3

The standard meaning of this notation is that you're using $n$-ary encoding of the indices. In this case $n=3$, so \begin{align}11 \sim 1\\ 12 \sim 2 \\ 13 \sim 3 \\ 21 \sim 4 \\ 22 \sim 5 \\ 23 \sim 6 \\ 31 \sim 7 \\ 32 \sim 8 \\ 33 \sim 9 \end{align} The matrix element $M_{1,6}$ is then represented as $M_{1,23}$ which is given by $M_{213}$.


2

Yes. Any mixed state $\rho$ is a convex combination of pure states, that is $$ \rho = \sum_i \lambda_i |\phi_i\rangle\langle\phi_i| $$ where $\lambda_i >0$, $\sum_i\lambda_i=1$. The partial trace is linear, so that $$ \rho_A = \sum_i \lambda_i \text{Tr}_B(|\phi_i\rangle\langle\phi_i|). $$ Every $\lambda_i\text{Tr}_B(|\phi_i\rangle\langle\phi_i|)$ is a ...


2

Bob does not know the outcome of Alice's measurement. All he knows is that Alice would have obtained $|\psi\rangle$ with probability $\frac12$ and $|\psi^\perp\rangle$ with probability $\frac12$ for some orthonormal basis $|\psi\rangle$, $|\psi^\perp\rangle$. Therefore, his state is a mixture $$ \rho_B = \frac12|\overline\psi\rangle\langle\overline\psi|+\...


2

Let us recall the Ekert-91 (E91) protocol. Initially Alice and Bob each independently receive respective elements of a plurality of Bell pairs (for example, generated by spontaneous parametric down-conversion; Upon deciding to generate a secret key, Alice and Bob each independently measure their pairs in one of three bases, and publicly announce their ...


1

Grover’s search uses an equal superposition of all basis states in the search space both as its starting state and as a part of “reflection about the mean” step. So if you're searching among these 4 states, you need to have an operation that prepares this superposition from the $|0000\rangle$ state to start with, and then your "reflection about the mean&...


1

Cirq has a function cirq.sub_state_vector which can extract a single qubit's state from a full state vector. It doesn't just do the single qubit case, it can do arbitrary subsets of qubits. It will raise an exception if the subset you pick is entangled with other stuff. It's unfortunately a bit picky about error tolerances and input shape.


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