QM1 midterm, worked solutions

My worked solutions to the four problems on the first-year graduate quantum midterm at Columbia, fall 2021.

Exponential operators

Assume $A$, $B$ and $C$ are square matrices of the same rank and obey the commutation relations $[A,B] = C$ and $[C,A] = [C,B] = 0$. Given $e^{A+B} = e^A e^X e^Y$, find $X$ and $Y$.

Solution. Recall the Baker-Campbell-Hausdorff relation,

$$e^A e^B = e^{A+B+\frac{1}{2}[A,B]+\cdots}$$

where the higher terms are nested commutators. In the given question, $A$ and $B$ both commute with the commutator $[A,B] = C$, so the BCH relation reduces to

$$e^A e^B = e^{A+B+\frac{1}{2}C}$$

Now let $A+B = \Delta$ and $\frac{1}{2}C = \Gamma$, so the exponent is $\Delta + \Gamma$ with $[\Delta,\Gamma] = 0$. Applying BCH again in reverse, $e^{\Delta+\Gamma} = e^\Delta e^\Gamma$, so

$$e^A e^B = e^{A+B}\,e^{\frac{1}{2}C} \implies e^A e^B e^{-\frac{1}{2}C} = e^{A+B}$$

and we read off $X = B$, $Y = -\frac{1}{2}C$.

Orthonormality

Given three degenerate eigenfunctions $|u_1\rangle, |u_2\rangle, |u_3\rangle$ that are linearly independent but not necessarily orthogonal, find three linear combinations that are orthogonal to one another and normalized. Are the new combinations eigenfunctions, and if yes, are they still degenerate?

Solution. By the Gram-Schmidt process we construct three new orthogonal vectors. Writing $\langle u_i|u_i\rangle = \|u_i\|^2$ and $\langle u_i|u_j\rangle = S_{ij}$ for $i \neq j$,

$$\begin{gathered} |b_1\rangle = |u_1\rangle \\ |b_2\rangle = |u_2\rangle - \frac{S_{12}}{\|u_1\|^2}|u_1\rangle \\ |b_3\rangle = |u_3\rangle - \frac{S_{13}}{\|u_1\|^2}|u_1\rangle - \frac{\langle b_2|u_3\rangle}{\|b_2\|^2}|b_2\rangle \end{gathered}$$

and each is normalized by dividing by its norm, $|b_i'\rangle = |b_i\rangle / \|b_i\|$. The new combinations are still eigenfunctions, since any linear combination of degenerate eigenfunctions is an eigenfunction with the same eigenvalue, and for that reason they are still degenerate.

The limit as ℏ goes to zero

Given operators $A(\hat{x},\hat{p})$ and $B(\hat{x},\hat{p})$ expressible as power series in $x$ and $p$, evaluate and simplify

$$\lim_{\hbar\to 0}\frac{1}{i\hbar}[A,B]$$

Solution. Suppose our operators are of the form $A = \sum_{mn} a_{mn}\,x^m p^n$ and $B = \sum_{\alpha\beta} a_{\alpha\beta}\,x^\alpha p^\beta$. For a commutator of the general form $[AB, CD]$ we can expand

$$[AB,CD] = A[B,C]D + C[A,D]B + [A,C]BD + CA[B,D]$$

which reduces the problem to the elementary commutator $[x^m, p^\beta]$. Peeling off one power of $x$ at a time,

$$[x^m, p^\beta] = \beta\, i\hbar \sum_{k=0}^{m-1} x^k p^{\beta-1} x^{m-k-1}$$

After dividing by $i\hbar$, every term left inside the double sum still contains reordered products of $x$ and $p$. Reordering any of them costs a further commutator proportional to $\hbar$, and since we eventually take $\hbar \to 0$, those corrections vanish. In the limit we are free to move $x$ and $p$ past each other as if they commute, and the $k$-sums simply count $m$ and $n$ factors. Recognizing

$$\sum_{mn} a_{mn}\, m\, x^{m-1}p^n = \frac{\partial A}{\partial x}$$

we obtain

$$\lim_{\hbar\to 0}\frac{1}{i\hbar}[A,B] = \frac{\partial A}{\partial x}\frac{\partial B}{\partial p} - \frac{\partial B}{\partial x}\frac{\partial A}{\partial p}$$

which is the classical Poisson bracket.

Density matrices

Suppose we have a beam of photons whose polarization state is unknown to us. It might be pure or mixed, and we would like to perform measurements that determine the state.

(a) For a pure state, prepare two sets of the same beam. Let the beam direction be the $z$ axis and pass the copies through two separate polarizers oriented along $x$ and $y$. The ratio of the two measured intensities, if both are nonzero, yields the tangent of the polarization angle with respect to the $x$ axis. If the $x$ polarizer yields zero intensity the light is $y$ polarized and vice versa, and if the two intensities are equal the light is polarized at $\pi/4$ from the $x$ axis. The minimum number of measurements is thus 2. If the beam is in a mixed state, the “mixed nature” cannot be determined experimentally, as elaborated in (b).

(b) An unpolarized beam is a mixed state with equal probabilities on $|\psi_i\rangle\langle\psi_i|$. In the Cartesian basis the density matrix is

$$M = \frac{1}{2}\begin{pmatrix}1&0\\0&0\end{pmatrix} + \frac{1}{2}\begin{pmatrix}0&0\\0&1\end{pmatrix} = \frac{1}{2}\begin{pmatrix}1&0\\0&1\end{pmatrix}$$

There is, however, no measurement that guarantees the light is unpolarized. For a pure state $c_1|\psi_1\rangle + c_2|\psi_2\rangle$ the interference terms mixing $c_1$ and $c_2$ average out over the total phase, so if $|c_1|^2 = |c_2|^2 = 1/2$ the measured distribution is identical to that of unpolarized light, and the two cannot be distinguished.

(c) For a mixed state of 50% light polarized along $\hat{x}$ and 50% right circularly polarized, take $|x\rangle = (1\ 0)^\dagger$, $|y\rangle = (0\ 1)^\dagger$, and $|R\rangle = \frac{1}{\sqrt{2}}(1\ {-i})^\dagger$. The density matrix is

$$M = \frac{1}{2}|x\rangle\langle x| + \frac{1}{2}|R\rangle\langle R| = \frac{1}{4}\begin{pmatrix}3 & -i\\ i & 1\end{pmatrix}$$

To find two orthogonal states giving the same density matrix, diagonalize $M$. The eigenvectors

$$\underline{v}_1 = \begin{pmatrix}-i(1+\sqrt{2})\\ 1\end{pmatrix}, \qquad \underline{v}_2 = \begin{pmatrix}i(-1+\sqrt{2})\\ 1\end{pmatrix}$$

are orthogonal (their inner product vanishes). Building $M' = f_1\,\underline{v}_1\underline{v}_1^\dagger + f_2\,\underline{v}_2\underline{v}_2^\dagger$ and matching the element $M_{22} = 1/4$ admits $f_1 = f_2 = 1/8$, which indeed recovers $M$. Since probabilities must sum to 1, we instead take $f_1 = f_2 = 1/2$ and rescale the vectors by $1/2$, so that

$$\underline{v}_1 = \frac{1}{2}\begin{pmatrix}-i(1+\sqrt{2})\\ 1\end{pmatrix}, \qquad \underline{v}_2 = \frac{1}{2}\begin{pmatrix}i(-1+\sqrt{2})\\ 1\end{pmatrix}$$

are two orthogonal states that give the same density matrix, each with probability $1/2$.