WhatIAm
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Commutators involving functions
13 votes

A more general relation than $\left[q^n, p \right] = i\hbar nq^{n-1}$ is $$\left[f(A), B\right] = \left[A, B \right]\frac{\partial f}{\partial A}$$ if $\left[A, \left[A, B \right]\right] =0$. In this ...

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Fourier Transform and Lattices
Accepted answer
6 votes

$f(X)$ is a periodic function that has a Fourier series. In other words, $f(X)$ is a periodic function and so its Fourier transform has a spectrum at only discrete values of q. Still the transformed ...

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How do photons affect accessible microstates?
2 votes

All you need is the statistical definition of temperature. \begin{align*} \frac{1}{T} := k\frac{\partial \ln \Omega(E)}{\partial E} \\ \frac{1}{T} = k\frac{1}{\Omega(E)}\frac{\partial \Omega(E)}{\...

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Faraday's law question: wire falling over two rails
2 votes

I think you should re-examine your expression for $\frac{\partial \phi}{\partial t} $. Might it be that it is not a function of time?

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Electromagnetic wave propagation through two lossless dielectrics
2 votes

$$ E_1 = Re\left[\left(E_{i0}e^{-j\beta z} + \Gamma E_{i0}e^{+j\beta z}\right)e^{j\omega t}\right]$$ $$ E_1 = E_{i0}\cos(\omega t - \beta z) + \Gamma E_{i0}\cos(\omega t +\beta z) $$ $$ E_1 = E_{...

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How groups act on fields in QFT?
1 votes

To answer the last question first, yes, this is about group representations. There are ways to transform the quantum field that obey the laws of particular groups. These are called representations of ...

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Electric Field Inside of a Conductor
0 votes

Well it goes with the assumption that it's a charge distribution in equilibrium and there isn't any current. It's important to note that electric field is not zero in a conductor carrying current. But ...

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Use of ladder operators
0 votes

The ladder operators for the energy will correspond to the particular Schrodinger equation you have. So they will be different for different potentials and different systems. The presence of bound ...

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Number of modes in cavity radiation
0 votes

Each linearally independent mode counts as a unique micro state. Notice $$\sin(kx) = -\sin(-kx)$$ so in this example $k$ and $-k$ are not unique modes. One is equal to the other with negative ...

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How come you see what happens to the energy of a system when 2 free linear plane waves cancel out?
0 votes

Waves which perfectly cancel each other out are equivalent to no waves to start with. So the energy of the perfect cancellation will be zero. This does not violate conservation of energy. Note that ...

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