Infinite square well energy eigenstates

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

[@aliciaroberts3965]

thank you! this video helped a lot with my finals studying. we use the MIT textbook so im glad there are well-done lectures that go over the material :)

[@mayukbasak1429]

Please don't edit out the discussions between the professor and the students. That's a part of learning !

[@finn9000]

I think it's a privacy issue

[@jeetsharma9892]

Thank you so much for this informative and understandable video

[@user-rg1nt9lf4s]

very good content sir. thank you .. sir kindly make a video on Bound
States for Potential Wells with no rigid walls.

[@Sk-bp6ji]

Thanks MIT and thank sir your video is very helpful

[@miffyn1737]

Very satisfying explanation, thanks sir

[@not_amanullah]

Thanks ❤️🤍

[@not_amanullah]

This is helpful ❤️🤍

[@saikatmaji2917]

If a is rational , lets say a = p/q , then for n multiple p the eigenstate vanishes. Hence we cant take a rational here.

[@anamikasrivastava8714]

Thanks sir....❣️

[@AT-zf2xf]

It is not clear why n=1,2,3,... here. He argues that for the circle also n

[@anmolsubba7394]

Andrea Tononi the |wave function | square for a circle will have different values for +n and -n, different momentum but energy will be same ,,,, for infinite square well for + n and - n wave function will have |waves function| square same indicating same probability . Therefore we can take negative integers

[@GaneshGunaji]

You could pretend n can be negative and continue to solve the problem. What would happen is you would eventually find a way to group the eigenfunctions corresponding to the -n terms and the eigenfunctions corresponding to the +n terms so they can be expressed solely as a Fourier series of +n terms using the identity sin(-nx)=-sin(nx). When solving for the coefficients, you would see that sin(-npix/a) and sin(npix/a) are not orthogonal on [0,a], and you would end up grouping them as a single sin(npix/a) term, taking n to be a positive integer. (Zero is excluded as an eigenvalue because of the normalization requirement that there is a particle in the box. If Psi were 0, we couldn't have that the integral of Psi times its complex conjugate from 0 to a is 1.)

[@thomaslupo382]

What about the derivative of the outer and inner functions at the boundary. The derivative of sin is not 0. Should inner function be 1-cos(2*pi/a*x)

[@negasonicteenagewarhead]

Sin(nπ)=0

[@skya6863]

I'm 2 years late, but posting in case somebody else wonders about this.
No the wave function is correct, and it's true that you should expect the derivative to be continous everywhere for the wave function. But the problem here lies in the potential energy function, V(x). You can show that the derivative of the wave function is continous in all places except for when V(x) makes an infinitely large jump. In nature, there is no such infinite jump but for this theoretical square well we see a discontinuity in the derivative on the boundaries

[@doublecross8323]

is N^2 is the maximum value of probability in the graph?

[@michaelwagner6877]

Holy, effin, shite. Why was this so much easier to compartmentalize? The 1/2 of the integral of sin was something wizard that makes way too much sense when pointed out like this.

[@debanjan7883]

❤

[@GB3770]

Oh MIT why you can't record audio correctly?
Should some 16 year old sound engineers tell your professors how to do it? :)

[@michaelterrell5061]

Yes they should.

[@medonis5913]

wawawawawa.....shut up