Recent content by uxioq99

I'm quite embarrassed. The values are on the off-diagonal. I apologize for asking this in the first place. I've been studying too long and now am just making stupid mistakes. Even by my standards, this was an incredible mistake -- sorry.

@TSny Thank you, dividing by ##i## introduces the negative sign. I make more mistakes when I type things in Latex. Is the solution still supposed to be exponential, or should I have instead looked for a oscillatory behavior? I don't see how that is possible, given that the matrix defining the...

@hutchphd Sorry, I realized when I typed "\exp(-\omega/2t)" it produced ##\exp(-\omega/2t)## instead of ##\exp(-\frac{\omega}{2} t)## Wouldn't the units of ##\omega## cancel those of time given that it is an angular frequency? Would ## \begin{pmatrix} \Psi_1(x,t) \\ \Psi_2(x,t) \end{pmatrix} =...

Thanks, I added it back into the equation, right at the end of the editing window.

By the statement of the question, a solution must take the form ##\begin{pmatrix} \Psi_1 \\ \Psi_2 \end{pmatrix}## and the energy operator will be as per usual ##\hat{E} = i\hbar \frac{\partial}{\partial t}##. I am confused by the fact that ##S_y = \frac{\hbar}{2} \begin{pmatrix} 0 & -i \\ i & 0...

@TSny Thank you, I forgot that they didn't commute. My brain was still operating in "elementary mode".

By considering the power series for ##e^x##, I assert that ##N=e^{-\lambda^2/2}## and that ##a\Psi_\lambda = \lambda \Psi_\lambda##. Because the Hamiltonian may be written ##\hbar \omega(a^\dagger a + 1/2)##, ##\langle E \rangle = \hbar \omega(\langle a \Psi_\lambda, a \Psi_\lambda \rangle +...

@vanhees71 Thank you so much for the insight. I didn't mention this originally, but the question asked me to solve it in position space as a way to build my mathematical stamina. I agree that it would have been nicer in momentum space. @anuttarasammyak Sorry, that I forgot the factor of ##t##...

##\begin{align} \langle E \rangle &= \int_{\mathbb R^3} \int_{\mathbb R^3} \int_{\mathbb R^3} \int_{\mathbb R^3} g^\dagger (\tilde K) g(K) |\psi_0(x)|^2 \left(E_0 +\frac{\hbar^2 |K|^2}{2m}\right) e^{i(K-\tilde K)\cdot X -\frac{i}{\hbar} \left(\frac{\hbar^2 |K|^2}{2m}-\frac{\hbar^2...

Mine is a simple question, so I shall keep development at a minimum. If a particle is moving in the absence of a potential (##V(x) = 0##), then ##\frac{\langle\hat p \rangle}{dt} = \langle -\frac{\partial V}{\partial x}\rangle=0## will require that the momentum expectation value remains...

I have been self-studying the MIT 8.04 Introduction to Quantum Mechanics course. This question is not graded, so I have no reservation asking about it on the internet. Imagine an electron bound by tritium (Z=1). One of the two neutrons undergoes beta decay and becomes a proton, causing the...