28 votes

How do I show that a two-qubit state is an entangled state?

A two qudit pure state is separable if and only if it can be written in the form $$|\Psi\rangle=|\psi\rangle|\phi\rangle$$ for arbitrary single qudit states $|\psi\rangle$ and $|\phi\rangle$. ...
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12 votes

Composing the CNOT gate as a tensor product of two level matrices

The whole point is that CNOT cannot be written in the form $A\otimes B$. This is absolutely essential because if we only ever had operators of the form $A\otimes B$, states of the form $|\psi\rangle\...
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  • 46.1k
11 votes
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What role does the non-commutativity of the tensor product play in experimental quantum computation?

For many questions that appear on this site, and about quantum information and computation in general, it is possible to ask a completely classical version of the question, and often the (sometimes ...
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  • 4,413
9 votes
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Is there an algorithm for determining if a given vector is separable or entangled?

Imagine you have a vector that can be written in the form $$ |\psi\rangle=\sum_{i=0}^{d_A-1}\sum_{j=0}^{d_B-1}c_{ij}|i\rangle|j\rangle. $$ The coefficients can be arranged as a $d_A\times d_B$ matrix $...
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  • 46.1k
8 votes
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Partial trace over a product of matrices - one factor is in tensor product form

The equation at the top of the question is not correct: there is a missing factor of $1/d$ on the right-hand side. Let's eliminate this factor from the left-hand side to make it simpler, so that the ...
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  • 4,413
8 votes

Prove that $\|p^{\otimes n} - q^{\otimes n}\| \leq n \|p-q\|$ for density operators $p,q$

Marsl is correct, and his "hint" is really more a sketch of a solution than a hint. Rather than viewing the question or its solution as just formal algebra, you can also approach his same solution ...
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7 votes
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What circuit or operation corresponds to the tensor product?

I think that you misunderstood the concept of tensor product here. There is no need of a Clifford gate in order to have a multi-qubit system. The fact that a multi-qubit system is described by the ...
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7 votes

Partial trace over a product of matrices - one factor is in tensor product form

Here the important fact is that the maximally mixed state is in fact an identity matrix. Let me rewrite the expression on the left in index notation (the summation sign is omitted according to the ...
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7 votes

Confusion regarding projection operator

Does $|0\rangle\langle0|$ represent a tensor product or is it just matrix multiplication? You can think of $|0\rangle\langle0|$ as tensor product of $|0\rangle$ and $\langle0|$, or equivalently as ...
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  • 18.1k
7 votes
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Prove that $\|p^{\otimes n} - q^{\otimes n}\| \leq n \|p-q\|$ for density operators $p,q$

Hint: To make your induction work, write $$\eqalign{p^{\otimes n} - q^{\otimes n} & = & \left(p^{\otimes(n-1)}\otimes p \right)-\left(q^{\otimes (n-1)} \otimes q\right)\\ & = & \left(...
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  • 779
7 votes
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Stinespring dilation: Size of environment

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{...
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7 votes

Writing state $ |\Psi⟩ =\frac{1}{\sqrt{2}}|00⟩+\frac{i}{\sqrt{2}}|01⟩$ as separate qubits (qiskit textbook)

As is the case with ordinary multiplication, tensor product distributes over addition, so we can pull $|0\rangle$ on the first qubit out in front $$ \begin{align} |\Psi⟩ &= \frac{1}{\sqrt{2}}|\...
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7 votes

In Stinespring dilation, can we always use a mixed state as the ancilla?

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 ...
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6 votes

What do we mean by the notation $\lvert \mathbf{x}, 0\rangle$?

Yes, $|\mathbf{x},0\rangle$ is a shorthand for $|\mathbf x\rangle\otimes |0\rangle$. Note that $|\mathbf x\rangle$ itself, with $\mathbf x = x_1x_2\dots x_N$ a bit string, is just a shorthand for $$ |...
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6 votes
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Symmetry of tensor product w.r.t. Vazirani 2-qubit video

The tensor product is not commutative, i.e. in the computational basis, $$X\otimes Z=\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \...
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  • 2,334
6 votes
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Proof of no-cloning

For step (116), the equivalence between both of them is proved by \begin{equation} (\langle\psi_1|\otimes\langle0|)C^\dagger C(|\psi_2\rangle\otimes|0\rangle) = (\langle\psi_1|\otimes\langle0|)(|\...
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6 votes
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What does the notation $U(B,\beta) = \prod_{j =1}^n e^{-i \beta \sigma_j^x} $ mean in the context of QAOA?

I think there are two ways that you could denote the same thing. The first is what is done here: $$ \prod_{j =1}^n e^{-i \beta \sigma_j^x} $$ The second is $$ \bigotimes_{j-1}^ne^{-i \beta \sigma^x}, $...
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6 votes
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Quantum tensor product closer to Kronecker product?

The tensor product of two objects with $m$ and $n$ components is an object with $mn$ components that consists of the pairwise products of the components of the inputs. The Kronecker product and the $v ...
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  • 639
5 votes

Why is the state of multiple qubits given by their tensor product?

Perhaps it helps to take a step back and start with something simpler: why do we tabulate probability amplitudes for state vectors and unitaries? For a single quantum system with $d$ distinct ...
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  • 46.1k
5 votes

What role does the non-commutativity of the tensor product play in experimental quantum computation?

When you say $\neq$ I presume you are talking about the implied basis in usual ordering like (00, 01, 02, 10 etc). Otherwise you would have the isomorphism of Hilbert spaces vs an equality statement. ...
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  • 3,503
5 votes
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What is the $\left| 22\right>$ state?

It is worth emphasising that the stuff that you write inside a ket is completely arbitrary. It's just a label you're attaching to something, so it should have a proper definition somewhere. Now, ...
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5 votes
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How to keep track of entanglements when emulating quantum computation?

It is certainly sufficient to always calculate the full $2^n\times 2^n$ unitary matrix, and then apply it to the $2^n$-entry state vector. If that's what you choose to do, that's all you have to do as ...
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  • 46.1k
5 votes

Composing the CNOT gate as a tensor product of two level matrices

Any two qubit Controlled-U gate is not decomposable as $A\otimes B$ in general. The whole point is that a conditional operator has a non-local nature and it gives rise to non-trivial correlations when ...
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5 votes
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How to interpret $-\rvert1\rangle \otimes \rvert1\rangle = -\rvert11\rangle$?

First of all note that, strictly speaking, quantum states are defined up to multiplication by complex scalars (that is, they are elements of the associated projective space $\mathbb{CP}^n$). This ...
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  • 18.1k
5 votes

Can every bipartite state be written as $\rho_{AB} = \sum_{ij} c_{ij}\sigma_A^i\otimes \omega_B^j$?

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 $\rho$ ...
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5 votes
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If you apply a unitary transformation to an entangled state, is it still entangled?

The state $|\psi \rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$ is a maximal entangled state that can be created by applying the unitary transformation $U_1 = CNOT \cdot (H \otimes I) $ to the ...
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  • 12.3k
5 votes
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Can we express $\mathrm{tr}_A((A\otimes B)\rho_{AB})$ in terms of $A$, $B$, $\rho_A$ and $\rho_B$?

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 ...
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5 votes

Negative Probability — Reality vs Description

"I understand that quantum physics supports the concept that the probability of a qubit collapsing into (say) 1, can be negative or positive [...] If that is wrong, please say so; I understand ...
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  • 11.9k
4 votes

Symmetry of tensor product w.r.t. Vazirani 2-qubit video

I'll add a bit to the other answer. State of a two qubit system is written as $|\psi_1\rangle\otimes|\psi_2\rangle$ where $|\psi_1\rangle$ is the state vector of the first qubit and $|\psi_2\rangle$ ...
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