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Definition A two-qubit state $|\psi\rangle \in \mathbb{C}^4$ is an entangled state if and only if there not exist two one-qubit states $|a\rangle = \alpha |0\rangle + \beta |1\rangle \in \mathbb{C}^2$ and $|b\rangle = \gamma |0\rangle + \lambda |1\rangle \in \mathbb{C}^2$ such that $|a\rangle \otimes |b\rangle = |\psi\rangle$, where $\otimes$ ...

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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 $|\phi\rangle$. Otherwise, it is entangled. To determine if the pure state is entangled, one could try a brute force method of attempting to find satisfying states $|\psi\rangle$ and $... 11 For many questions that appear on this site, and about quantum information and computation in general, it is possible to ask a completely classical version of the question, and often the (sometimes obvious) answer that one finds in the more familiar classical setting translates directly to the quantum setting. In this case, a reasonable classical version of ... 10 The whole point is that CNOT cannot be written in the form$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\otimes|\phi\rangle$would always remain separable. There's be no entanglement, and all quantum circuits would be easy to simulate on a classical computer. As ... 8 Marsl is correct, and his "hint" is really more a sketch of a solution than a hint. Rather than viewing the question or its solution as just formal algebra, you can also approach his same solution more conceptually. The conceptual reasoning is really identical to the algebra, just phrased differently. You can rely on the following two facts: 1) Trace ... 7 Does$|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 the matrix multiplication (more precisely, Kronecker product) of the vectors representing them. Also, I thought that we must always be able to write a ... 7 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$C$, with the elements$c_{ij}$(in your special case, you're talking about setting$d_A=d_B=\sqrt{m}$). Now, if you calculate$\rho_A=CC^\dagger, this is ... 7 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(p^{\otimes(n-1)}-q^{\otimes (n-1)} \right)\otimes p+\left(q^{\otimes (n-1)} \right) \otimes (p-q)} Then, use triangle inequality and finally the fact that ... 6 The tensor product is not commutative, i.e. in the computational basis, $$X\otimes Z=\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ \end{array} \right)$$ while $$Z\otimes X=\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\... 6 Here the important fact is that the maximally mixed state is in fact an identity matrix. Let me rewrite the expression on the left in index notation (the summation sign is omitted according to the Einstein convention):$$ Tr(\rho^{AB} (\sigma^A \otimes I/d)) = [\rho^{AB}]_{ijkl} [\sigma^A]_{ji} [I/d]_{lk} $$But [I/d]_{lk} = \frac1d \delta_{lk}, ... 6 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 tensor produt of their state vectors comes from the fourth postulate of quantum mechanics, refer to the 94^{th} page of Nielsen and Chuang to see the formulation ... 5 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 states, labelled 0 to d-1, we associate a complex number a_i with the probability amplitude for being in state i. For a unitary, we associate a probability ... 5 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. That is, AB implies a certain ordered basis and BA a different one. The experiment has it's observables on the combined system in a basis independent way. If ... 5 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, usually, we're talking about quantum spins with a certain size of Hilbert space, say d. Probably here you're talking d\geq 3, and perhaps specifically d=3. ... 5 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 all the entanglement information is contained in that vector. A quick and easy way to see if a particular qubit is entangled is to take the partial trace of ... 5 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$$ |\mathbf x\rangle \equiv |x_1\rangle \otimes |x_2\rangle \otimes\cdots |x_N\rangle\ . $$5 Any two qubit Controlled-U gate is not decomposable as A\otimes B in general. The whole point is that a conditional operator has a non-local nature and it gives rise to non-trivial correlations when operated on a state. The Physics behind is that; imagine a Controlled operation which can be factored as A\otimes B, and the sub-systems (or the two qubits)... 4 I'll add a bit to the other answer. State of a two qubit system is written as |\psi_1\rangle\otimes|\psi_2\rangle where |\psi_1\rangle is the state vector of the first qubit and |\psi_2\rangle is the state vector of the second qubit (it is upto you to decide which one will be called "first" and which one will be called "second"). The order is ... 4 The tensor product is not a gate, but rather a way for us as humans to model the behavior of a quantum system. Whenever we're using multiple qbits, we can look at them in two ways: in their product state (a complex vector of size 2^n for n qbits) or their individual state (|\psi_0\rangle \otimes \ldots \otimes |\psi_{n-1}\rangle). We can usually switch ... 4 The equation at the top of the question is not correct: there is a missing factor of 1/d on the right-hand side. Let's eliminate this factor from the left-hand side to make it simpler, so that the equation we want is this:$$ \text{Tr}\bigl(\rho^{AB} \bigl(\sigma^A \otimes I\bigr)\bigr) = \text{Tr}\bigl(\rho^A \sigma^A\bigr). $$To see why this is true, ... 4 Here i and j are bit strings of size n. Correspondingly, |i\rangle, |j\rangle are some basis vectors in 2^n-dimensional space, that corresponds to n-qubit register. Those controlled operations C and V act on (n+m)-qubit space. You can consider first n qubits as control register and last m qubits as target register. Now, C_n^j(U_{... 4 First of all note that, strictly speaking, quantum states are defined up to multiplication by complex scalars (that is, they are elements of the associated projective space \mathbb{CP}^n). This means that it is not true that "-\lvert1\rangle is just \lvert1\rangle but time or space delayed in time". The vectors -\lvert1\rangle and \lvert1\rangle ... 4 The reason is relatively straightforward. Consider an m dimensional vector space V with basis \lbrace \vert v_1 \rangle,...,\vert v_m \rangle \rbrace, and an n dimensional vector space W with basis \lbrace \vert w_1 \rangle,...,\vert w_n \rangle \rbrace. As your intuition suggests, we can naturally express any element A \in V \otimes W in the ... 4 As you say, start by expanding e^{-iS\Delta t}=\cos(\Delta t)I-i\sin(\Delta t)S, so you'd be calculating$$ (\cos(\Delta t)I-i\sin(\Delta t)S)\rho\otimes\sigma(\cos(\Delta t)I+i\sin(\Delta t)S). $$If you multiply out all the terms, then the \cos^2(\Delta t) comes from the two I terms, leaving you with \rho\otimes\sigma. If you trace out the first ... 4 I think there are two ways that you could denote the same thing. The first is what is done here:$$ \prod_{j =1}^n e^{-i \beta \sigma_j^x} $$The second is$$ \bigotimes_{j-1}^ne^{-i \beta \sigma^x},$which I imagine is what you're thinking of. In the first expression, note the subscript on the Pauli matrix. This means that it's an operator over all$n$... 4 The idea of a tensor product is to link two Hilbert spaces together in a nice mathematical fashion so that we can work with the combined system. Normally, these two Hilbert spaces each consist of at least one qubit, and sometimes more. Let's say we have a qubit, which we label$a$, and a qubit which we label$b$. These qubits 'live' in the Hilbert spaces of$...

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Annoyingly, the answer is "it depends on what you mean by $|\psi\rangle \otimes |\phi\rangle$." A tensor product of vectors from a collection of Hilbert spaces over the same field is simply a choice of one vector from each Hilbert space, with some equivalence relations modded out. (A tensor product of Hilbert spaces is more complicated.) Let $\... 3 For step (116), the equivalence between both of them is proved by $$(\langle\psi_1|\otimes\langle0|)C^\dagger C(|\psi_2\rangle\otimes|0\rangle) = (\langle\psi_1|\otimes\langle0|)(|\psi_2\rangle\otimes|0\rangle) = \langle\psi_1|\psi_2\rangle\otimes\langle0|0\rangle=\langle\psi_1|\psi_2\rangle\langle0|0\rangle,$$ where in the ... 3 The order in the tensor product is a convention and has nothing to do with experiments. As an example, if I have a cavity (with photons in it,$H_A$) and an atom (with internal states,$H_B\$), it is clear which is the atom and which is the cavity, regardless of the order ones chooses for their Hilbert spaces in the tensor product when describing the setup ...

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