Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The LSZ reduction formula

  1. Aug 4, 2009 #1

    haushofer

    User Avatar
    Science Advisor

    Hi, a small question about the LSZ reduction formula, as treated in Srednicki's QFT book (chapter 5).

    He argues that in a QFT in a scalar theory with interactions you would like that

    [tex]
    <0|\phi(x)|0> = <0|\phi(0)|0> = 0
    [/tex]

    and

    [tex]
    <p|\phi(0)|0> = 1
    [/tex]

    For the first he argues that were this NOT the case, then the creation operators [itex] a^\dagger (\pm\infty)[/itex] would create a linear combination of a single particle state and the ground state.

    For the second he argues that the second condition accounts for the fact that the creation operators create a correctly normalized one-particle state.

    How does one see these two statements? Thanks in forward! :)
     
  2. jcsd
  3. Aug 4, 2009 #2
    I think the general idea is that at times t=-infinity and infinity, the interacting field turns into the free field. So you can calculate [tex]<p|\phi(0)|0> [/tex] using the free-field for [tex]\phi[/tex] (if you're wondering what [tex] \phi(0)[/tex] has to do with infinity, the formula is really [tex]<p|\phi(0)|0>e^{-ipx} [/tex] and at infinity this should be e^{-ipx}).

    When calculating this, Srednecki chooses the convention that the integration measure for [tex]\phi[/tex] is proportional to d^3k/E, where E is the energy. This is for Lorentz invariance, but then states must be normalized proportional to E. This convention is interesting because in a lot of textbooks the integration measure is proportional to d^3k/E^(1/2), which seems to make all the formulas look nicer.

    Sometimes [tex]<p|\phi(0)|0> [/tex] can't equal one and you'll have to correctly normalize creation and annihilation operators by 1/|<p|\phi(0)|0>| which happens in MS renormalization.

    As for [tex]<0|\phi(0)|0> [/tex] the same idea applies - just calculate it in the free-field. Usually this is zero unless there is spontaneous symmetry breaking. For Dirac fields this is zero by Lorentz symmetry.
     
  4. Aug 5, 2009 #3

    haushofer

    User Avatar
    Science Advisor

    Thanks for the answer! So, at [itex] t \rightarrow \pm\infty [/itex] I can use the free field mode expansion. But if I calculate this concretely, I get

    [tex]

    <0|\phi(x)|0> = <0| \int\frac{d^3 k}{(2\pi)^3 2E} [a(k)e^{ikx} + a^{\dagger}(k) e^{-ikx}] |0>

    [/tex]

    which for x=0 becomes

    [tex]
    \int\frac{d^3 k}{(2\pi)^3 2E} <0| [a(k) + a^{\dagger}(k)]|0> = \int\frac{d^3 k}{(2\pi)^3 2E}[<k|0> + <0|k>]
    [/tex]

    Via [itex] <k|q> = (2 \pi)^3 2E \delta^3 (k-q)[/itex], equation 5.5 in Srednicki's book, I then get

    [tex]
    2 \int\frac{d^3 k}{(2\pi)^3 2E} (2\pi)^3 2E \delta^3(k) = 2 \int d^3 k \delta^3(k) = 2
    [/tex]

    I'm obviously overlooking something. I really don't see the connection between this "lineair combination of the ground state and a one-particle state", the creation\annihilation operators at [itex] t \rightarrow \pm\infty [/itex] and the conditions I mentioned in my first post.
     
  5. Aug 5, 2009 #4

    haushofer

    User Avatar
    Science Advisor

    I would say that [itex]<0|a(k)|0> = <0|a^{\dagger}(k)|0> = 0 [/itex], so the demand that the creation/annihilation operators create a one-particle state at [itex]t=\pm\infty [/itex] is basically the same as demanding that your interaction goes to zero at these times (or, relativistically, the interaction has compact support in space-time). So then I wonder: how can I calculate explicitly the statement "then the creation operators would create a linear combination of a single particle state and the ground state"? Which expression for the field/mode expansion should I take?
     
  6. Aug 5, 2009 #5
    Instead of equation 5.5 I think you have to look at equation 5.3

    Basically any single operator between vacuum states, such as [tex]<0|a^{\dagger}|0>[/tex] or [tex]<0|a|0>[/tex] or [tex]<0|a^{\dagger}+a|0>[/tex] is zero.

    So in order for [tex]<0|\phi(x)|0>[/tex] to be nonzero, [tex]\phi(x)[/tex] operated on the vacuum would have to produce a multiple of the vacuum. This can be accomplished for example if the expansion for [tex]\phi(x)[/tex] has a constant term [tex]\nu [/tex]: [tex]


    \phi(x) = \nu + \int\frac{d^3 k}{(2\pi)^3 2E} [a(k)e^{ikx} + a^{\dagger}(k) e^{-ikx}]


    [/tex]

    Then putting [tex]\phi(x)[/tex] between two vacuums yields [tex]<0|\nu|0>=\nu [/tex] since everything else with just one operator between two vacuums is zero. So [tex]\phi(x)[/tex] has to create a linear combination of the vacuum when acting on the vacuum for the result to be nonzero. In this case, [tex]\nu|0>=\nu|0> [/tex] is the lienar combination of the vacuum.
     
  7. Aug 6, 2009 #6

    haushofer

    User Avatar
    Science Advisor

    Yes, obviously the normalization 5.5 doesn't count for k=0 or k'=0.

    I see your point now. What I wanted to do is to write down the expression in the interaction-picture explicitly and show that without 5.17 (a non-vanishing vev), the creation operator would create a one-particle state and a vacuum state. But in this picture you already assume that the fields at infinite time are the same as the free fields so that's kind of stupid. So in your example, you think of v being part of the creation operator? (because the text explicitly says "we would like the creation operator when acting on |0> to create a single particle state). Thanks for your help!!
     
  8. Aug 7, 2009 #7
    I think if you look at eqn. (5.2), and ask what would happen if you added 'v' to [tex]\phi[/tex], then you can see that there would be an extra term added to the creation operator [tex]a^{\dagger}[/tex] that, acting on the vacuum state, produces a multiple of the vacuum state.

    The example of adding a constant 'v' (which is the identity operator times a scalar) to the field is just one example of how you can construct an operator that acted on the vacuum, produces the vacuum. You can also imagine the operator [tex] a(k)a^{\dagger}(k)[/tex]. I chose to show the constant 'v' instead because that is something that actually comes up, although not till chapter 30 of Srednicki.
     
  9. Aug 8, 2009 #8

    haushofer

    User Avatar
    Science Advisor

    Yes, I now see completely your point. Thanks again for the clarification! I did a lot with GR in the past, and our courses on QFT weren't really advanced, so lately I try to catch up.
     
  10. Aug 10, 2009 #9

    haushofer

    User Avatar
    Science Advisor

    I have another small question about Srednick's book;it's about ultraviolet cutt-off. In eq. (9.22) Srednicki makes the replacement

    [tex]
    \Delta(x-y) \rightarrow \int\frac{d^4k}{(2\pi)^4}\frac{e^{ik(x-y)}}{k^2 + m^2 - i\epsilon} \Bigl(\frac{\Lambda^2}{k^2 + \Lambda^2 - i\epsilon}\Bigr)^2
    [/tex]
    instead of cutting the integral explicitly of at [itex]\Lambda[/itex]. Are there any arguments besides Lorentz invariance why such a particular convergent replacement makes sense?
     
  11. Aug 10, 2009 #10
    I have no clue as to why Srednicki chose that method (Pauli-Villars) instead of the cutoff method.

    I don't even get the comment about Lorentz invariance, as either choice should retain Lorentz invariance, but I guess the Pauli-Villars choice makes Lorentz invariance obvious, while with the cutoff method you would have to insert a step-function in the integrand that vanishes after the cutoff in order to call something the Fourier transform from -infinity to infinity.

    The rest of Srednicki doesn't use the cutoff method or Pauli-Villars but something else entirely (dimensional reguarlization).

    The cutoff method still should be obviously Lorentz invariant as the step-function is a scalar function.

    So I don't really get it either.
     
    Last edited: Aug 10, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: The LSZ reduction formula
  1. LSZ representation (Replies: 5)

  2. LSZ formula help! (Replies: 3)

  3. Why the LSZ formula? (Replies: 7)

Loading...