Please don't edit out the discussions between the professor and the students. That's a part of learning !
@finn900010 ай бұрын
I think it's a privacy issue
@aliciaroberts3965 Жыл бұрын
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 :)
@user-rg1nt9lf4s5 жыл бұрын
very good content sir. thank you .. sir kindly make a video on Bound States for Potential Wells with no rigid walls.
@jeetsharma98924 жыл бұрын
Thank you so much for this informative and understandable video
@miffyn17373 жыл бұрын
Very satisfying explanation, thanks sir
@Sk-bp6ji6 жыл бұрын
Thanks MIT and thank sir your video is very helpful
@saikatmaji29172 жыл бұрын
If a is rational , lets say a = p/q , then for n multiple p the eigenstate vanishes. Hence we cant take a rational here.
@AT-zf2xf4 жыл бұрын
It is not clear why n=1,2,3,... here. He argues that for the circle also n
@anmolsubba73944 жыл бұрын
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
@GaneshGunaji3 жыл бұрын
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.)
@anamikasrivastava87143 жыл бұрын
Thanks sir....❣️
@thomaslupo3823 жыл бұрын
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)
@negasonicteenagewarhead2 жыл бұрын
Sin(nπ)=0
@skya68632 ай бұрын
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
@doublecross832310 ай бұрын
is N^2 is the maximum value of probability in the graph?
@michaelwagner68779 ай бұрын
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.
@debanjan78834 жыл бұрын
❤
@GB37703 жыл бұрын
Oh MIT why you can't record audio correctly? Should some 16 year old sound engineers tell your professors how to do it? :)