6

Typically this is a slight abuse of notation. One can have a unitary operator $U$ chosen from some Haar measure, such as the circular unitary ensemble. Then, taking some fiducial state $|\psi_0\rangle$, a "Haar-random state" would be $|\psi_U\rangle=U|\psi_0\rangle$ for $U$ randomly chosen according to the Haar measure. If you want to find the ...


6

There's a very quick proof if you can use the properties of the Choi-Jamiołkowski isomorphism. Define a map that acts on subsystem $B$ as $$\Lambda(\rho_B) = \mathrm{Tr}(\rho_B) |B| I_B - \rho_B.$$ The Choi operator of this map is $J(\Lambda)_{BB'} = |B| I_{BB'} - |B| \Phi^+_{BB}$, where $\Phi^+$ is the maximally entangled state. It follows that $J(\Lambda)\...


5

We have, $$\begin{align}\begin{aligned}\newcommand{\th}{\frac{\theta}{2}}\\\begin{split}Ry(\theta) = \exp\Big(-i \th Y\Big) = \begin{pmatrix} \cos{\th} & -\sin{\th} \\ \sin{\th} & \cos{\th} \end{pmatrix}\end{split}\end{aligned}\end{align}$$ If the input state is $|0\rangle$, then the probability of getting $|0\rangle$ as a ...


5

Summary: The expression you're looking for is: $$ \frac{1}{4} \left[ (III + IZZ + ZIZ + ZZI) + (XXX - XYY - YXY - YYX)\right] $$ where Pauli string notation like $XYX$ denotes $\sigma_1 \otimes \sigma_2 \otimes \sigma_1$, for example. To start, we need to write the $n$-qubit GHZ state as an operator, namely \begin{equation} |\psi^{(n)}\rangle\langle\psi^{(n)...


4

A direct while brute-force way is by writing state $|\psi\rangle$ as density matrix $\rho$. Then by noticing that $\rho=\frac{1}{2^3}\sum_{ijk}t_{ijk}\sigma_{i}\otimes\sigma_{j}\otimes\sigma_{k}$ for $i,j,k=1,2,3,4$ stands for Pauli matrices and identity matrix, find all the coefficients $t_{ijk}$ by $Tr(\rho\sigma_{i}\otimes\sigma_{j}\otimes\sigma_{k})$. If ...


3

In order to apply Nyquist-Shannon sampling theory, we need to know the maximum frequency that will be present in the signal we intend to measure. We will do this by rewriting a time-dependent observable in terms of the frequencies present in the Hamiltonian $H$. Consider an arbitrary observable $O$ which will time evolve under application of $H$ in the ...


3

It's impossible to do it in polylogarithmic depth, because for a modular multiplication (or even just an increment!) the output value of the most significant bit is a function of every other input bit. One spot is affected by all the other spots. So your runtime has to be at least as long as the longest path between that spot and any one of the other used ...


3

Let $\mathcal{H}_A$ denote the Hilbert space of qubits in partition $A$ and similarly for $\mathcal{H}_\bar{A}$. Define the operator $P:=Q|0^n\rangle\langle 0^n|Q^\dagger$ and write its Schmidt decomposition $$ P = \sum_a R_a\otimes Q_a\tag1 $$ where $R_a$ are operators on $\mathcal{H}_A$ and $Q_a$ are operators on $\mathcal{H}_\bar{A}$. See for example the ...


2

Applying a reset to a qubit is equivalent to measuring it, and then applying a bit flip to it conditioned on the measurement result. def reset(qubit): if measure(qubit) == ON: X(qubit) For example, in this Quirk circuit, you can see that the post-reset state matches the state you'd get when conditioning on a measurement-via-ancilla+bit-flip of ...


2

Bob does not know the outcome of Alice's measurement. All he knows is that Alice would have obtained $|\psi\rangle$ with probability $\frac12$ and $|\psi^\perp\rangle$ with probability $\frac12$ for some orthonormal basis $|\psi\rangle$, $|\psi^\perp\rangle$. Therefore, his state is a mixture $$ \rho_B = \frac12|\overline\psi\rangle\langle\overline\psi|+\...


2

The term "non-local states" most likely refers to states which allow to violate a Bell inequality. They give rise to statistics that cannot be explained with a local hidden variable model.


2

You don't state this explicitly, but I'm guessing this is the crucial part: How do $|A_i\rangle$ relate to $A$? I assume that $|A_i\rangle$ correspond to normalized rows, $i$ of matrix $A$, while $\|A_i\|$ is the weight of the row $i$ such that $\|A_i\||A_i\rangle$ would have all the elements corresponding to the $i^{th}$ row of matrix $A$. In other words, $$...


2

The claim does not specify what protocols for distinguishing quantum states are acceptable. In particular, it does not state whether we are allowed to err or reserve judgment. Below, we note success probability for protocols allowed to err and compute success probability for an error-free protocol. The success probability of the former is not bounded by $4\...


2

A single qubit is indeed different from a classical coin in a probabilistic state, for the reasons that the other answers have articulated. But at a deeper level, you are correct: a single qubit can indeed be thought of as a classical system, because there exists a local hidden-variable model that exactly reproduces its statistics. The only important ...


1

This is because of these lines of code: # bob reverse the initialization gate inverse_init_gate = init_gate.gates_to_uncompute() qc.append(inverse_init_gate, [2]) These two lines add the gates that set $q_2$ back to $|0\rangle$. Just remove them and the output should look like: {'1 1 1': 266, '1 0 0': 261, '1 1 0': 240, '1 0 1': 257}


1

Heuristically, when $\sigma$ and $\rho$ are "close," we can write $$\sigma=(1-\epsilon)\rho+\epsilon\varrho$$ for some small positive number $\epsilon$ and some other normalized state $\varrho$. Then $$\sigma-\rho=\epsilon(\varrho-\rho),$$ so $T$ is the projector onto the positive eigenspace of $\varrho-\rho$. If we take the extreme case that $\rho$...


1

We can transform the second expression as follows $$ \begin{align} \bigotimes_{l=1}^{n} \frac{1}{\sqrt 2}\left(|0\rangle + e^{2\pi i k/2^l} |1\rangle\right) &=\frac{1}{\sqrt{2^n}}\bigotimes_{l=1}^{n}\left(e^{2\pi i\cdot 0 \cdot k/2^l}|0\rangle + e^{2\pi i\cdot 1 \cdot k/2^l} |1\rangle\right)\tag1\\ &=\frac{1}{\sqrt{2^n}}\bigotimes_{l=1}^{n}\sum_{m=0}^...


1

Yes, probabilities shown in a histogram are squares of absolute values of coefficients in a state vector (i.e. probability amplitudes). Note that the amplitudes are generally complex numbers, hence we square the absolute values and not the amplitudes themselves.


1

Yes, the reasoning is correct. In fact, it can be generalized beyond pure states. By definition, every mixed quantum state $\rho$ is a positive semidefinite operator with unit trace. Since every positive semidefinite operator is Hermitian, we may interpret $\rho$ as an observable. In this case, the expectation of observable $\rho$ in state $\sigma$ $$ \...


1

(Case of pure states) Let $\rho=|\psi\rangle\!\langle\psi|$ be a pure bipartite state, suppose the underlying space is $\mathbb C^n\otimes\mathbb C^m$, and write as $|\psi\rangle=\sum_{k=1}^r \sqrt{p_k}(|u_k\rangle\otimes|v_k\rangle)$ the Schmidt decomposition of $|\psi\rangle$, with $r\le \min(m,n)$. Then $\rho$ has a single nonzero eigenvalue equal to $1$, ...


1

An easy (but perhaps cheating) answer: The relative entropy $S(A||B)$diverges when $A$ has support over the kernel of $B$ (e.g., wikipedia). Now, the kernel of $B=|0\rangle^{\otimes n}\langle 0|^{\otimes n}$ is everything other than $|0\rangle^{\otimes n}$, i.e., $\mathbb{I}-|0\rangle^{\otimes n}\langle 0|^{\otimes n}$! For the inequality to be meaningful, ...


1

If you run the following code you will see the output statevector plotted in Bloch sphere. qc = QuantumCircuit(1) qc.ry(3 * np.pi/4, 0) sim = Aer.get_backend('statevector_simulator') result = execute(qc, sim).result() output_state = result.get_statevector(qc) plot_bloch_multivector(output_state) Now from the picture you can see that the state lies near ...


1

Repetition codes are great for helping to demonstrate some understanding of how things work. But they are not good quantum codes. In fact this is part of the reason why they are so good for demonstrating some of the ideas - they help bridge the gap between the classical intuition that we're used to and the quantum world which often feels a bit less ...


1

Cirq has a function cirq.sub_state_vector which can extract a single qubit's state from a full state vector. It doesn't just do the single qubit case, it can do arbitrary subsets of qubits. It will raise an exception if the subset you pick is entangled with other stuff. It's unfortunately a bit picky about error tolerances and input shape.


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