3
$\begingroup$

I've been working through Scott Aaronson's "Introduction to Quantum Information Science" notes / problem sets on his blog, and I'm running into a little bit of trouble with an exercise on Homework 3. The problem is stated as follows:

No-Communication Theorem: Suppose Alice and Bob share the entangled state $\sum_{i=1}^{N -1}\sum_{j=1}^{N -1} \alpha_{ij} |i\rangle | j \rangle$.

  1. What is Bob's local density matrix?
  2. Show that Bob's local density matrix is unchanged if Alice measures her subsystem in the standard basis $\{|0 \rangle, |1 \rangle, \cdots |n\rangle \}$.
  3. Show that Bob's local density matrix is also unchanged if Alice applies any $N \times N$ unitary matrix $U$ to her subsystem.

My work is as follows:

For part 1, Bob's local density matrix has entries given by: $$\rho^B_{jj'} = \sum_{i=0}^{N-1} \alpha_{ij} \alpha_{ij'}^*$$ I do not know a more specific way to express this matrix. If there is a better way, then please let me know.

For part 2, when Alice measures her part of the system, she measures outcome $|i \rangle$ with probability $p_i = \sum_{j=1}^{N-1}|\alpha_{ij}|^2$. Thus, when Alice measures her system, since Bob doesn't know the outcome, from his perspective the new quantum system is the mixed state: $$\sum_{i=0}^{N-1} p_i \left [ |i\rangle \otimes \frac{\sum_{j=0}^{N-1} \alpha_{ij} |j\rangle}{\sqrt{|\alpha_{i,0}|^2 + \cdots + |\alpha_{i,N-1}|^2}} \right ]$$ But by the bi-linearity of the tensor product, we can pull the denominator out, which cancels out $p_i$, yielding the mixed state: $$\sum_{i=0}^{N-1} \left [ |i\rangle \otimes \sum_{j=0}^{N-1} \alpha_{ij} |j\rangle \right ]$$

So calculating an entry of Bob's local density matrix yields: $$\rho^B_{jj'} = \sum_{i=0}^{N-1} \alpha_{ij} \alpha_{ij'}^*$$ This is the same as before, so Alice taking a measurement doesn't change Bob's local density matrix.

For part 3, I am completely lost. I thought about writing $U|i\rangle$ in the standard basis as the sum $\sum_k \beta_{ik} |k \rangle$, but when calculating Bob's local density matrix, the triple sum hasn't worked out nicely. It's very possible that I've made an algebraic mistake somewhere, but I feel like there must be a better way. Perhaps there's a better way of If anyone could point me in the right direction, I would appreciate it. Thanks!

$\endgroup$
  • $\begingroup$ Part 1 and 2 are good. For part 3, it seems the good way, and do not forget the effect of unitarity of $U$ on the $\beta_{ik}$. $\endgroup$ – Frédéric Grosshans Sep 12 '18 at 19:01
  • $\begingroup$ @FrédéricGrosshans Are you sure that part 2 is good? I think I made a mistake in treating the mixed state as a sum. I'm working on fixing it now. $\endgroup$ – Joe Sep 12 '18 at 19:20
3
$\begingroup$

For part 1, you can also concisely write it as $\rho^B_{jj'}=(\alpha^\dagger \alpha)_{j'j}$, thinking of $\alpha$ as a matrix. Equivalently, $$\rho^B= \sum_{jk} \left(\sum_i \alpha_{ij}\alpha_{ik}^*\right)\lvert j\rangle\!\langle k\rvert.$$

Part two looks also correct. Here is how I would have done it: start from the shared state $\lvert\Psi\rangle=\sum_{ij}\alpha_{ij}|i,j\rangle$. If Alice measures her state and finds the $i$-th outcome, the state of Bob is $|\phi_i\rangle\equiv \sum_j \alpha_{ij}|j\rangle/\sqrt{p_i}$, with $p_i=\sum_j \lvert\alpha_{ij}\rvert^2$.

Because Bob does not know what outcome Alice observed, he must describe his state as a convex combination of the various possible states: $$\sum_i p_i \lvert\phi_i\rangle\!\langle\phi_i\rvert= \sum_{ijk} \alpha_{ij}\alpha^*_{ik} \lvert j\rangle\!\langle k\rvert = \rho^B.$$

For the third part you don't really need to do any more work. Notice that applying $U$ locally to Alice's system means to consider the state $$\lvert\Psi'\rangle=\sum_{ij}\alpha_{ij} (U\lvert i\rangle)\otimes\lvert j\rangle = \sum_{ijk} \alpha_{ij}U_{ki}\lvert k,j\rangle = \sum_{jk}\left(\sum_i \alpha_{ij}U_{kj}\right)\lvert k,j\rangle = \sum_{kj}\alpha'_{kj}\lvert k,j\rangle.$$ Thus, applying $U$ is equivalent to considering a state with a different matrix of coefficients $\alpha'$. However, note how the argument of part 2 holds regardless of what $\alpha_{ij}$ is. The conclusion is thus immediate.

$\endgroup$
  • $\begingroup$ This is a much more concise way of looking at this proof. Thank you very much. $\endgroup$ – Joe Sep 19 '18 at 17:11
  • $\begingroup$ similar question on qc: quantumcomputing.stackexchange.com/q/4325/55 $\endgroup$ – glS Oct 3 '18 at 10:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.