Recent content by isochore

Makes sense - thanks!

I was using ##\psi^*## to be ##\langle \psi \vert##, but that's probably not good notation. So ## P = (c_1^* \langle \psi_1 \vert + c_2^* \langle \psi_2 \vert) (c_1 \vert \psi_1\rangle + c_2 \vert \psi_2\rangle) = c_1^* c_1 \langle \psi_1 \vert \psi_1\rangle + c_2^* c_2 \langle \psi_2 \vert...

I guess I can expand it, like ## P = (c_1^* \psi_1^* + c_2^* \psi_2^*) (c_1 \psi_1 + c_2 \psi_2) = c_1^* c_1 \psi_1^* \psi_1 + c_2^* c_2 \psi_2^* \psi_2 + c_1^* c_2 \psi_1^* \psi_2 + c_2^* c_1 \psi_2^* \psi_1 = \vert c_1 \vert^2 \vert \psi_1 \vert^2 + \vert c_2 \vert^2 \vert \psi_2 \vert^2 +...

Makes sense. My initial thought for getting something like ##P = \vert c_1 \vert ^2 \vert \psi_1 \vert ^2 + \vert c_2 \vert ^2 \vert \psi_2 \vert ^2 + ## (more terms involving ## \psi_1 \psi_2##) was by using the triangle identity, but I don't see a way to get the mixed terms. Is there a...

Yeah, I have, and it makes sense that it would matter. This is all the info I got with the problem (my professor just asked us to consider it, rather than it being a formal problem), so I guess they're normalised/orthogonal?

I guess I'm not following here?

I think they're just supposed to be wave functions corresponding to some experiment, and ## c_1 ## and ##c_2 ## are supposed to be scalars.

I am not sure what I can do with the equation. I realize that ## \vert c_1 \vert ^2 = \vert c_2 \vert ^2 = \frac{1}{2} ## does not mean that ## c_1 ^2 = c_2 ^2 = \frac{1}{2} ## or that ## c_1 = c_2 ##, so I don't know how to use it. I think ideally I might have something like ##P = \vert c_1...

Ahh, not sure why I did not think of that... So you can expand starting with the original expression ## \langle l, m \vert \exp((a+bi)L_z) \vert l, m \rangle ##, then since ##L_z = \hbar m \vert l, m \rangle## you can simply compress back into something like ## \exp (\hbar m (a+bi)) ##?

I am struggling to figure out how to calculate the expectation value because I am finding it hard to do something with the exponential. I tried using Euler's formula and some commutator relations, but I am always left with some term like ##\exp(L_z)## that I am not sure how to get rid of.

Thanks! I was trying something complicated with the Schrödinger equation because I thought I would get something similar but not quite the final formula if I just used the straight Hamiltonian. Turns out all you need is a few messy algebraic substitutions to get the two sides equivalent :)

I can show that ##\frac{d}{dt} \langle \psi (t) \vert X^2 \vert \psi (t) \rangle = \frac{1}{m} \langle \psi (t) \vert PX+XP \vert \psi (t) \rangle##. Taking another derivative with respect to time of this, I get ##\frac{d^2}{dt^2} \langle \psi (t) \vert X^2 \vert \psi (t) \rangle = \frac{i}{m...