Minimum Momentum and Energy from Schrödinger Equation

  • Thread starter erok81
  • Start date
  • #1
464
0

Homework Statement



I have a 2D square well with sides a.

What is the minimum momentum and energy for this well?

Homework Equations



Minimum momentum given by Δp≈ℏΔx and Δp≈ℏΔy
Minimum energy given by ΔE≈ℏΔt

This might be right. It says to consider the Schrödinger Equation in 2D. I've never seen this in 2D, so I don't know if this is correct.

[tex]\frac{- \hbar^{2}}{2m} \left( \frac {\partial ^{2}}{\partial x^{2}} + \frac {\partial ^{2}}{\partial y^{2}} \right) \Psi (x,y,t) + U(x,y,t) \Psi (x,y,t) = i \hbar \frac{\partial}{\partial t} \Psi (x,y,t)[/tex]

The Attempt at a Solution



So I first though I would separate the variables into a spatial piece and a time dependent piece and then relate those to the uncertainty principles above. But then realized that might not work since I am in 2D. Then again....can I split it into two spatial (one for x and one for y) set each of those in the uncertainty principle since they aren't axis dependent and then do the same for time.

Although I don't think that is right either since that doesn't give me a Δx,y,t.

Basically I am stuck at the beginning it seems. :redface:
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,044
1,624
Are you supposed to use the uncertainty relations? It's pretty straightforward to find the solutions to the 2D equation.
 
  • #3
464
0
That's where I am not sure. That was my part. I don't have to use them. I just assumed since we were finding the momentum and energy minimum they might be used.

Other than that, I am not sure how what I have above relates to momentum.
 
Last edited:
  • #4
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,044
1,624
You can get estimates using the uncertainty principle, but if you need exact answers, you need to solve the SE.
 
  • #5
464
0
So in this case I can apply separation of variables to get an x only, y only, and t only SE. Then I can solve from there.

The only thing I am not sure is how they relate to momentum and energy. We haven't gone into a huge amount of depth on actually solving them.
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,044
1,624
When you solve the SE, you're finding the energy eigenstates and eigenvalues, so to find the minimum energy, you just need to figure out what the lowest eigenvalue is.

The energy eigenstates are not eigenstates of the momentum operators, but it turns out they are easy to express as linear combinations of momentum eigenstates. This might have been discussed for the case of the 1D square well. Are you familiar with that?

I'm still wondering whether the intent of the problem is to have you solve the SE or to find estimates using the uncertainty relations. On the one hand, when a problem asks you for the minimum momentum, it typically means to estimate it using the uncertainty principle. On the other hand, you said the problem said to consider the SE in 2D, which suggests you're supposed to solve it. Have you asked your instructor for clarification?
 
  • #7
464
0
Well in class we haven't said anything about eigenstates/values. The only thing we've talked about that deals with minimum momentum is the uncertainty principle.

I haven't had a chance yet. I was going to try after class tomorrow, but of course I have a test in another class right after.

If one was to solve using uncertainty principles, how would I relate them using the SE? For the problem of course, but I'd also like to know in general. I still get confused when using it so any chance I get to use it more, the better.

Thanks for all the help so far.
 
  • #8
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
15,044
1,624
In terms of the momentum operators

[tex]\begin{align*}
\hat{p}_x &= \frac{\hbar}{i}\frac{\partial}{\partial x} \\
\hat{p}_y &= \frac{\hbar}{i}\frac{\partial}{\partial y}
\end{align*}[/tex]

the time-independent Schrodinger equation is

[tex]\left(\frac{\hat{p}_x^2 + \hat{p}_y^2}{2m}\right)\psi(x,y) + V(x,y)\psi(x,y) = E\psi(x,y)[/tex]

Inside the well, the potential is 0, so the energy is all kinetic, and if you have an estimate of the magnitude of the momentum, you can get an estimate of the particle's kinetic energy.
 
  • #9
464
0
I read ahead into the next chapter last night and found the energy eigenstates and eigenvalues. It is in an "advanced" section of my text book (thankfully spring break is next week so I can study this chapter!) so I don't think solving it will require those.

Also after you posted those operators I remembered learning that in class last time, but hadn't had a chance to use them. So I can use them..the only problem now is I don't have a wave function. Without that I don't think I can do much besides write out the equations.
 

Related Threads on Minimum Momentum and Energy from Schrödinger Equation

Replies
13
Views
2K
Replies
1
Views
2K
Replies
1
Views
43K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
3
Views
643
Replies
2
Views
7K
Replies
3
Views
22K
Top