Energy eigenstates for particle on a circle
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- Опубликовано: 30 июл 2017
- MIT 8.04 Quantum Physics I, Spring 2016
View the complete course: ocw.mit.edu/8-04S16
Instructor: Barton Zwiebach
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
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Use left audio only.
Right audio is mostly background noise.
left audio channel is fine, right audio is messed up. please just export in mono audio. there is no option to watch on youtube left audio only
You can go on your device settings and change it mono audio for your device
Energy operator is also hermitian so the eigenvectors can also be chosen orthonormal for the E operator
even though there are degenerate states?
"A teacher affects eternity; he can never tell where his influence stops" ---HENRY ADAMS
He forgot a psi in the differential equation in the beginning
Easy and hard at the same time.
kind of superposed voice of professor is coming XD
Why are we allowed to assume that the wavefunction of the particle on a string is stationary?
It doesn't have to be, but from stationary wavefunctions you can always construct any wavefunction by superposition. That's a standard procedure in QM, you first find the stationary ones and then you construct any other by summing them. Hope this helps.
Whenever the V(x) in the Schrödinger equation is not time-dependent, you can separate psi(x,t) into a function of x and a function of t and you get the time-independent SE. Which means you only need to find stationary solutions and can then simply multiply by exp(-iEt/hbar)
@@chrisr9320 yes but multiplication by e^-iet/h bar is only for Hamilton operator right?
If you have a different operator the time dependence will look different
How come kL= 2πn ?
Solve e^ikx = e^ikx e^ikL
since e^ikL=1, using eulers eq cos(kL)+isin(kL)=1 which means sin(kL)=0 and cos(kL)=1 Therefore kL has to be 2pi*n!!