# Tag Info

6

Given $\rho$ and a fixed ensemble $\{ |\psi_i \rangle \}$ it might not be possible to write $\rho$ as $\sum_i p_i |\psi_i \rangle \langle \psi_i |$. For example, let $| + \rangle = \frac{1}{\sqrt{2}} (| 0 \rangle + | 1 \rangle )$. Then the state $|+\rangle \langle + |$ cannot be expressed as a convex combination in the ensemble $\{ | 0 \rangle, |1\rangle \... 5 Choi operator of a linear map$\mathcal{E}$is defined as $$J(\mathcal{E}) = \sum_{ij} \mathcal{E}(|i\rangle\langle j|)\otimes |i\rangle\langle j|.\tag1$$ Substituting$\mathcal{E}(\rho)=\sum_k E_k\rho E_k^\dagger$into$(1), we have \begin{align} J(\mathcal{E}) &= \sum_{ijk} \left(E_k|i\rangle\langle j| E_k^\dagger\right)\otimes |i\rangle\langle j|... 4 We know that \begin{gather} |0\rangle = \frac{|+\rangle+|-\rangle}{\sqrt{2}} \\ |1\rangle = \frac{|+\rangle-|-\rangle}{\sqrt{2}} \end{gather} $$Thus, we can rewrite the GHZ state as$$ \begin{align} |GHZ\rangle &= \frac{1}{\sqrt{2}}\left(|0\rangle|00\rangle+|1\rangle|11\rangle\right) \\ &=\frac{1}{2}\left(|+\rangle|00\rangle+|-\rangle|00\rangle+... 3 LetM\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$be some linear operator whose input and output spaces are both$\mathcal Y\otimes\mathcal X$, for some pair of finite-dimensional Hilbert spaces$\mathcal X,\mathcal Y$. Moreover, suppose$M$is positive semidefinite:$M\ge0$. It being positive semidefinite implies it admits a decomposition of the form$M=\...

3

Consider the state $|\Psi\rangle$. This has a Schmidt decomposition $$|\Psi\rangle=U_A\otimes U_B\sum_i\lambda_i|ii\rangle.$$ Its reduced density matrix is $$\rho_A=U_A\left(\sum_i\lambda_i|i\rangle\langle i|\right)U_A^\dagger.$$ It must be that if $|\Phi\rangle$ has the same reduced density matrix, the density matrices have the same spectrum and hence $|... 2 As per N&C, fidelity is "analogous to the probability of doing the decompression correctly" (emphasis added). The goal is to do the operation correctly with 100% probability, which means the probability is 1. This is the desired limit of fidelity, so no error means the fidelity is 1. 2 There are many demos on https://pennylane.ai/qml/demonstrations.html. You could perhaps get some inspiration from there. 2 This is due to how the$\mathbf{A}$matrix was defined; from that same tutorial page we have: $$\tag{1} \mathbf{A} = \sum_{n} c_n A_n$$ where each$A_n$is unitary and$c_n$is complex (and in the original VQLS paper they further impose$\lVert {\mathbf{A}}\rVert<1$and bounded condition number) but$\mathbf{A}$is never required to be unitary. Therefore,... 1 An intuitive way to think about it is that$E[M]=E[X_1 \otimes Z_2]=E[X_1 \otimes \mathbb{1}]E[\mathbb{1} \otimes Z_2]$If we only think about$E[\mathbb{1} \otimes Z_2]$, it is just the expectation value of$Z_2$on the second qubit. Consider that our second Qubit in the entangled state$\frac{| 00\rangle + | 11\rangle}{\sqrt{2}}$is measured to be$\frac{+\...

1

Taking the last two terms of last expression you gave, we can do the following \begin{align} M \left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) &= X_1\otimes Z_2\left(\frac{|00\rangle+|11\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{X_1|0\rangle \otimes Z_2|0\rangle+X_1|1\rangle \otimes Z_2|1\rangle}{\sqrt{2}}\right) \\ &= \left(\frac{|1\... 1 Although it is not explained up to that point in the Qiskit textbook, the quantum toss is in reality applying the Hadamard gate, denoted H. In matrix form, this operator looks like: H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} $$Now, we express the basis states in column form as follows:$$ \begin{gather} |0\rangle = \...

1

Quantum (pure) states are, by definition, defined up to a scalar complex factors. That means that a state that we write as $|\psi\rangle$, should really be understood as the full set of vectors (an equivalence class if you will) $\{\lambda|\psi\rangle : \lambda\in\mathbb C\}$. The more formal way to put this is to say that quantum states are elements in the ...

1

When we consider an uniform (or equal, as stated in Nielsen and Chuang) superposition, that is, a state that can be written as: $$|\psi\rangle=\frac{1}{2^n}\sum_x|x\rangle,$$ it is quite common not to write the normalisation constant $\frac{1}{2^n}$. Similarly, when the amplitutes of all vectors on the superposition are equal, we omit the normalisation ...

1

First of all, if we write down $\left|\psi_1\right\rangle$, we get: $$\left|\psi_1\right\rangle=\frac{1}{\sqrt{2}^n}\sum_x|x\rangle\left[\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right].$$ Applying $f$ on this state gives us: $$\left|\psi_2\right\rangle=\frac{1}{\sqrt{2}^n}\sum_x|x\rangle\left[\frac{|f(x)\rangle-|1\oplus f(x)\rangle}{\sqrt{2}}\right].$$ Note that ...

1

$P$ is acting on the space $V$, projecting onto the subspace $W$. Yes, if it only acted on the subspace $W$, it would be identity, but it is acting on a larger space. For example, think about a qubit, where $V$ is spanned by the basis states $\{|0\rangle,|1\rangle\}$. You can define a projector $P=|0\rangle\langle 0|$ which projects onto the space $W$ which, ...

1

In a similar way to how the global phase difference of a state makes no physical difference, neither does amplitude of a state. We normalise states to have unit magnitude for mathematical convenience in the same way we don't carry around an $e^{i\phi}$ factor for arbitrary $\phi$ with all our states. This is because having unit vectors means we don't need to ...

Only top voted, non community-wiki answers of a minimum length are eligible