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
Let $M\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$ 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
Find expectation value of the observable $X_1\otimes Z_2$ for a maximally entangled two-qubit system
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
Find expectation value of the observable $X_1\otimes Z_2$ for a maximally entangled two-qubit system
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 ...
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