Recent content by flyusx

I see that I did make a typo, for which I listed the radial integral as having the value of its square. I've reworked it from scratch, and it appears that aside from accidentally dropping the squared magnitude on the integral, the rest of the algebra works out (probably since the radial integral...

Thanks! I'm still not entirely sure why this is needed since I summed over ##m_l## already in resolving the Kronecker deltas, and because the solved problem doesn't include an overall ##\frac{1}{3}## factor when using the dipole expansion (epsilon method). I'll keep experimenting and see what I...

I know that to find the intensity, I must first calculate the transition rate. I started from the transition rate equation for spontaneous emission. $$W_{i\to f}=\frac{4\omega_{f\to i}^{3}}{3\hbar c^{3}}\left\vert\boldsymbol{d}_{f\to i}\right\vert^{2}=\frac{4e^{2}\omega_{f\to i}^{3}}{3\hbar...

Thanks!

I am a bit confused in Part (b) in calculating the first-order perturbative first excited state energies. I know that the total wavefunction must be antisymmetric because it is describing a system of fermions. This requires that an antisymmetric spatial wavefunction ##\phi## be paired with a...

I was reviewing some undergraduate mechanics when I found something I hadn't realised before. Consider a damped driven oscillator governed by the differential equation $$\ddot{x}+2\beta\dot{x}+\omega^{2}x=F_{d}\cos\left(\omega_{d}t\right)$$ where ##\omega## is the system's natural frequency and...

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 Schrödinger 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...