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!
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.
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
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.
@@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 :)
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.