Recent content by JTFreitas

Of course, this makes a lot of sense. It's been a minute since I've done linear algebra for real, so I forget the eigenvectors pertain to how the matrix is structured. And this was the one thing I was missing. For some reason I was thinking each energy state had a different wave function, and...

Thanks for the response. I see what you're saying. And I suppose that we did not change the Hamiltonian anywhere, so throughout the algebra naturally the system did not become different. Meaning that I am expecting the results for the usual oscillator. That said, I think the discretization is...

##x## can be discretized as ##x \rightarrow x_k ## such that ##x_{k + 1} = x_k + dx## with a positive integer ##k##. Throughout we may assume that ##dx## is finite, albeit tiny. By applying the Taylor expansion of the wavefunction ##\psi_n(x_{k+1})## and ##\psi_n(x_{k-1})##, we can quickly...

Thank you for taking a look at it. I just double-checked, and according to my textbook, it is defined with ##g_{ij}## as is.

I want to show that ##[C, a_{r}] = 0##. This means that: $$ Ca_{r} - a_{r}C = \sum_{i,j} g_{ij}a_{i}a_{j}a_{r} - a_{r}\sum_{i,j} g_{ij}a_{i}a_{j} = 0$$ I don't understand what manipulating I can do here. I have tried to rewrite ##g_{ij}## in terms of the structure...

Okay, got it. Possibly you mean the ##u^{k}_{n+1}## as a phase factor times ##u^{k}_{n}##? Oh, seeing how you altered the LHS index to ##l##, I think I understand why we need to change the summation index. The k-index corresponds to the k-th function of the n-th particle coordinate, but when I...

Is it strictly necessary to change the last index? The original differential equation was for ##q_n##, so the corresponding ##k## ought to be the same, or am I misunderstanding something? And with regards to re-expressing the basis functions, I am wondering if maybe we can change the index of...

In terms of a clear definition, ##a_k## is just the coefficient of expansion of ##q(t)## as a discrete Fourier transform, and it carries the time dependence. In essence, I introduce the new definition of ##q_n(t)## because my initial equation is coupled (I have ##q_n, q_{n-1}, q_{n+1} ## in a...

Hello, I hope the equation formatting comes out right but I'll correct it if not. So far, I have attempted to write ##\ddot{a}_k(t) = \sum_{n}(u^{k}_n)^*\ddot{q}_n(t) ##. Then I expand the right hand side with the original equation of motion, and I rewrite each coordinate according to its own...

Once again, thanks for the answer, this is really helpful. Then, this does mean that both kernels can be deduced based on the very physical properties of the system. As in, the zero-kernel of the raising operator, and the non-zero kernel of the lowering operator are necessary conditions for the...

Okay, that makes sense. Thanks for your answer. I suppose there is no direct mathematical way of establishing the kernels of these particular operators then? Other than writing them out in their matrix representation, finding the inverse, solving ##\hat{a}\mathbf{x} = 0##, etc etc etc...?

First time posting in this part of the website, I apologize in advance if my formatting is off. This isn't quite a homework question so much as me trying to reason through the work in a way that quickly makes sense in my head. I am posting in hopes that someone can tell me if my reasoning is...

Perfect, thank you so much! We can close the thread now

God it makes complete sense now. I've been at this for days. My professor's notes weren't exactly very clear, but pulling out the ##\partial x'^{\sigma}## made everything work out nicely and made complete sense. Thank you so much for the response, that was really helpful! A quick follow-up...

Apologies in advance if I mess up the LaTeX. If that happens I'll be editing it right away. By starting off with ##\nabla^{'}_{\mu} V^{'\nu}## and applying multiple transformation laws, I arrive at the following expression $$ \frac{\partial x^{\lambda}}{\partial x'^{\mu}} \frac{\partial...