quantum(1) - hachem.wtf

name

quantum mechanics - angular momentum, perturbation theory, and operator identities

synopsis

quantum [--angular-momentum] [--pauli-rotations] [--harmonic-oscillator] [--perturbation-theory] [--operator-identities] [--density-matrix]

description

the results i double-check instead of trusting my memory, like exact coefficients, sign conventions, and the multi-term identities. $\hbar$ is kept explicit.

angular momentum

the ladder commutators are $[J_z, J_\pm] = \pm\hbar J_\pm$ and $[J_+, J_-] = 2\hbar J_z$
the exact action, which is the bit people misremember, is $J_\pm|j,m\rangle = \hbar\sqrt{j(j+1) - m(m\pm1)}\,|j, m\pm1\rangle$
addition goes $j_1\otimes j_2 = (j_1+j_2)\oplus\cdots\oplus|j_1-j_2|$ , mixed with the clebsch–gordan coefficients $\langle j_1m_1\,j_2m_2|JM\rangle$ , which are nonzero only if $M = m_1+m_2$

pauli & rotations

the product identity is $(\mathbf{a}\cdot\boldsymbol{\sigma})(\mathbf{b}\cdot\boldsymbol{\sigma}) = (\mathbf{a}\cdot\mathbf{b})\,I + i\,(\mathbf{a}\times\mathbf{b})\cdot\boldsymbol{\sigma}$
the exponential, for a unit $\hat{\mathbf{n}}$ , is $e^{i\theta\,\hat{\mathbf{n}}\cdot\boldsymbol{\sigma}} = \cos\theta\,I + i\sin\theta\,(\hat{\mathbf{n}}\cdot\boldsymbol{\sigma})$
the rotation operator is $R(\theta,\hat{\mathbf{n}}) = e^{-i\theta\,\hat{\mathbf{n}}\cdot\mathbf{J}/\hbar}$ , and for spin-½ that's $e^{-i(\theta/2)\,\hat{\mathbf{n}}\cdot\boldsymbol{\sigma}} = \cos(\theta/2)\,I - i\sin(\theta/2)\,(\hat{\mathbf{n}}\cdot\boldsymbol{\sigma})$

harmonic oscillator

the actions are $a|n\rangle = \sqrt{n}\,|n-1\rangle$ , $a^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle$ , and $N|n\rangle = n|n\rangle$
the operators are $x = \sqrt{\hbar/2m\omega}\,(a + a^\dagger)$ and $p = i\sqrt{\hbar m\omega/2}\,(a^\dagger - a)$
a coherent state has $a|\alpha\rangle = \alpha|\alpha\rangle$ and $|\alpha\rangle = e^{-|\alpha|^2/2}\sum_n\frac{\alpha^n}{\sqrt{n!}}|n\rangle$ . it's minimum-uncertainty and not orthogonal

perturbation theory

nondegenerate, with $H = H_0 + \lambda V$ .

the energy is

E_n = E_n^{(0)} + \lambda\langle n|V|n\rangle + \lambda^2\sum_{k\neq n}\frac{|\langle k|V|n\rangle|^2}{E_n^{(0)} - E_k^{(0)}} + \cdots

the state is

|n\rangle = |n^{(0)}\rangle + \lambda\sum_{k\neq n}\frac{\langle k|V|n\rangle}{E_n^{(0)} - E_k^{(0)}}\,|k^{(0)}\rangle + \cdots

in the degenerate case, diagonalize $V$ inside the degenerate subspace, and its eigenvalues are the first-order splittings
fermi's golden rule (time-dependent) is $\Gamma_{i\to f} = \frac{2\pi}{\hbar}|\langle f|V|i\rangle|^2\rho(E_f)$

operator identities

the hadamard lemma gives $e^A B e^{-A} = B + [A,B] + \frac{1}{2!}[A,[A,B]] + \frac{1}{3!}[A,[A,[A,B]]] + \cdots$
bch gives $e^A e^B = \exp\big(A + B + \tfrac12[A,B] + \tfrac{1}{12}([A,[A,B]] - [B,[A,B]]) + \cdots\big)$
glauber's formula, when $[A,B]$ commutes with both $A$ and $B$ , is $e^A e^B = e^{A+B}\,e^{\frac12[A,B]}$

density matrix

a state is pure when $\rho^2 = \rho \iff \operatorname{tr}\rho^2 = 1$ , and mixed when $\operatorname{tr}\rho^2 < 1$
the von neumann entropy is $S = -\operatorname{tr}(\rho\ln\rho) = -\sum_i\lambda_i\ln\lambda_i$ , where $\lambda_i$ are the eigenvalues

author

hachem <im@hachem.wtf>