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Single Photon Interference and the Hong-Ou-Mandel (HOM) Effect in an Interferometer

In our current era of technology, this is pretty straightforward. Two Photon Interference: As you mention the HOM effect, I have the original 1987 paper they presented - still relevant today. Two ...
DrChinese's user avatar
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How can a QFT field act on particle states in Fock space?

You had encountered exactly the same situation already in elementary quantum mechanics, but maybe you did not recognize the analogies. Consider a single harmonic oscillator with angular frequeny $\...
Hyperon's user avatar
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How can a QFT field act on particle states in Fock space?

The field operators do not represent particles, they are operators that, if acting on a state (a vector in fock space which represents a state with or without particles), can change the state of the ...
Wolphram jonny's user avatar
-1 votes

Single Photon Interference and the Hong-Ou-Mandel (HOM) Effect in an Interferometer

We can tune both arms of an MZ interferometer so that no photons go thru. So now we have 0 photon interference! Photons will not travel the arms if the path length between mirrors is lambda x n + 0.5 ...
PhysicsDave's user avatar
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Why are the angular momentum raising and lowering operator coefficients real?

Yes, OP is right: Normalized eigenstates $|j, m\rangle$ are in principle defined modulo an arbitrary phase factor. However, it is customary to chose the phase factors of neighboring eigenstates $|j, m\...
Qmechanic's user avatar
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Deducing the ground state from a known first excited state

For b=1, your problem reduces to the well-known PΓΆschl–Teller potential, shifted by a unit, so your excited state has $$ \psi''= \left (1-{2(2+1)\over \cosh^2(x)}\right )\psi, $$ alright. And then the ...
Cosmas Zachos's user avatar
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What is the distinction between a ket and a state in quantum mechanics?

I think all other answers so far do not address the problem directly. The definition of the OP regarding kets is a quite common, though not the most used one, see e.g. references 1 and 2. So I don't ...
Tobias Fünke's user avatar
-1 votes

Why are particles in Quantum Mechanics indistinguishable?

Identical vs Indistinguishable A particle is not a point in space in QM, it has a spread of its probability. Many answers assume that indistinguishability = Identical, that is not what we are talking ...
quanta's user avatar
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Why we cannot transfer information using a spin entangled singlet?

In quantum mechanics, the information you can get from a system is given by the expectation values of its observables. If you have two spatially separated systems $S_1$ and $S_2$, the expectation ...
alanf's user avatar
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5 votes

What is the distinction between a ket and a state in quantum mechanics?

This is really two questions in one, namely what is a ket and what is a state. Both questions are bound to cause some degree of controversy between physicists of different mathematical pedigrees. I ...
J. Murray's user avatar
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Why we cannot transfer information using a spin entangled singlet?

There are a number of issues with your description of entanglement, including the question title. You acknowledge that FTL (nonlocal) communication is not feasible using entanglement, and even supply ...
DrChinese's user avatar
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What is the distinction between a ket and a state in quantum mechanics?

A ket is an element of the Hilbert space, $| \psi \rangle \in \mathcal{H}$, i.e., it is an element (a vector) of the complex inner-product space. I think this is well-accepted terminology (introduced ...
Ben H's user avatar
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What is the distinction between a ket and a state in quantum mechanics?

Besides the correct introduction to the bra-ket notation done by @AndrewSteane, a cleaner mathematical notation may help to understand the situation and how manipulations go. Kets are elements of the ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
3 votes

What is the distinction between a ket and a state in quantum mechanics?

A ket is an element of a Hilbert space. A ket may also be called a state vector. An operator acting on a ket produces a ket. Suppose we have a two-dimensional Hilbert space and a Hermitian operator $\...
Andrew Steane's user avatar
1 vote

Probability distribution for the momentum of a quantum harmonic oscillator

...point me towards the... probability distribution for the momentum of a quantum harmonic oscillator in the canonical ensemble. To establish some notation, I note that the distribution you are ...
hft's user avatar
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The contradiction between Gell-mann Low theorem and the identity of Møller operator $H\Omega_{+}=\Omega_{+}H_0$

Actually, the original Gell-Mann-Low Theorem(GLT) can be deduced by the Adiabatic Theorem(AT), which requires no degeneracy and discrete spectrum. However, in more general cases, say, degenerate ...
Sakana's user avatar
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What will be wave function after application of operator?

No, the state vector after a position measurement is not given by $X |\psi\rangle$. Suppose your detector can decide whether the position of the particle is inside oder outside a certain (space) ...
Hyperon's user avatar
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Can it be disproven that an electron is a wave packet of photons?

As mentioned in the comments, photons are neutral, so no collection of photons can have charge $-1$; moreover, they are spin 1, so no collection of photons can have spin $\frac 1 2$. Meanwhile equal ...
JEB's user avatar
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Can it be disproven that an electron is a wave packet of photons?

In physics we prove things by testing them experimentally. You already seem to know what the experiments say. Our current (very accurate) model of quantum electrodynamic phenomena doesn't support your ...
John Doty's user avatar
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What is the difference between local operations and global operations for quantum systems?

Typically in a quantum circuit context, people aren't binary about whether an operation is "local" or "global", since, for example, in a 4-qubit system you could have an operation ...
KandC's user avatar
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4 votes
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Energy levels of a translating quantum harmonic oscillator

OP seemingly wants to understand whether or not there is a contribution to the energy from the overall translation of the system with velocity $v$. Of course there should be, and we should expect ...
hft's user avatar
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Energy levels of a translating quantum harmonic oscillator

$\hat H_𝑣=\exp(βˆ’π‘–π‘£π‘‘\hat p/ℏ)~\hat H\exp(𝑖𝑣𝑑\hat p/ℏ)$, so the two hamiltonians are canonically equivalent and have the same spectrum. You may also see this from $[\hat x -vt,\hat p]=i\hbar$, so ...
Cosmas Zachos's user avatar
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Why does this not break the security of quantum teleportation

I am going to assume in this answer that you're concerned about the security of quantum cryptography not teleportation. You say that Eve gives Alice and Bob the following state: $$|\Psi_{ABE}^-\rangle=...
alanf's user avatar
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Time reversal in momentum space

According to a theorem by Wigner (see E.P. Wigner, Gruppentheorie, Vieweg, Braunschweig, 1931, pp. 251-254), a symmetry transformation in quantum mechanics is either represented by a unitary or an ...
Hyperon's user avatar
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Maximum number of partial waves and matching point

The picture shows is a plot of magnitude of S-Matrix square against angular momentum L. When the coulomb excitatian potential is absent (No pot in legend) but only imaginary potential is present the ...
SAKhan's user avatar
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2 votes
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What does $f(x)$ satisfies the given equation means?

Now the solution just requires one to prove that $\partial^2/\partial x^2=\partial^2/\partial (-x)^2$. Yes, if you can "just" prove the above-quoted equality you are basically "done.&...
hft's user avatar
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What does $f(x)$ satisfies the given equation means?

As you note, the form of the equation doesn't change, which implies that $\psi(-x)$ is just as much a solution as $\psi(x)$ is. Similarly, one can take any linear combination $\phi(x)=c_1\psi(x)\pm ...
Albertus Magnus's user avatar
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Why are these unbounded operators (essentially) self-adjoint?

Let me expand on my comment: For every operator you mentioned, I assume that $A$ is a certain function of $X$ and $P$ and possibly their powers with their usual definitions in $L^2 (\mathbb{R})$ which ...
DanielC's user avatar
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Why does classical light always result in super-Poissonian statistics?

For a single-mode thermal state, yes, we can, in general, show that the photon-counting statistics is super-Poisson using the method you proposed (as @glS has done it nicely). However, for a multi-...
Omid's user avatar
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1 vote

"Derivation" of group velocity and "sharply" peaked wave functions

The point is of course, to talk about the group velocity of a free particle with a somewhat well defined momentum. If you use a Dirac delta function then you throw any concept of "group" or ...
Albertus Magnus's user avatar
-1 votes

Spin-orbit interaction and retarded potential

The spin and magnetic moment of the electron come out of the relativistic Dirac equation for the electron. Their appearance is unrelated to the velocity of the electron, and does not require high ...
Jerrold Franklin's user avatar
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Spin-orbit interaction and retarded potential

Spin-orbit interaction is not directly related to the velocity of the electron (in whichever way such a velocity is defined). However, the velocity can serve as a general indicator of the importance ...
dennismoore94's user avatar
2 votes
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Square of the density operator for a mixed state

Assume, for example, that the states $|\psi_k\rangle$ are mutually orthogonal. In an orthonormal basis including the $|\psi_k\rangle$, $\rho$ is diagonal and its elements are the $p_k$. To obtain $\...
Sancol.'s user avatar
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2 votes
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What happens if two people have different knowledge about a state in a quantum mechanical system?

Suppose you have an ensemble of identical quantum systems prepared in a pure state $\psi$. Next a measurement occurs of an observable $A$ with two possibles outcomes $a$ and $a’$. A measurement is an ...
Valter Moretti's user avatar
3 votes

Are Hermitian operators Hermitian in any basis?

An operator is not Hermitian "in a basis". An operator $A$ on an inner product space is defined to be Hermitian iff it equals its own adjoint, $A = A^\dagger$. This condition has nothing to ...
tparker's user avatar
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4 votes
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Are Hermitian operators Hermitian in any basis?

A linear operator on a Hilbert space $A:\mathcal H\to\mathcal H$ is Hermitian if for all $v,w\in\mathcal H$, $$\langle Av, w\rangle = \langle v,Aw\rangle$$ This definition makes no reference to any ...
Er Jio's user avatar
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What happens if two people have different knowledge about a state in a quantum mechanical system?

Working in the basis of the 1st measurement, the electron is in a pure state: $$ \chi_0 = \left[\begin{array}c a \\ b\end{array}\right]_1$$ with $||a||^2 + ||b||^2 = 1$. You measure, and it goes to a ...
JEB's user avatar
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1 vote

What happens if two people have different knowledge about a state in a quantum mechanical system?

Measuring is how people acquire this knowledge. One person measures the spin of the electron with respect to a vertical axis. This returns an up or down value, and puts the electron in a spin up or ...
mmesser314's user avatar
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4 votes

Difference between real operators and Hermitian operators in quantum mechanics

In order to avoid unnecessary mathematical complications, let us consider a finite-dimensional complex Hilbert space $\mathcal H$, i.e. a complex vector space with dimension $n\lt \infty$ and a (...
Hyperon's user avatar
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Deduction of Kinetic energy operator in quantum mechanics

Actually mathematical part of kinetic energy derivation is Ok : $$ \begin{align} \hat T &= \frac {\hat p^2}{2m}\\ &=\frac {1}{2m}\left(\hat p \cdot \hat p\right)\\ &=\frac {1}{2m}\left(-i\...
Agnius Vasiliauskas's user avatar
-1 votes

How to determine parameters such that the state $|\psi\rangle=\frac1{\sqrt2}|+\rangle|+\rangle+a|+\rangle|x+\rangle+b|-\rangle|-\rangle$ is separable?

So I started with rewriting state $|x+\rangle$ in z-representation: $$ |\psi \rangle = \frac{1}{\sqrt{2}} |+\rangle|+\rangle + a|+\rangle \Bigl( \frac{1}{\sqrt{2}} \bigl( |+\rangle + |-\rangle \bigr) \...
hft's user avatar
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How to determine parameters such that the state $|\psi\rangle=\frac1{\sqrt2}|+\rangle|+\rangle+a|+\rangle|x+\rangle+b|-\rangle|-\rangle$ is separable?

The best way to quantify the entanglement for a 2-qubit state is something called the "concurrence"; it goes to zero for a separable state and 1 for a maximally entangled state (like a Bell ...
Ken Wharton's user avatar
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What happens to entangled particles at a Black Hole?

Anything you could learn from an observation on one particle of a maximally entangled particle pair, you could still learn even after the other enters a black hole. To understand why this is true: ...
DrChinese's user avatar
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2 votes
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Difference between real operators and Hermitian operators in quantum mechanics

A real linear operator is one whose matrix elements are real. For example, given an orthonormal basis $\{|\psi_k\rangle\}$,and an operator $\hat O$, suitably defined on a Hilbert space $\mathcal H$; ...
Albertus Magnus's user avatar
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How to solve for a particle in a triangular well?

Here's how I would approach it. I'll use units where $m = q\epsilon = \hbar = 1$ for notational simplicity. First, use a temporary normalization where $a = 1$. We'll fix the overall normalization ...
Michael Seifert's user avatar
2 votes
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Deduction of Kinetic energy operator in quantum mechanics

In operatorial language for any operator $A$, $A^2$ means $A \circ A$, that is $A$ composed with itself. In the particular case of $A = \hat{p} = -i\hbar \frac{d}{dx}$, indeed $$\hat{p}^2 = -\hbar^2 \...
lucabtz's user avatar
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Deduction of Kinetic energy operator in quantum mechanics

Once a differential operator represents the momentum $p$, $p^2 f$ should be intended as the operator $p$ acting on $pf$. Therefore, the second expression is the correct one. Edit after some exchange ...
GiorgioP-DoomsdayClockIsAt-90's user avatar
1 vote

Do the length of electrons appear to be shorter when they travel close to the speed of light?

Note also that relativistic velocities affect the shape of the electric field in space created by a moving charge, since that field propagates at c.
niels nielsen's user avatar
2 votes

Planck Length vs Size of objects that exhibit quantum behaviour

The idea that the electron is mach larger than the Planck scale is not a scientific notion. Science involves knowledge that we obtained with the aid of experimental observations in support of out ...
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