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

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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

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

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

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

but what do I do now? I'm not sure how to go about normalizing this.
 
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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
[tex]\int_{-\infty}^{+\infty}dE\;|E\rangle\langle E| = I,[/tex]
where I is the identity operator.
 
Thanks!
It's a bit much too write everything as tex but I ended up with
[tex]A = \left( \frac{1}{2 \pi \hbar f} \right)^{1/2}[/tex]
Now, how do I get a propagator out of this!
 
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The propagator is <p'|e^(-iHt/hbar)|p>. Can you think of a use for the completeness statement?
 
[tex] \langle p'| e^{-iHt/ \hbar} |p\rangle = \int_{-\infty}^{+\infty}dE \langle p'|E \rangle \langle E| e^{-iHt/ \hbar} |p\rangle = \int_{-\infty}^{+\infty}dE \langle p'|E \rangle \langle E|p-ft\rangle[/tex]
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.
 
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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[itex]^{-iEt/h}[/itex]|p> ?
 
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