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!

  • $\begingroup$ 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 $\beta_{ik}$. $\endgroup$ Commented Sep 12, 2018 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
    Commented Sep 12, 2018 at 19:20

1 Answer 1


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.

  • $\begingroup$ This is a much more concise way of looking at this proof. Thank you very much. $\endgroup$
    – Joe
    Commented Sep 19, 2018 at 17:11
  • $\begingroup$ similar question on qc: quantumcomputing.stackexchange.com/q/4325/55 $\endgroup$
    – glS
    Commented Oct 3, 2018 at 10:46

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