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QM: changing basis of quantum states

  • Thread starter doublemint
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  • #1
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Hello,

I am trying to express a given wavefunction through different basis, momentum and position. Look at 5.2(b) and (c) through the link: http://qis.ucalgary.ca/quantech/443/2011/homework_five.pdf" [Broken]

I complete part (b) by doing the following:
[tex]\int\left\langle{x}\left|e^{\frac{-i\hat{p}a}{h}}\left|{x'}\right\rangle\left\langle{x'}\left|\psi\right\rangle dx' = \psi(x+a)[/tex]
If I did part (b) correctly, then for part (c), I did the same thing to find the state in the momentum basis:
[tex]\int\left\langle{p}\left|e^{\frac{-i\hat{p}a}{h}}\left|{p'}\right\rangle\left\langle{p'}\left|\psi\right\rangle dp' = \int e^{\frac{-ip'a}{h}}\psi(p')dp'[/tex]
Now I do not know where to go..and I do not know if i did the right steps.
So if anyone can help, that would be great!
Thank You
Doublemint

Edit:
I have looked at 5.3(b) and the integration is horrible...
Given that: [tex]\tilde{\psi}(p) = \frac{1}{\sqrt{2\pi\hbar}}\int\psi(x)e^{\frac{-ipx}{\hbar}}dx = \frac{1}{\sqrt{2\pi\hbar}}\int Axe^{\frac{-k^{2}x^{2}}{2}}e^{\frac{-ipx}{\hbar}}dx[/tex]. Any one got ideas on this one as well?
 
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Answers and Replies

  • #2
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I think the link you gave doesn't work. I can't help you anyway, I'm just curious :P
 
  • #3
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thanks Telemachus, Ill get that fixed.
 
  • #4
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I did the same thing to find the state in the momentum basis:
[tex]\int\langle p |e^{\frac{-i\hat{p}a}{h}} |p'\rangle \langle p'|\psi \rangle dp' = \int e^{\frac{-ip'a}{h}}\psi(p')dp'[/tex]
You made a mistake here. You need to use:

[tex]
\langle p |e^{\frac{-i\hat{p}a}{h}} |p'\rangle = e^{\frac{-i p' a}{\hbar}} \langle p | p' \rangle = e^{\frac{-i p' a}{\hbar}} \delta(p - p')
[/tex]
 
  • #5
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Hey matonski,
So you are saying it should be like this?
[tex]\int e^{\frac{-ip'a}{\hbar}}\delta(p-p')\psi(p')dp' = e^{\frac{-ipa}{\hbar}}\psi(p)[/tex]
Okay, so then how do I use the Fourier Transformations to verify part (b)?
 
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  • #6
vela
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The problem wants to you to take the explicit expressions you have for the wave functions [tex]\tilde{\psi}(p)[/tex] and [tex]\tilde{\psi}'(p)[/tex] and Fourier transform them to find their representation in the position basis. In the position basis, those two wave functions should have the relationship you figured out in part (b).

Edit:
I have looked at 5.3(b) and the integration is horrible...
Given that: [tex]\tilde{\psi}(p) = \frac{1}{\sqrt{2\pi\hbar}}\int\psi(x)e^{\frac{-ipx}{\hbar}}dx = \frac{1}{\sqrt{2\pi\hbar}}\int Axe^{\frac{-k^{2}x^{2}}{2}}e^{\frac{-ipx}{\hbar}}dx[/tex]. Any one got ideas on this one as well?
Complete the square in the exponent.
 
  • #7
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Alright I figured out 5.3. Now for 5.2..
okay so for [tex]\tilde{\psi}(p) = Ae^{\frac{-b^{2}p^{2}}{2}[/tex]
i would sub it into:
[tex]\psi(x)= \frac{1}{\sqrt{2\pi\hbar}}\int\tilde{\psi}(p) e^{\frac{-ipx}{\hbar}}dp = \frac{Ae^{\frac{-1x^{2}}{2b^{2}\hbar^{2}}}}{ib\sqrt{\hbar}}[/tex]
So far so good?
Then to find the momentum basis of [tex]\left|\psi'\right\rangle = e^{\frac{-i\hat{p}a}{\hbar}\left|\psi\right\rangle[/tex]
I would just do what I did before which was
[tex]\int e^{\frac{-ip'a}{\hbar}}\delta(p-p')\psi(p')dp' = e^{\frac{-ipa}{\hbar}}\psi(p)[/tex]
But I do not see the consistency compared to 5.2(b)..
 
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  • #8
vela
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I would just do what I did before which was
[tex]\int e^{\frac{-ip'a}{\hbar}}\delta(p-p')\psi(p')dp' = e^{\frac{-ipa}{\hbar}}\psi(p)[/tex]
But I do not see the consistency compared to 5.2(b)..
You have

[tex]\tilde{\psi}'(p) = \langle p \vert \psi' \rangle = e^{-ipa/\hbar}\tilde{\psi}(p) = e^{-ipa/\hbar}Ae^{-b^2p^2/2}[/tex]

Now take its Fourier transform to find the wave function [itex]\psi'(x)[/itex] for the state [itex]\vert \psi' \rangle[/itex] in the position basis.

What does 5.2(b) say about how [itex]\psi(x)[/itex] and [itex]\psi'(x)[/itex] should be related?
 
  • #9
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You have

[tex]\tilde{\psi}'(p) = \langle p \vert \psi' \rangle = e^{-ipa/\hbar}\tilde{\psi}(p) = e^{-ipa/\hbar}Ae^{-b^2p^2/2}[/tex]

Now take its Fourier transform to find the wave function [itex]\psi'(x)[/itex] for the state [itex]\vert \psi' \rangle[/itex] in the position basis.

What does 5.2(b) say about how [itex]\psi(x)[/itex] and [itex]\psi'(x)[/itex] should be related?
[tex]\tilde{\psi}'(x) = \frac{Ae^{\frac{-1(a+x)^{2}}{2b^{2}\hbar}}}{b\sqrt{\hbar}}[/tex]
So 5.2(b) should be describing a shift between [itex]\psi(x)[/itex] and [itex]\psi'(x)[/itex], right?
 
  • #10
vela
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Right. That's what 5.2(b) showed. The effect of the operator [itex]\exp(-i\hat{p}a/\hbar)[/itex] was to shift the state by a to the left. So when you took the two momentum-space wave functions and Fourier-transformed them back to the position basis, you should find that one is the shifted version of the other if everything is consistent.
 
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  • #11
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Cool! Thanks alot vela!
 

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