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Klein Gordon & spacelike/timelike virtual photon

  1. Feb 16, 2014 #1
    1. The problem statement, all variables and given/known data

    See attachment.

    2. Relevant equations

    3. The attempt at a solution

    So I have shown that the plane wave sol'n satisfies the Klein-Gordon equation by subbing in and reducing the equation to:

    [tex]E^2 = p^2c^2 + m^2c^4[/tex]

    which reduces to:

    [tex]E = pc[/tex]

    for an m = 0 particle, however I don't really know how to show that this would correspond with the wave travelling at the speed of light.

    Next I am asked to act on the wave function with the Hamiltonian and momentum operators, I assume I should act with the relativistic Hamiltonian so:

    [tex]\hat{H} = (-\hbar^2\frac{\partial^2 }{\partial x^2}c^2 + m^2c^4)^{1/2}[/tex]

    So I get:

    [tex]\hat{H}\psi = \sqrt{4\hbar^2c^2k^2 + m^2c^4}(e^{i(wt-kx)})[/tex]


    [tex]E = \sqrt{4\hbar^2c^2k^2 + m^2c^4}[/tex]

    [tex]h\nu = \sqrt{4\hbar^2c^2k^2 + m^2c^4}[/tex]

    Reduces to:

    [tex]-3h^2\nu^2 = m^2c^4[/tex]

    So I'm not sure what I have done wrong,
    now for the momentum:

    [tex]\hat{p}\psi = -i\hbar\frac{\partial }{\partial x}(e^{i(wt-kx)})[/tex]

    Reduces to:

    [tex]= -\hbar k(e^{i(wt-kx)})[/tex]

    [tex]\frac{h}{\lambda} = -\hbar k[/tex]

    [tex]\frac{h}{\lambda} = -\frac{h}{\lambda}[/tex]

    Once again something has gone wrong, not sure what. . .

    On the very last bit 2.(a) I basically don't know how to show that it does not satisfy the energy-mass-momentum relation. I have tried separating components and solving for E3 with,

    [tex](m_ec,\bar{0}) = ((|\bar{p}_2|^2 + m_e^2c^2)^{1/2},\bar{p}_2) + (E_3,\bar{p}_3)[/tex]

    to no avail, so i'm pretty lost on this one any general advice or direction would help.

    Thanks in advance!

    Attached Files:

  2. jcsd
  3. Feb 18, 2014 #2
    Hi, hope it's not too late to help..
    So first, you can't use the relativistic Hamiltonian the way you do because you can't ignore the square-root. in this particular example, you can start with the relation:
    H = pv – L = pv + mc2/γ,
    then express v in terms of p and E (γmc2) to get a Hamiltonian you can work with. I get the right relation at the end.

    Then, i think obtaining p = – h/λ is no big deal because |p| = h/λ, which is the significant relation.

    For (2), i would start with expressions in terms of ω and k for the electrons to obtain E and p for the photon and show that the usual relation doesn't hold.
  4. Feb 18, 2014 #3
    Awsome, no you're not too late so I will take your advice and retry but I didn't think I ignored the square root because I set [tex]e^{i(wt-kx)}=\sqrt{e^{2i(wt-kx)}}[/tex], thought that was mathematically consistent but I guess not anyway thanks again, will report back if I still run into any problems.
  5. Feb 18, 2014 #4
    Yeah, operators and state-functions don't work this way...
    Also, in the first question you're not required to find a relation in terms of E, p and m like you apparently did: once you substitute the wave solution into the Klein-Gordon equation you just obtain a relation between w and k. Then once you set m = 0 and remember that v = w/k you get the speed of your wave, which should be the desired result for that question.
  6. Feb 22, 2014 #5
    Ok, so only just got round to carrying out your advice due to project work taking up my time anyway, converting the Hamiltonian using the relativistic Lagrangian H = pv – L and then applying the new Hamiltonian I get:

    [tex] E = h\nu + \frac{mc^2}{\gamma} [/tex]

    You definitely end up with just [tex] E = h\nu [/tex]

    Also on Q2 when you say use the expressions in terms of ω & k do you mean the relativistic energy in terms of ω & k and sub that into the 4-momentums of the electrons?
  7. Feb 23, 2014 #6
    Ok, so first:
    in special relativity you have: v = pc2/E (and E = γmc2),
    Take E as the energy eigenvalue, so H = pv – L = p2c2/E +m2c4/E.
    Looks familiar, right? now see what you get when you apply the Hamiltonian to ψ.
    What did you obtain in the first question (relation between k and ω)? Plug it in and see how it simplifies...

    Q2: I read it quickly the first time and thought it was related to the first question, but apparently it's not. So forget what i said, but take this hint: you can use a reference frame where the first electron is at rest...
  8. Feb 23, 2014 #7
    Ahh.. in Q2 you did that already but the way you put it is confusing. Separate the equations: first the energy conservation then momentum, plug the second into the first and everything becomes clear.
  9. Feb 23, 2014 #8
    So heres my working for hamiltonian in Q1:

    [tex]H = \vec{p}\cdot \vec{v} - L[/tex]

    I end up with,

    [tex]H = \frac{|\vec{p}|^2c^2}{\gamma mc^2}+\frac{mc^2}{\gamma}[/tex]


    [tex]\hat{H} = \frac{1}{\gamma mc^2}(-\hbar^2\frac{\partial^2 }{\partial x^2}c^2+m^2c^4)[/tex]

    [tex]\hat{H}\psi = \frac{1}{\gamma mc^2}(\hbar^2k^2c^2+m^2c^4)e^{-i(\omega t-kx)}[/tex]

    [tex] E = \frac{1}{\gamma mc^2}(\hbar^2k^2c^2+m^2c^4)[/tex]

    using the relation [tex]\omega = 2\pi \nu[/tex] &
    [tex]k = \frac{2\pi}{\lambda}[/tex] & that [tex]\frac{h\nu}{mc^2} = \gamma[/tex]

    i end up with

    [tex] E = h\nu + \frac{mc^2}{\gamma}[/tex]

    Notes: the wave solution should be [tex]e^{-i(\omega t-kx)}[/tex] apparently the one on the sheet is missing a minus sign, typo.
    Last edited: Feb 23, 2014
  10. Feb 23, 2014 #9
    No!! use the relation you've found in the very first question and express k in terms of ω...
  11. Feb 23, 2014 #10
    Aha, yep sorry about that, it does work, thanks! As for Q2 I do what you say but I get [tex]m_e^2-2E_3m_e+E_3^2/c^2=p_2^2+m_e^2c^2[/tex] and I can't really solve for E_3 here? Or am I not supposed to be solving for that to show that it doesn't satisfy the relation :S?
  12. Feb 23, 2014 #11
    Almost done: the last c^2 on the right shouldn't be there, so two of your terms cancel...
    then what did you obtain for the momentum conservation equation? you should be able to use it and get an equation only in terms of the photon's energy and momentum (and the constants m and c). Does it satisfy the usual relation?
  13. Feb 23, 2014 #12
    ah ok sweet, should have it from here thanks for all the help! :D
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