# Tag Info

9

The existence of the inverse of a linear map is independent of the way the map affects the trace. Moreover, if an invertible map preserves a property then its inverse necessarily also preserves the property. Since depolarizing channel preserves the trace, so does its inverse. Inverse of depolarizing channel We can derive the formula for the inverse $\mathcal{... 8 The Pauli matrices form an orthogonal basis of$\mathcal{M}_2$, this vector space can be endowed with a scalar product called the Hilbert-Schmidt inner product $$\langle A,B\rangle=\mathrm{Tr}(A^\dagger B)$$ since the Pauli matrices anticommute, their product is traceless, and since they are Hermitian this implies that they are orthogonal with respect to ... 8 No dimension-independent bound is possible. Consider states$\rho_A$and$\sigma_A$that are close in$p$-norm (for$p>1$) but have relatively low fidelity. Specifically, assume $$\|\rho_A - \sigma_A\|_p = \varepsilon$$ and $$\operatorname{F}(\rho_A,\sigma_A) = \bigl\|\sqrt{\rho_A}\sqrt{\sigma_A}\bigr\|_1 = \delta,$$ where$\varepsilon$is small and$...

8

Transposing a matrix is trace preserving since for $\rho = \sum_{a,b} \rho_{a,b} | a \rangle \langle b |$: $$\text{Tr}(\rho)= \sum_c \langle c| \big( \sum_{a,b} \rho_{a,b} | a \rangle \langle b | \big) | c \rangle = \sum_{a,b,c} \rho_{a,b} \delta_{a,c} \delta_{b,c} = \sum_c \rho_{c,c}$$ $$\text{Tr}(\rho^T)= \sum_c \langle c| \big( \sum_{a,b} \rho_{a,b} | b \... 7 No. The minimal size of the environment is just the rank of the Choi matrix of \mathcal E, call it J(\mathcal E). Since J(\mathcal E^{\otimes n}) = \big(J(\mathcal E)\big)^{\otimes n} and \text{rank}(A \otimes B) = \text{rank}(A)\text{rank}(B), the minimal size of the environment is just \text{rank}\big(J(\mathcal E)\big)^n. 7 Trace preservation The trace must be preserved in the transpose of a matrix, because the trace is the sum of the diagonal elements. When transposing a matrix, you do not change the diagonals at all! Only the off-diagonals change when you transpose a matrix. "Positive" preservation "A positive matrix is a matrix in which all the elements are ... 7 No, that doesn't work. It's fine to use an arbitrary pure state because the unitary U can always be used to take it to any pure state you want. This argument doesn't work for a mixed state, as unitaries cannot take mixed states to pure states. As a concrete example, consider the CPTP map$$ \Lambda(\rho) = |0\rangle\langle 0| \operatorname{tr}(\rho), $$... 6 First, note that |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} and |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} and therefore \langle 0 |1\rangle = \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix}= 0  and similarly,  \langle 1|0\rangle = 0. Also note that |00\rangle = |0\rangle \otimes |0\rangle, |01\rangle = |0\... 6 In the general case I think it's easier to consider the matrix in the form$$ M = \sum_{i_1,\dots,i_n, j_1, \dots j_n} c_{i_1,\dots,j_n} |i_1 \dots i_n\rangle \langle j_1 \dots j_n|, where the i_1, \dots, i_n,j_1, \dots, j_n are all binary and the c with the awful index are the elements of the matrix. Now we know the transformation rules so it's not ... 6 No such (orthonormal) basis can exist. An orthonormal basis \{|\psi_i\rangle\} requires \langle \psi_i | \psi_j \rangle = 0 for i\neq j, and so clearly \begin{align} [\rho_i, \rho_j] &= |\psi_i\rangle \langle \psi_i | \psi_j\rangle \langle \psi_j | - | \psi_j\rangle \langle \psi_j |\psi_i\rangle \langle \psi_i | \\ &= 0 \end{align} So to get a ... 5 First, we can start with U = P_+ - P_-, since the Hermitian is the sum of the projection operators of the eigenspaces scaled by their eigenvalues. If U^2 = I, that means I = (P_+ - P_-)(P_+ - P_-) = P_+P_+ - P_+P_- - P_-P_+ + P_-P_-. Since the two projectors correspond to orthogonal eigenspaces, operating on themselves doesn't change them and one ... 5 Your suspicion is correct, even when A=B. Consider the Hilbert space of two qubits and let ^{T_A} denote the partial transpose with respect to one of them. Suppose that A=B=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}. $$Then$$ A^{T_A}=B^{T_A}=\begin{...

5

From linear algebra, if $v_1, \dots, v_n$ is a basis of the vector space $V$ then every vector $v\in V$ can be written as a linear combination $$v = a_1 v_1 + \dots + a_n v_n\tag1$$ where the coefficients $a_k$ belong to the underlying scalar field. Moreover, if $V$ is an inner product space and $v_1, \dots, v_n$ is an orthonormal basis then the ...

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 $$\mathrm{tr}_A((A\otimes B)\rho_{AB}) \ne \mathrm{tr}_A((A\otimes B)\sigma_{AB}).$$...

5

The answer is no. Define X=[[0,1,0],[0,0,1],[1,0,0]] Z=[[1,0,0],[0,w,0],[0,0,w^2]], w^3=1 Then the Pauli group is generated by X and Z and is of order 27. With H being your matrix, you can check that H'XH and H'ZH are not in the group. Calculations like this are easy to do in gap The dim=3 counterpart of the Hadamard gate is the 3 dimensional Fourier ...

4

With the chosen structure of $U$, i think it's even possible to prove the stronger statement: $$\langle z| \rho|z \rangle = \langle z| \sigma_\rho|z \rangle, \hspace{0.2em} \text{where} \hspace{0.2em} \sigma_\rho = \mathbb{E}_U \big[U\rho U^\dagger\big] \text{and} \hspace{0.3em} |z\rangle \hspace{0.3em} \text{a computational basis vector.}$$ You may ...

4

Quantum channels are foremost, linear operators. So given a basis for the Hilbert-Schmidt operator space (for example the states $\{|0\rangle\langle 0|,|0\rangle\langle 1|,|1\rangle\langle 0|,|1\rangle\langle 1|\}$ that you've chosen above), where density matrices reside, it acts linearly on the basis elements. Perhaps, the easiest way to see it is to write ...

4

This is an application of the following identity $$Be^AB^{-1} = e^{BAB^{-1}}\tag1$$ where $A$ is any $n\times n$ real or complex matrix and $B$ is any invertible $n\times n$ real or complex matrix. Proof of $(1)$. First, recall that the matrix exponential of $A$ is defined as $$e^A = \sum_{k=0}^\infty \frac{1}{k!}A^k.$$ Next, note that for any integer $k$...

4

The reason is when you make measurement on the state $|\psi \rangle = \alpha |0\rangle + \beta |1\rangle$, the state $|\psi\rangle$ will collapsed onto whether the state $|0\rangle$ or $|1\rangle$. If it collapsed to the state $|0\rangle$ then it is indeed belongs to the $+1$ eigenspace, and you record it as an event occurred in that space. If instead it ...

4

No. If you take $\sigma=\tfrac1{d_A}\mathbb I$, you will have that for $\rho=\tfrac1{d_S}\mathbb I$, $$\mathrm{tr}_A(U^\dagger(\rho\otimes\sigma)U)=\tfrac1{d_S}\mathbb I\ ,$$ and thus, it will not be possible to implement any channel for which $$\Lambda(\tfrac1{d_S}\mathbb I)\ne \tfrac1{d_S}\mathbb I\ ,$$ such as any channel mapping all inputs to a ...

4

I do not know what the state-of-the-art for this problem is, but here is my attempt at it. First, I doubt whether the required unitary $U$ can be found for every matrix $A$. The most glaring issue occurs when $b\in\ker A$. In this case, $|Ab\rangle$ is ill-defined, so it is not clear what $U|b\rangle$ should be approximately equal to. A slightly more subtle ...

4

TL;DR: These inner products are equal to the amplitudes and therefore the squares of their magnitudes sum to one. By Pythagoras theorem $\sin^2\theta + \cos^2\theta = 1$, so if one of the amplitudes is $\cos\theta$ then the magnitude of the other must be $|\sin\theta|$. Since $\{|\psi_1\rangle, |\psi_1^\perp\rangle\}$ is a basis, we can expand $|\psi_2\... 4 If you mean by this sentence: we are able to get whole solution to know all the components of vector$x$, then I'm afraid you are wrong. Given a sparse$N \times N$matrix with a condition number$\kappa$, HHL algorithm has a runtime$O(\log(N)\kappa ^{2})$. This offers an exponential speedup over the best known classical algorithm, which has a runtime$O(...

4

The eigenvalues of a matrix are independent of the choice of basis in which we represent it. This remains true for choices of bases that are not orthogonal. Consider then a matrix $A=\sum_k P_k$, where $P_k\equiv\lvert u_k\rangle\!\langle u_k|$, and $\{|u_k\rangle\}$ is a set of normalized (not necessarily orthogonal) vectors. Observe that $A=UU^\dagger$, ...

3

To avoid long computation of all pair-wise inner products in Bell basis, you can think as follow. Switching from computational basis to the Bell basis is done by transformation $$CNOT(H \otimes I)$$ As any operation on quantum computer (measurement and reset being exceptions), this operation is unitary. A unitary operation preserve angles among vectors and ...

3

The most basic but laborious way of checking that Bell states are orthonormal is to carry out the calculations for all sixteen inner products such as $\langle\Phi^+|\Psi^-\rangle$. One way to do this is to switch from Dirac notation to standard linear algebra by replacing the kets and bras with appropriate column and row vectors. After this conversion you ...

3

I'm writing an alternate proof because it uses some interesting tools, computes the value of these expressions, and gives some insights into how we can interpret the quantities in consideration. The first term is $\mathbb{E}_{\mathrm{Haar}} \left[ \left\langle z | \rho | z \right\rangle \right]$, where $\rho := U | \psi_{0} \rangle \langle \psi_{0} | U^{\... 3 An adjoint to partial trace is just tensoring by$I$, i.e.$\text{Tr}_2^\dagger(\sigma) = \sigma \otimes I$. So in this case we can write $$N^\dagger(\sigma) = V^\dagger (\sigma \otimes I_E)V.$$ It still can be written as$\text{Tr}_{E'}(U\sigma U^\dagger)$for an isometry$U:B \rightarrow AE'$, but I don't think there is a simple expression for$U$in ... 3 Yes, in fact there exits Stinespring dilations such that $$\frac{\|N_1-N_2\|_{cb}}{\sqrt{\|N_1\|_{cb}}+\sqrt{\|N_2\|_{cb}}}\leq \|V_1-V_2\|\leq \sqrt{\|N_1-N_2\|_{cb}}$$ where the distance between the isometries is the in terms of the operator norm and$N_1,N_2$are completely positive maps and$V_1,V_2$are their Stinespring isometries. See https://arxiv.... 3 After @Rammus answer explaining that the given inequality does not hold in general, i'll try to prove a weaker statement. Define$ \Delta = V - \tilde{V} $. The assumption is equivalent to$\$ \text{Tr}[I - \tilde{V}^{\dagger} V] \leq \epsilon \cdot |D| \implies \text{Tr}[V^{\dagger}V - \tilde{V}^{\dagger} V] \leq \epsilon \cdot |D| \implies \text{Tr}[\...

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