CSE 599Q: Homework #5

Quantum probability and information

Due: Sun, Dec 11 @ 11:59pm
Gradescope Link

1. Positive semidefinite matrices [12 points]

    A Hermitian matrix \(M \in \C^{d \times d}\) is said to be positive semidefinite, denoted \(M \succeq 0\) if it holds that \(\bra{u}M\ket{u} \geq 0\) for all \(\ket{u} \in \C^d\).

  1. [3 pts] Prove that \(M \succeq 0\) if and only if \(\bra{u}M\ket{u} \geq 0\) holds for all unit vectors \(\ket{u} \in \C^d\).

  2. [3 pts] Let \(M \in \C^{d \times d}\) be a diagonal matrix. Verify that \(M\) is Hermitian if and only if all its diagonal entries are real and \(M \succeq 0\) if and only if all diagonal entries is nonnegative.

  3. [3 pts] Consider any matrix \(A \in \C^{d \times d'}\). Show that \(A^* A\) is Hermitian and \(A^*A \succeq 0\).

  4. [3 pts] Recall the Frobenius inner product on matrices \(A,B \in \C^{d\times d}\):

    \[\langle A,B\rangle \seteq \tr(A^*B).\]

    Prove that if \(A \succeq 0\) and \(B \succeq 0\), then \(\langle A,B\rangle \geq 0\).

    You may use the fact that every Hermitian matrix \(M \in \C^{d \times d}\) can be written as \(M = \sum_{i=1}^d \lambda_i \ket{u_i}\bra{u_i}\) where \(\lambda_1,\ldots,\lambda_d\) are the real eigenvalues of \(M\) and \(\ket{u_1},\ldots,\ket{u_d}\) is an orthornormal basis of eigenvectors for \(M\).

2. Purification of quantum states [10 points]

    Suppose that \(\rho^A\) is a $d \times d$ density matrix. Write $\rho^A$ in its eigenbasis:

    \[\rho^A = \sum_{j=1}^n \lambda_j \ket{v_j}\bra{v_j}\,.\]

    Let \(\mathbb{C}^B\) denote a \(d\)-dimensional Hilbert space with basis \(\ket{1},\ket{2}\,\ldots,\ket{d}\) and define

    \[\begin{align*} \ket{u^{AB}} & \seteq \sum_{j=1}^n \sqrt{\lambda_j} \ket{v_j} \ket{j} \\ \rho^{AB} &\seteq \ket{u^{AB}} \bra{u^{AB}} \end{align*}\]

    Show that \(\rho^A = \tr_B(\rho^{AB})\). In other words, the mixed state \(\rho^A\) can be seen as arising from taking a joint system in the pure state \(\ket{u^{AB}}\) and then discarding the \(B\)-part of the space.

3. Quantum uncertainty splits evenly [10 points]

    Recall that if $X$ is a classical random variable such that \(p_i \seteq \Pr[X=i]\) for \(i=1,2,\ldots,d\), then the Shannon entropy of $X$ is defined by

    \[H(X) \seteq \sum_{i=1}^d p_i \log \frac{1}{p_i}.\]

    This is a measure of the uncertainty of \(X\) measured in bits (or measured in “nats” if we use the natural logarithm).

    Define the von Neumann entropy of a \(d \times d\) density matrix \(\rho\) by

    \[\mathcal{S}(\rho) \seteq \sum_{j=1}^d \lambda_j \log \frac{1}{\lambda_j}\,.\]

    Suppose that \(\rho = \ket{u^{AB}} \bra{u^{AB}}\) is a pure state and \(\rho^A = \tr_B(\rho), \rho^B = \tr_A(\rho)\). Show that

    \[\mathcal{S}(\rho^A) = \mathcal{S}(\rho^B)\,.\]

    Note that the two states don’t necessarily have the same dimension, so they could each have a different number of eigenvalues.

    [ Hint: Show first that if \(U\) is a \(d \times d\) matrix, then \(UU^*\) and \(U^*U\) have the same non-zero eigenvalues. ]

4. Negative conditional entropy [10 points]

  1. [5 pts] In classical probability theory, if \(A\) and \(B\) are two random variables, one defines the entropy of \(A\) conditioned on \(B\) by the formula

    \[H(A \mid B) \seteq \sum_{x} \Pr[B=x] \cdot H(A \mid \{B=x\})\,,\]

    where \(A \mid \{B=x\}\) is the random variable \(A\) condition on \(X\). This quantity is nonnegative because the entropy \(H(A \mid \{B=x\})\) is always nonnegative.

    Prove that

    \[H(A \mid B) = H(A,B) - H(B)\,.\]

    In particular, right-hand side is always nonnegative, and therefore

    \[H(B) \leq H(A,B)\,.\]

    This asserts the relatively obvious fact that the pair of random variables \(\{A,B\}\) has more uncertainty than the single random variable \(B\). In other words, it is easier to predict $B$ than to simultaneously predict both $A$ and $B$.

  2. [5 pts] You will show that this fails dramatically in the quantum setting where the conditional entropy can be negative! Using Problem 2, Show that there is a state \(\rho^{AB}\) with

    \[\mathcal{S}(\rho^B) > \mathcal{S}(\rho^{AB})\,,\]

    where \(\rho^B \seteq \tr_A(\rho^{AB})\). In other words, the entropy of the subsystem is actually bigger than the entropy of the full system.