$(v,w) \geq 0$, 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. $H_{prod} = \Sigma_{ia} \Phi_{ia} e_{i} \otimes e_{a}$)
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
The set of all rotation matrices for a group (ie. they satisfy group properties)
Identity: $\delta_{ij}$
The determinant can be $\pm 1$ (plus is called SO(3), minus is called O(3))
In the complex plane, we replace Orthogonal, real matrices with unitary matrices
For an infinitesimal complex rotation, we can represent a rotation as $U = \approx 1+\frac{i \omega_{ij}}{2}J_{ij}$, where $\omega$ is an anti-symmetric matrix (taken from rotation in real plane) and J is the generator (using Einstein notation for summations)
We need this to remain unitary. Using equation $U^{\dag}U=1$ yields that $J_{ij}$ must also be unitary (ie. a unitary peturbation)
This also implies that vectors transform like $(1-\frac{\omega_{ij}J_{ij}}{2}) v_{k} (1+\frac{\omega_{ij}J_{ij}}{2}) = v_{k}+\omega_{ij} v_{j}$
We know that $[J_{i}, J_{j}] = i\epsilon_{ijk} J_{3}$ (fundamental algebra of angular momentum)
We also know that we can write $J = L+S$, where L is the angular momentum of the particle, and S is the spin
Spin commutes with position and momentum, while angular momentum doesn’t
We know that $[J^{2}, J_{i}]=0$ for i=0,1,2. What are the eigenvalues of J?
We can define raising and lowering operators $J_{\pm} = J_{1}+iJ_{2}$. We can use these to generate a ladder of eigenstates for $J_{3}$
For the z component
$J_{3} \psi_{m} = m\psi_{m}$
Act raising and lowering operators on both sides, use fundamental commutator to see that $J_{3}(J_{+} \psi_{m}) = m (J_{+} \psi_{m})$ (same for minus)
We can rewrite $J^{2} = J_{-}J_{+}+J_{3}+J_{3}^{2}$ (use the fact that $J^{2} = J_{1}^{2}+J_{2}^{2}+J_{3}^{2}$, plus the commutator relationship). acting this onto $\psi$, which is an eigenstate of $J_{z}$ and $J^{2}$ yields $J^{2}\psi = (j+j^{2}) \psi$
Acting $J_{\pm}$ on $\psi_{j}$ (eigenstate of $J_{3}$) yields $(j\pm 1) \psi_{j\pm 1}$
Similar argument shows that there is a lower bound of j’=-j (can represent $J^{2}$ by swapping plus and minus around, and futzing with the sign of $J_{3}$)
What values can j take?
We know $J_{3}(J_{+}^{N})\psi_{-j} = (-j+N) \psi_{-j} = -j$, which implies that j must be an integer or a half integer
What is the normalization of $J_{\pm}$?
$J_{+}\psi_{m} = \alpha(j,m) \psi_{m+1}$
Take the norm of this (remembering that $J_{+}^{\dag} = J_{-}$), you can show that $J_{\pm} \psi_{m} = \sqrt{j(j+1)-m^{2}\mp m} \psi_{m \pm 1}$
In words, we can represent a state of total angular momentum j and total z-spin m as a linear combination of states of definite z-spin m for each hilbert space
We have two selection rules on this system:
$C_{j_{1}j_{2}}(jm;m_{1},m_{2}) = 0$ if $m_{1}+m_{2}\neq m$
$C_{j_{1}j_{2}}(jm;m_{1},m_{2}) = 0$ if $|j_{1}-j_{2}| \leq j \leq j_{1}+j_{2}$
You could in theory construct these coefficients via recursion
note that $\psi_{jj}^{j_{1}j_{2}} = \psi_{j_{1}j_{2}}^{j_{1}j_{2}}$ for $m=j$ and $m_{1}=j_{1}$ and $m_{2}=j_{2}$ for physical reasons (it’s the only possible way of doing this addition)
Imagine that you lower the z angular momentum by 1. The only way this couple happen is if either particle 1 reduces m by 1, or particle 2 reduces m by 1. You can then generate these states and their normalizations via lowering operators.
This procedure then recurses: you reason about what states you could generate by changing z angular momentum by 1, then generating their normalizations
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
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
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
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