A collection of hints and solutions to the exercises I pose on this blog. I’m not trying to be paternalistic, but I do put in the 30 second cooldown for viewing full solutions as a stop-gap. On the internet, there is not enough friction between us and choices not good for us. If you would like to view a solution, it should be a considered choice and not one made in less than a second.
Exercise 1: Show that this is the correct state space by showing that for any two states $\ket{\psi}\neq \ket{\phi}$, there exists a unitary $U$ such that measuring after applying $U$ to each of these states yields different probability distributions.
Hint: Construct a $U$ mapping $\ket{\psi}$ to $\ket{1}$. What the probability of measuring $1$ with respect to $\ket{\psi}$ vs. $\ket{\phi}$?
Please wait 30 seconds.
Solution: One can construct a $U$ mapping $\ket{\psi}$ to $\ket{1}$ by extending $\ket{\psi}$ to an orthonormal basis $\ket{\psi_i}$, $\psi_1 = \psi$, and having $U = \sum_i \ket{i}\bra{\psi_i}$ with respect to this basis. After applying $U$ and then measuring we get $1$ with probability $1$ when applied to $\ket{\psi}$. One gets $1$ with probability $\bra{\psi}\phi\rangle < 1$ when applied to $\ket{\phi}$.
Exercise 2: Show that this is true for the following distributions on one qubit: $(\ket{0}, \ket{1})$, and $(\ket{0}+\ket{1},\ket{0}-\ket{1})$ where both of these probability distributions are 50-50 chances for the two respective quantum states.
Exercise 3: Given two non-equal mixed states, $D_1, D_2$, show that there is some $i,U$ such that $\lambda_{i,U}(D_1) \neq \lambda_{i,U}(D_2)$.
Hint: Note that $D_i$ are self-adjoint matrices $D_i^\dagger = D_i$. That is, their difference is also self-adjoint. Look at the spectral decomposition of the difference.
Please wait 30 seconds.
Solution: As $\Delta D = D_1 - D_2$ is self adjoint, we can find an orthonormal eigenbasis $\ket{\psi_i}$ for $\Delta D$ with real eigenvalues $e_i$. Pick $i$ with a nonzero eigenvalue $e_i$. If we measure with respect to this basis, which is equivalent to measuring to the standard basis after applying $U = \sum_i \ket{i}\bra{i}$, the probability of observing $\psi_i$ from mixed-state $D_j$ is $\bra{\psi_i}D_j\ket{\psi_i}$. Notice these probabilities are different as $\bra{\psi_i}D_1\ket{\psi_i} - \bra{\psi_i}D_2\ket{\psi_i} = \bra{\psi_i}\Delta D\ket{\psi_i} = \lambda_i$. Otherwise stated, $\lambda_{i,U}(D_1) - \lambda_{i,U}(D_1) = e_i\neq 0.$