Time evolution operator on momentum basis (1 Viewer)

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Hey i was wondering how to express the time evolution operator U(t,to) to a momentum eigen state |p> for a particle moving in the xdirection under a zero potential, V= 0. The reason i need this is that iam told the only way to get the matrix element of the time evolution operator using position eigen states <x|U(t,to)|x2> is to first apply the Time evolution operator on a momentum basis and use a linear transformation to position eigenstates, Iam not quite sure how to apply the time evolution operator on the momentum eigenstate though.

Thanks any help would be appreciated

My attempt:

I thought of trying to expand the exponential time evolution operator using

e^{x}=\\sum_n x^n/n!

but then your just left with the exact same problem which is i can't seem to be able to use the operator on the momentum eigenstate, not quite sure how to begin...
 
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fzero

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Write down the Hamiltonian. The reason I said to use the momentum basis is that the momentum eigenstates are also eigenstates of the free-particle Hamiltonian.
 
I see i think iam starting to get the idea of how to solve it, but what i have been struggling with is just the math of it all, i am not sure how to take the hamiltonian using a momentum basis.
 

fzero

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The Hamiltonian, expressed in terms of operators, does not change with the basis. Write down the Hamiltonian in terms of the momentum operator and try to compute

[tex] \hat{H} | p\rangle.[/tex]
 
[tex]
\hat{H} | p\rangle = {E}| p\rangle
[/tex]

so you would just have to find the constant eigenvalue and since the potential is zero wouldn't you just end up with a second order ODE and solve for the eigenfunction (momentum). But iam not sure how that would apply to the time operator in terms of momentum eigenstate.
 

fzero

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What is

[tex] \hat{p}|p\rangle ?[/tex]

How is [tex]\hat{H}[/tex] expressed in terms of [tex]\hat{p}[/tex]? Is there another way to write [tex] \hat{H}|p\rangle [/tex] using these 2 pieces of information?
 
[tex]
\hat{H} = (ih/2m)(d/dx) hat{p}
[/tex]


Are you saying that since the hamiltonian generates the time evolution of eigenstates, we can express the time evolution operator in terms of eigenstates |p> using the hamiltonian?
 

fzero

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You can write it entirely in terms of [tex]\hat{p}[/tex] as

[tex] \hat{H} = \frac{1}{2m} \hat{p}^2.[/tex]

Now try to answer the other questions I asked.
 
Maybe iam just struggling with dirac notation, but wouldn't the first just give you an eigenstate proportional to the momentum operator by a scaler, eigenvalue. Is it the eigenvalue you are interested in or the eigenstate |p>.
 

fzero

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Maybe iam just struggling with dirac notation, but wouldn't the first just give you an eigenstate proportional to the momentum operator by a scaler, eigenvalue. Is it the eigenvalue you are interested in or the eigenstate |p>.
You sort of have it. Don't be afraid to write down an equation if you're not 100% sure. I'm trying to lead you on to think a bit for yourself, explain what you understand or don't and then we'll continue back and forth. To put what you're saying into a formula, we have that

[tex] \hat{p} |p\rangle = p |p\rangle .[/tex]

That is, the states [tex]|p\rangle [/tex] are eigenstates of the momentum operator [tex]\hat{p}[/tex] so acting with [tex]\hat{p}[/tex] gives us the eigenvalue times the eigenstate.

So try to compute

[tex]\hat{H}| p \rangle[/tex]

in terms of the momentum eigenvalue [tex]p[/tex]. You probably have enough to try to compute

[tex] U(t,t_0) | p\rangle[/tex]

as well now.
 
[tex]
hat{H}| p>
[/tex]


so would that just be the same as the case with the momentum operator but now its (h^2)/2m multiplied by the momentum eigenstate squared? so the question is asking to find the eigenvalue associated with the momentum eigenstate when you use the time evolution operator on the momentum eigenstate?

I.e ( i replaced the hamiltonian with Energy)

e^(iEt/h) |p> = U|p>

where u is the eigenvalue

if it is how would i find it?
 

fzero

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[tex]
hat{H}| p>
[/tex]


so would that just be the same as the case with the momentum operator but now its (h^2)/2m multiplied by the momentum eigenstate squared?
The momentum eigenvalue will appear squared in the answer, not the eigenstate. It's not quite what you wrote. Since

[tex]\hat{p} | p\rangle = p | p \rangle,[/tex]

you should be able to convince yourself by direct calculation that

[tex]\hat{p}^2 | p\rangle = p^2 | p \rangle.[/tex]

Then

[tex]\hat{H} | p\rangle = \frac{1}{2m} \hat{p}^2 | p\rangle =\frac{1}{2m} p^2 | p \rangle. [/tex]

So the energy eigenvalue can be written in terms of the momentum eigenvalue as

[tex] E = \frac{p^2}{2m}. [/tex]



so the question is asking to find the eigenvalue associated with the momentum eigenstate when you use the time evolution operator on the momentum eigenstate?

I.e ( i replaced the hamiltonian with Energy)

e^(iEt/h) |p> = U|p>

where u is the eigenvalue

if it is how would i find it?
Yes, because the momentum eigenstates are energy eigenstates, they're also eigenstates of the time evolution operator. I would express the eigenvalue in terms of the momentum eigenvalue. When you have the explicit result, you can go on to computing the matrix element

[tex] \langle p' | \hat{U}(t,t_0) | p'' \rangle [/tex]
 
how would i find the time evolution eigenvalue though and then the matrix element? that always seems to be where i struggle
 

fzero

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how would i find the time evolution eigenvalue though and then the matrix element? that always seems to be where i struggle
Well you should go back and verify that

[tex] \hat{p}^2 | p\rangle = p^2 | p \rangle .[/tex]

In fact, earlier in your course or text, you've almost certainly seen the result that a general function of an operator applied to an eigenstate of that operator returns the same function evaluated on the eigenvalue of the operator,

[tex] f(\hat{p})| p\rangle = f(p) | p \rangle .[/tex]

The simplest way to verify this is to consider a function [tex]f(u)[/tex] that has a power series expansion and then consider the individual terms.

The idea here is that we used this result to compute

[tex]\hat{H}|p\rangle[/tex]

and want to use it for the time evolution operator as well.
 
Oh I see so its similar to my orginal plan which was to use the power series expansion

e^x = sum(1 + x...) (i'll only use first 2 terms)

therefore you should get:

1 - iHt/h for your eigenvalue

which is what i had previously and doesn't seem to make much sense
 

fzero

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Oh I see so its similar to my orginal plan which was to use the power series expansion

e^x = sum(1 + x...) (i'll only use first 2 terms)

therefore you should get:

1 - iHt/h for your eigenvalue

which is what i had previously and doesn't seem to make much sense
Essentially, but the point is that after you express the series in terms of the eigenvalue p, you can resum the series into the original exponential form.
 
Essentially, but the point is that a
fter you express the series in terms of the eigenvalue p, you can resum the series into the original exponential form.


Ah i see, so the point was to show that you can express the hamiltonian in terms of the momentum eigenvalue (by showing that the hamiltonian operator can be expressed in terms of momentum operator) so that you can find the time evolution operator (essentially due to the hamiltonian) using a momentum basis:


U(t,to)|p> = e^((-itp^2)/(2m))|p>

where you can expand the exponential using power series.

So when iam trying to find the matrix element <x|U(t,t0)|x'> would i try to do the same thing and express the hamiltonion operator in terms of position eigenvalues?
 

fzero

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Ah i see, so the point was to show that you can express the hamiltonian in terms of the momentum eigenvalue (by showing that the hamiltonian operator can be expressed in terms of momentum operator) so that you can find the time evolution operator (essentially due to the hamiltonian) using a momentum basis:


U(t,to)|p> = e^((-itp^2)/(2m))|p>

where you can expand the exponential using power series.
That's right, but to be precise I think you want [tex]t-t_0[/tex] in the exponent instead of just t.

So when iam trying to find the matrix element <x|U(t,t0)|x'> would i try to do the same thing and express the hamiltonion operator in terms of position eigenvalues?
No, that won't work well since the position eigenstates are not eigenstates of the Hamiltonian. However, we can express the position space matrix element in terms of that in momentum space by

[tex] \langle x | \hat{U}(t,t_0)|x' \rangle = \int dp \int dp' \langle x | p\rangle \langle p| \hat{U}(t,t_0)| p' \rangle \langle p'|x' \rangle . [/tex]

To evaluate this, we need to use the fact that the position and momentum eigenstates are related by the Fourier transform, so

[tex] \langle x | p \rangle = \frac{1}{\sqrt{2\pi \hbar}} e^{i p x/\hbar} . [/tex]

You still have another step to do in evaluating

[tex] \langle p| \hat{U}(t,t_0)| p' \rangle , [/tex]

which is straightforward if you review some more facts about the Dirac notation.
 
Ah i see so for the matrix element using momentum basis and momentum eigenstate i get:


<p|u|p'> = U*<p|p'>


U is just the eigenvalue found earlier which is just the exponential in terms of momentum eigenvalue

therefore when you find the matrix element using position, you have to use compltness relation to cancel out the <p|p'> term (essentially just delta function) thus p = p' and you cancel out both the exponentials in <x|p> and <p'|x'>



i still get the double integral evalulated over the time evolution (comlex conjugate though) eigen value:


e^(ip^2)(t-to)/2m multiplied by 1/pi*h

which when i try to solve using gaussian integrals from the back of my book i get squareroot i which doesn't seem plausible
 

fzero

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Ah i see so for the matrix element using momentum basis and momentum eigenstate i get:


<p|u|p'> = U*<p|p'>


U is just the eigenvalue found earlier which is just the exponential in terms of momentum eigenvalue

therefore when you find the matrix element using position, you have to use compltness relation to cancel out the <p|p'> term (essentially just delta function) thus p = p' and you cancel out both the exponentials in <x|p> and <p'|x'>
Be careful here. You don't cancel the exponentials since one involves px and the other involves p'x'.

i still get the double integral evalulated over the time evolution (comlex conjugate though) eigen value:


e^(ip^2)(t-to)/2m multiplied by 1/pi*h

which when i try to solve using gaussian integrals from the back of my book i get squareroot i which doesn't seem plausible
The result will have the form

[tex] A \frac{1}{\sqrt{t-t_0}} e^{i\alpha \frac{(x-x')^2}{t-t_0}} , [/tex]

so you sound like you're on the right track. If your answer is really off from this, show some more work so we can find the error.
 

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