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 ...

$f(X)$ 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 ...

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)}{\...

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?

$$E_1 = Re\left[\left(E_{i0}e^{-j\beta z} + \Gamma E_{i0}e^{+j\beta z}\right)e^{j\omega t}\right]$$ $$E_1 = E_{i0}\cos(\omega t - \beta z) + \Gamma E_{i0}\cos(\omega t +\beta z)$$ $$E_1 = E_{... View answer 1 votes To answer the last question first, yes, this is about group representations. There are ways to transform the quantum field that obey the laws of particular groups. These are called representations of ... View answer 0 votes Well it goes with the assumption that it's a charge distribution in equilibrium and there isn't any current. It's important to note that electric field is not zero in a conductor carrying current. But ... View answer 0 votes The ladder operators for the energy will correspond to the particular Schrodinger equation you have. So they will be different for different potentials and different systems. The presence of bound ... View answer 0 votes Each linearally independent mode counts as a unique micro state. Notice$$\sin(kx) = -\sin(-kx) so in this example $k$ and $-k$ are not unique modes. One is equal to the other with negative ...