Energy eigenstates for particle on a circle

MIT 8.04 Quantum Physics I, Spring 2016
View the complete course: ocw.mit.edu/8-0...
Instructor: Barton Zwiebach
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu

[@zokalyx]

Use left audio only. Right audio is mostly background noise.

[@not_amanullah]

Wow

[@cafe-tomate]

Energy operator is also hermitian so the eigenvectors can also be chosen orthonormal for the E operator

[@rahilshaik1603]

even though there are degenerate states?

[@zokalyx]

@@rahilshaik1603 yes, when there is degeneracy you can choose eigenvectors with some degrees of freedom such that they are orthogonal

[@nicktohzyu]

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

[@varunshrivastav8876]

You can go on your device settings and change it mono audio for your device

[@sipraneye70]

"A teacher affects eternity; he can never tell where his influence stops" ---HENRY ADAMS

[@yeahyeah54]

He forgot a psi in the differential equation in the beginning

[@not_amanullah]

Thanks ❤️🤍

[@aide1326]

Easy and hard at the same time.

[@michielsnoeken5596]

Why are we allowed to assume that the wavefunction of the particle on a string is stationary?

[@kyubey3166]

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.

[@chrisr9320]

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)

[@pixelberrychoicespodcast5861]

@@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

[@goopyt267]

kind of superposed voice of professor is coming XD

[@wondererasl]

How come kL= 2πn ?

[@bendiknyheim6936]

Solve e^ikx = e^ikx e^ikL

[@한두혁]

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!!