Recent content by flyusx

I agree. I wrote a note in the book margins about including the singlet state a while back and have reached time-independent perturbation theory now. I just wanted to double check. Thanks!

Hey, sorry for the late reply! Zettili says the second excited state is ##\phi_{2,0,0}\left(r_{1}\right)\phi_{2,0,0}\left(r_{2}\right)\chi_{\text{singlet}}##. If it's of any help, he says the ground state energy is ##2E_{1}=-27.2Z^{2}## eV, the first excited energy is ##E_{1}+E_{2}=-17Z^{2}##...

Thanks for your reply. I had wrote down the additional solutions when I originally solved it, but I thought I may have been wrong after being confused in the 9th chapter. It's nice to get some additional confirmation.

I had read up on identical particles and their associated symmetric/antisymmetric wavefunctions a while back and solved a few problems. It seems like I'm still confused on some fronts. I have picked here two solved problems from Zettili's QM book (Edition 3) that I believe illustrate the part I...

This is a solved problem from Zettili Chapter 7 Problem 7.8(b) but I am having troubles reproducing some of the quantities he produces. Zettili approaches this problem by describing ##\textbf{r}## using a spherical basis: a product between a radial and angular part...

I was able to derive the energy eigenvalue equations $$\tan(k(a-b))\tanh(\tilde{k}b)=-\frac{k}{\tilde{k}}\text{, even wave function}$$ and $$\tan(k(a-b))\coth(\tilde{k}b)=-\frac{k}{\tilde{k}}\text{, odd wave function}$$ As TSny said, it doesn't seem like I can find numerical energies...

Good catch, I wouldn't like to confuse it for energy.

So if I look at the entire setup (instead of looking at half of it) and write $$\begin{cases}\frac{\text{d}^{2}\psi_{1}}{\text{d}x^{2}}+k^{2}\psi_{1}(x)=0&-a<x<-b\\\frac{\text{d}^{2}\psi_{2}}{\text{d}x^{2}}-\tilde{k}\psi_{2}(x)=0&-b\leq x\leq...

Sorry, I'm still a bit confused. It seems I have more confusion with solving the Schrodinger equation than I thought. I can accept that even functions have zero slope at ##x=0##, but if I plot a function like ##\sin(k_{2}\vert x\vert)+\cos(k_{2}x)## (where I arbitrarily set ##A=B=1##), I get an...

So it is generally better to have the wave function vanish at the necessary 'vanishing points' than impose the conditions after writing them like I did. In hindsight, I suppose it would have been better to write the arbitrary wave function of form ##A\sin(k_{2}x)+B\cos(k_{2}x)## instead of with...

I have been stuck on part (a) for a few days and would appreciate a nudge in the right direction. This is Zettili Exercise 4.4. I know this is a symmetric potential so that the (non-degenerate) bound states will have definite parity. I have tried to take advantage of this symmetry by focusing...

Reading a few (or quite a few) more posts in the past few hours has given me the impression that the wave could transmit through the barrier, 'bleed into' it and then get reflected. All of the wave will eventually get reflected so ##R=1## and ##T=0##. One could hypothetically measure the...

Looking from that perspective, it makes sense how why the transmission coefficient is zero in this case; the wave function (and hence its probability density) is static for ##x\geq0##. On the wikipedia page for transmission coefficients, I read the following: "In non-relativistic quantum...

Wouldn't exponential decay for ##x\geq0## still suggest ##\vert\psi\vert^{2}\neq0## and hence the probability density is non-zero for this region (the particle can tunnel into the potential barrier)? If so, how would one go about calculating this probability for the transmission coefficient is...

This isn't really homework but rather me working through my quantum textbook and coming across something I don't understand. Consider the potential function $$V(x)=\begin{cases}0&x<0\\V&x\geq0\end{cases}$$ where ##V## is a constant. If the energy of the incident particle ##E## is less than...