Commutator Relations: [x,p]=ih, Proof of p=-iħ∂/∂x+f(x)

alisa
Messages
3
Reaction score
0
given that [x,p]=ih, show that if x=x, p has the representation p=-iħ∂/∂x+f(x) where f(x) is an arbitrary function of x
 
Physics news on Phys.org
alisa, you're supposed to show an attempted solution at the problem. That goes for your other threads as well.

alisa said:
given that [x,p]=ih, show that if x=x,

What do you mean " if x=x". x=x by definition.
 
Tom Mattson said:
What do you mean " if x=x". x=x by definition.

It's the coordinate representation in which the Hilbert space is L^{2}(\mathbb{R},dx). The "x" operator is realized by a multiplication by "x". She's asked to prove that the most general representation of the momentum operator in this Hilbert space is the one written there.
 
OK, so then it should read something like "If \hat{x}|\psi>=x|\psi>...", right?
 
Last edited:
Exactly. That's the spectral equation, but nonetheless, yes.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top