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Time dependent annihilation op

  1. Feb 17, 2010 #1

    In the free theory [tex] \mid k \rangle=a^{\dag} (\vec{k}) \mid 0 \rangle [/tex]. Then in Srednicki chapt 5, he defines time-independent operator that he says in free theory creates a particle localized in momentum space about [tex] \vec{k_1} [/tex] as:

    [tex] a^{\dag}_1 \equiv \int f_1 (\vec{k}) a^{\dag}(\vec{k}) \,d^3 x [/tex]


    [tex]f_1 (\vec{k}) \propto \exp{ -\frac{(\vec{k}-\vec{k_1})^2} {4\sigma^2} } [/tex]

    I have a few questions about this:

    Firstly why is [tex] \mid k \rangle=a^{\dag} (\vec{k}) \mid 0 \rangle [/tex], creating a one particle state only valid in the free theory?

    Secondly what is the point in this time independent operator? Is it something along the lines of a particle you create shouldn't have absolutely definite momentum k, as would be the case using the first creation operator, it must have a smeared momentum range about some value, to account for then uncertainty relation?

    I'm just finding it weird because in a previous QFT course I studied (not using the path integral approach), I never seen this relation, and everything just seemed to work fine with only [tex] \mid k \rangle=a^{\dag} (\vec{k}) \mid 0 \rangle [/tex].

    Finally how does this time independent operator create a particle localized in position space near the origin?

    Last edited: Feb 17, 2010
  2. jcsd
  3. Feb 17, 2010 #2
    Also I'm having issues seeing why if we act with this time independent operator on the vacuum [tex]a^{\dag}_1 \mid 0 \rangle [/tex] the wave packet will propopagate and spread out (presumably in space, not momentum?). This sounds intuitive but I can't see it mathematically, but I think that that is related to my final question above.
  4. Feb 18, 2010 #3
    hmm, perhaps my questions where too vague? Sorry, I'm finding hard to express exactly what it is I don't understand here.
    Perhaps my final concrete question would be the solid one to ask: "how does this time independent operator create a particle localized in position space near the origin?"

    thanks for any help
  5. Feb 19, 2010 #4
    Disclaimer: My QFT is self-taught. Always willing to be corrected! Here's my understanding:

    Srednicki uses the creation operator defined in terms of the field operator:
    [tex]a^{\dag}(k) = -i \int d^3 x e^{ikx} \stackrel{\leftrightarrow}{\partial_0} \phi(x)[/tex]
    The operators defined in this way only obey the algebra for creation&annihilation operators if the field is a free field. What you hope in the interacting theory is that the interaction doesn't spoil the algebra as long as the particle wave packets are well separated, which allows you to define in and out scattering states as (approximately) free particle states.

    He then defines a new creation operator that creates a wave packet state. [tex]f_1 (x) \propto e^{-(k-k_1)^2 / 4\sigma}[/tex] is just the Fourier transform of the wave packet in position space, which is centered at the origin and has average momentum [tex]k_1[/tex].

    The reason for the wave packet is that in the derivation coming up (eq 5.10 on pg 50 in my copy of the text) he needs to integrate by parts and the wave packet allows you to rigorously drop the boundary terms. You can drop the wave packet after that. The physical reason for the wave packets is to ensure that the particles are well separated in the distant past and distant future, meaning the interactions will be weak and there is a chance that the approximation in terms of free particle in and out states is meaningful.

    The result is the LSZ formula which connects transition amplitudes (what you want to compute cross sections etc) to time ordered correlation functions (what you compute using path integrals/Feynman diagrams etc).

    Of course renormalization (i.e., quantum corrections) screws everything and you then have to go back and check that everything in the LSZ derivation goes through, which leads to things like wave function renormalization and rules for "amputating" Feynman diagrams.

    Hope this helps.
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