I did this at University, about 30 years ago. Nice to see it still makes sense after all these years.
@-alto10 жыл бұрын
Brilliant videos, helped a ton studying for my QM final and supplemented the Griffiths book well to read ahead for QMII. Thank you!
@grantmaybe4 жыл бұрын
4:34 my name is Grant. He said that at a really weird time, I didn't realize it was the video, I thought someone was responding to me.
@Salmanul_4 жыл бұрын
Haha
@07carlsberger10 жыл бұрын
Shows how much of a geek I am. My brain almost exploded when he put the inner products as zero haha!
@kalyanjyotikalita45628 жыл бұрын
me tooo.. :-D
@scitwi91647 жыл бұрын
Yeah, it was like a function could be orthogonal to itself :q
@lineakristensen18215 жыл бұрын
Yeah I had the same feeling when he forgot to square y and z in the second derivatives. And he didn't even correct it. How will I sleep tonight? 😂
@Introvertrains4 жыл бұрын
I was just freaking out
@vaishnavipal2298 Жыл бұрын
He made one more error I hope you can figure that out that's some simple arithmetic error.
@fidgetspinner10506 жыл бұрын
7:44 Shouldn't ^p*^p be (h/2pi)^2*[nabla operator]? You said ^p is defined as -i(h/2pi)*[nabla operator] and i*i = -1.
@Libservative799 жыл бұрын
You forgot the squared term on dy and dz at 8:38 :)
@hammietime840410 жыл бұрын
Thank you so much for these videos. Very helpful.
@albertliu25994 ай бұрын
Check your understanding: . . . . 1. [x, Py] = 0 2. So we could measure x and Py as precise as we want. 3. [Px, Py] = 0 also. So we could measure Px and Py as precise as we want.
@2000freefuel5 жыл бұрын
it amuses me how many people don't realize that Schrodinger created this now historical thought experiment as a piss take on the Heisenberg uncertainty principle!
@MohammadHassan-ud8iq3 жыл бұрын
Thanks for going over the seperable solutions for the time dependent Schrodinger equation. Griffiths doesn't do that.
@paulooliveiracastro7 жыл бұрын
At 12:50, after dividing by RT, why did he write VR? Shouldn't it be only V (since he divided VRT by RT)?
@quenteijnvancooten55706 жыл бұрын
The division of R is present in front of the bracket
@ifrazali30524 ай бұрын
Because R is being operated on by Potential operator
@gerontius17266 жыл бұрын
@ 8:55 in the Laplacian he forgot to square the delta x and delta z denominators.
@mohammed-090z_aljuboory3 жыл бұрын
Thank you very much 🌹🌹🌹
@pascal34582 жыл бұрын
حصل
@vaishnavipal2298 Жыл бұрын
Shouldn't it be V instead VR since u divided both sides by RT and this V term had a multiple RT
@ifrazali30524 ай бұрын
No, Because R is being operated on by Potential operator and Laplacian.
@Albeit_Jordan5 жыл бұрын
So would cubing the three-dimensional wave-function give us a three-dimensional probability distribution? Edit: I'm not gonna get an answer, am I? - because the video was uploaded in 2013.. xD
@TheALANARDO5 жыл бұрын
If you solve for each individual dimension and multiply those solutions together, you get a psi equation that is already parameterized for three dimensions, cubing a one-dimensional solution does not give you probability. To get the probability, the function first has to be normalized to one, and then squared to get rid of the negative values.
@Albeit_Jordan5 жыл бұрын
@@TheALANARDO I know cubing a one-dimensional solution won't give you a probability distribution, I actually asked about cubing a three-dimensional solution - tangentially I was wondering if the squared normalized one-dimensional solution was time independent (thereon if the three-dimensional solution should be the same.) But thank you, I do appreciate your response :)
@Salmanul_4 жыл бұрын
Why dot product and not cross product?
@MiguelGarcia-zx1qj3 жыл бұрын
I think (without knowing more about QM than this course, up to this chapter) that there is a better way to expand the kinetic energy operator than appealing to the square of the momentum operator p. Because, in fact, there is more than one way to multiply vectors (besides the cross product, you also can do the tensor product). The rationale to obtain the final expression with nabla squared in it is that, in classical mechanics, the kinetic energy encompasses the sum of the three squared components of the velocity vector; by analogy, each velocity component becomes a coordinate derivative: multiply twice by one velocity component (to get a square) equals to derive two times with respect to that coordinate. The 3D kinetic energy (in QM) is what it is because of Physics, not Mathematics.