6

Here the important fact is that the maximally mixed state is in fact an identity matrix. Let me rewrite the expression on the left in index notation (the summation sign is omitted according to the Einstein convention): $$ Tr(\rho^{AB} (\sigma^A \otimes I/d)) = [\rho^{AB}]_{ijkl} [\sigma^A]_{ji} [I/d]_{lk} $$ But $[I/d]_{lk} = \frac1d \delta_{lk}$, ...


4

The equation at the top of the question is not correct: there is a missing factor of $1/d$ on the right-hand side. Let's eliminate this factor from the left-hand side to make it simpler, so that the equation we want is this: $$ \text{Tr}\bigl(\rho^{AB} \bigl(\sigma^A \otimes I\bigr)\bigr) = \text{Tr}\bigl(\rho^A \sigma^A\bigr). $$ To see why this is true, ...


3

Let me give you the structure for how you go about answering this question. Let $|\Psi\rangle$ be the pure state shared between Alice and Bob. Bob measures his system in an orthonormal basis $\{|\phi_i\rangle\}$. Thus, Bob gets an answer $i$ with probability $p_i$, and the overall system is left in the state $|\psi_i\rangle|\phi_i\rangle$. Now, Alice does ...


3

Let's start with a general state $$ \rho\otimes\rho_0=\sum_{x,y\in\{0,1\}}\langle x|\rho|y\rangle|x\rangle\langle y|\otimes |0\rangle\langle 0|. $$ If we apply the controlled-not, we have $$ \rightarrow\rho_{\text{final}}=\sum_{x,y\in\{0,1\}}\langle x|\rho|y\rangle|x\rangle\langle y|\otimes |x\rangle\langle y|. $$ Now we want to take the partial trace over ...


2

With further understanding coming from the expanded question, I'm entirely revising my answer, but the original version is kept below in case it's useful. The point, clearly, is to show how to simulate a probabilistic classical computation. So, we will store a classical distribution by using a diagonal density matrix: $$ \rho=\sum_ip_i|i\rangle\langle i|. $$...


2

I think it helps here to write things explicitly. Suppose $\mathcal E(\rho)=\operatorname{Tr}_E[U(\rho\otimes|\mathbf e_0\rangle\!\langle\mathbf e_0|)U^\dagger]$. Pick a basis for the environment in which $|\mathbf e_0\rangle$ is the first element. Note that here $U$ is a unitary matrix in a bipartite system. The operator before taking the partial trace ...


2

$|e_k\rangle$ is the basis of the environment. Taking the sum of projections onto an orthonormal basis of one subsystem is the definition of the partial trace over that subsystem.


2

Consider a bipartite state $|\psi\rangle=\sum_{ij}\psi_{ij}|i\rangle\otimes|j\rangle$. In the following, I will work directly on the matrix elements of the objects involved. Tracing out the second space amounts to the following mapping $$\psi_{ij}\rightarrow \rho_{ii'}\equiv\sum_j \psi_{ij}\bar\psi_{i'j}.\tag A$$ Now forget about the partial trace, and ...


2

Yes. Choi's Theorem a priori uses different Hilbert spaces of potentially different dimensions $d_1$ and $d_2$. Then $d_1=d_2$ is a corollory. The proof is included there.


2

As pointed out in a comment, what you wrote as $\rho$ should more precisely be written as $\rho\otimes\mathbb 1$ (although the Kraus operator decomposition can be obtained similarly with any initial ancilla state, in which case you have $\rho\otimes|\phi\rangle\!\langle\phi|$). The standard algebraic properties of tensor product spaces then tell you that $$(...


1

I've never thought about this before, and certainly not done any in-depth calculations, but see if this gets you started.... Obviously you want to do two different things: show that there are $n$-sharable states, and show that these states are not $m$-shareable for some $m>n$ (and hopefully get $n$ and $m$ as close as possible). Let $$ |\psi\rangle=\...


1

The answer is yes, you need $d_1 d_2$ operators, as already pointed out in the other answer. Here I'll show explicitly why this is the case. Let $\Phi\in\mathrm{T}(\mathcal X,\mathcal Y)$ be a CPTP operator sending maps in $\mathcal X$ into maps in $\mathcal Y$. The spaces $\mathcal X$ and $\mathcal Y$ have dimensions $d_1$ and $d_2$, respectively. Short ...


1

There are two possible answers. Let's say the universe evolves from $t=0$ to $t_f$ then the unitary evolution $U$ from $0$ to $t_f$ induces a CP evolution on the subsystem. To see this, note that the composition of CP maps is CP. Now, the reduced (system) evolution is $Tr_E U\rho_s\otimes\rho_E U^\dagger$ which is a composition of the map $\rho_s\...


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