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Simple Problem in One Dimension

  1. Jul 14, 2007 #1
    1. The problem statement, all variables and given/known data

    Exercise 5.1.1 : Show that Eq (5.1.9( may be rewritten as an integral over E and a sum over the +/- index as

    [tex]U(t) = \sum_{a=\pm}\int^{\infty}_{0}|E,a><E,a|e^{-iEt/\hbar}dE[/tex]

    You will find at http://books.google.com/books?id=2z...sig=dZ4WX7ruNL46QZzNSbqIExbz3qw#PRA1-PA152,M1 equation 5.1.9 and the background of this problem.


    2. Relevant equations



    3. The attempt at a solution

    I am trying to use the substitution rule starting with 5.1.9 which requires that we tack on a 2m/sqrt(2mE) to get and integrand that looks like
    [tex]\frac{2m}{(2mE)^{1/2}}|p><p|e^{-iEt/\hbar}dE[/tex]

    I am not sure how you deal with the limits of integration, how we get rid of the 2, and how you deal with the eigenvectors. Do we need to change anything for the eigenvectors since they are the same for both p and E?

    Please just give me hints not the entire solution. Thanks.
     
    Last edited: Jul 14, 2007
  2. jcsd
  3. Jul 14, 2007 #2

    nrqed

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    I would start by breaking into two pieces the integral over p [itex] \int_{- \infty}^{\infty} dp = \int_0^{\infty} dp + \int_{-\infty}^0 dp [/itex]

    Now, you replace each in terms of the Energy states (corresponding respectively to positive and negative values of p). Then you do a change of variable [itex] E \rightarrow -e [/itex] in the second integral and it will look like the first one. You add the two back together. I haven't checked all the steps but I think that should work out. Post again if you run into problems.


    Patrick
     
  4. Jul 14, 2007 #3

    StatusX

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    The change of variables is a little tricky here. Note that at t=0 the propagator should be the identity operator. That is, we should have:

    [tex] 1 = \int_{-\infty}^{\infty} |p><p| dp [/tex]

    and:

    [tex] 1 =\sum_{a=\pm}\int^{\infty}_{0}|E,a><E,a|dE[/tex]

    which is just the continuous version of the discrete formula that says that [itex]|\phi_n>[/itex] is a complete set of states iff:

    [tex] \sum_n |\phi_n><\phi_n|=1 [/tex]

    Here it's assumed the states are normalized. In the continuous case, the states are unnormalizable, and this means they are ambiguous up to a scalar multiple.

    In fact, it's clear from the above two integrals that this must be the case, since if |E,a>=|p(E,a)> exactly, not just up to a scalar, the two integrals would have different dimensions. So evidently, we have:

    [tex] |E,a> = \sqrt{\frac{dp}{dE}} |p(E,a)> [/tex]

    then use nrqed's advice about changing the regions of integration.
     
  5. Jul 14, 2007 #4

    George Jones

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    Yes, this follows directly from the two (E and p) completeness relations that you wrote down.
     
  6. Jul 14, 2007 #5
    ehrenfest: sorry for the slight hijack, but since I happen to be struggling with the same problem: could some kind person please explain the last step to:
    [tex] |E,a> = \sqrt{\frac{dp}{dE}} |p(E,a)> [/tex] in a bit more detail. By the looks of it you're comparing the two completeness relations, then getting rid of the integrals by evaluating at p=..., but then why can you go from |p><p| to a square root?
    Thank you.
     
    Last edited: Jul 14, 2007
  7. Jul 14, 2007 #6

    George Jones

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    The two completeness relations require [itex]\left< p | p' \right> = \delta \left( p - p' \right)[/itex] and [itex]\left< E | E' \right> = \delta \left( E - E' \right).[/itex] To see, this multiply the [itex]p[/itex] completeness relation by [itex]\left| p' \right>[/itex]. This gives

    [tex]
    \begin{equation*}
    \begin{split}
    \left| p' \right> &= \int \left| p \right> \left< p | p' \right> dp\\
    &= \int \left| p \right> \delta \left( p - p' \right) dp\\
    &= \left| p' \right>.
    \end{split}
    \end{equation*}
    [/tex]

    Now, suppose [itex]\left| E \right> = \alpha \left| p \right>.[/itex] Then, [itex]\left< E | E' \right> = \left| \alpha \right|^2 \left< p | p' \right>.[/itex] But,

    [tex]
    \begin{equation*}
    \begin{split}
    \left< E | E' \right> &= \delta \left( E - E' \right)\\
    &= \delta \left( \frac{p^2}{2m} - \frac{p'^2}{2m} \right).
    \end{split}
    \end{equation*}
    [/tex]

    Now, use properties of the delta function to find [itex]\alpha.[/itex]
     
    Last edited: Jul 14, 2007
  8. Jul 14, 2007 #7
    Maybe I just don't know this property of the delta function...

    Its not this property,

    [tex]\delta'(x-x') = \frac{d}{dx}\delta(x-x') = -\frac{d}{dx'}\delta(x-x')[/tex],

    is it?
     
  9. Jul 14, 2007 #8

    olgranpappy

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    umm... but there are two roots to the equation in the argument of the delta function...

    ehrenfest, the property that you are expected to use is apparently
    [tex]
    \delta(f(x)) = \sum_r \frac{\delta(x-x_r)}{|f'(x_r)|}
    [/tex]
    where the x_r are roots of f(x). E.g. [tex]\delta(ax)=\frac{1}{|a|}\delta(x)[/tex].

    Consider this explaination:

    [tex]
    1=\int dp |p><p|=\int dE \frac{dp}{dE}|p(E)><p(E)|=\int dE \sqrt{\frac{dp}{dE}}|p(E)><p(E)|\sqrt{\frac{dp}{dE}}
    [/tex]
    and also
    [tex]
    1=\int dE |E><E|
    [/tex]

    so comparing the above equation with the last line of the equation above it we see that we might like to make the identification
    [tex]
    |E>=\sqrt{\frac{dp}{dE}}|p(E)>
    [/tex]
    and so we do.

    Also, n.b., the ket |p> is *not* the same thing as the ket |p(E)>.
     
  10. Jul 15, 2007 #9

    George Jones

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    Not if the region of integration in equation 5.1.9 has been spit into two, as Patrick and StatusX have said. And if the region hasn't been split,then [itex]\left|E\right> = \alpha \left|p\right>[/itex] can't, in general, be written as Shankar makes clear in the link given in ehrenfest's original post, i.e., in my previous post, it's actually [itex]\left|E,a>\right> = \alpha \left|p\right>[/itex], with [itex]a=-[/itex] in one region, and [itex]a=+[/itex] in the other.
     
    Last edited: Jul 15, 2007
  11. Jul 15, 2007 #10

    olgranpappy

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    Silly me. I guess I should have read all the posts and clicked on all the links before responding. Oh well. Cheers.
     
  12. Jul 15, 2007 #11
    Almost there. So we have

    [tex]\int_{-\infty}^{\infty}|p><p|exp(-i p^2 t /2 m \hbar) [/tex]

    [tex]= \sum_{\pm \alpha}\int_{-\infty}^{0}|E,\alpha><E,\alpha| \sqrt{dp/dE}exp(-i E t /2 m \hbar) + \sum_{\pm \alpha}\int_{0}^{\infty} |E,\alpha><E,\alpha| \sqrt{dp/dE}exp(-i E t /2 m \hbar)[/tex]

    [tex]= \sum_{\pm \alpha}\int_{-\infty}^{0}|E,\alpha><E,\alpha| \sqrt{m/(2Em)^{1/2}}exp(-i E t /2 m \hbar) + \sum_{\pm \alpha}\int_{0}^{\infty} |E,\alpha><E,\alpha| \sqrt{m/(2Em)^{1/2}}exp(-i E t /2 m \hbar)[/tex]

    I'm not sure what you mean nrqed, when you said change the variable E -> -e. If you switch the limits of integration in one of these integral isn't the sum just 0? Where did I go wrong?

    Also,

    This should really be |p(E,alpha)> right?
     
    Last edited: Jul 15, 2007
  13. Jul 15, 2007 #12

    olgranpappy

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    Here's how it goes. Call [tex]E=\frac{p^2}{2m}[/tex]. This thing is always positive, right. Yes, it is. Also the sqrt
    sign will always mean the positive square root so that [tex]\sqrt{x}[/tex] is positive and [tex]-\sqrt{x}[/tex] is negative.
    Now, we also label states by this letter E, but for a given E there are two p's which have that value of E.

    Anyways, let's just go thru the formalism without the factor of [tex]e^{itp^2/2m}[/tex] in there:

    [tex]
    \int_{-\infty}^{\infty}dp|p><p|
    =
    \int_{-\infty}^0dp|p><p|
    +
    \int_{0}^{\infty}dp|p><p|
    [/tex]
    now I'm going to change variables from p to -p in the first integral there. then I will change variables from p to E.
    [tex]
    =\int_{0}^{\infty}dp|-p><-p|+\int_{0}^{\infty}dp|p><p|
    =\int_{0}^{\infty}dE\frac{dp}{dE}|p=-\sqrt{2mE}><p=-\sqrt{2mE}|
    +\int_{0}^{\infty}dE\frac{dp}{dE}|p=+\sqrt{2mE}><p=+\sqrt{2mE}|
    [/tex]
    You see that the integration limits are the same. So I can write
    [tex]
    =\int_{0}^{\infty}dE\frac{dp}{dE}\left(|p=-\sqrt{2mE}><p=-\sqrt{2mE}|+|p=+\sqrt{2mE}><p=+\sqrt{2mE}|\right)
    =\sum_\alpha\int_{0}^{\infty}dE\frac{dp}{dE}|E,\alpha><E,\alpha|
    [/tex]

    Everything goes thru exactly the same with the factor of e^(itp^2/2m) inside there too. It just becomes e^{itp(E)^2/2m}=e^{iEt}
     
    Last edited: Jul 15, 2007
  14. Jul 15, 2007 #13
    You did not even use the fact that

    ,

    right?
     
  15. Jul 15, 2007 #14

    olgranpappy

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    right. but, see that last far RHS of my last equation... you see that I could factor that into
    [tex]
    \sum\int dE \sqrt{\frac{dp}{dE}}|E,\alpha><E,\alpha|\sqrt{\frac{dp}{dE}}
    [/tex]
    if I wanted to...

    But why should one do that? You don't have to.
     
  16. Jul 15, 2007 #15

    olgranpappy

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    and, again, n.b. your |E,a> is not the same as Shankar's... apparently. But dont worry about it. it's notation. physicists are always using the same notation for different functions (and kets)
     
  17. Jul 15, 2007 #16

    StatusX

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    Because otherwise it will hold that:

    [tex] \int |p><p|dp = 1 [/tex]

    But:

    [tex] \sum_a \int |E,a><E,a| dE \neq 1[/tex]

    Why should p be special? There's no priveleged definition of these states - as I mentioned above they are unnormalizable, and so intrinsically ambiguous up to a scalar multiple. This scalar is picked depending on the variable used to index the states so that the above relations hold. This might not be necessary, but it's certainly the most natural thing to do
     
    Last edited: Jul 15, 2007
  18. Jul 16, 2007 #17

    olgranpappy

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    I didn't need to explicitly make that identification. It is a pointless identification, really; it's just a roundabout way to stick in the Jacobian.
     
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