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8

A mixed state represents a lack of knowledge on the system (maybe caused by the observer, maybe more fundamental). It is a notion in my opinion closely related to the Bayesian interpretation of quantum theories (seen as non-commutative probability theories). Quantum states are (non-commutative) probabilities, i.e. they encode all the information on ...


8

A unitary operator $U$ can only take a pure state $|\psi \rangle$ to a pure state. Let $U | \psi \rangle = | \psi' \rangle$. Then acting the unitary $U$ on the pure state $| \psi \rangle$ yields $$ U \left( | \psi \rangle \langle \psi | \right) U^\dagger = | \psi' \rangle \langle \psi' |,$$ which is a pure state. Another way to see this is that a (...


3

By an incoherent state (relatively to a basis, and it must be specified), they simply mean a mixed state described by a diagonal density matrix (in this basis). The word "coherence" refers to the usual thing in the discussion of "decoherence" (indeed, it has no simple relationship with the coherent states of harmonic oscillators). Coherence is the ...


2

It seems that by "operator" you mean a time evolution operator $\exp\left(\frac{i}{\hbar}\,H\,t\right)$ where $H$ is a quantum system's Hamiltonian, and such an operator by definition always maps (acts on) a pure quantum state to another pure quantum state. Unitary evolution is what happens whenever quantum measurement doesn't. So your statemtent "I assume ...


2

I think the most interesting approach in this direction is Caticha's entropic dynamics, for example in his "Entropic Dynamics, Time and Quantum Theory", arxiv:1005.2357. "Quantum mechanics is derived as an application of the method of maximum entropy. No appeal is made to any underlying classical action principle whether deterministic or stochastic. ... Both ...


2

This experiment has been done, first by Birgit Dopfer in 1998, then later by Dr. John Cramer of the University Of WA. In Dopfer's experiment, there was a "coincidence detector" which is basically an AND gate to filter out only the entangled pairs. By moving the detector in the beam of photons not going to the double-slit, the information about the photon's ...


2

If Alice observes an electron from a maximally entangled EPR pair, the expectation value of the spin is always $$\langle \vec S \rangle = 0,$$ not the cosine indicated by the OP. It's because the EPR pair always has the 50% probability for one value of the spin and 50% probability for the opposite value of the spin. These odds are true regardless of the axis ...


2

By "Copenhagen interpretation", I assume that you mean the interpretation with instantaneous "collapse" one usually encounters in an introductory quantum theory course. Such collapse is a useful rule to do calculation but it is only a fiction. What typically happens is that the quantum system is correlated with the macroscopic measurement device and other ...


2

The first term (sum) in $\bar H$ obviously commutes with all $\sigma_x$ variables because it's a function of $\sigma_x$ only and they commute with each other. The second term (sum) in $\bar H$ also commutes with the product of all $\sigma_x$ because the first term in the summand is a $c$-number and the second term $\sigma_z^j \sigma_z^{j+1}$ anticommutes ...


2

The given ansatz includes two assumptions. (1) The approximate ground states is seeked on a subspace of the Hilbert space consisting of rotated versions of single constant vector. (This subspace is a $2$-sphere parametrized by a unit vector in $\mathbb{R}^3$ (2) The value of the spin projection in the direction of the unit vector is half of the number of ...


2

Let $[G,H]=0$, and consider an eigenstate $|\psi\rangle$ of $H$, $$ H|\psi\rangle = E\psi\rangle\ . $$ Then, $|\psi'\rangle := G|\psi\rangle$ is also an eigenstate of $H$ with the same eigenvalue, since $$ H|\psi'\rangle = HG|\psi\rangle = GH|\psi\rangle = EG|\psi\rangle = E|\psi'\rangle\ . $$ Thus, also $|\chi\rangle = \tfrac{1}{\sqrt{2}}(|\psi\rangle + |\...


2

1) The reduction they are referring to is explain in the middle of page 13: "We now write (4.5) in the invariant sector as a sum of $n/2$ commuting $2×2$ Hamiltonians that we can diagonalize." 2) They are "reducing" the difficult problem of diagonalizing a very large matrix to the much easier problem of separately diagonalizing $n/2$ different $2 \times 2$ ...


2

When don't you condition on the result, measurement of a qubit can only decrease its purity (you end up with less information than you started with). When you do condition on the result, measurement of a qubit will make it 100% pure but there are two possible results. One possible result is along the measurement axis you measured. The other is against the ...


2

\begin{equation} b_{j} = \frac{1}{\sqrt{n}}\sum_{p}e^{-i\pi pj/n}\beta_{p}\qquad b_{j+1} = \frac{1}{\sqrt{n}}\sum_{p}e^{-i\pi q(j+1)/n}\beta_{q} \end{equation} Then \begin{equation} b_{j}^{\dagger}b_{j+1}^{\dagger} = \frac{1}{n}\sum_{p,q}e^{\pi i(p+q)j/n}e^{\pi iq/n}\beta_{q}^{\dagger}\beta_{p}^{\dagger} = \sum_{p} e^{-\pi ip/n}\beta_{-p}^{\dagger}\beta_{p}^{...


1

Yes. We can do this for any number of qubits using N dimensional spherical coordinates. For two qubits if we can write a general density matrix as a linear combination of direct products of Pauli matrices and identity , $$ \rho = \sum_{ij=0}^3 a_{ij}~ \sigma_{i} \otimes \sigma_{j}$$. Here $\sigma_{0} = I$, and the rest are the usual Pauli matrices. For ...


1

A valid density operator is any Hermitian, trace 1, matrix (with complex entries) and all eigenvalues between 0 and 1. Any two qubit system may be represented therefore by a Hermitian, trace 1 4x4 matrix. Your qubit representation could be rewritten, more suggestively as: \begin{align} \rho &= \frac{1}{2}\left(\operatorname{I} + a_1 \sigma_x + a_2 \...


1

I don't think identical purity by itself suffices to make $\rho$ and $\rho'$ equivalent, which implies identical spectrum. Assuming this is what you are looking for and both states are known, one general approach is to diagonalize both $\rho$ and $\rho'$, then find the basis transformation ...


1

Your example matrix $A$ isn't unitary. The right-most column has a length of 2 instead of 1, and the left-most column is only perpendicular to the right-most column if $a=-1$. You also seem to have computed $\rho_{\text{new}} = A \rho$ instead of $\rho_{\text{new}} = A \rho A^\dagger$. You can tell your $\rho_{\text{new}}$ must be wrong at a glance because ...


1

To see that there are two sector, corresponding to the eigenvalues of $G$ note that $G^2=1$ since $$ G^2 = (\prod^n_{j=1}\sigma^{(j)}_x)^2 = \prod^n_{j=1}(\sigma^{(j)}_x)^2 = 1$$ Thus there are two eigenvalues to $G$ that are $\pm1$. These sectors need not be of the same size. Consider just 2 spins in theire singlet and triplet configurations. The singlet ...


1

Yes, one particle can be in a mixed state when there is classical uncertainty present. Or at least it is the cleanest visualization one can make of a quantum state with classical uncertainty to think of it as mixed. The classic example here is the state of the particle ("cat") in the Wigner Friend Thought Experiment as Wigner would calculate it if he had to ...


1

Yes, it can. "Ensemble" may mean many experiments with one particle, not obligatory many particles at the same experiment.


1

Yes, they're realizable. All unitary operations on qubits (or qudits) are physically realizable (with negligible error) (unless you're working with an extremely limited gate set). The 'shift' operator is just adding a constant number to the little-endian number represented by the qubits (as if they were bits). Here's a blog post about efficiently ...


1

Both of the forms you propose, $$\sigma^{(1)}_x \otimes \sigma^{(2)}_x \ldots \otimes \sigma^{(n)}_x \tag1$$ and $$\left(\sigma_x \otimes \mathbb{I}_{n-1}\right)\cdot \left(\mathbb{I}\otimes \sigma_x\otimes \mathbb{I}_{n-2}\right) \cdot \left(\mathbb{I}_2\otimes \sigma_x\otimes \mathbb{I}_{n-3}\right)\cdot \ldots \cdot \left(\mathbb{I}_{n-1}\otimes \sigma_x\...


1

Take for example the two particle wave function $$|\Psi\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle_1 \otimes |1\rangle_2 + |1\rangle_1 \otimes |0\rangle_2\right)$$ The probability to measure 0 for the first particle is given by $$ |\langle 0_1|\text{Tr}_2(|\Psi\rangle)|^2 = \frac{1}{2} $$ The wave function after the collapse if 0 was measured is $$ |\Psi_c\...


1

First of all, let's clarify things: In Quantum Mechanics decomposing the wavefunction into a linear combination of eigenstates is not the same as decomposing the system into single particle basis. Say you have a unidimensional system of two particles, the wavefunction in terms of their position is $\psi(x_1,x_2)$ can be decomposed into a linear ...


1

A book that I highly recommend which covers the said topics is "Lectures on Quantum Theory: Mathematical and Structural Foundations" by Chris Isham. His treatment is more thorough than most textbooks and has some interesting insights. It is also well suited for someone without a deep quantum background. Two books which also covers these topics and are ...


1

You are misunderstanding (and thus misquoting) the book of Nielsen and Chuang. What they do is to first define the reduced density matrix $\rho_A$ of a bipartite state $\rho_{AB}$ as $$\rho^{A} =\mathrm{tr}_B(\rho^{AB})$$ Then, they explain what a partial trace is: It is the linear map defined by its action $$ \mathrm{tr}_B \left( |a_1 \rangle \langle a_2| \...


1

For context, any general quantum operation $\Phi$ on a bipartite system $AB$ with finite dimensional Hilbert space $H_{A} \otimes H_{B}$ has a Kraus representation $\Phi(\rho) = \sum_j K_j \rho K_j^{\dagger}$ where the "Kraus operators" $K_j$ are linear operators on $H_{A} \otimes H_{B}$ such that $\sum_j K_j^{\dagger} K_j = I_{AB}$. The Kraus ...



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