac{\kappa exp(-\kappa r)}{exp(-\kappa r)} = \frac{1}{r-a}$. In the limit where R« a, we have that $\kappa = \frac{1}{a}$ The binding energy of this bound state then becomes $-E_{bound} = \frac{\hbar^{2}\kappa^{2}}{2m} \approx \frac{\hbar^{2}}{2ma^{2}}$ Resonances In some systems, certain spherical waves experience cause a peak in the cross section. These are called resonances because of the contrifugal term, potentials can look like attractive potentials followed by a repulsive barrier at larger distances. Hence, a particle can get trapped in the attractive portion (ie. a quasi-bound state) with some chance of tunnelling through the barrier Look at the scattering phase shift of this situation. $\delta_{l}$ rises through the value $\frac{\pi}{2}$ as the incident energy rises through that of the quasi-bound state ($\frac{\pi}{2}$ comes from the $\sin \delta_{l}$ in the cross section formula) Recall that $f_{l}(k) = \frac{1}{k\cot \delta_{l}-ik}$. At resonance, $\cot(\delta_{l})=0$. If we Taylor expand around $E_{R}$ (ie. the resonance energy), w have that $\cot \delta_{l} = -c(E-E_{R})$ we define $\frac{d \cot \delta_{l}}{dE} = -c = -\frac{2}{\Gamma}$ where $\Gamma$ can be though of as the FWHM of the cross section around the resonance Assuming that one spherical wave dominates the cross section, we have that $\sigma_{l} = \frac{4\pi}{k^{2}}\frac{(2l+1)(\frac{\Gamma}{2})^{2}}{(E-E_{r})^{2}+\frac{\Gamma^{2}}{4}}$ This is called the Breit-Wigner formula

Logistics Principle of “Least” Action Symmetries Gauge Invariance Example Reduce to Quadratures Hamiltonian Mechanics Phase Portraits Poisson Brackets Properties Jacobi Proof Sketch Poisson’s theorem Canonical Transformations Why We Care Liouville’s Theorem Canonical Invariants Entropy Tangent Infinitesimal CTs Hamiltonian Noether’s Theorem Hamilton Jacobi Equation Action Variables Multiperiodic motion Planetary motion Liouville’s Integrability Theorem Cannonical Peturbation Theory Adiabatic Motion Quantum Mach Zender Hilbert Space Linear Algebra Review Time Evolution Heisenberg Formulation Compatible Observables Symmetries Uncertainty Relationship Phase Space Interpretation Entanglement And Mixed States Mixed States Von Neumann Entropy Partial Trace (“Tracing Out”) Infinite Spaces Translation Operator Scattering Probability Current Delta Function Path Integral Free Particle Saddle Point Approximation Path Integral Proof WKB Logistics Matthew Klebon [email protected] Office Hours: Grading: TBD TA: [email protected]. Room 943 Principle of “Least” Action N particles in 3D There are 6N coordinates We want to find $\vec{q}(t)$ of the system where q represents the coordinates in position for each particle $q = {x_{0},x_{1},… x_{n}}$ As long as the equations of motions are 2nd order, there exists a unique solution for the system Let’s examine a simple system: the double pendulum Naively, one would expect 6 equations of motions for the 6 coordinates. However, there are 4 constrains of the system $z_{1} = z_{2} = 0$ $x_{1}^{2}+y_{1}^{2} = l_{1}^{2}$ $x_{2}^{2}+y_{2}^{2} = l_{2}^{2}$ With these constraints, you can reduce the system to a function of two coordinates: $\theta_{1}$ and $\theta_{2}$ holonomic: when the constraints of a system are solely of a position $f_{k}(q_{1}…q_{n}) = 0$ Not all constraints are holonomic: Rolling without slipping, inequalities etc. To incorporate these constraints, you can either use Lagrange multipliers, or use the constraints to eliminate degrees of freedom Assuming no dissipative forces, you can describe the system by a Lagrangian $\mathcal{L(q, \dot{q}, t)}$ This is typically $\mathcal{L} = T-V$ In polar coordinates, we have that $\dot{r}^{2}+(r\dot{\theta})^{2}$ In spherical coordinates, we have that $\dot{r}^{2}+(r\dot{\theta}\sin \phi)^{2} + (r\dot{\phi})^{2}$ From this Lagrangian, you can define something called the action of the path: $S(q_{i}(t)) = \int_{t_{1}}^{t_{2}} \mathcal{L}(q(t), \dot{q(t)}, t) dt$ This is a functional (ie. it’s arguments are other functions) Units of Plank’s constant and angular momentum You want to find the stationary points of the action function as a function of time. Pretend that the endpoints of the path are fixed at two points in time: $q(t_{1}) == q(t_{2})$ This implies that $\delta q_{i}(t_{1}) = \delta q_{i}(t_{2}) = 0$ Taking the total derivative of the action, use integration by parts, utilize the boundary conditions and rearrange to get $\delta S = \int_{t_{1}}^{t_{2}} dt (\frac{\partial L}{\partial q_{1}}-\frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{q_{i}}})\delta q_{i}) = 0$ This must hold for all possible $q_{i}$, so if we imagine placing delta functions everywhere, this implies that the bracketed term must be 0 these are the Euler-Lagrange equations: $\frac{\partial L}{\partial q_{1}}-\frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{q_{i}}}) = 0$ Sometimes, it’s easier to plug in $q+\delta q$ directly into the action and reduce up to first order terms You can add a total derivative $\frac{d}{dt}f(q,t)$ (note lack of dependence on $\dot{q}$) to a Lagrangian and leave the equations of motion unchanged Simply do variational principle with additional term The converse is not necessarily true The principle of least action is invariant under coordinate changes Symmetries When there is a symmetry, there is a conserved quantity (Noether’s Theorem) For instance, if the Lagrangian is time independent (ie. $L(q,\dot{q},t) = L(q,\dot{q})$), then via the Euler Lagrange equations, energy conservation follows Taking the total derivative of the Lagrangian w.r.t. time, applying the time-independent constraint, and then rearranging yields that that the following is conserved. This is the energy (usually)! $\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}\dot{q}-L) = 0$ A trivial symmetry is when $\frac{\partial L}{\partial q} = 0$. This is called a cyclic coordinate Another example is $L(q_{i}+\delta q_{i}) = L(q_{i})$. If q is a position, then the associated momentum is also conserved Yet another is conservation of angular momentum: $L = \frac{1}{2} m \dot{r}^{2}- V(|\vec{r}|)$ $\frac{\partial \vec{r}}{\partial \theta_{i}} = \hat{n_{i} \times \vec{r}} = \epsilon_{jkl} n_{k} r_{l}$ Can reduce this to $P_{\theta_{i}} = \epsilon_{ilj} r_{l} (m\dot{r}_{j})$, which is just $\vec{r}\times \vec{p}$ This can be done by utilizing the fact that $\hat{n_{ik}} = \delta_{ik}$ that the Levi-Civita symbol is invariant under cyclic permuations of it’s indicies the definition that $p_{\theta} = \frac{\partial L}{\partial \dot{r_{j}}}\frac{\partial \dot{r_{j}}}{\partial \dot{\theta_{i}}}$ More formally, a transformation $q_{i} => f(\epsilon, \bar{q}); f_{i}(0) = q_{i}$ (where $\epsilon$ is the implicit variable and $\bar{q}$ represents all of the coordinates) is a symmetry of the system if the Lagrangian is unchanged (ie. $\delta L = \frac{d}{dt}(\frac{\partial L}{\partial q_{i}} \delta q_{i}) = 0$) If the symmetry holds, then the conserved quantity obeys the following equation $\frac{\partial L}{\partial q_{i}}\delta q_{i} = \epsilon \frac{\partial}{\partial q_{i}}\frac{\partial f_{i}}{\partial \epsilon}|_{\epsilon = 0}$ Gauge Invariance We know that $L’ = L + \frac{d\Phi}{dt}$ holds in general for some scalar field (gauge invariance) This modifies Noether’s therom: $\frac{d}{dt}(\frac{\partial L}{\partial q_{i}}\dot{q_{i}}-\Phi) = 0$ Hence $\frac{\partial L}{\partial q_{i}}\dot{q_{i}}-\Phi$ is conserved Example Let’s couple a charge particle with some fields: $L = \frac{1}{2} m\vec{\dot{x}^{2}}-qV+\frac{q}{c}\vec{A}\cdot \vec{x}$ There is a gauge invariance from the transformation $\vec{A} => \vec{A}+c\nabla \Lambda$ and $V => V - \frac{\partial}{\partial t} \Lambda$ Plugging in this transformation yields that $L’ = L + \frac{\partial}{\partial t}(q\Lambda)+\frac{\partial}{\partial x_{i}}(q\Lambda) \vec{\dot{x_{i}}} = L + \frac{d}{dt} \Phi$ Reduce to Quadratures Instead of solving a PDE to find equations of motion, you can solve a different PDE in terms of conserved quantities For instance: $L = \frac{1}{2}m\dot{x}^{2} -V(x)$ has $h = \frac{1}{2}m\dot{x}^{2} +V(x)$ as a conserved quantity You could solve $m\ddot{x} = 9\frac{\partial V}{\partial x}$ You could also solve $(\frac{dx}{dt})^{2} = \frac{2}{m} (h-V(x))$ Hamiltonian Mechanics The Legendre transform of the Lagrangian is the Hamiltonian Graphically, you can take the slope of the initial function, and project that onto the y-axis. In symbols $v \frac{\partial L}{\partial v}-L = H(p)$ where v is the x axis coordinate and p is the y axis coordinate This assumes that $\frac{dL}{dv} = p(v)$ We assume that p is invertible (ie. p(v) implies v(p)) $L’’(v) \neq 0$ This one allows H to be a well defined function With the above in mind, we define the Hamiltonian as $H(q_{i},p_{i},t) = \Sigma_{i=1}^{3} p_{i}\dot{q_{i}}-L(q_{i},\dot{q_{i}},t)$ $p_{i} = \frac{\partial L}{\partial \dot{q_{i}}}$ Taking the total derivative of the Hamiltonian, applying the equation $p_{i} =\frac{\partial L}{\partial \dot{q_{i}}}$ and applying the EL equations yields $H = \dot{q}dp_{i}-\dot{p_{i}}dq_{i}-\frac{\partial L}{\partial t} dt$ $\frac{\partial H}{\partial q_{i}} = -\dot{p_{i}}$ $\frac{\partial H}{\partial p_{i}} = \dot{q_{i}}$ $\frac{\partial H}{\partial t} = \frac{\partial L}{\partial t}$ You solve double the number of equations, but they are 1st order now Phase Portraits Can plot momentum and position of a particle. Closed loops are called librations Unbounded paths are called rotations The boundary between the two is called the separatrix Poisson Brackets Suppose that you have some observable $f(q_{i}, p_{i},t)$ Utilizing the total time derivative and Hamilton’s equations, you show $\frac{df}{dt} = [f,H]+ \frac{\partial f}{\partial t}$ $[A,B] = \frac{\partial A}{\partial q_{i}}\frac{\partial B}{\partial p_{i}}- \frac{\partial A}{\partial p_{i}}\frac{\partial B}{\partial q_{i}}$ This is the Poisson bracket Properties $[\alpha f_{1} + \beta f_{2}, g] = [\alpha f_{1}, g] + [ \beta f_{2}, g]$ $[f,g] = -[g,f]$ $[f,[g,h]]+[h,[f,g]]+[g,[h,f]] = 0$ (Jacobi identity) $[f,gh]= [f,g]h + g[f,h]$ (Leibniz’s rule) Any object which satisfies the above properties is called a Lie Algebra. ...

Compilation of notes for Quantum I class for Spring 2023. The Basics The Current Density Operator (probability current) Erenhfest’s Principle Uncertainty Princple Time Independent SE Interference of Stationary States Solving TISE $V(x)=V_{0}$ Infinite Square Well Properties of Eigenfunctions Finite Square Well Dirac Delta Function Free Particle Fourier Transform Definition Scattering States SHO Commutation relation of $\hat{x}$ and $\hat{p}$ Ladder Operators Hermetian Operators Momentum Eigenvalues Fourier transforms Generalized Uncertainty Relationship Dirac Notation Solution to Spherically Symmetric Schrodinger’s Equation Angular Momentum Spherical Coordinates Spin Electron in Magnetic Field Addition of Angular Momenta The Basics The Wavefunction Written in 1 spatial dimension as $\Psi (x,t)$. Can be complex Will suppress the inputs from now on, unless needed It’s evolution is given by the Time Dependent Schrodinger equation $i \hbar \frac{\partial \Psi}{\partial t} = \frac{-\hbar^{2}}{2m}\frac{\partial^{2} \Psi}{\partial^{2} x}+V\Psi$ Born’s Statistical Interpretation: The probability of finding the particle between a and b is given by $\frac{\int_{a}^{b} |\Psi|^{2}dx}{\int_{-\infty}^{\infty} |\Psi|^{2}dx}$ Denominator is typically 1 because functions we care about are typically normalized Classically: Assume there is some $P(x,t)$ that gives the probability of a particle appearing at x at time t $ <x(t)> = \int_{\infty}^{\infty} xP(x,t)dx$ $\int_{-\infty}^{\infty} P(x,t) dx = 1$ Suppose we have some operator Q. The expectation value: $ <Q(t)> = \int_{\infty}^{\infty} QP(x,t)dx$ Quantum: We are given a wave function $\Psi(x,t)$. You then get: $ <x(t)> = \int_{\infty}^{\infty} x|\Psi|^{2}dx$ $\int_{-\infty}^{\infty} |\Psi|^{2} dx = 1$ Suppose we have some operator Q. The expectation value: $ <Q(t)> = \int_{\infty}^{\infty} \Psi^{*}Q\Psi dx$ Velocity/momentum $<x(t)> = \int_{\infty}^{\infty}x|\Psi|^{2}dx$ $v = \frac{d}{dt}<x(t)> = \int_{\infty}^{\infty}x \frac{\partial}{\partial t}|\Psi|^{2}dx$ $\frac{\partial}{\partial t}|\Psi|^{2} = \Psi^{*}\frac{\partial \Psi}{\partial t}+\Psi\frac{\partial \Psi^{*}}{\partial t}$ From Schrodinger’s equation, can replace: $\frac{\partial \Psi}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^{2}}{\partial x^{2}}\Psi-\frac{i}{\hbar}V\Psi$ Taking the conjugate of Schrodinger’s equation (assuming V is real): $\frac{\partial \Psi^{*}}{\partial t} = \frac{i\hbar}{2m}\frac{\partial^{2}}{\partial x^{2}}\Psi^{*}-\frac{i}{\hbar}V\Psi^{*}$ Subbing into above yields: $\frac{\partial}{\partial t}|\Psi|^{2} = i\frac{\hbar}{2m}(\Psi^{*}\frac{\partial^{2} \Psi}{\partial x^{2}} - \Psi\frac{\partial^{2}}{\partial x^{2}}\Psi_{*}) = \frac{\partial}{\partial x}(\frac{i\hbar}{2m}(\Psi^{*}\frac{\partial \Psi}{\partial x}-\Psi\frac{\partial \Psi^{*}}{\partial x}))$ Hence: $\frac{d}{dt}<x(t)> = \int x\frac{\partial}{\partial x} (\frac{i\hbar}{2m}(\Psi^{*} \frac{\partial \Psi}{\partial x} - \Psi \frac{\partial \Psi^{*}}{\partial x}))$ Integrate by parts , and utilize the boundary conditions that $|\Psi(x)|^{2} \rightarrow 0$ and $|\Psi^{*}(x)|^{2} \rightarrow 0$ at $\pm \infty$ $v = \frac{d}{dt}<x> = \int \Psi^{*} (\frac{-i\hbar}{m} \frac{\partial}{\partial x}) \Psi$ We call operator $\hat{p} = -i\hbar \frac{\partial}{\partial x}$ The Current Density Operator (probability current) probability current J is defined as $\hat{J}(x,t) = \frac{i\hbar}{2m}(\Psi\frac{\partial \Psi^{*}}{\partial x}-\Psi^{*}\frac{\partial \Psi}{\partial x})$ We can think of this as the “flow” of probability across a boundary Can be derived by finding $\frac{\partial}{\partial t}|\Psi|^{2}$ rewrite this time derivative using Schrodinger’s equation Use conservation of probability: $\frac{\partial}{\partial t}|\Psi|^{2} = -\frac{\partial J}{\partial x}$ Rationale: Imagine a wavefunction as a plane wave: $\Psi(x,t)=Ae^{i(kx-\omega t)}$ $P=|\Psi|^{2}=|A|^{2}$ $J = \frac{\hbar k}{m} |A|^{2}$ $\frac{\hbar k}{m}$ has units of speed, and $|A|^{2}$ is a probability density, so J can be interpreted as a probability flux Erenhfest’s Principle P1: $<v> = \frac{d}{dt}<x>$ Namely, that quantum mechanic predictions approach the classical predictions P2: The equivalent of Newton’s Law in Quantum Mechanics $\frac{d}{dt}<p > = -<\frac{\partial V}{\partial x}>$ Write out the formula for the expectation value of $p = -i\hbar\frac{\partial}{\partial x}$ Expand out derivatives and utilize equivalence of mixed partials to swap $\frac{\partial^{2}}{\partial t \partial x}$ to $\frac{\partial^{2}}{\partial x \partial t}$ Use Schrodinger’s equation to substitute first order time derivatives with second order spatial derivatives (Don’t forgot to conjugate for $\Psi^{*}$ term!) Use integration by parts twice in a row to eliminate $\frac{\partial^{3}}{\partial x^{3}}$ term (remember the boundary conditions $\Psi(\pm \infty) \rightarrow 0$ and $\Psi^{*}(\pm \infty) \rightarrow 0$) Expand all the partials and things should cancel to P2 of Erenhfest’s Theorem Uncertainty Princple $\sigma_{i} = \sqrt{<i^{2}>-<i>^2}$ for a coordinate and its associated momentum: $\sigma_{x}\sigma_{p_{x}} \geq \frac{\hbar}{2}$ For a coordinate with another coordinate: $\sigma_{x}\sigma_{y} \geq 0$ For a momentum with a momentum: $\sigma_{p_{x}}\sigma_{p_{y}} \geq 0$ For a coordinate with a momentum that isn’t it’s own: $\sigma_{x}\sigma_{p_y} \geq 0$ Time Independent SE $\Psi(x,t) = \psi(x)\phi(t)$ is only possible iff $V(x,t) = V(x)$ Making this assumption, use seperation of variables, take the partials. You will notice that the LHS only depends on time and the RHS only depends on space. So each side must equal some constant (call this constant E) The time equation yields $\phi(t) = \phi(0)exp(\frac{-i Et}{\hbar})$ $\omega=\frac{E}{\hbar}$ as shorthand $\phi(0)$ is a constant at $t=0$ The space equation (called the Time Dependent Schrodinger Equation) is: $\frac{-\hbar}{2m}\frac{d^{2}\psi}{dx^{2}}+V(x)\psi(x) = E\psi(x)$ Alternatively: $\hat{H}\psi = E\psi$ This is an eigenvalue problem in an infinite dimensional space So $\Psi(x,t) = \psi(x)exp(\frac{-iEt}{\hbar})$ Observe that $|\Psi(x,t)|^{2}$ is independent of time the expectation value of any time-independent operator is time-independent Importantly, the expectation value of the energy is independent of time and has a definite value ($\sigma_{E} = 0$). Corresponds to the Hamiltonian $\hat{H} = \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)$ Also, all other expectation values of operators are time independent Interference of Stationary States Suppose that we are dealing with the time-independent Schrodinger Equation. The general time evolution of the wavefunction in a stationary potential is: $\Psi(x,t)=\Sigma_{n=1}^{\infty}c_{n}\psi_{n}(x)exp(\frac{-iE_{n}t}{\hbar})$ $\Sigma|c_{n}|^{2}=1$ If we want to Fourier decompose a function $f(x)$ in terms of the eigenfunctions $\Psi_{n}$, we can write the $c_n$ as $c_{n} = \int \Psi_{n}(x)^{*} f(x) dx$ Suppose that you have the state: $\Psi(x,t)=c_{1}\psi_{1} exp(\frac{-iE_{1}t}{\hbar})+c_{2}\psi_{2} exp(\frac{-iE_{2}t}{\hbar})$ This implies (assuming that the stationary states are real eigenfunctions and you have real coefficients): $|\Psi(x,t)|^{2}=c_{1}^{2}\psi_{1}^{2}+c_{2}^{2}\psi_{2}^{2}+2c_{1}c_{2}\cos(\frac{(E_{1}-E_{2})t}{\hbar})$ Looking at the expectation value of $\hat{Q}$, we get $<Q> = |c_{1}|^{2}Q_{11}+|c_{2}|^{2}Q_{22}+ {c_{1}}^{*}c_{2}Q_{12} + {c_{2}}^{*}c_{1}Q_{21}$ $Q_{ij} =\int_{\infty}^{\infty}\psi_{i}^{*}\hat{Q}\psi_{2} dx$ called matrix elements Solving TISE Remember the boundary conditions: $\Psi(\pm\infty)\rightarrow 0$ At each interface: The wavefunction is continuous ($\psi_{l}=\psi_{r}$) The derivative of the wavefunction is continuous (usually, there are some exceptions like the delta function) ($\frac{d\psi_{l}}{dx}=\frac{d\psi_{r}}{dx}$) You can also use symmetry arguments if appropriate $V(x)=V_{0}$ $E>V_{0}$ $\frac{\partial^{2}\psi}{\partial x^{2}} = \frac{-2m}{\hbar}(E-V_{0})\psi(v)$ Let $k^{2}=\frac{2m}{\hbar}(E-V_{0})$ $\psi(x)=Ae^{ikx}+Be^{-ikx}=C\cos(kx)+D\sin(kx)=A\sin(kx+\delta)$ $E<V_{0}$ Let $k^{2}=\frac{2m}{\hbar}(V_{0}-E)$ $\frac{\partial^{2}\psi}{\partial x^{2}} = k^{2}\psi(v)$ $\psi(x)=Ae^{kx}+Be^{-kx}=C\cosh(kx)+D\sinh(kx)$ Classically forbidden Infinite Square Well The well has a width of $a$ The walls of infinite height occur at $x=0$ and $x=a$ Energies: $E_{n}=\frac{\hbar^{2}n^{2}\pi^{2}}{2ma^{2}}$ Wavefunctions: $\psi_{n}=\sqrt{\frac{2}{a}}\sin(\frac{n\pi x}{a})$ Properties of Eigenfunctions Symmetric potentials have symmetric eigenfunctions All eigenfunctions are muthually orthogonal $\int \psi_{m}^{*} \psi_{n} = \delta_{mn}$ For the infinite square well, useful orthogonality relations include: $\int_{0}^{a} sin(\frac{n\pi x}{a})sin(\frac{m\pi x}{a}) = \frac{a}{2}\delta_{mn}$ $\int_{0}^{a} cos(\frac{n\pi x}{a})cos(\frac{m\pi x}{a}) = \frac{a}{2}\delta_{mn}$ $\int_{0}^{a} cos(\frac{n\pi x}{a})sin(\frac{m\pi x}{a}) = 0$ Any other function can be represented as a sum of the eigenfunctions (complete set of state) This means that you can write the time evolution of a wavefunction $f(x)$ as $\Psi(x,t) = \Sigma_{n=1}^{\infty}c_{n}\psi_{n}(x)exp(\frac{-iE_{n}t}{\hbar})$ The coefficients can be found via $c_{n} = \int \psi^{*}_{n}(x)f(x)dx$ Probability must be conserved though ($\Sigma_{n}c_{n}^{2}=1$) Finite Square Well $V(x) = -V_0$ between -a to a, and equals $0$ otherwise $\psi(x) = Be^{kx}$ for $x \leq -a$, $\psi(x) = C\sin(lx)+D\cos(lx)$ for $-a \leq x \leq a$, and $\psi(x) = Fe^{-kx}$ Match the functions and derivatives at the boundary to yields $k = l\tan(la)$ $k=\sqrt{\frac{2m(E)}{\hbar^{2}}}$ $l = \sqrt{\frac{2m(E-V_{0})}{\hbar^{2}}}$ Solve for the value of E that solves this equation (transcendental) To do this, the above can be rewritten in terms of $z=la$ and $z_0 = \frac{a}{\hbar}\sqrt{2mV_0}$, and then solve for z $tan(z) = \sqrt{(\frac{z_0}{z})^2-1}$ For a wide and deep well ($z_0$ is big), we get an infinite square well with only the odd states and offset by $V_0$ For a shallow and narrow well, we approach the delta well You can find the transmission coefficient $T = \frac{|F|^2}{|A|^2}$ and see that periodically, the well becomes transparent ($T=1$) Dirac Delta Function $V(x) = -\alpha\delta(x)$ if $x=0 \rightarrow \delta = \infty$ if $x\neq 0 \rightarrow \delta = 0$ $\int_{-\infty}^{\infty}\delta(x)dx = 1$ $\int_{0}^{-\infty}f(x)\delta(x-a)dx = f(a)$ On the LHS and RHS, you just get growing exponentials on the left and decaying exponentials on the right. Call the parameter $\kappa = \frac{\sqrt{2mE}}{\hbar}$ From the continuity of the wavefunction, we know the coefficients of the LHS and RHS match (call this coefficient B) We can’t assume that the derivative is continuous. Taking a step back, let’s look at the TISE: $\frac{-\hbar}{2m}\frac{d^{2}\psi}{dx^{2}}+V(x)\psi(x) = E\psi(x)$ Integrate w.r.t. x from $-\epsilon$ to $\epsilon$ (ie. in a small region around zero) to get $\frac{-\hbar}{2m}(\frac{d\Psi_{L}}{dx}\rvert_{-\epsilon}-\frac{d\Psi_{R}}{dx}\rvert_{\epsilon})+\int_{-\epsilon}^{\epsilon}V(x)\psi(x) = \int_{-\epsilon}^{\epsilon}E\psi(x)$ $\Psi_{L}$ and $\Psi_{R}$ is the wavefunction on the left and the right Let $\epsilon$ go to zero. The RHS goes to zero, and if V(0) is finite then the 2nd term on the LHS is zero. Hence $\Psi_{L}=\Psi_{R}$ If V(0) is infinite, then you need to take into account the value of the 2nd LHS term. This also means that $\Psi_{L}$ and $\Psi_{R}$ are NOT equal to each other. For the Dirac delta, we know what $\Psi_{L}$ and $\Psi_{R}$ and we know $\int_{-\epsilon}^{\epsilon}-\alpha\delta(x)\psi(x) = -\alpha B$ Doing the algebra yields $\kappa = \frac{m\alpha}{\hbar^{2}}\rightarrow E = \frac{-m^{2}\alpha^{2}}{2\hbar^{2}}$ So we only have a single bound state with energy $E = \frac{-m\alpha^{2}}{2\hbar^{2}}$ with eigenfunction $\Psi_{L}=\frac{\sqrt{m\alpha}}{\hbar}e^{-\frac{m\alpha|x|}{\hbar^2}}$ The scattering states have any energy $E>0$ Both sides of the equation have sinusoidal solutions. Matching continuity of function, taking into account discontinuity of the first derivative, and assuming only incidence of particles from the left ($G=0$), we find that (with $\beta=\frac{m\alpha}{\hbar^2k}$) $B = \frac{i\beta}{1-i\beta}A$ $F = \frac{1}{1-i\beta}A$ Transmission and reflection coefficients can be found from F and B respectively Free Particle $\frac{-\hbar^{2}}{2m}\frac{d\psi^{2}}{dx^{2}} = E\psi$ Solutions are $\Psi = Ae^{ikx}+Be^{-ikx}$ where $k = \frac{\sqrt{2mE}}{\hbar}$ We can add on the energy dependent phase to see that the two states represent right and left moving waves respectively Not normalizable We can take sums of various k values to construct normalizable functions Define a right moving wave: $\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{=\infty}^{\infty} \phi(k)exp(i(kx-\frac{\hbar^{2}k^{2}}{2m}t))$ $\Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{=\infty}^{\infty} \phi(k)exp(ikx)$ Fourier Transform Definition Define the Fourier Transform as: $f(x) = \frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}F(k)exp(ikx)$ $F(k) = \frac{1}{\sqrt{2\pi}}\int_{\infty}^{\infty}f(x)exp(-ikx)$ If we make the correspondence: $f(x)\rightarrow \Psi(x,0)$ and $F(k) \rightarrow \phi(k)$, then we can calculate the Fourier components of any moving wave packets (ie. decompose any wave packet in terms of complex exponentials of various magnitude) We also have the problem that the speed of the packet is twice the classical speed. This is because of the distinction between the group velocity ($v_{g} = \frac{d\omega}{dk}$) and the phase velocity ($v_{p} = \frac{\omega}{k}$) Scattering States Flux of a particle is $k|C|^{2}$, or in words, it is the momentum of the particle times the probability density Define a potential such that $V(x) = V_{1}$ for $x<0$,$V(x) = 0$ for $0<x<a$,$V(x) = V_{2}$ for $x>a$ For scattering problems, $R = \frac{k|B|^{2}}{k|A|^{2}}$ and $T = \frac{k^{’}|F|^{2}}{k|A|^{2}}$ To remember which k is assigned to each probability, look at the region where the wave exists in and choose this momentum For a symmetric potential well, at widths of integer multiples of half wavelength, you get perfect transmittance SHO Define $[A,B] = AB-BA$ to be the commutator where A and B are operators The SHO (simple harmonic oscillator) has the potential of the form $V(x) = \frac{(m\omega x)^{2}}{2m}$ $\omega^2 = \frac{k}{m}$ The central idea of solving the harmonic oscillator via ladder operators is to rewrite the Hamiltonian $H = \frac{1}{2m}[p^{2}+(m\omega x)^{2}]$ in terms of ladder operators Commutation relation of $\hat{x}$ and $\hat{p}$ $[\hat{x},\hat{p}] = i\hbar$ You utilize this ad nauseum for Angular momentum operators and, in this case, ladder operators Ladder Operators Let $a^{+} = \frac{1}{\sqrt{2\hbar\omega m}}(-i\hat{p}+m\omega\hat{x})$ be the creation/raising operator Let $a = \frac{1}{\sqrt{2\hbar\omega m}}(i\hat{p}+m\omega\hat{x})$ be the annihilation/lowering operator Alternatively, this can go the other way around where you rewrite $\hat{x}$ and $\hat{p}$ in terms of the raising and lowering operators $\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}(a_{+}+a)$ $\hat{p} = i\sqrt{\frac{\hbar m \omega}{2}}(a_{+}-a)$ H can be rewritten as $H = \hbar\omega aa^{+}-\frac{\hbar\omega}{2} = \hbar\omega a^{+}a+\frac{\hbar\omega}{2} $ $[a,a^{+}] = 1$ Applying the Hamiltonian to $a^{+}\psi$ and utilizing the commutators shows that $H(a^{+}\psi) = (E+\hbar\omega)a^{+}\psi$ Analogously, we get that $H(a\psi) = (E-\hbar\omega)a\psi$ There exists a ground state to the system since $V(x) = \frac{1}{2}m\omega^{2}x^{2} \geq 0$. In words, you can’t lower the energy to a negative value Let $a\psi_{0} = 0$, where $\psi_{0}$ refer to the ground state. Applying a to this eigenstate kills the wavefunction Expand out what a is and solve the dif eq to get (after normalization) $\psi_{0} = (\frac{m\omega}{\pi\hbar})^{\frac{1}{4}}exp(\frac{-m\omega x^{2}}{2\hbar})$ To get the ground state energy, apply H to $\psi_{0}$ to get $E_{0} = \frac{\hbar\omega}{2}$ We can recursively apply $a^{+}$ to $\psi_{0}$ to generate the higher quantum level eigenfuctions and eigenvalues: $\Psi_{n} = \frac{1}{\sqrt{n!}}(a_{+})^{n} \Psi_0$ Alternatively, the general eigenfunctions are $\psi_n(x) = (\frac{m\omega}{\pi\hbar})^\frac{1}{4}\frac{1}{\sqrt{2^n n!}}H_n(\xi)e^{\frac{-\xi^2}{2}}$ where $\xi = \sqrt{\frac{m\omega}{\hbar}}x$ where $H_n$ are the Hermite polynomials Hermetian Operators $<Q> = \int \phi^{*} \hat{Q} \phi dx = <\phi|Q\phi>$ We define Hermetian operators such that $<f|\hat{Q}f> = <\hat{Q}f|f>$ All observables can be represented by hermetian operators. The expectation value of hermetian operators is real The hermetian conjugate (or the adjoint) of an operator is denoted $Q^{\dagger}$ such that $<f|\hat{Q}g> = <\hat{Q^{\dagger}}f|g>$ for all f and g An operator is hermetian if $Q = Q^{\dagger}$ The eigenvalues and vectors of a hermetian Q satisfy $Q\psi = q\psi$ All eigenvalues of hermetian operators are real All eigenfunctions of distinct eigenvalues of hermetian operators are orthogonal Momentum Eigenvalues Let $f_{p} = \frac{1}{\sqrt{2\phi \hbar}} e^{\frac{ipx}{\hbar}}$ be the eigenvector of the equation $-i\hbar \frac{d}{dx} f_p = p f_p$ Satisfies the orthogonality relation $<f_{p’}|f_{p}> = \delta(p’-p)$ Any function can then be written as $\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} c(p) exp(\frac{ipx}{\hbar})$ where $c(p)$ is given by the Fourier trick $c(p) = <f_{p’}|f>$ Fourier transforms $\phi(p,t) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} exp(\frac{-ipx}{\hbar}) \psi(x,t) dx$ $\psi(x,t) = \frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^{\infty} exp(\frac{ipx}{\hbar}) \psi(p,t) dx$ $phi(p,t)$ is the wave function in the position basis, and holds equivalent information to the wavefunction in position basis $x = i\hbar \frac{\partial}{\partial p}$ is position basis Generalized Uncertainty Relationship Let A and be B be Hermitian operators. Then $\sigma_A^{2} \sigma_B^{2} \geq |\frac{<[A,B]>}{2i}|^{2}$ This is derived using the definition of the variance $\sigma_A^2 = <A-<A>\Psi^{*}|A-<A>|\Psi >$ You then use the Schwartz Inequality $f^2g^2 \geq |<f|g>|^2$ where f and g are complex numbers, as well as the inequality that $|z|^2 \geq Im(z)^2$ where z is complex To find a minimum uncertainty, we want both of the inequalities to the equalities. Doing this, we find that $g(x) = ia f(x)$ where $a$ is real In particular, if we let $g(x) = p-<p>|\Psi>$ and $f(x) = x-<x>|\Psi>$, we get a differential equation of the form $\frac{d\Psi}{\Psi} = \frac{-ax}{\hbar}++\frac{a<x>}{\hbar}+\frac{i}{\hbar} <p> dx$, where the expectation values are constants. Solving this, we get that $\Psi(x) = Aexp(\frac{-ax^2}{2\hbar}+\frac{a<x>x}{\hbar}+\frac{i}{\hbar}<p>x) = A’ exp(\frac{-a(x-<x>)^{2}}{2\hbar})exp(\frac{i<p>x}{\hbar})$ This is a Gaussian Another important relation is $\Delta E\Delta t \geq \frac{\hbar}{2}$ Look at $\frac{d}{dt}<Q> = <\frac{\partial \Phi}{\partial t}|Q\Phi>+<\Phi|\frac{dQ}{dt}\Phi>+<\Phi|Q\frac{\partial \Phi}{\partial t}>$ From Schrodinger’s equation, we know that $\frac{\partial \Phi}{\partial t} = \frac{-i}{\hbar}H\Phi$. Plug this in above to get that $\frac{d}{dt}<Q> = \frac{i}{\hbar}<[H,Q]>+<\frac{\partial Q}{\partial t}>$ Combine the above with the Generalized Uncertainty Principle where A=H and B = Q, and define ($\frac{\sigma_Q}{\hbar<Q>} = \Delta t$), we can prove the energy time uncertainty Dirac Notation Have been using it throughout these notes The only weird thing is the projection operator $\hat{P} = |a><a|$ This operator picks out the portion of any other vector that lies along a $P|b> = <a|b>|a>$ $\Sigma_{n}|e_{n}><e_{n}| = 1$ is the identity operator. Can be inserted into any dirac notation expression with whatever basis you want Need to remember that $<p|x> = \frac{1}{\sqrt{2\pi\hbar}}exp(\frac{-ipx}{\hbar})$ Solution to Spherically Symmetric Schrodinger’s Equation The angular solutions take the form of spherical harmonics: $Y_{l}^{m}(\theta,\phi) = \sqrt{\frac{(2l+1)(l-m)!}{4\pi (l+m)!}}e^{im\phi}P^m_l(\cos\theta)$ $P_{l}(\cos\theta)$ are the associated Legendre polynomials These are orthonormal functions The radial equation becomes $\frac{-\hbar^2}{2m}\frac{d^2 u}{dr^2}+(V+\frac{\hbar^2}{2m}\frac{l(l+1)}{r^2})u = Eu$ $u(r) = rR(r)$ For a general hydronic atom $E_{n} = -(\frac{\mu}{2\hbar^2}(\frac{(Ze^2)}{4\pi \epsilon_{0}})^2)\frac{1}{n^2}$ We also define the Bohr Radius as $a = \frac{4\pi \epsilon_{0}\hbar^2}{\mu Ze^2}$ The final normalized wavefunctions are $\Phi_{nlm} = \sqrt{(\frac{2}{na})^3\frac{(n-l-1)!}{2n(n+l)!}}e^{-\frac{r}{na}}(\frac{2r}{na})^{l}L^{2l+1}_{n-l-1}(\frac{2r}{na})Y_l^m(\theta,\phi)$ $L_q(x) = \frac{e^x}{q!}(\frac{d}{dx})^q(e^{-x}x^q)$ Angular Momentum NOTE: A general rule of commutators: $[AB,C] =A[B,C]+[A,C]B$ $[L_x,L_y] = i\hbar L_z$ Cyclic permutations hold (x->y->z) Anticommutative $L^{2}$ commutes with $L_{x}$,$L_{y}$ and $L_{z}$ This means that $L^2$ can have simultaneous eigenstates with a direction. Let’s take $L_{z}$ to be said direction $L^2f = \lambda f$ and $L_{z}f = \mu f$ Define the ladder operators $L_{\pm} = L_{x}\pm L_{y}$ $[L_{z},L_{\pm}] =\pm \hbar L_{\pm}$ $[L^2,L_{\pm}] =0$ This means that $L_{\pm}f$ is also an eigenfunction of $L_{z}$ with a new eigenvalue $\mu \pm \hbar$. Let $L_{z}f = \hbar l f$ where l is an integer $L^{2} = L_{\pm}L_{\mp}+L_z^2\mp \hbar L_{z}$ $L_{\pm}|lm> = \hbar \sqrt{l(l+1)-m(m\pm 1)}|s(m\pm 1)>$ $L^{2}|lm> = \hbar l(l+1)|lm>$ $L_{z}|lm> = m\hbar |lm>$ $l$ can take either half or integer values Spherical Coordinates $L_{x} = -i\hbar (-\sin\phi \frac{\partial}{\partial \theta}-\cos\phi\cot\theta \frac{\partial}{\partial \phi})$ $L_{y} = -i\hbar(\cos\phi\frac{\partial}{\partial \theta}-\sin\phi\cot\theta\frac{\partial}{\partial \phi})$ $L_{z} = -i\hbar \frac{\partial}{\partial \phi}$ $L^{2} = -\hbar^2[\frac{1}{\sin\theta}\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial}{\partial \theta})+\frac{1}{\sin^{2}\theta}\frac{\partial^2}{\partial \phi^2}]$ The only reason you would ever use these forms is to calculate the explicit eigenfunctions of these operators Spin Has the same commutator relationships as angular momentum In matrix form, for spin 1/2, we have $S^{2} = \frac{3}{4} \hbar^2 \begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}$ $S_{+} = \hbar \begin{pmatrix} 0& 1\\0&0 \end{pmatrix}$ $S_{-} = \hbar \begin{pmatrix} 0& 0\\1&0 \end{pmatrix}$ $S = \frac{\hbar}{2} \sigma$ $\sigma_{x} = \begin{pmatrix} 0 & 1\\1&0 \end{pmatrix}$ $\sigma_{y} = \begin{pmatrix} 0 & -i\\i&0 \end{pmatrix}$ $\sigma_{z} = \begin{pmatrix} 1 & 0\\0&-1 \end{pmatrix}$ Electron in Magnetic Field The magnetic moment is defined as $\mu = \gamma S$ where $\gamma$ is the gyromagnetic ratio The Hamiltonian of the system becomes $-\mu\cdot B = -\gamma B\cdot S$ Suppose that we have a B field configured like $B = B_{0} \hat{k}$. For an electron, we have 2 possible energies whose eigenfunctions are the same as $S_{z}$ Writing down the time evolution of the system and taking the expectation value of $S_{x}$, $S_{y}$ and $S_{z}$, we find that the spin vector makes an angle of $\alpha$ to the z axis and precesses around the z-axis with frequency $\omega = \gamma B$ Addition of Angular Momenta Suppose that you have two particles in states $|s_{1}m_{1}>$ and $|s_{2}m_{2}>$ where the notation $|sm>$ denotes a particle of spin s with quantum humber m Denote the composite states as $|s_{1}s_{2}m_{1}m_{2}>$, or alternatively as $|s_{1}m_{1}>|s_{2}m_{2}>$ The new z quantum number is just $m = m_{1}+m_{2}$ The new total spin is kind of indeterminate. By this, I mean that the composite particle has the potential to have a variety of total spins (ie. a superposition of total spins) For instance, combining two spin 1/2 particles could yield either a spin 0 particle or a spin 1 particle In general, given a spin $s_{1}$ and a spin $s_{2}$ particle, the possible total spins range from $s_{1}+s_{2}$ all the way down to $|s_{2}-s_{1}|$ Let’s look at combining two spin 1/2 particles. Start with the composite state $|\frac{1}{2}\frac{1}{2} m_{1}m_{2}>$. You can repeatedly apply the normalized raising and lowering operators $S^{\pm}$ on the state with the highest possible spin (ie. 1) Once You exhaust all the possible spin 1, move onto spin 0 You get the following for spin 1 (triplet): $|11> = |\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{1}{2}>$ $|10> = \frac{1}{\sqrt{2}}(|\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{-1}{2}>+|\frac{1}{2}\frac{1}{2}\frac{-1}{2}\frac{1}{2}>)$ $|1-1> = |\frac{1}{2}\frac{1}{2}\frac{-1}{2}\frac{-1}{2}>$ And the following form spin 0 (singlet): $|00> = \frac{1}{\sqrt{2}}(|\frac{1}{2}\frac{1}{2}\frac{1}{2}\frac{-1}{2}>-|\frac{1}{2}\frac{1}{2}\frac{-1}{2}\frac{1}{2}>)$ In general, you can state that $|sm> \Sigma_{m_1+m_2=m} C_{m_1 m_2 m}^{s_1 s_2 s} |s_1 s_2 m_1 m_2>$ And conversely $|s_{1}s_{2}m_{1}m_{2}> = \Sigma_{s} C_{m_1 m_2 m}^{s_1 s_2 s} |sm>$ for ($m = m_1 +m_2$) The constants are called the Clebsch-Gordan coefficients and they are a fancy look up table NOTE: Take the square root of the absolute value in the cell, then apply the negative sign as needed