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How to get from <P> to probability density P(k)?

  1. Jun 22, 2014 #1
    1. The problem statement, all variables and given/known data
    Knowing the momentum operator -iħd/dx , the expectation value of momentum and the Fourier transforms how can I prove that <p> = ∫dk [mod square of ψ(k)] h/λ. From this, mod square of ψ(k) is defined to be equal to P(k) right?

    2. Relevant equations
    Momentum operator, p: -iħd/dx
    Expectation value of p: <p> = ∫ψ(x)*pψ(x)dx
    Fourier transform (not going to type here assuming it is known since it is a lengthy pair of eqs)

    3. The attempt at a solution
    I have tried various "tactics" but the closest I got was through these steps. I used the fourier transforms of ψ(x) and ψ*(x) to get ψ(k) and ψ*(k) somewhere in the eq for <p>. I tried to integrate by parts to pull both ψ(k) out in front so it would form mod square ψ(k). Basically i get as constants ∫[h/(2piλ)][mod square of ψ(k)]*[a large integral]dk. Here is where i got stuck from the solution the integral should equal 2pi which it looks like it doesn't.

    Thank you for your time. This problem seems important to know but I just can't find any solutions online.
     
  2. jcsd
  3. Jun 22, 2014 #2

    vanhees71

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    2016 Award

    It's much easier in steps:

    (a) Write the position-representation wave function, [itex]\psi(t,x)=\langle x|\psi,t \rangle[/itex] in terms of the momentum-representation wave function, [itex]\tilde{\psi}(t,p)=\langle{p}|\psi,t \rangle[/itex].

    (b) Use the result to evaluate [itex]\hat{p} \tilde{\psi}(t,p)[/itex] from the momentum operator in position representation, [itex]\hat{p} \psi(t,x)=-\mathrm{i}/\hbar \partial_x \psi(t,x)[/itex].

    (c) Finally write [itex]\langle p \rangle=\langle \psi,t|\hat{p}|\psi,t \rangle[/itex].

    Above, I assumed the Schrödinger picture for the time dependence (eigen vectors of position and momentum time-independent state ket time dependent).
     
  4. Jun 22, 2014 #3
    How on earth can anybody check your work if you don't post it? Don't just describe your calculation. Show it. How can we tell if an integral is equal to 2pi if you don't show the integral?
     
  5. Jun 22, 2014 #4
    Thanks Vanhees but I have not learned this notation yet which i believe will be the next chapter in the book. Once I get a little bit further ill revisit this calculation to see how this notation might simplify things.

    Sorry about not posting the work it is a lot to write without knowing latex. Ill give it a try:

    [tex] |\psi(x)|^{2}=P(x) = \psi^*(x)\psi(x)[/tex]
    [tex] <p>=\int\psi^*(x)(-i\hbar\frac{\partial}{\partial x})\psi(x) dx[/tex]

    Knowing Fourier transforms:
    [tex] \psi(x) = \frac{1}{\sqrt {2\pi}}\int_{-\infty}^{\infty}\tilde{\psi}(k)e^{ikx} dk[/tex]

    Using the first transform and its complex conjugate we get:
    [tex] <p>= \int (\frac{1}{\sqrt {2\pi}}\int_{-\infty}^{\infty}\tilde{\psi^*}(k)e^{-ikx} dk(-i\hbar\frac{\partial}{\partial x})\frac{1}{\sqrt {2\pi}}\int_{-\infty}^{\infty}\tilde{\psi}(k')e^{ik'x}dk') dx[/tex]

    I think here is where i messed up since the complex conjugate of a integral might not be the integral of the complex conjugate, but i proceeded anyways

    [tex] \int (\frac{k\hbar}{2\pi}\int_{-\infty}^{\infty}\tilde{\psi^*}(k)e^{-ikx} dk \int_{-\infty}^{\infty}\tilde{\psi}(k)e^{ikx}dk) dx[/tex]

    This is where I got stuck since integration by parts didn't help me at all. Also did some "illegal" movement of integrals


    Edit: I believe I have solved it so ill just post up my solution for anyone that needs it in the future

    [tex] \int (\frac{k\hbar}{2\pi}\int_{-\infty}^{\infty}\tilde{\psi^*}(k)e^{-ikx} dk \int_{-\infty}^{\infty}\tilde{\psi}(k')e^{ik'x}dk') dx[/tex]

    [tex] \int (\frac{k\hbar}{2\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\tilde{\psi^*}(k)\tilde{\psi}(k')e^{ix(k'-k)}dkdk'dx[/tex]

    By [tex]\int_{-\infty}^{\infty} e^{ix(k'-k)} = 2\pi\delta(k'-k)[/tex]
    This was what I was missing

    We get:
    [tex]k\hbar\int_{-\infty}^{\infty}\tilde{\psi^*}(k)\tilde{\psi}(k')\delta(k'-k)dkdk'[/tex]
    [tex]k\hbar\int|\psi(k)|^2dk[/tex]
     
    Last edited: Jun 22, 2014
  6. Jun 23, 2014 #5

    vanhees71

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    You started very well! It's important to keep different labels for the [itex]k[/itex] and [itex]k'[/itex] integrals, before you do the [itex]x[/itex] integration. At the end, of course, the [itex]k[/itex] must be under the [itex]k[/itex] integral to get the average momentum. As for any distribution, the expectation value is given by (using [itex]p=\hbar k[/itex])
    [tex]\langle p \rangle=\int_{\mathbb{R}} \mathrm{d} k \; \hbar k \; \tilde{P}(k).[/tex]
     
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