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2 votes

Is a fermionic boson possible?

If a wavefunction is symmetric under exchange of the particle labels, then it's a perfectly good bosonic state. If this just happens to arise as the product of antisymmetric spatial and spin states, ...
march's user avatar
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2 votes

Basic doubt in quantum mechanics

Are entities like electrons which are considered point particles in classical Mechanics have actually have a definite position at a particular time (irrespective of it can be measured or not)? The ...
hft's user avatar
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3 votes
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Basic doubt in quantum mechanics

This is a question that haunts anyone who is interested about the foundations of quantum mechanics. In standard textbook quantum mechanics, the answer would be no, the electron does not have a ...
Chang Hexiang's user avatar
4 votes

Basic doubt in quantum mechanics

In substance the answer is no. If you take electrons and pass them through an interference experiment, then the result in general can't be explained by assuming it just went down one of the paths in ...
alanf's user avatar
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3 votes
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Can any meaning be given to a path integral with no fixed end point?

Intead of a $\psi(x_{\rm final})$ number, it is a function $\psi $of the unstated endoint. When an $x$ is chosen you get the wavefunction $\psi(x,t)= \langle x|\psi(t)\rangle$. In other words, the ...
mike stone's user avatar
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2 votes

A simple question in quanum mechanics on position and momenum eigenstates

First of all $\phi(x,t) = \sin(kx - \omega t)$ is not a momentum eigenstate. Its a super position of of $\exp(i(kx - \omega t))$ and $\exp(-i(kx - \omega t))$, which are momentum eigenstates for $+k$ ...
Zaph's user avatar
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0 votes

Does QM recognise empty waves?

It is the EM field that you are interested in here ... not just any field. The EM field propagates many virtual fields even before the photon is created. Virtual fields are your "empty waves&...
PhysicsDave's user avatar
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0 votes

Does QM recognise empty waves?

In QM the wave function is the fundamental physical property describing your system. In your case, before you disturb the system by the measurement, it will be an extended object reaching threw the ...
Zaph's user avatar
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1 vote
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Does QM recognise empty waves?

In the MWI what is happening in reality is described by quantum equations of motion. In the pilot wave theory reality is described by the same equations of motion with point particles sprinkled on top....
alanf's user avatar
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0 votes

Does QM recognise empty waves?

I'm glad you called the photon a particle and that's the main thing you need to remember. There is no wave unless your talking about millions of individual coherent photons radiating from a common ...
Bill Alsept's user avatar
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2 votes

Ordinarily continuous function of the wave function

In regard to your second question, I believe Griffiths is stating that, in general, the second integral is zero, and $d\psi/dx$ is continuous. However, when $V(x) \propto \delta(x)$, this statement no ...
Alejandra Mendez's user avatar
0 votes

Would a high energy Hydrogen atom start emanating electromagnetic radiation?

The physics others said in their answers are bang on. But we can also get to the error by saying that the Correspondence principle doesn't say what you say it does. That is because there is no single ...
Dr. Nate's user avatar
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1 vote
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Scattering Matrix and the Lippmann-Schwinger equation in QM

However, is there a formula that relates the S-matrix elements and these states $\psi^{(+)}$ and $\psi^{(-)}$? You have not provided a sufficiently detailed description of what exactly the $\psi^{(+)}...
hft's user avatar
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7 votes
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Question about Griffiths' proof that $\Psi$ stays normalized

The proof by Griffiths relies in fact on stronger (implicit) hypotheses about the behaviour of $\psi$ at infinity. These hypotheses could be justified by requiring (as in Giorgio's answer of the other ...
Valter Moretti's user avatar
1 vote
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Orthogonality of wavefunction and overlap integral

Not sure I fully understand what you mean by overlap. Specifically, support in real space or inner product? But hopefully I can answer anyway. To be clear, if you have two H atoms, the full ...
just a phase's user avatar
1 vote

Doubts on Particle in a box model

For 1) and 2), we can ignore phases (including negative signs) as there is only one particle in the box. When we measure a particles position we do not measure $\phi(x)$ but we instead measure $|\phi(...
psychgiraffe's user avatar
8 votes
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When do two state functions represent the same quantum state?

The set of wavefunctions which corresponds to the same state as $\psi$ is just the set of multiples of $\psi$ by a non-zero complex number (respectively, a complex number of absolute value $1$, if you ...
SolubleFish's user avatar
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1 vote

What is the difference between a vector and a representation of a vector in QM?

The state vector $|\Psi\rangle$ "encodes" everything about a quantum state, because you can choose to expand $|\Psi\rangle$ in any orthonormal, complete basis corresponding to some ...
BioPhysicist's user avatar
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9 votes
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Why is probability outside the infinite square well zero?

If the well is sufficiently deep to bind the particle, the wavefunction decays exponentially outside the well. The rate of exponential decay depends on the depth of the well. For an infinitely deep ...
John Doty's user avatar
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2 votes

A theoretical issue in the mathematical description of the Aharonov-Bohm experiment

All the talk about extending the formulation of QM (which is a good thing to understand) is in this case a red herring. Even if you insist on an unsophisticated view of the wavefunction as a class of ...
HTNW's user avatar
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2 votes

Gaussian wave packet with complex coefficients

Given your integral: $$ \langle p'|\alpha\rangle = \int dx' \frac{1}{\sqrt{2 \pi \hbar}} \exp\left(-\frac{i p'x'}{\hbar}\right) N \exp\left[-(a+ib)x'^2 +(c+id)x'\right] $$ You can complete the square ...
Adversing's user avatar
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1 vote

Is complex integration needed when normalizing a wave function?

The only complex quantity you might have here is $\phi_0$ (which phase is irrelevant, by the way), so you only have to integrate \begin{equation} \lVert \phi(k) \rVert = \int_{-\infty}^{\infty}|\phi(k)...
Gabriel Ybarra Marcaida's user avatar
7 votes
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Minimization over a function is equivalent to the problem of finding the minimum energy eigenstate in an infinite potential well?

The "Schrödinger equation" of your problem is the eigenvalue equation $$- \frac{d^2 \tilde{g}(\mu)}{d\mu^2} = \lambda \tilde{g}(\mu), \qquad \tilde{g}(a)=\tilde{g}(b)=0.$$ Obviously, the ...
Hyperon's user avatar
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1 vote
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Doubt obtaining the expected value of $x^2$ of a bidimensional harmonic oscillator

You need to calculate the expectation value in the position basis: \begin{align} \langle x^2 \rangle &= \langle \phi | x^2 | \phi \rangle = \int \mathrm{d}x \int \mathrm{d}y \; \phi^*(x,y) x^2 \...
WillHallas's user avatar
1 vote

Orthogonality of wavefunction and overlap integral

The support of an atomic wavefunction is never compact. This implies, for example, that the overlap of the wavefunctions of two hydrogen atoms, even very far apart, is never exactly zero. Thus, two s-...
GiorgioP-DoomsdayClockIsAt-90's user avatar
1 vote

Normalizing the wave-function

To see that the function is normalized take its inner product with itself: $$\begin{align} |\Psi(x,0)|^2&={1\over 5}\int_{0}^{+a}(2\psi_2(x)^{*}-\psi_3(x)^{*})(2\psi_2(x)-\psi_3(x))\;dx\\ &={1\...
Albertus Magnus's user avatar

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