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 cannonical transformations are an important part of classical Hamiltonian mechanics. What is the analog in quantum?
    • We want the probability to be conserved under a cannonical 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 $(\phi_{j’’}^{m’’}O_{j}^{m}\psi_{j’}^{m’}) = C(j’’m’’;j’m’) (\phi || P|| \psi)$
    • 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

  • 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
    • $c_{n}(t) \approx c_n{0}-\frac{i}{\hbar}\Sigma_{m} c_{m}(0) \int_{0}^{t} dt’H_{nm}’(t’) exp(i\frac{(E_{n}-E_{m})t’}{\hbar})$
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

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-relativistical 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:
    • $\sigma = \frac{4\pi}{k} Im(f_{k}(\theta=0))$

More General Scattering

  • Suppose our Hamiltonian now looks like $H_{0}+V(\vec{x})$
  • 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 $\frac{\sigma(\hat{x},\vec{k_{0}})}{d\Omega} = |f_{k}|^{2}$

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 $q = |\vec{k}-k\hat{x}|$

The S-Matrix

  • Suppose that we have some Hamiltonian $H = H_{0}+V$ where V becomes negligible as the particles move away from each other. At $t = \pm \infty$, we can treat the Hamiltonian as free:
    • $H_{0} \Phi_{\alpha} = E_{\alpha} \Phi_{\alpha}$ where $\alpha$