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I know the question is already answered, but there was some question on my comment and I wanted to elaborate on that. First, let us consider one system only. The $\mathbb{R}$-span of all states $\rho$ is the space of Hermitian operators. Indeed, by the spectral decomposition, already the set of pure states is enough. This also implies that the $\mathbb{C}$-...


4

Note that: $$ X |0\rangle = |1\rangle \hspace{1 cm} X|1\rangle = |0\rangle $$ but $$ Y |0\rangle = i|1\rangle \hspace{1 cm} Y|1\rangle = -i|0\rangle $$ So $X \neq iY$. In fact, the set $\{I, X, Y, Z\}$ is an orthogonal basis set for $2 \times 2$ matrices. They are not just some factor off from each other. They are independent from one another. As for ...


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

$\frac{\sqrt{2}}{2}(1+i) $ = $\frac{1}{\sqrt{2}}(1+i)$. To see how this is the case, multiply the numerator and denominator of $\frac{1}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ = $1$. $\frac {\sqrt{2}}{\sqrt{2}} \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{{2}}$ . The result doesn't differ; only how it's displayed in comparison to your manual calculation does.


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

If all $c_{ij}$ are non-negative, this is the definition of a separable state. But you can argue this by giving a protocol that lets you construct, using LOCC, the state $\rho_{AB}$. You probably could use this description to cover entangled states if you let the $c_{ij}$ be negative. Here's a simple argument for bipartite states of qubits: the Pauli ...


3

Note that $$ |0\rangle\langle 1| = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix}= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ and similarly $$ |1\rangle\langle 0| = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$ and therefore $$ ...


3

First of all, your question should be more carefully formulated, since it is not even possible to always find a non-trivial subsystem (not subspace) of $\mathcal H_A$, see also my comment here Can a single qutrit in superposition be considered entangled? Thus, let us assume that $\mathcal H_A$ does not have prime dimension, $d_A=\dim\mathcal H_A$. The ...


3

Consider a vector space $V$ with an inner product and a linear operator $A:V\rightarrow V$. The definition of support that you have posted from Wikipedia can be a bit confusing. It says $\text{supp}(A) = \{u\in V|Au\neq 0\}$. This is the complement of the kernel where the kernel of $A$ is $\text{ker}(A) = \{v\in V| Av = 0\}$. However, this definition of the ...


3

I'd like to add to Angelo Lucia's answer slightly. It is not very surprising that $S(\rho \| \sigma)$ can take the value $+\infty$ once we realise that the relative entropy is a generalization of the Kullback-Liebler divergence $D(p \| q)$ between probability distributions $p$ and $q$. Formally, given two distributions $p,q$ over some finite set $\mathcal{X}$...


3

If $\sigma$ is not full rank, then the correct way to interpret the quantum relative entropy formula you wrote is to assign it the value of $+\infty$ when the support of $\rho$ is not included in the support of $\sigma$. Wikipedia has a nice explanation of how to interpret this, but you can think that the reason for which the quantum relative entropy is ...


3

For a single CNOT operation, the simplest approach is to think in terms of Boolean logic. Mathematically, this is usually represented by modular addition, which gives the action of CNOT as $$\vert A, B \rangle \rightarrow \vert A, A \oplus B \rangle,$$ where $\vert A, B \rangle$ is the tensor product of $\vert A \rangle$ and $\vert B \rangle$, and $\oplus$ ...


3

For $ n = 2 $, it is known that the Pauli matrices together with the identity matrix $ I $ form a basis. Now observe that we can write: $ I = |0 \rangle \langle 0| + |1 \rangle \langle 1| $ $ \sigma_z = 2 \cdot |0 \rangle \langle 0| - I $ $ \sigma_x = 2 \cdot |+ \rangle \langle +| - I $ $ \sigma_y = 2 \cdot |+i \rangle \langle +i| - I $ This means ...


3

Is it clear how the syndrome measurement for $3$ qubit bit-flip code is done that is described in the same source with this circuit? If yes then is it clear how this can be done for $9$ qubit code with this circuit for bit-flip error? If no here are some ideas that might help. A toy model example about what is happening in these codes. Imagine we have a ...


3

The complex conjugation flips the sign of the imaginary part of a complex number. Transposition exchanges the row and column co-ordinates of a value in a matrix. A vector can be thought of as a matrix with 1 column and a certain amount of rows. The conjugation of this takes it to it's row form, which for a vector $|\psi\rangle$, becomes $\langle\psi|$. Now ...


3

Part 1 Monotonicity under channels is sometimes also referred to as satisfying a data-processing inequality. One way to prove this is to use a variational formula for the fidelity function, see Theorem 3.17 and the subsequent discussions in TQI - Watrous. This is slightly cheating as you first need to prove the variational formula is correct but in my ...


2

You can consider X and Z gates as "inversion" in computational basis and circular and Hadamard bases, respectively. Lets start with X. It holds that $$ X|0\rangle = |1\rangle\,\,\,\,\,\,\ X|1\rangle = |0\rangle, $$ so X is analogical to classical negation, i.e. it converts 0 to 1 and conversely. Instead of computational basis $\{|0\rangle, |1\...


2

I guess that part of the confusion is that you defined the matrix $$ M = [m_{x,y}] ,\quad 1 \leq x \leq |A|, 1 \leq y \leq |B| $$ which is actually the transpose of the common matrix representation of a linear operator. This later on means that composition of the linear operators $ V M $ is not the usual matrix multiplication. All this to say, that you ...


2

You seem to have got, pretty successfully (I won't claim to have checked all the fine details), to the point of showing that you need $$ MM^\star=M'(M')^\star=\rho. $$ However, you then assume $M'=VM$. You cannot do this as what you're trying to prove is that the only option is for $M'=VM$. What you could do is assume a singular value decomposition of both $...


2

Amplitudes $\alpha$ and $\beta$ in the description of Alice's state are complex numbers; to describe them precisely, you're going to need infinite number of bits of information. If you're only using a finite number of bits, you get an approximation of the state, which can be a very good one but still not an exact representation. (Why? There are uncountably ...


2

At the beginning of $40$ page of the same book an arbitrary spin state is written as: $$c_1| \uparrow \rangle + c_2|\downarrow \rangle$$ where $|c_1|^2 + |c_2|^2 = 1$, $c_1$ and $c_2$ are in general complex numbers, but how I understand from the parts of the book the author uses only real $c_1$ and $c_2$ here for some (maybe more pedagogical) reasons. There ...


2

$|+\rangle$ and $|-\rangle$ represented by $|0\rangle$ and $|1\rangle$ basis vectors: $$|+\rangle = \frac{1}{\sqrt{2}} (|0\rangle + |1\rangle) \qquad |-\rangle = \frac{1}{\sqrt{2}} (|0\rangle - |1\rangle)$$ By substituting these expressions that are written in $|0\rangle$ and $|1\rangle$ basis will prove what is wanted: $$\frac{1}{\sqrt{2}} (|+\rangle + |-\...


2

Let $P$ be the projector onto the symmetric subspace. We want to find $$ \max|\langle\gamma|\phi,T\rangle| $$ for $|\gamma\rangle\in \mathcal{V}$. This is equivalent to $$ \max|\langle\gamma|P|\phi,T\rangle|, $$ so if we calculate $P|\phi\rangle|T\rangle$, then $|\gamma\rangle$ will be the normalised state parallel to that. Now, if we project $|\phi\rangle|T\...


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

I follow Wilde's notation here. The coherent information of a channel $N:A' \rightarrow B$ is given by $$Q(N) \equiv \max_{\phi_{A A^{\prime}}} I(A\rangle B)_{\rho},$$ where $\rho_{AB}=N_{A^{\prime} \rightarrow B}\left(\phi_{A A^{\prime}}\right)$. Notice that the channel only acts on the $A'$ register. The $A$ register is used to purify the input to the ...


1

First, note that the circuit construction using classical condition like you have is not executable on IBM hardware at the moment. Devices like Honewywell Trapped ion allows you to do such thing (I think). However, thanks to principle of deferred measurement, we can push the measurement all the way back to the end of the circuit. See here. Essentially, your ...


1

For a 3 qubit system, you might get results like 000, 010, 011 etc. but the leftmost bit of the readout will always be zero. This is how we know that the teleport has worked.


1

Hope this helps you get started, If two vectors are orthogonal, then their inner product is zero. The inner product is a measure of how similar two vectors are; or, the degree to which the first vector lies along the second. So you know that |𝜙⟩|𝑇⟩~ 0


1

I will do my best to answer from my understanding of your question: I am assuming you mean something like Unitary Coupled Cluster variational form where it is implemented the exponential to the initial state $|\psi \rangle$ $$ e^{T - T^\dagger} $$ here $T$ is the cluster operator, which, when acting on $|\psi$, produces a linear combination of excited ...


1

If i understand your question correctly, no, it's not possible. Take for example $$ \rho = | 0 \rangle \langle 0|^{\otimes |A|} \hspace{0.2em} \in D(\mathcal{H_A}) $$ For every choice $ \mathcal{H_B}, \mathcal{H_C} $ such that $ \mathcal{H_A} = \mathcal{H_B} \otimes \mathcal{H_C} $ we have $$ \text{Tr}_C [ \rho ] = | 0 \rangle \langle 0|^{\otimes |B|} $$ so ...


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