# QM Course Notes

Here is a conglomerate of notes gathered for the graduate QM classes 221A and 221B.

## 221A Notes

Fundamentals

Unitary transformations:

Unitary transformations are really Fourier Transforms, at least for transforms between momentum and position.

$UU^\dagger = U^{\dagger} U = 1$ $U^\dagger = U^{-1}$ $U |n\rangle = e^{i \phi}|n\rangle \text{ , i.e. eigenvalues are phases.}$

Projectors:

$P^2 = P$

Show that $T(a) = e^{-\frac{iapx}{\hbar}}$ is unitary and has eigenvalue of unity up to a phase:

State two ways in which a wavefunction can evolve (i.e. change) in time according to the Copenhagen interpretation:

Without measurement wavefunction evolves as $|\psi(t)\rangle = e^{-\frac{iHt}{\hbar}} | \psi(t=0)\rangle$. With measurement, wavefunction evolves by reduction to a value according to $|\psi\rangle = P(a) |\psi \rangle$.

Standard Boundary Conditions for $\psi$:

1. $\psi$ is continuous over a boundary

2. $\frac{\partial \psi}{\partial x}$ is continuous over a boundary, except for delta function boundaries where this term goes to infinity; for delta boundaries split up $\psi(x)$ and take moment of SE: $\lim_{\epsilon \to 0} \int_{a-\epsilon}^{a+\epsilon} \left( \frac{\partial^2 \psi(x) }{\partial x^2} - \frac{2m}{\hbar} \delta (x-a) \psi(x) \right) dx = \lim_{\epsilon \to 0} \int_{a-\epsilon}^{a+\epsilon} \frac{-2mE}{\hbar} \psi(x) dx \approx 0 \Rightarrow \\ \Rightarrow \frac{\partial \psi_{I}}{\partial x} \mid_{x=a} -\frac{\partial \psi_{II}}{\partial x} \mid_{x=a} = \frac{2m}{\hbar} \psi(a)$

Schwartz Inequality:

$|\langle x,y\rangle| \leq \|x\| \cdot \|y\|.\,$

Hilbert space:
General HUP:

$\sigma_{A}=\sqrt{\langle{A}^{2} \rangle-\langle{A}\rangle ^{2}}$ $\sigma_{B}=\sqrt{\langle{B}^{2} \rangle-\langle{B}\rangle ^{2}}$ $\sigma_{A}\sigma_{B} \geq \left| \frac{1}{2i}\langle[A,B]\rangle \right| = \frac{1}{2}\left|\langle[A,B]\rangle \right|.$

Harmonic Oscillators

Commutation relations, Hamiltonian, G.S. SHO wavefunction:

$[a, a^\dagger] = 1 \\ [a^\dagger, a] =- 1$ $a^\dagger a |n\rangle = N |n \rangle = n |n\rangle,\text{ where N is the number operator } N = a^\dagger a$ $H = \hbar \omega ( a^\dagger a + \frac{1}{2}) = \hbar \omega ( N + \frac{1}{2})$ $\text{Using the identity }[A, BC] = B [A, C] + [A, B] C \text{ we find: } \\ [N, a] = -a \\ [N, a^\dagger] = a^\dagger$

$|n\rangle = \frac{1}{\sqrt{n!}} (a^\dagger)^n | 0 \rangle$ $|0\rangle = \left ( \frac{m \omega}{\pi \hbar}\right )^{\frac{1}{4}} e^{-\frac{m \omega}{2 \hbar} x^2}$ Note: The G.S. wavefunction is unique due to the lower bound $a |0\rangle = 0$.

Pictures

Schrödinger Picture vs. Heisenberg picture:

Schrödinger picture: - time dependent state, constant operator $i \hbar | \dot{\psi} \rangle = H \psi\rangle$ - [x, H] = 0, in principle

Heisenberg picture: - time-dependent operators, constant states - yields Heisenberg equation of motion: $\frac{d x}{d t} = \frac{[x, H]}{i \hbar}$

Heisenberg picture:
Interaction picture:

Charges in Magnetic Fields

Origin of Canonical Momentum:

$p = \left( p - \frac{e}{c} A \right), \text{ where p inside the brackets is the quantum mechanical momentum } p = -i \hbar \nabla \text{ and A is the gauge invariant vector potential.}$ Canonical momentum is not the momentum fired out of a cannon, but rather the momentum that satisfies $p = \frac{\partial \mathcal{L}}{ \partial \dot{q}}$, where $\mathcal{L}$ is the Lagrangian of the system. $H=\frac{1}{2m} \left(p-\frac{qA}{c}\right)^2 +q\phi$

Angular Momentum

Momentum Commutation Relations:

$[L_i, L_j] = i \hbar \epsilon_{ijk} L_k$

How does one find the matrices $J_i$ of a single particle state with angular momentum j:

Use raising and lowering operators $J_{\pm}= J_x \pm i J_y$ $J_x = \frac{J_{+} + J_{-}}{2}$$J_y = \frac{J_{+} - J_{-}}{2i}$

Express $J^2$ in terms of $J_z, J_{\pm}$:

$J^2 = J_z^2 + J_+J_- -\hbar J_z$, where $J_+ J_- = J_x^2 + J_y^2 - i [J_x, J_y] = J_x^2 + J_y^2 + \hbar J_z$

### Exercises

How do you find the simultaneous eigenvectors of two commuting matrices, A and B?

Non-degenerate: easy. Degenerate: Find relationship between eigenvector coordinates of A and then plug this into eigenvector equation for B. [1]

## 221B Notes

Perturbation Theory

Wigner-Eckart Theorem (W.E.T.) selection rules:

$\langle n'j'm' | T_q^k | njm\rangle \ne 0$ if $m' = q+m \text{ and }|j-k| \le j' \le j+k$

Use W.E.T. to find $$\langle njm|x|njm\rangle$$ for Hydrogen:

From wikipedia: "This matrix element is the expectation value of a Cartesian operator in a spherically-symmetric hydrogen-atom-eigenstate basis, which is a nontrivial problem. However, using the Wigner–Eckart theorem simplifies the problem. (In fact, we could obtain the solution quickly using parity, although a slightly longer route will be taken.)

We know that x is one component of r, which is a vector. Vectors are rank-1 tensors, so x is some linear combination of T1q for q = -1, 0, 1. In fact,

$x=\frac{T_{-1}^{1}-T^1_1}{\sqrt{2}}\,,$

where we defined the spherical tensors T10 = z and $T^1_{\pm1}=\mp \frac{x \pm i y}{\sqrt{2}}$ (the pre-factors have to be chosen according to the definition of a spherical tensor of rank k. Hence, the T1q are only proportional to the ladder operators). Therefore $\langle njm|x|n'j'm'\rangle = \langle njm|\frac{T_{-1}^{1}-T^1_1}{\sqrt{2}}|n'j'm'\rangle = \frac{1}{\sqrt{2}}\langle nj||T^1||n'j'\rangle (C^{jm}_{1(-1)j'm'}-C^{jm}_{11j'm'})$ The above expression gives us the matrix element for x in the $$|njm\rangle$$ basis. To find the expectation value, we set n′ = n, j′ = j, and m′ = m. The selection rule for m′ and m is $$m\pm1=m'$$ for the $$T_{\mp1}^{(1)}$$ spherical tensors. As we have m′ = m, this makes the Clebsch-Gordan Coefficients zero, leading to the expectation value to be equal to zero."

Non-degenerate t-indep. PT, first order energy and state:

$E_n^{(1)} = \langle n|H_1|n\rangle$$|n\rangle^{(1)} = \sum_{m \ne n} \frac{\langle m|H_1|n\rangle}{E_n-E_m} |m\rangle$

Non-deg. t-indep. PT, second order energy:

$E_n^{(2)} = \sum_{n \ne m} \frac{|\langle m|H_1|n\rangle|^2}{E_n -E_m}$

Degenerate t-indep. PT:

Diagonalize matrix.

Time-dep. PT:

If the unperturbed system is in eigenstate $$|j\rangle$$ at time $$t = 0\,$$, its state at subsequent times varies only by a phase (we are following the Schrödinger picture, where state vectors evolve in time and operators are constant):

$|j(t)\rangle = e^{-iE_j t /\hbar} |j\rangle$

We now introduce a time-dependent perturbing Hamiltonian $$V(t)\,$$. The Hamiltonian of the perturbed system is

$H = H_0 + V(t) \,$

Let $$|\psi(t)\rangle$$ denote the quantum state of the perturbed system at time t. It obeys the time-dependent Schrödinger equation,

$H |\psi(t)\rangle = i\hbar \frac{\partial}{\partial t} |\psi(t)\rangle$

The quantum state at each instant can be expressed as a linear combination of the eigenbasis $${|n\rangle}$$. We can write the linear combination as

$|\psi(t)\rangle = \sum_n c_n(t) e^{- i E_n t / \hbar} |n\rangle$

where the $$c_{n}(t)\,$$s are undetermined complex functions of t which we will refer to as amplitudes (strictly speaking, they are the amplitudes in the Dirac picture). We have explicitly extracted the exponential phase factors $$\exp(- i E_n t / \hbar)$$ on the right hand side. This is only a matter of convention, and may be done without loss of generality. The reason we go to this trouble is that when the system starts in the state $$|j\rangle$$ and no perturbation is present, the amplitudes have the convenient property that, for all t, cj(t) = 1 and $$c_n (t) = 0\,$$ if $$n\ne j$$.

$c_n^{(1)}(t) = \frac{-i}{\hbar} \sum_k \int_0^t dt' \; \langle {n}|V(t')|k\rangle \, c_k(0) \, e^{-i(E_k - E_n)t'/\hbar}\\ \text{where the former is used when } V= H_1 = \text{ constant; } H = H_{total}$

Fermi's Golden Rule

In quantum physics, Fermi's golden rule is a way to calculate the transition rate (probability of transition per unit time) from one energy eigenstate of a quantum system into a continuum of energy eigenstates, due to a Perturbation theory perturbation.

We consider the system to begin in an eigenstate, $$\scriptstyle | i\rangle$$, of a given Hamiltonian, $$\scriptstyle H_0$$. We consider the effect of a (possibly time-dependent) perturbing Hamiltonian, $$\scriptstyle H'$$. If $$\scriptstyle H'$$ is time-independent, the system goes only into those states in the continuum that have the same energy as the initial state. If $$\scriptstyle H'$$ is oscillating as a function of time with an angular frequency $$\scriptstyle \omega$$, the transition is into states with energies that differ by $$\scriptstyle \hbar\omega$$ from the energy of the initial state. In both cases, the one-to-many transition probability per unit of time from the state $$\scriptstyle| i \rangle$$ to a set of final states $$\scriptstyle| f\rangle$$ is given, to first order in the perturbation, by $T_{i \rightarrow f}= \frac{2 \pi} {\hbar} \left | \langle f|H'|i \rangle \right |^{2} \rho,$ where $$\scriptstyle \rho$$ is the density of final states (number of states per unit of energy) and $$\scriptstyle \langle f|H'|i \rangle$$ is the matrix element (in bra-ket notation) of the perturbation $$\scriptstyle H'$$ between the final and initial states.This transition probability is also called decay probability and is related to mean lifetime.

Fermi's golden rule is valid when the initial state has not been significantly depleted by scattering into the final states.

Scattering

General Wavefunction for Scattering Problem:

$\psi(\mathbf{r}) = A[e^{ikz} + f(\theta)\frac{e^{ikr}}{r}] \;,$

Born Approximation $f(\theta)$:
 $f(\theta)= -\frac{m}{2\pi \hbar^2} \int_V e^{i\mathbf{(\mathbf{k}'-\mathbf{k}).\mathbf{r}}} V(\mathbf{r}) d^3\mathbf{r}$


, where k' is the incident beam direction.

Born Approximation $f(\theta)$ for central potential:
 $f(\theta)= -\frac{2m}{\hbar^2} \int_0^\infty V(r) \frac{sin(qr)}{q} r dr$


, where k' is the incident beam direction.

Unified coordinate wavefunction for scattering:
 $\psi(r, \theta) = A \sum_{\ell=0}^{\infty} i^\ell (2l+1)[j_\ell(kr) + ik a_\ell h_{\ell}^{(1)}(kr)]P_{\ell}(cos\theta) \\ \text{ This wavefunction is zero at the boundary of a hard sphere r=a, allowing mulitplication by a Legendre Polynomial}\\ \text{ and consequent calculation of partial amplitude as follows:}\\ a_\ell = \frac{-j_\ell(k a)} {ik h_{\ell}^{(1)}(ka)}$

Scattering Amplitude $f(\theta)$ in partial wave expansion:

In the partial wave expansion the scattering amplitude is represented as a sum over the partial waves, $f(\theta)=\sum_{\ell=0}^\infty (2\ell+1) a_\ell(k) P_\ell(\cos(\theta)) \;,$ where $$a_\ell(k)$$ is the partial amplitude and $$P_\ell(\cos(\theta))$$ is the Legendre polynomial.

Partial amplitude $a_\ell(k)$

$a_\ell = \frac{e^{2i\delta_\ell}-1}{2ik} = \frac{e^{i\delta_\ell} \sin\delta_\ell}{k} = \frac{1}{k\cot\delta_\ell-ik} \;.$

Bessel Functions $j_0$ and $j_1$:

$j_0(x) = \frac{sin(x)}{x}$ $j_1(x) = \frac{sin(x)}{x^2} - \frac{cos(x)}{x}$

General form of Bessel function and Neumann function and derivatives thereof:

$j_l(x) = (-x)^l \left(\frac{1}{x}\frac{d}{dx}\right)^l\,\frac{\sin(x)}{x} ,$ $n_l(x) = -(-x)^l \left(\frac{1}{x}\frac{d}{dx}\right)^l\,\frac{\cos(x)}{x}.$

$\left( \frac{1}{x} \frac{d}{dx} \right)^m \left[ x^\ell J_{\ell} (x) \right] = x^{\ell - m} J_{\ell - m} (x)$ $\left( \frac{1}{x} \frac{d}{dx} \right)^m \left[ \frac{J_\ell (x)}{x^\ell} \right] = (-1)^m \frac{J_{\ell + m} (x)}{x^{\ell + m}}.$ where J also denotes Y, H(1), or H(2).

Hence, $\frac{d J_{\ell}}{dx} = \frac{\ell J_{\ell}}{x} - J_{\ell+1}$

Exercises

Find $E_n^{(1, 2)}$ for $V= cx$ in SHO using P.T. and compare to the exact solution:

$E_n^{(1)} = c \langle n^{(0)} | a + a^{\dagger} | n^{(0)} \rangle = 0$ $E_n^{(1)} = - \frac{c^2}{2m \omega^2}$ $\text{Lol and behold, this is exact:}$ $H= \frac{p^2}{2m} + 1/2 m \omega^2 x^2 + cx = p^2/2m + 1/2 m \omega^2 (x + c/m \omega^2 )^2 - c^2/2m \omega^2$

Stark Effect, i.e. polarization of H-atom by E-field:

$H_1 = -qz E$

## General Identities

The inner product of a vector with itself is the square of its norm (magnitude): $\langle A | A \rangle = |A_x|^2 + |A_y|^2 + |A_z|^2 = || A ||^2$

BCH for non-cummuting [A,B], but [[A,B], A]= 0 and [[A, B], B] =0 : $e^A e^B = e^B e^A e^{[A,B]}$ $e^{A+B} = e^A e^B e^{-\frac{1}{2}[A,B]}$

$(AB)^\dagger= B^\dagger A^\dagger$

$[J_z, J_{\pm}]$:

$[J_z, J_{\pm}]] = \pm \hbar J_{\pm}$

$[J_+, J_-]$:

$[J_+, J_-] = 2 \hbar J_z$

$[V_i, J_j ]$:

$[V_i, J_j] = i \hbar \epsilon_{ijk} V_k$

$[J^{\pm}, V^0]$:

$[J^{\pm}, V^0] = \mp \hbar V^{\pm}$

$[J^{\pm}, V^{\pm}]$:

$[J^{\pm}, V^{\pm}] = 0$

$[J^{\pm}, V^{\mp}]$:

$[J^{\pm}, V^{\mp}] = \pm 2 \hbar V^0$

## Gallery

Lecture Notes

Quantum Mechanics by Prof. Eric D'Hoker[2].

Quotes

"The student who has read the book but cannot do the problems has learnt nothing." - J. J. Sakurai

Figures

## References

1. M.L. Boas, p.158.
2. http://www.pa.ucla.edu/content/eric-dhoker-lecture-notes, retrieved 17th April 2014.