# Tag Info

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

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

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, ...

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

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 $$\text{Tr}(\rho'_{AB}) = \text{Tr}\big(\sum_i A_i \otimes \Pi_BB_i\Pi_B\big) =$$ $$= \sum_i \text{Tr}(A_i)\text{Tr}(\Pi_BB_i\Pi_B) = ... 3 Let D_{\alpha}(\rho\|\sigma):= \frac{1}{\alpha - 1} \log \mathrm{Tr}[\rho^\alpha \sigma^{1-\alpha}] be the Petz-Rényi divergence for \alpha \in (0,1)\cup(1,\infty). Note that for \alpha \in (0,1)\cup(1,2] this quantity satisfies the data processing inequality$$ D_{\alpha}(\rho\|\sigma) \geq D_{\alpha}(\mathcal{E}(\rho) \| \mathcal{E}(\sigma)), $$... 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 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 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 ... 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|. $$Now we want to take the partial trace over ... 2 The answer is yes, you need d_1 d_2 operators, as already pointed out in the other answer. Here I'll show explicitly why this is the case. Let \Phi\in\mathrm{T}(\mathcal X,\mathcal Y) be a CPTP operator sending maps in \mathcal X into maps in \mathcal Y. The spaces \mathcal X and \mathcal Y have dimensions d_1 and d_2, respectively. Short ... 2 Yes. Choi's Theorem a priori uses different Hilbert spaces of potentially different dimensions d_1 and d_2. Then d_1=d_2 is a corollary. The proof is included there. 2 As pointed out in a comment, what you wrote as \rho should more precisely be written as \rho\otimes\mathbb 1 (although the Kraus operator decomposition can be obtained similarly with any initial ancilla state, in which case you have \rho\otimes|\phi\rangle\!\langle\phi|). The standard algebraic properties of tensor product spaces then tell you that$$(...

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

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

Firstly, just because an operator $M$ is defined on a composite space, it does not mean that the operator itself has a tensor product structure. You need $$M=\sum_{i,j,k,l}M_{ij,kl}|ij\rangle\langle kl|$$ Now let $$O=\sum_{i,k}O_{i,k}|i\rangle\langle k|.$$ With these in place, you just calculate the two sides of the equation and see if they're the same. (...

2

I think the presentation in N&C is a little confusing because $\rho$ is used in two contexts. I'll substitute one of those for a $\sigma$. You can define $$E_i=I\otimes\langle j|,$$ which will certainly achieve the effect stated in your first equation. This lets us define the quantum operation $$\mathcal{E}(\sigma)=\sum_iE_i\sigma E_i^\dagger$$ ...

2

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 ...

2

No, there's not a lot you can say. Consider these two cases, both with $\epsilon=0$. First, the obvious one, $\rho=\sigma=|\psi\rangle\langle\psi|\otimes |\psi\rangle\langle\psi|$. Clearly $\rho_r-\sigma_r=0$. Second, let $|\psi^\perp\rangle$ be orthogonal to $|\psi\rangle$. You can have $$\rho=(|\psi\rangle\langle\psi|\otimes |\psi^\perp\rangle\langle\psi^\... 1 On a quantum computer, there is no point in implementing a partial trace. Just ignore the qubits which you want to trace. 1 qiskit.quantum_info has a partial_trace method you can use for this. It takes as input either a Statevector or a DensityMatrix object. Both these classes implement the from_instruction routine that you can use with the QuantumCircuit object you already have. You would have to do something along the lines of: circuit2 = QuantumCircuit(4,3) ... # your ... 1 A way that's perhaps more compact (if less terse), we can write$$ \mathrm{Tr}_B(O_AM)=\mathrm{Tr}_B((O_A\otimes \mathbb 1_B)M)=\sum_i (\mathbb 1_A\otimes \langle i|_B) (O_A\otimes \mathbb 1_B)M(\mathbb 1_A\otimes |i\rangle_B)=\\=\sum_i O_A(\mathbb 1_A\otimes \langle i|_B) M(\mathbb 1_A\otimes |i\rangle_B)=O_A\mathrm{Tr}_B(M)$$The only nontrivial thing I ... 1 I would probably set this up as a semi-definite program and throw it at the computer. Basically, your problem is a linear problem in the coefficients of the matrix \delta_{ABC} except for the constraint that \delta_{ABC} is positive semi-definite. This is exactly what semi-definite programming is designed to do. Note: the calculation below uses a ... 1 I've never thought about this before, and certainly not done any in-depth calculations, but see if this gets you started.... Obviously you want to do two different things: show that there are n-sharable states, and show that these states are not m-shareable for some m>n (and hopefully get n and m as close as possible). Let$$ |\psi\rangle=\...

1

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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