6

There's a very quick proof if you can use the properties of the Choi-Jamiołkowski isomorphism. Define a map that acts on subsystem $B$ as $$\Lambda(\rho_B) = \mathrm{Tr}(\rho_B) |B| I_B - \rho_B.$$ The Choi operator of this map is $J(\Lambda)_{BB'} = |B| I_{BB'} - |B| \Phi^+_{BB}$, where $\Phi^+$ is the maximally entangled state. It follows that $J(\Lambda)\...


5

Yes; in fact, $\rho$ is both separable and pure. We can start by writing any state $\rho$ in its eigenbasis $$\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|,$$ where $p_i$ are probabilities (i.e., positive and sum to unity) and $|\psi_i\rangle$ are pure states that may or may not be entangled. If $\rho$ is bipartite, each eigenstate $|\psi_i\rangle$ is ...


5

In general, the knowledge of the marginals $\rho_A$ and $\rho_B$ and the operators $A$ and $B$ is insufficient to compute $\mathrm{tr}_A((A\otimes B)\rho_{AB})$. Indeed, we can find two different density matrices $\rho_{AB}$ and $\sigma_{AB}$ with the same marginals for which $$\mathrm{tr}_A((A\otimes B)\rho_{AB}) \ne \mathrm{tr}_A((A\otimes B)\sigma_{AB}).$$...


4

TL;DR Tracing out a subsystem corresponds to discarding it. Suppose Alice has subsystem $A$ and Bob has subsystem $B$ of the composite system $AB$ in state $\rho_{AB}$. Tracing out subsystem $B$ gives us $$ \rho_a=\mathrm{tr}_B\rho_{ab}=\sum_j \langle j_B|\rho_{ab}|j_B \rangle $$ which represents the state of Alice's subsystem in the absence of any ...


4

If $\rho_{AB}, \sigma_{AB} \ge 0$, i.e. they are positive semi-definite then yes (otherwise no). To prove it observe that if $\text{supp}(\rho_{AB})\subset \text{supp}(\sigma_{AB})$ then we can write that $$ \sigma_{AB} = \epsilon \cdot \rho_{AB} + \Delta, $$ for some sufficiently small $\epsilon > 0$ and some operator $\Delta \ge 0$. Partial trace is ...


4

The equality follows by hitting both sides of $$ X_A\mathrm{tr}_B(Y_{AB})=\mathrm{tr}_B((X_A\otimes I_B)Y_{AB})\tag1 $$ with the trace and setting $X_A=M$ and $Y_{AB}=\rho^{AB}$. We can establish $(1)$ for $Y_{AB}$ of the form $Y_{AB}=Y_A\otimes Y_B$ using $$ \mathrm{tr}_B(A\otimes B)=A\mathrm{tr}(B)\tag2 $$ as follows $$ \begin{align} X_A\mathrm{tr}_B(Y_{...


3

Sanity check: the statement is indeed true when $\rho$ is a pure state. We can start by finding the singular values of the combination of purified systems, which I will write as $|\phi\rangle$ and $|\Psi\rangle$. Given the Hermitian matrix $M=|\phi\rangle\langle\phi|-|\Psi\rangle\langle\Psi|$, we can look for eigenvectors of the form $\alpha|\phi\rangle+\...


3

One data point for the general case (that indicates it's not always possible): $\rho_1=|0\rangle\langle0|$, $\rho_2=|1\rangle\langle1|$, $p_1p_2\neq0$ while $\sigma_1=p_1\rho_1+p_2\rho_2$ and $q_1=1$. Note that the purifications of the left-hand side are separable, so $p_1|\psi_1\rangle\langle\psi_1|+p_2|\psi_2\rangle\langle\psi_2|$ is separable. Meanwhile, ...


3

Let $\newcommand{\rmD}{\mathrm{D}}\newcommand{\rmU}{\mathrm{U}}\newcommand{\calU}{\mathcal{U}}\newcommand{\CC}{\mathbb{C}}\rho\in\rmD(\CC^n),\sigma\in\rmD(\CC^m)$ be two arbitrary finite-dimensional quantum states, and let $\calU\in \rmU(\CC^n\otimes\CC^m)$ be a unitary in the total space. We have $$[U(\rho\otimes\sigma)U^\dagger]_{ij,k\ell} = \sum_{mnpq} U_{...


3

The key thing that it tells you is that the W-state is partially entangled. This is perhaps a little clearer to see if you trace over two qubits: $$ \text{Tr}_{AB}(|W\rangle\langle W|)=\frac23|0\rangle\langle 0|+\frac13|1\rangle\langle 1|. $$ The state is mixed, so the overall pure state is entangled, but it's not maximally mixed, so the overall state is not ...


3

For any $\theta\in\mathbb{R}$ and any operator $T$ such that $T^2=I$ we have $$ \exp(i\theta T) = I\cos\theta + i T\sin\theta $$ (c.f. exercise 4.2 on p.175 in Nielsen & Chuang). Therefore, $$ \exp(i\theta S_{AB}) = I \cos\theta + i S_{AB} \sin\theta $$ and we have $$ \exp(-i\theta S_{AB}) (\rho_A \otimes \sigma_B) \exp(i\theta S_{AB}) = \\ \rho_A\otimes\...


3

Using the fact $ e^{i\theta S} = \text{cos}(\theta) \cdot I + i \cdot \text{sin}(\theta) \cdot S $, we calculate \begin{align*} e^{-i\theta S} (\rho \otimes \sigma) e^{i\theta S} & = (\text{cos}(\theta) \cdot I - i \cdot \text{sin}(\theta) \cdot S) \big(\rho \otimes \sigma \big) (\text{cos}(\theta) \cdot I + i \cdot \text{sin}(\theta) \cdot S) \\ &...


3

You could try solving this numerically using semidefinite programming. We know the trace norm of an operator $X$ can be formulated as $$ \begin{aligned} \|X\|_1 &= \min_{Y,Z}\quad \frac12\mathrm{Tr}[Y+Z] \\ &\quad \mathrm{s.t.} \quad \begin{pmatrix} Y & X \\ X^* & Z \end{pmatrix} \geq 0. \end{aligned} $$ Furthermore, we can write your problem ...


3

No - take $\rho$ one of the four Bell states, which all have the same marginals. Then the trace you give will evaluate to $\mathrm{tr}[APBP]$, with $P$ one of the four Paulis (including $I$), which is not only a function of $A$ and $B$ (as it depends on the Pauli, which cannot be inferred from the reduced states).


3

In the absence of the constraint on the marginal, it is true that there exists $U_{ABC}$ such that $U_{ABC}|\rho_{ABC}\rangle = |\sigma_{ABC}\rangle$. Indeed, extend $|\rho_{ABC}\rangle = |\rho_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\rho_{ABC}^{(k)}\rangle$ and $|\sigma_{ABC}\rangle = |\sigma_{ABC}^{(1)}\rangle$ to an orthonormal basis $|\sigma_{ABC}^{...


3

Minimalist formal proof (I'll use $\mu_a\equiv \mu(a)$): $\textrm{(A)}\Rightarrow\textrm{(B)}:$ Let $\Gamma\ge0$. Then, $$ (\Phi_A\otimes I_B)(\Gamma_{AB}) = \sum (\sigma_a)_A\otimes\mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ , $$ which is a separable decomposition, since $\mathrm{tr}_A[((\mu_a)_A\otimes I_B)\,\Gamma_{AB}]\ge0$ because it describes ...


3

It's easy to see that it's true for $\rho^{AB} = A \otimes B$ for any matrices $A,B$ (even when $\rho^{AB}$ is not a state, but just a matrix). Any matrix (not only states) is a linear combination of such products, that is $\rho^{AB} = \sum_i A_i \otimes B_i$, where $A_i,B_i$ are some matrices. Thus $tr(M\rho^A)=tr((M\otimes I_B)\rho^{AB})$ since both sides ...


3

If you're tracing over two systems, $A$ and $B$, you can split this into two steps $$ \text{Tr}(Q_{AB})=\text{Tr}\left(\text{Tr}_B(Q_{AB})\right) $$ (To see this, let the basis you use for taking the first trace be the standard basis, $|ij\rangle$. All I'm doing here is separating out the sums over $i$ and $j$.) So, if you let $Q_{AB}=\tilde M\rho^{AB}$, ...


2

When we trace out system b, what we are doing is basically reducing the system down to as if we had just measured system a Its as if you had just measured or discarded system $b$. Otherwise yes, the probability distribution over the computational basis states described by $\rho_a$ on $\mathcal{H}_a$ is precisely the marginal of the distribution described ...


2

That (A) implies (B) should be obvious from the physical intuition behind (A): A channel of the form (A) can be interpreted as performing a POVM measurement with elements $\mu_a$, and on obtaining outcome $a$ preparing the state $\sigma_a$. It should be obvious that this breaks any entanglement, since it (destructively) measures the input. (Note that also ...


2

That is indeed some weirdly written exposition with typos, but the result is correct. Let $\Phi(\rho) = \sum_k R_k \text{Tr}(F_k\rho)$ and $\Phi_k(\rho)=R_k \text{Tr}(F_k\rho)$. For $\Gamma = \rho_1 \otimes \rho_2$ we have $$ (I \otimes \Phi_k)(\Gamma) = \rho_1 \otimes \Phi_k(\rho_2) = \rho_1 \otimes R_k\text{Tr}(F_k\rho_2) = $$ $$ = \rho_1\text{Tr}(F_k\...


2

Yes, the conjugate is $\langle 0|_A\langle 1|_B$. This is also other times written as $\langle 0_A|\langle 1_B|$, or $\langle 0_A|\otimes \langle 1_B|$, or just $\langle 01|$, or similar ways. These are all just notational differences. They are all equivalent as long as one knows what is being discussed.


2

No, this is not the case. Consider the situation where $$ \rho_{AB}=\frac12(|0\rangle\langle 0|\otimes |0\rangle\langle 0|+|1\rangle\langle 1|\otimes |1\rangle\langle 1|). $$ So, we have that $\rho_B=\text{Tr}_A(\rho_{AB})=\frac12(|0\rangle\langle 0|+|1\rangle\langle 1|)$. Now let $\Pi_A=|0\rangle\langle 0|$, which means that $\sigma_{AB}=|00\rangle\langle ...


2

DaftWullie's answer is correct. The key identity they are using is $$ \mathrm{tr}_B(\rho_A\otimes\sigma_{BC}) = \rho_A \otimes (\mathrm{tr}_B\sigma_{BC})\tag1 $$ which says that we can pull out tensor factors that do not act on the system being traced over. Using $(1)$ and the symbols defined in the linked question and answer, we have $$ \begin{align} \...


2

The two equations are part of the measurement postulate of quantum mechanics which states that probability of the outcome $m$ in a measurement described by operators $M_m$ on a state $\rho$ is $$ p(m) = \mathrm{tr}(M_m^\dagger M_m \rho)\tag1 $$ (c.f. $(2.159)$ in Nielsen & Chuang) and the post-measurement state is $$ \frac{M_m\rho M_m^\dagger}{\mathrm{tr}...


2

Computationally, the easiest way to do this is probably as follows: Let your state be $$ |\psi\rangle=\sum_{i,j,k,l}c_{ijkl}|ij\rangle_A|kl\rangle_B $$ Rewrite this as a matrix $$ C=\sum_{i,j,k,l}c_{ijkl}|ij\rangle\langle kl| $$ Effectively, you just have to reshape your numpy array. Then, you can calculate $$ \rho_A=CC^\dagger $$ or $$ \rho_B=C^\dagger C. $$...


2

I'm not sure exactly what the question is, but I can expand a bit about these states. The states you mention are sometimes referred to as "one-way quantum-classical correlated states" (eg here and arxiv version) to refect the properties you describe. They differ from "strictly classical-classical states" of the form $\sum_j p_j|j\rangle \...


2

Another interesting example is a single-photon state in a superposition of two different spatial (or any other type of degree of freedom really) modes: $$\frac{1}{\sqrt2}(a_1^\dagger + a_2^\dagger)|0\rangle\equiv \frac{1}{\sqrt2}(|1\rangle+|2\rangle).$$ This type of states is sometimes not considered "entangled", as there is only a single particle ...


2

It absolutely depends on the subdivision of the spaces. Take the 3-qubit system (qubits A, B and C) in a state $$ |0\rangle_A(|00\rangle+|11\rangle)_{BC} $$ We can partition these qubits in various different ways. Clearly the $A|BC$ partition has no entanglement across the partition, while the $AB|C$ partitioning is maximally entangled. Note that if you go ...


2

The question amounts to whether there are nice enough expressions for the partial traces of powers of an operator in a bipartite (finite-dimensional) space. That's where the result with the standard trace comes from: just observe that $\frac{\partial}{\partial X_{ij}}\operatorname{Tr}(X^n)=n (X^{n-1})_{ji}$ and apply it to the series expansions of the given ...


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