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Path Integral example, analysis of integral limits for K-G theory

  1. Jan 24, 2012 #1
    In general I find in books that the path integral approach is an equivalent alternative of the hamiltonian approach for QFT (and for QT in general, but my concern is with QFT). There I usually find that this method is usually developed in a formal way and used to derive Feynmann rules, gauge identities and so on.

    However, I did not find lots of explicit applications of this method to QFT (I found the typical example of the path integral representation of the 2 slit experiment but that is non relativistic one particle QT and not QFT).

    In this sense, my doubts are related with how amplitudes of QFT should be computed with this method.

    Lets put an example: Klein Gordon non interacting theory. I want to calculate the amplitude of finishing with a final 2-particle state with given energy and momentum (lets say (E2a;p2a) y (E2b;p2b)) given that we started with an initial 1 particle state with given energy and momentum (lets say (E1;p1)). (I know that the answer should be 0 because when there is no interaction, we can not go from a 1-particle state to a 2 particle state, but I would like to arrive to this result through path integral formalism.

    My questions are:
    1) What are the limits of the integral?
    2) Are they “fi(0,x)=exp(E1*t+p1*x) and fi(T,x)=exp(E2a*t+p2a*x)+exp(E2b*t+p2b*x)”?
    3) If my guess is correct, what would the final limit “fi(T,x)=exp(E2a*t+p2a*x)+0,5*exp(E2b*t+p2b*x)” mean? That we end we half a particle?
    4) If that is what that mean, this amplitude should give 0 as a result, isn´t it? Is this the result that we arrive when we compute this amplitude

    PS1: Thanks in advance for everything you can do?
    PS2: Sorry for my English, I speak Spanish
    PS3: Sorry for my physics, I am an actuary!
    PS4: I know that there is no point in calculating this amplitude through this method. I just want to know how this amplitude should be calculated, if I had a computer where I can put 1 millon “for i=1 to n”, in order to understand properly what this method exactly mean.
  2. jcsd
  3. Jan 24, 2012 #2
    Note that the usual Feynman diagrams will give you the answer here, because you can't draw any diagram that takes a 1-particle state to a 2-particles state in a free theory, so the amplitude is zero. And the diagrams do come pretty directly out of the path integral.

    If you want to think about how this could in principle be done "directly" using the path integral, I suggest considering a 1-D harmonic oscillator in regular quantum mechanics first, which as you may know can be considered as a 0+1 dimensional QFT.

    In the regular QM of one particle, the "straightforward" thing to calculate with the path integral is the propagator: the amplitude to from a position x1 to a position x2 in a time interval T. You do this by integrating over all paths x(0<t<T) with the limits x(0) = x1 and x(T) = x2.

    The easier equivalent of the QFT question "what is the amplitude to go from a 1-particle state to a 2-particle state" is the QM SHO question "what is the amplitude to go from the n = 1 state of the oscillator to the n = 2 state of the oscillator." But you will notice that these oscillator energy states are not position eigenstates. So the apply the path integral we should probably write them as superpositions of position eigenstates:

    [itex]|\psi_1 \rangle = \int dx \psi_1(x) |x \rangle [/itex]

    and similarly for the n = 2 state |psi2>. Then we can compute the amplitude for psi1 to go to psi2:

    [itex]\langle \psi_2 | e^{-iHT} | \psi_1 \rangle = \int dx \psi_2^*(x) \int dy \psi_1(y) \langle x | e^{-iHT} | y \rangle[/itex]

    where now the matrix element is just a propagator that we know how to calculate using the path integral.


    Now go to QFT. In QFT the path integral is over field configurations phi(x,t). The field variable phi(x) is the equivalent of the position variable x in QM. In the QFT path integral, the easy thing to do is specify boundary conditions for the path as phi(x,0) = phi1(x), phi(x, T) = phi2(x) where phi1 and phi2 are some initial and final field states. However, just as the energy eigenstates of the harmonic oscillator are not position eigenfunctions, particle states (the energy eigenstates of the field theory) are not field-eigenstates! The one-particle and two-particle states are each superpositions of infinitely many field configurations. If you like you can describe them by "wave functionals"

    [itex]\psi[f(x), t] = \langle f(x) | \psi(t) \rangle[/itex]

    Where here |psi(t)> is a general time-dependent state vector, and for any function f(x), |f(x)> is the "field-eigenstate," defined to satisfy the "eigenvalue equation"

    [itex]\hat{\phi}(x)|f(x) \rangle = f(x)| f(x) \rangle[/itex]

    where the hat on phi is emphasizing that here it as a field operator. psi[f(x), t] a is the probability amplitude to find the field phi(x) in the configuration f(x) at the time t. It is the equivalent of the QM wave function psi(x, t).

    The point is that the wave functional of a particle state is non-trivial (but I think analytically calculable for free theories). Once you find the wave functionals of your initial and final state, you can write them as superpositions of field-eigenstates, and then calculate the amplitude to go from one to the other in terms of the transition amplitudes between field-eigenstates, which is what the path integral explicitly calculates.
  4. Jan 24, 2012 #3
    Ok, thank you very much for such a large response. It will take me some time to get the idea but surely it will help. I work and so I have only the nights to take some time to think about this things. Nevertheless, I think Im getting what you say. I will tell you my thoughts and doubts about what you said when I make some time to think about it.

    Anyone else who has another answer would also be very appreciated.
  5. Jan 25, 2012 #4
    Well, for the 1 particle->2 particles question, besides what the Duck said, a somewhat more direct approach is to use LSZ reduction formula, which states that S-matrix element can be identified with the coefficients of the poles of the correlation function. For 1 particle->2 particles scenario we need to calculate the 3-point correlation function, which can be shown to be 0 for the free KG field(c.f. Peskin chapter 9), thus no poles and no transition probability.
  6. Jan 26, 2012 #5
    Ok The Duck, Im getting the idea. My mistake was not taking into account that an state could be defined as quantum superpositions of position field eingenstates (I dont know if I said it right but it is more or less the same words that you used). In this sense, I should express the one or the two particle eigenstate as a quantum superposition of position eigenstate.

    Now, my new problem is that I dont know which is the function that expresses the "x position" eigenstate (that is, the function that I should obtain when I perform phi(x)|0> and that should be in a limit in the path integral formalism). I think the answer is delta(x), is it? (QUESTION NUMBER 1)

    If this is right there is an equation in Mark Srednicki's Quantum Field Theory that expresses momentum creation operators (a(k)) as superposition of position creation operators (phi(x)). Something like:
    triple integral of (exp(-i*k*x)*i*timederivative of phi(x) + w*phi(x)).
    If what I say is right then:
    a(k)|0> = triple integral of (exp(-i*k*x)*i*timederivative of phi(x) + w*phi(x)) | 0>
    is my quantum superposition of position eigenstates that Im looking for. Am I right? (QUESTION NUMBER 2).

    If all this is right (or not too false), I just need to understand what is the time derivative of phi(x) and how to calculate it (QUESTION NUMBER 3) and then think a little bit how to extend this to the 2 particle momentum state (QUESTION NUMBER 4) (I have to recognize that I havent give this last question enough time to think about it).

    Ps: Thanks for everything you can do to help me with these 4 questions
    Ps2: Thanks for everything you have already done
    Ps3: Sorry for my English again (and for my physics)
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