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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 ...


6

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 ...


0

$\renewcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ The original post leaves out the strength of the Hamiltonian, which is confusing, so I'm going to put it back in. We write $H/\hbar = \Omega \sigma_x$. Following the original post we have \begin{align} \exp(-i H t / \hbar) = \exp \left( -i \Omega t \sigma_x \right) &= \left( \begin{array}{cc} ...


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 ...


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| ...


0

A mixed state just describes our ignorance about a particular system I don't think you can call our inability to access a pure state of any system an ignorance about a particular system. Because I think of pure state as a mathematical abstruction that can only be related to reality by application of Born rule - wich either reflects our fundamental ...


5

The answer to your question is NO. The simplest counter-example is the identity matrix. There is no solution to $|a\rangle\langle b| = I$. For example, try to solve $\begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} z & t \end{bmatrix} = \begin{bmatrix} 1&0 \\0&1 \end{bmatrix}$. Since $xt=0$ we know that $x=0$ or $t=0$. But clearly $x$ ...


7

Not every tensor is a simple tensor. The simple tensors of the form $a\otimes b\in A\otimes B$ span the tensor product, which means that a general element $t$ of the tensor product is a linear combination of these simple tensors, i.e. $$ t = \sum c_{ij} a_i\otimes b_j$$ for some basis $a_i,b_j$ of $A$ and $B$ respectively. If $A$ itself is a space of ...


6

No. Counterexample: choose a basis in 2D quantum space, so the basis states are written $\left(\begin{array}{c}1\\0\end{array}\right)$ and $\left(\begin{array}{c}0\\1\end{array}\right)$. Now think of a mixture, with a state's having probability $p$ to be in state 1. The density matrix is then $\mathrm{diag}(p,\,1-p)$, which is not singular for ...


2

It seems I have figured out an answer for 2 terms in the original state. Suppose that the state is $$\rho = a |\alpha \rangle \langle \alpha | + (1-a) |\beta\rangle \langle \beta|$$ We need to write it in a basis, which is $$|+\rangle = \frac{|\alpha\rangle + |\beta\rangle}{\sqrt{2}}; \quad |-\rangle = \frac{|\alpha\rangle - |\beta\rangle}{\sqrt{2}}. $$ ...


1

The basis states are represented as four-dimensional vectors, like so: \begin{align} |00\rangle&=[1, 0, 0, 0]^T \\ |01\rangle&=[0, 1, 0, 0]^T \\ |10\rangle&=[0, 0, 1, 0]^T \\ |11\rangle&=[0, 0, 0, 1]^T \end{align} You then simply apply the matrix transformation to these vectors. For example, a quantum state may be expanded as a linear ...


2

The answer is superposition. Let's take one classical bit. It can be in two states: $|0\rangle$ or $|1\rangle$. Now let's consider a qubit: its general state will be $$a |0\rangle + b |1\rangle$$ where $a,b$ are complex numbers with the constraint $$|a|^2+|b|^2=1$$ Now you should be able to see the difference. A state like $$\frac{|0\rangle + ...


-1

Lets not use technical terms and understand this bit by bit, by example. In classical terms if a cat is kept with a radioactive poison inside a box, without any type of knowledge of what is going inside. A person with common sense would say that the cat is either alive "OR" dead. The word OR here is emphasized, because of some reason which we will come to ...


1

You can use circuit simulators to help with understanding small circuits like this, and to check your work. For example, here's your circuit in Quirk: The key things you need to understand for this circuit are: How to apply an operation on paper to an entangled state (for the H after the CNOT). Group by the uninvolved bits, and apply the operation ...


0

You seem very confused. You can work this one out without using density matrices and density matrices would only add complexity in this case, so you should avoid them in this particular problem. To figure out the state, you consider what each gate does in turn. And in this case one of the gates, CNOT, involves an interaction between the two qubits, so you ...


2

The title of the paper is "Optimum Unambiguous Discrimination Between Linearly Independent Symmetric States". Random guessing is an ambiguous discrimination procedure. It can return the wrong answer without telling you that it failed. More generally, unambiguous discrimination procedures are less likely to succeed because removing any chance of accidental ...


0

Wave Function collapse is a reconciliation between what quantum mechanics states should occur, and what we classically observe occurring. It is a process in which the quantum superposition of states becomes less important, and a single observed state becomes more important. Nothing in the equations of quantum mechanics demand the observer effect. You can ...


1

A full quantum mechanical description of the interferometer is complicated because it's not an isolated system. But we can do a thought experiment where we imagine it to be made out of mirrors that are floating in free space. Then as Anna has explained, the photon will interact with the entire system, it will not excite vibrational modes of the lattice. This ...


2

This hand waving assumption you are making is the crux: At the mirrors A and B (and also at the half-transparent ones not considered) the photon interacts with one or more electrons of the mirror, transferring momentum The photon is not interacting with one or more electron on its way, it is interacting with the lattice of atoms. This means that the ...


0

The author shows that since $UN$ equivalent to $UNU^\dagger U$, that it is possible to change the order of the operations, if $N$ is replaced by $M=UNU^\dagger $. In that case, $UN$ = $MU$. The remaining logic follows from this. Note that $ M $ is just $ N $ in the $ U $ basis; the choice of $ U $ uniquely determines how $ N $ changes to $ M $ for this ...


4

The operations aren't faster, they're more flexible. From that flexibility comes power. For example: in a classical computer there is no single-bit operation that, when applied twice, flips a bit. There's no boolean function $f$ such that $f(f(x)) = \overline{x}$. But quantum computers do have such an operation, represented by the matrix $M = \frac{1}{2} ...


4

No, $|\phi\rangle$ doesn't need to be orthogonal to $|0\rangle$. Neither does $|\psi\rangle$. They only need to be orthogonal to each other. Example Suppose you know that you will be given a state $|\omega\rangle$ that will either equal $|\phi\rangle = \frac{1}{\sqrt 2}|0\rangle + \frac{1}{\sqrt 2}|1\rangle$ or equal $|\psi\rangle = \frac{1}{\sqrt ...


1

Particles are described by quantum fields, and the quantum field determines the mass, spin and charge. So for example all electrons (and positrons) have the same mass, spin and (magnitude of) charge because they are all excitations of the electron quantum field. Individual electrons can have different energy and momenta, but I'm guessing you wouldn't ...



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