Stationary states infinite cubic well

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For a state to be stationary in an infinite cubic well, it must be time independent, meaning its absolute square remains constant over time. The discussion revolves around finding the values of c that eliminate time dependency in the wave function. The eigenfunction provided is a sine function multiplied by an exponential decay term, with the eigenvalue calculated as 12ħ²/(8mL²). However, the approach taken in the calculations is questioned, suggesting a need for a review of the fundamentals of operators and wave functions. Clarification is sought on the role of the operator L_z in this context.
happyparticle
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Homework Statement
For which values of c the state ##e^{c \cdot L_z} |2,2,2>## is stationary for the infinite cubic well Hamiltonian.
Relevant Equations
##\psi (r,t) = \sqrt{\frac{8}{l^3}} sin(\frac{2 \pi x}{l}) sin(\frac{2 \pi u}{l}) sin(\frac{2 \pi z}{l}) e^{-iEt/\hbar}##
##E = \frac{6}{m} (\frac{\pi \hbar}{l})^2##
For a state to be stationary it must be time independent.

Naively, I tried to find the values of c where I don't have any time dependency.

##e^{c \cdot L_z} \psi (r,t) = e^{c L_z} \sqrt{\frac{8}{l^3}} sin(\frac{2 \pi x}{l}) sin(\frac{2 \pi u}{l}) sin(\frac{2 \pi z}{l}) e^{-iEt/\hbar}##

##e^{c \cdot L_z -iEt/\hbar} ##

Thus
##c = \frac{L_z - iEt}{\hbar}##

##c = \frac{L_z - i\frac{6}{m} (\frac{\pi \hbar}{l})^2t}{\hbar}##

I'm wondering if this is correct.

Thanks
 
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happyparticle said:
Homework Statement:: For which values of c the state ##e^{c \cdot L_z} |2,2,2>## is stationary for the infinite cubic well Hamiltonian.
Relevant Equations:: ##\psi (r,t) = \sqrt{\frac{8}{l^3}} sin(\frac{2 \pi x}{l}) sin(\frac{2 \pi u}{l}) sin(\frac{2 \pi z}{l}) e^{-iEt/\hbar}##
##E = \frac{6}{m} (\frac{\pi \hbar}{l})^2##

For a state to be stationary it must be time independent.

Naively, I tried to find the values of c where I don't have any time dependency.

##e^{c \cdot L_z} \psi (r,t) = e^{c L_z} \sqrt{\frac{8}{l^3}} sin(\frac{2 \pi x}{l}) sin(\frac{2 \pi u}{l}) sin(\frac{2 \pi z}{l}) e^{-iEt/\hbar}##

##e^{c \cdot L_z -iEt/\hbar} ##

Thus
##c = \frac{L_z - iEt}{\hbar}##

##c = \frac{L_z - i\frac{6}{m} (\frac{\pi \hbar}{l})^2t}{\hbar}##

I'm wondering if this is correct.

Thanks
A stationary state itself is not time independent. Its absolute square is time independent.
That won't get you very far though. You have to look at something else, I would say.
 
The eigenfunction is ##
\sqrt{\frac{8}{l^3}} sin(\frac{2 \pi x}{l}) sin(\frac{2 \pi u}{l}) sin(\frac{2 \pi z}{l}) e^{-iEt/\hbar}
##
The eigenvalue is ##\frac{12 \hbar^2}{8mL^2}##
Maybe ##e^{c L_z} = e^{\frac{12 \hbar^2}{8mL^2}}##
 
I assume ##L_z## is the operator corresponding to the z-component of orbital angular momentum, so your attempts don't make mathematical sense. You should go back and review the basics of operators and wave functions.

If it's not supposed to represent that operator, what is it supposed to be?
 
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...