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Virtual particle energy-momentum

  1. Mar 10, 2008 #1
    A mentioning about virtual particle problem in my other thread just reminded me of some thoughts, which I now succeeded putting together.

    When calculating cross sections in QFT, we encounter terms like this

    [tex]
    \langle 0|a_{\textbf{k}'} a_{\textbf{p}'} a^{\dagger}_{\textbf{p}_1} a^{\dagger}_{\textbf{p}_2} a'_{\textbf{k}_1} a^{'\dagger}_{\textbf{k}_2} a_{\textbf{p}_3} a_{\textbf{p}_4} a^{\dagger}_{\textbf{k}} a^{\dagger}_{\textbf{p}}|0\rangle,
    [/tex]

    where we start with some excitation, operate on it with annihilation operators, create some intermediate particle, annihilate the intermediate particle, create two final particles, and then consider the inner product with some fixed outcome.

    As it turns out, the calculations lead into some expressions that contain propagator of the intermediate particle, and the energy-momentum in the propagator is not on shell. This is usually interpreted as sign of the intermediate particle being off shell, but this doesn't fully make sense to me.

    Firstly, the calculation is quite abstract. I don't see any obvious reason to interpret the resulting propagator as actually describing the propagation of the intermediate particle.

    Secondly, isn't it impossible to create off shell particles with the usual creation operators? I mean, if you operate on vacuum with [itex]a^{'\dagger}_{\textbf{p}}[/itex], you get a particle whose energy momentum is [itex](\sqrt{|\textbf{p}|^2 + (m')^2}, \textbf{p})[/itex], right? Here [itex]m'[/itex] is the constant mass, characteristic to the particular particle type.

    So, what I hear everyone explaining, is this:

    "Since the energy-momentum is conserved in the vertexes, the intermediate particle is clearly off shell. This, however, is not problematic, since it is a virtual particle that exists only for short period of time, and cannot be observed directly."

    But why not like this:

    "Since the intermediate particle is on shell, clearly the energy-momentum is not conserved in the vertexes. This, however, is not problematic, since the state, at which energy-momentum conservation is violated, exists only for a short period of time, and cannot be observed directly."?
     
    Last edited: Mar 10, 2008
  2. jcsd
  3. Mar 10, 2008 #2

    tiny-tim

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    It's a trick!

    The operators in [tex]
    \langle 0|a_{\textbf{k}'} a_{\textbf{p}'} a^{\dagger}_{\textbf{p}_1} a^{\dagger}_{\textbf{p}_2} a'_{\textbf{k}_1} a^{'\dagger}_{\textbf{k}_2} a_{\textbf{p}_3} a_{\textbf{p}_4} a^{\dagger}_{\textbf{k}} a^{\dagger}_{\textbf{p}}|0\rangle,
    [/tex]
    are all for real, not virtual, particles.

    From that, we get the position Feynman diagrams, in which the integration is over [tex]d^3p[/tex].

    (The momentums in the position Feynman diagrams are still all on-shell, since they still represent real operators for real particles.)

    Those integrals include a step function.

    Then a Fourier trick is used to convert that step function into a fourth integral, which combined with the [tex]d^3p[/tex] gives a convenient Lorentz-friendly four-variable integral (see, for example, Weinberg QFT page 276).

    The four-vectors in this, and in the associated momentum Feynman diagram, have nothing to do with the real momentums in the position diagram - they only look similar. It's a good trick, isn't it? :smile:

    It is convenient to call them off-shell four-vectors, and to think of them as representing the energy-momentum of "virtual particles". :smile:
     
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