# Tag Info

9

No, not necessarily. For example, take $\rho$ to be a GHZ state and let $\sigma$ be the completely mixed state of one qubit. We then have $\lambda=4$ and $\mu=2$.

9

Measurement average Measurement average $\langle M \rangle_\rho$ of observable (a Hermitian operator) $M$ on the state $\rho$ is the average of measurement outcomes $m$ in the limit of infinite number of measurements. It is straightforward to show that $\langle M \rangle_\rho = \mathrm{tr}(M\rho)$. Physical consistency Suppose we have a theory which provides ...

8

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, ...

7

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 $[I/d]_{lk} = \frac1d \delta_{lk}$, ...

6

The $1$-norm decreases under partial trace and so we have an upper bound of $1$ when the states are normalized, $$\|\mathrm{Tr}_B[|\psi_1\rangle \langle \psi_2|]\|_1 \leq \||\psi_1\rangle \langle \psi_2|\|_1 = 1.$$ This bound cannot be improved upon without extra information about the states. Here is a counterexample. Take $|\psi_1 \rangle = |00\rangle$ ...

6

A permutation of the qubits is a unitary operation. The trace distance is invariant under unitaries (https://en.wikipedia.org/wiki/Trace_distance#Properties). Thus, statement 1 is true.

5

I'd like to add a small addition to the answer of @DaftWullie about why you would expect this operationally to be true -- without knowing permutations correspond to unitary matrices. It boils down to the Holevo-Helstrom Theorem (HHT) that says the trace distance between two states characterizes operationally the probability that we can distinguish the two ...

5

Let's start with the second question. There is nothing special about an extension $\sigma_{AR}^{\ast}$ that allows it to be optimal for the right-hand side of (1); any extension $\sigma_{AR}$ of $\sigma_A$ could happen to be optimal for the right choice of $\rho_{AR}$. For example, if we suppose that $\sigma_{AR}$ is any given extension of $\sigma_A$, and we ...

5

The point of physical consistency is about how we can define the operator $M$ as acting on system $\rho^A$, or we can define the operator $M \otimes \mathbb{1}_B$ as acting on the system $\rho^{AB}$, and both must 'do the same thing', as it relates to expected value and uncertainty, I believe this is what 'measurement average' is referring to. So since we ...

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

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 ...

4

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 ...

4

No. Just take two Bell states. They have identical reduced density matrices yet are orthogonal, that is, as distant from each other as it gets.

4

They are exactly the same. Remember that you can write $$\rho=\sum_{i,j}\rho_{ij}\sigma_i\otimes\sigma_j.$$ If you take the partial trace, you have $$\rho_A=\sum_{i,j}\rho_{ij}\sigma_i \text{Tr}(\sigma_j).$$ $\text{Tr}(\sigma_j)=0$ unless $j=0$, i.e. the identity operator. Thus, we can write $$\rho_A=\sum_i2\rho_{i0}\sigma_i,$$ and of course we can ...

4

Let $D_{\alpha}(\rho\|\sigma):= \frac{1}{\alpha - 1} \log \mathrm{Tr}[\rho^\alpha \sigma^{1-\alpha}]$ be the Petz-Rényi divergence for $\alpha \in (0,1)\cup(1,\infty)$. Note that for $\alpha \in (0,1)\cup(1,2]$ this quantity satisfies the data processing inequality $$D_{\alpha}(\rho\|\sigma) \geq D_{\alpha}(\mathcal{E}(\rho) \| \mathcal{E}(\sigma)),$$ ...

4

Just notice that $$\text{Tr}(\rho'_{AB}) = \text{Tr}(\Pi_B\rho_{B}).$$ One way to see this is to consider any decomposition $$\rho_{AB} = \sum_i A_i \otimes B_i,$$ where $A_i, B_i$ just some matrices, not states. Then $$\text{Tr}(\rho'_{AB}) = \text{Tr}\big(\sum_i A_i \otimes \Pi_BB_i\Pi_B\big) =$$ $$= \sum_i \text{Tr}(A_i)\text{Tr}(\Pi_BB_i\Pi_B) = ... 4 \frac{s}{2}|0\rangle\langle1|\otimes|0\rangle\langle1|+\frac{s}{2}|1\rangle\langle0|\otimes|1\rangle\langle0| should disappear when you take the trace over them, as \langle0|1\rangle and \langle1|0\rangle = 0 Edit: To give a simpler example, if your traced out the second qubit of$$\frac{1}{\sqrt{2}}|00\rangle+|11\rangle$$, then \frac{1}{2}|00\rangle\... 4 If \rho_{AB}, \sigma_{AB} \ge 0, i.e. they are positive semi-definite then yes (otherwise no). To prove it observe that if \text{supp}(\rho_{AB})\subset \text{supp}(\sigma_{AB}) then we can write that$$ \sigma_{AB} = \epsilon \cdot \rho_{AB} + \Delta, $$for some sufficiently small \epsilon > 0 and some operator \Delta \ge 0. Partial trace is ... 4 TL;DR Tracing out a subsystem corresponds to discarding it. Suppose Alice has subsystem A and Bob has subsystem B of the composite system AB in state \rho_{AB}. Tracing out subsystem B gives us$$ \rho_a=\mathrm{tr}_B\rho_{ab}=\sum_j \langle j_B|\rho_{ab}|j_B \rangle $$which represents the state of Alice's subsystem in the absence of any ... 3 The Bell states are orthogonal and their partial trace is equal - the maximally mixed states - and thus indistinguishable. So yes, this is the maximum you can achieve. Another extremal example could be a tensor product |a\rangle\otimes|b_i\rangle with two orthogonal states |b_i\rangle. 3 The SWAP operator has been widely used to linearize'' polynomial functions of the density matrix. To understand this carefully, consider the following setup: Let \mathcal{H} = \mathcal{H}_{A} \otimes \mathcal{H}_{A'}, where \mathcal{H}_{A} \cong \mathcal{H}_{A'}, that is, we take two copies of the Hilbert space. The SWAP operator acting on this doubled ... 3 No, you wouldn't find 0.9 again. To make the partial trace calculation simpler you can note that the state |\psi'\rangle is separable under the bipartition a_1b_1 | a_2 b_2, i.e. |\psi'\rangle = |00\rangle \otimes (\sqrt{a} |00\rangle + \sqrt{1-a} |11\rangle). So irrespective of the value of a we have \operatorname{Tr}_{a_2b_2}[|\psi'\rangle\... 3 mixed-product property of the Kronecker product: (A \otimes B)(C \otimes D) = (AC \otimes BD) definition of the partial trace: \text{Tr}_1(A\otimes B) = \text{Tr}(A)B cyclic property of the trace: \text{Tr}(AB) = \text{Tr}(BA) Your 3 guesses actually follow from the cyclic property. \text{Tr}(|v_1\rangle\langle v_2|) = \text{Tr}(\langle v_2|\cdot|... 3 Let \newcommand{\rmD}{\mathrm{D}}\newcommand{\rmU}{\mathrm{U}}\newcommand{\calU}{\mathcal{U}}\newcommand{\CC}{\mathbb{C}}\rho\in\rmD(\CC^n),\sigma\in\rmD(\CC^m) be two arbitrary finite-dimensional quantum states, and let \calU\in \rmU(\CC^n\otimes\CC^m) be a unitary in the total space. We have$$[U(\rho\otimes\sigma)U^\dagger]_{ij,k\ell} = \sum_{mnpq} U_{...

3

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

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 $$(... 3 If perturbations are sufficiently small and \rho_{AB} has sufficiently broad support then a desired global state \rho_{AB}' exists. Define$$ \rho_{AB}' = \rho_{AB} + (\rho_A' - \rho_A) \otimes \rho_B + \rho_A \otimes (\rho_B' - \rho_B) \tag1. $$Note that \rho_{AB}' is Hermitian and trace one, but may not be positive. However, \rho_{AB}' is positive ... 3 Yes, the trace distance can only decrease under partial trace. One can see this via the variational characterization of the trace norm$$ \|\rho\|_1 = \max_{-I \leq M \leq I} \mathrm{Tr}[M\rho] $$where M is some hermitian operator satisfying the two operator inequalities M \leq I and M \geq - I. This is sometimes also known as the duality between ... 3 It is a variant of the formula for the average \langle A\rangle_\rho of an operator A measured on a state \rho$$ \langle A\rangle_\rho = \mathrm{tr}(A\rho).\tag1 $$In this case we are measuring an operator acting on the first subsystem, so \rho = \mathrm{tr}_B(|\psi\rangle\langle\psi|). Substituting into (1), we get$$ \begin{align} \langle A\...

3

You haven't really specified what you're happy with as a starting point. I suppose $$\langle A\rangle_\psi=\langle\psi|A\otimes 1_B|\psi\rangle?$$ To see that the other expressions are equivalent, I usually find it more helpful to work backwards. Let's say we start from $$\text{Tr}(A\otimes 1|\psi\rangle\langle \psi|),$$ well mathematically, we can take ...

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