1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Free relativistic particle (wave function)

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data

    The hamiltonian of a free relativistic particle moving along the x-axis is taken to be [tex]H=\sqrt{p^2c^2+m^2c^4}[/tex] where [tex]p[/tex] is the momentum operator. If the state of the wave function at time [tex]t=0[/tex] is described by the wave function [tex]\psi_0(x)[/tex] what is the wave function at time [tex]t>0[/tex] Hint: solve the time-dependent Schrödinger equation in momentum space. The answer can be left in the form of an integral.

    2. Relevant equations



    3. The attempt at a solution
    In momentum space [tex] \psi(x)=\frac{1}{\sqrt{2\pi}} \int_k \phi(k) e^{i k x} [/tex]
    does this mean that [tex] \psi_0(x)=\frac{1}{\sqrt{2\pi}} \int_k \phi_0(k) e^{i k x}[/tex]
    and how do i know what [tex] \phi_0(k)[/tex] is?

    Is the right answer something in form of [tex] \psi_0(x)=\frac{1}{\sqrt{2\pi}} \int_k \phi_0(k) e^{i k x}e^{i E t/\hbar}[/tex] where i just kind of write down the usual derivation of the time-dependent schrödinger equation?
     
    Last edited: Oct 28, 2009
  2. jcsd
  3. Oct 28, 2009 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    You can get [itex]\phi_0(k)[/itex] from [itex]\psi_0(x)[/itex] by taking its Fourier transform.

    Your final answer will involve a double integral, and come from

    [tex]\psi(x,t)=\frac{1}{\sqrt{2\pi}}\oint\phi(k,t)e^{ikx}dx[/tex]

    You will need to determine [itex]\phi(k,t)[/itex] from [itex]\phi_0(k)[/itex] by solving the time-dependent Schroedinger equation.
     
  4. Oct 28, 2009 #3
    So i get [tex]\phi_0(p)=\frac{1}{\sqrt{2\pi}} \int_k \psi_0(x) e^{i p x}[/tex] ?

    Then i solve the time dependent as following:
    [tex]H\phi_0 = E \phi_0 \,[/tex] where [tex]H=\sqrt{p^2c^2+m^2c^4} [/tex]


    Is it so, that:
    [tex] \phi(p,\,t)= A(t) \phi_0(p) \ [/tex]
    which leads to
    [tex] \Phi(p,\,t) = \phi_0(p) e^{-i{E t/\hbar}} \,[/tex]
    and then
    [tex]\psi(x,t)=\frac{1}{\sqrt{2\pi}}\int dpe^{-i{E t/\hbar}}e^{ip x} \frac{1}{\sqrt{2\pi}} (\int \psi_0(x) e^{i p x} dx)[/tex]
     
  5. Oct 28, 2009 #4
    Is this correct? It seems to be ok to me.
     
  6. Oct 28, 2009 #5

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    No, that's the time-independent Schroedinger equation...
     
  7. Oct 28, 2009 #6
    Of course. So the time dependent looks like
    [tex]i \hbar{\partial \over \partial t} \phi(p,\,t) = \hat H \phi(p,\,t)[/tex]
    and for
    [tex] \hat H=\sqrt{p^2c^2+m^2c^4} [/tex]
    it gives
    [tex]\phi(p,\,t) = \phi(p,\,0)e^{-i\sqrt{p^2c^2+m^2c^4}t/\hbar}[/tex]
    which would lead to (by the previous reasoning)
    [tex]\psi(x,t)=\frac{1}{2\pi}\int dpe^{-i\sqrt{p^2c^2+m^2c^4}t/\hbar}e^{ip x} (\int \psi_0(x) e^{i p x} dx)[/tex]
    is this the answer?
     
  8. Oct 28, 2009 #7

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Good:approve:

    Be careful, [itex]k=\frac{p}{\hbar}[/itex]...so when you rewrite your Fourier transforms in terms of [itex]p[/itex] instead of [itex]k[/itex], you should get some [itex]\hbar[/itex]s in there somewhere.
     
  9. Oct 28, 2009 #8
    one more try

    [tex]\psi(x,t)=\frac{1}{2\pi\hbar}\int dpe^{-i\sqrt{p^2c^2+m^2c^4}t/\\hbar}e^{ip x/\hbar} (\int \psi_0(x) e^{i p x/\hbar} dx)[/tex]
    which shold be correct if
    [tex] k=\frac{p}{\hbar}[/tex]
     
  10. Oct 28, 2009 #9

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Looks good to me!:approve:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook