Hi! around @7:48 the subtitle says "and real exponential solutions depend on the sines" 'sines' should be 'signs' @15:21 it says "know the sine" but sine -> sign @22:29 it says "8x psi plane" it should be "eta-xi plane"
@@TonalWorks you think that was annoying there is someone in my one lecture that constantly snorts his snot, like the snoring noise, during the first test they were going at it every like 20 or maybe less seconds for an hour and twenty minutes
They are all unit-free. [ V0 ] = [ E ] = [ M L^2 T^-2 ] [ hbar ] = [ E T ] = [ M L^2 T^-1 ] [ m ] = [ M ] [ a ] = [ L ] k^2 = (2m) (hbar^-2) (V0 - |E|) => [ k^2 ] = [ (M) (E^-2 T^-2) (E)] = [ (M) (E^-1) T^-2 ] = [ M (M^-1 L^-2 T^2) T^-2 ] = [ L^-2 ] => => [ η ] = [ ka ] = [ (L^-2)^(1/2) L ] = [ 1 ] thus unitless ,same with η^2 and we get the same result for κ (kappa) and by extension ξ and ξ^2, since the only difference between k and κ(kappa) is the replacement of the term (V0 - |E|) with |E| which we already know have the same units. Now using that [ k^2 ] = [ L^-2 ] ,from the end of my first line of calculation we get: z0^2 = (2m hbar^-2 V0) (a^2) => [ z0^2 ] = [ k^2 a^2 ] = [ L^-2 L^2 ] = [ 1 ] and thus is also unitless.
Does any force exist in this potential well ? Both inside and outside the well the derivative of potential energy with respect to position is 0 therefore zero force. So what is the point to have 2 different potentials ?
Fantastic lectures. I had studied QM before but this presentation makes it so comprehensible
My left ear understood better than my right ear. Thanks prof
@@MsMyrule ikr
Hi! around @7:48 the subtitle says "and real exponential solutions depend on the sines" 'sines' should be 'signs'
@15:21 it says "know the sine" but sine -> sign
@22:29 it says "8x psi plane" it should be "eta-xi plane"
Thanks for the feedback! The subtitles have been updated. :)
Awesome thanks prof 😅 I am watching from Algeria I am a theoretical physicist and qm is crucial for us , or for all modern physics in general 😅
Was searching up the finite curve song and stumbled upon this
If you’re able to learn quantum mechanics you can learn it here
You are really wonderful Sir. 😊😊😊😊😊😊😊😊😊😊
ok i think some mistakes are there we can t ignore the negative signs because wave function changes ..
"You'd better do something about that cough"
very clear !!!! ty for lecturing
I have a question sir. In the case where |x|
great lecture. Whoever that was coughing like a person on their deathbed, take some cough drops man. That was so distracting
Or even better would be to stay at home, and so preventing the whole lecture hall from getting infected. Annoying and also inconsiderate
@@TonalWorks you think that was annoying there is someone in my one lecture that constantly snorts his snot, like the snoring noise, during the first test they were going at it every like 20 or maybe less seconds for an hour and twenty minutes
at 19:51 the kappa square + k square = some quantity , But left hand side is unit less but right hand side have some units how it is possible
They are all unit-free.
[ V0 ] = [ E ] = [ M L^2 T^-2 ]
[ hbar ] = [ E T ] = [ M L^2 T^-1 ]
[ m ] = [ M ]
[ a ] = [ L ]
k^2 = (2m) (hbar^-2) (V0 - |E|) => [ k^2 ] = [ (M) (E^-2 T^-2) (E)] = [ (M) (E^-1) T^-2 ] = [ M (M^-1 L^-2 T^2) T^-2 ] = [ L^-2 ] =>
=> [ η ] = [ ka ] = [ (L^-2)^(1/2) L ] = [ 1 ] thus unitless ,same with η^2 and we get the same result for κ (kappa) and by extension ξ and ξ^2, since the only difference between k and κ(kappa) is the replacement of the term (V0 - |E|) with |E| which we already know have the same units.
Now using that [ k^2 ] = [ L^-2 ] ,from the end of my first line of calculation we get:
z0^2 = (2m hbar^-2 V0) (a^2) => [ z0^2 ] = [ k^2 a^2 ] = [ L^-2 L^2 ] = [ 1 ] and thus is also unitless.
i greatly appreciate the presentation. i like it
sir, what is the meaning of energy is negative?
Waw that was a blast!
My professor did her post-doc at MIT and she cannot teach quantum mechanics at all. I had to go to the source to find a better instructor xD
Sir what is the difference between particle in potential well and particle in a box?
The 'box' is a specific potential well.
The box is 3-dimensional and so you will have partial derivates wrt the 3 coordinates of position (instead of just d/dx)
Thanks 🤍❤️
14:06 if we are trying to find energy eigenstates , then they are supposed to be orthonormal anyways. why try to normalize at all?
This is helpful ❤️🤍
Can you do something about the noise from the audience. It is very hard to listen to the instructor with all the background noise.
Does any force exist in this potential well ? Both inside and outside the well the derivative of potential energy with respect to position is 0 therefore zero force. So what is the point to have 2 different potentials ?
Why k and kappa need unit-free??
ZiPan Huo
Como el picoo, no se entiende nada
I'm so glad I didn't go to MIT
not good ,could have been more understandable
why is it always the Indian complaining, i have seen your quality of lectures on RUclips lol, completely incomprehensible