Recent content by bluepilotg-2_07

Hello, thank you for the response! I should have known better, but wasn't considering even and oddness. I realize now that the first integral goes to zero since it is odd and being integrated over symmetric boundaries and the rest is trivial. Thanks again!

##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi...

Hi, thanks for responding! Just after posting, I finally found a resource that explained what I wanted to know. It feels as if it should have been obvious now that I have read about it. In response to the question you posited, we should be able to swap the summations and integral because the...

I am self-studying quantum mechanics from Griffiths' textbook and some other sources. I have come across this derivation shown in the photo. I've taken all three major calculus courses for physics, linear algebra, ODE, PDE, Complex Analysis, etc. However, I do still struggle with rules for...

I believe so. My derivation implies terms from -5 to 0 are all zero, so everything I wrote are the non-zero terms.

I believe it is expanded about zero.

Since ##|z| > 1## we can rewrite as ##|\frac{1}{z}|<1##. I rewrite the function as $$\frac{1}{z} \frac{1}{z^2(1-\frac{1}{z})^2}$$. For a typical geometric series: $$\frac{1}{1-\frac{1}{z}}=\sum_{k=0}\left(\frac{1}{z}\right)^k$$. The derivative is...

If my boundary conditions are ##\psi(x+L)=\psi(x) \Rightarrow \frac{1}{\sqrt{L}}e^{ikL}=\frac{1}{\sqrt{L}}##, then allowed values of k should be ##k=\frac{2\pi}{L}n##.

Hello, thanks for responding I assumed periodic boundary conditions when solving the Schrodinger equation for the 1D system. At zero temperature the number of occupied states is equal to the number of electrons N. The electrons exist along some line of length L in the 1D system. If I have period...

Fermi energy is given by $$\epsilon_F = \frac{\hbar ^2 k_F ^2}{2m}$$ ##N = \frac{2k_F L}{2\pi} \Rightarrow \frac{k_F L}{\pi}## the factor of two in the numerator comes from the electrons having two spins. $$E=\frac{2L}{2\pi} \int_{0}^{k_F} \frac{\hbar^2 k^2}{2m}\, 2\,dk$$ The two in front of the...

I've been working on this problem for the past 3 days. I have other papers with different ways of tackling the problem. However, I just cannot get to the answer (change in entropy = 2Nlog(2)).