How Do You Solve for the Propagator in Momentum Space Under Constant Force?

Click For Summary

Homework Help Overview

The discussion revolves around finding the propagator in momentum space for a particle subjected to a constant force, as presented in a problem from R. Shankar's "Principles of Quantum Mechanics." The original poster attempts to derive the propagator and normalize their wave function, leading to questions about the correct application of quantum mechanical principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the Hamiltonian formulation and the normalization of wave functions. There are attempts to clarify the relationship between eigenstates and the Hamiltonian, as well as the use of completeness in deriving the propagator. Questions arise regarding the evaluation of integrals and the correctness of assumptions made during the process.

Discussion Status

The discussion is active, with participants providing insights and corrections to each other's reasoning. Some guidance has been offered regarding the normalization process and the use of completeness, but there is no explicit consensus on the evaluation of the propagator or the assumptions made in the calculations.

Contextual Notes

Participants are navigating the complexities of quantum mechanics, particularly in the context of propagators and eigenstates. There are indications of potential misunderstandings regarding the application of operators and the evaluation of integrals, which remain unresolved.

WarnK
Messages
30
Reaction score
0

Homework Statement


(this is from R. Shankar, Principles of Quantum Mechanics, 2nd ed, exercise 5.4.3)
Consider a particle subject to a constant force f in one dimension. Solve for the propagator in momentum space and get

U(p,t;p',0) = \delta (p-p'-ft) e^{ i(p'^3-p^3)/6m\hbar f }

Homework Equations


The Attempt at a Solution



I write a hamiltonian H = p^2/2m + fx, plug that into H|p>=E|p>, with the x operator in momentum space being ih d/dp, it's all nice and seperable and I get

\psi(p) = A exp \left( i \frac{p^3-6mEp}{6m\hbar f} \right)

but what do I do now? I'm not sure how to go about normalizing this.
 
Physics news on Phys.org
First of all, you don't really mean H|p>=E|p>; |p> is an eigenstate of the momentum operator, but not of H; you mean H|E>=E|E>, where <p|E> is your momentum-space wave function.

You want to normalize it so that you can write a completeness statement in the form
\int_{-\infty}^{+\infty}dE\;|E\rangle\langle E| = I,
where I is the identity operator.
 
Thanks!
It's a bit much too write everything as tex but I ended up with
A = \left( \frac{1}{2 \pi \hbar f} \right)^{1/2}
Now, how do I get a propagator out of this!
 
Last edited:
The propagator is <p'|e^(-iHt/hbar)|p>. Can you think of a use for the completeness statement?
 
<br /> \langle p&#039;| e^{-iHt/ \hbar} |p\rangle = \int_{-\infty}^{+\infty}dE \langle p&#039;|E \rangle \langle E| e^{-iHt/ \hbar} |p\rangle = \int_{-\infty}^{+\infty}dE \langle p&#039;|E \rangle \langle E|p-ft\rangle<br />
and I can plug in my expression for psi, do the integral and out comes the given answer.
But how come e^(-iHt/hbar)|p> = |p-ft>? I just guessed it from experience of doing lots of integrals like these.
 
Last edited:
You don't need to guess it, and in fact it's not correct; evaluating your final expression does not yield the given result (so you made a mistake somewhere when you evaluated it).

In your middle expression, you can replace H with E, since H is sitting next to one of its eigenstates. Then the integral over E will generate the given answer.
 
Hi, i have a quick question on middle part of the integral. How do you evaluate <E|e^{-iEt/h}|p> ?
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
3K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 2 ·
Replies
2
Views
365