Schrodinger equation in 3d

[@craigfowler7098]

I did this at University, about 30 years ago. Nice to see it still makes sense after all these years.

[@-alto]

Brilliant videos, helped a ton studying for my QM final and supplemented the Griffiths book well to read ahead for QMII. Thank you!

[@hammietime8404]

Thank you so much for these videos. Very helpful.

[@grantmaybe]

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_]

Haha

[@07carlsberger]

Shows how much of a geek I am. My brain almost exploded when he put the inner products as zero haha!

[@kalyanjyotikalita4562]

me tooo.. :-D

[@scitwi9164]

Yeah, it was like a function could be orthogonal to itself :q

[@lineakristensen1821]

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? 😂

[@Introvertrains]

I was just freaking out

[@vaishnavipal2298]

He made one more error I hope you can figure that out that's some simple arithmetic error.

[@fidgetspinner1050]

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.

[@Libservative79]

You forgot the squared term on dy and dz at 8:38 :)

[@paulooliveiracastro]

At 12:50, after dividing by RT, why did he write VR? Shouldn't it be only V (since he divided VRT by RT)?

[@quenteijnvancooten5570]

The division of R is present in front of the bracket

[@ifrazali3052]

Because R is being operated on by Potential operator

[@gerontius1726]

@ 8:55 in the Laplacian he forgot to square the delta x and delta z denominators.

[@albertliu2599]

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.

[@2000freefuel]

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!

[@mohammed-090z_aljuboory]

Thank you very much 🌹🌹🌹

[@pascal3458]

حصل

[@MohammadHassan-ud8iq]

Thanks for going over the seperable solutions for the time dependent Schrodinger equation. Griffiths doesn't do that.

[@vaishnavipal2298]

Shouldn't it be V instead VR since u divided both sides by RT and this V term had a multiple RT

[@ifrazali3052]

No, Because R is being operated on by Potential operator and Laplacian.

[@Salmanul_]

Why dot product and not cross product?

[@MiguelGarcia-zx1qj]

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.

[@Albeit_Jordan]

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

[@TheALANARDO]

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_Jordan]

@@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 :)

[@nunoteixeira2129]

who the fuck disliked this

[@gibsonlespaul5709]

Nuno Teixeira The Amish