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Srednicki Path Integrals eq 6.22

  1. Aug 19, 2009 #1
    Hi everyone,

    I was reading through the section on path integrals in Srednicki's QFT book. I came across equation 6.22

    [tex]\langle 0|0\rangle_{f,h} = \int\mathcal{D}p\mathcal{D}q\exp{\left[i\int_{-\infty}^{\infty}dt\left(p\dot{q}-H_{0}(p,q)-H_{1}(p,q)+fq+hp\right)\right]}[/tex]

    [tex]= \exp{\left[-i\int_{-\infty}^{\infty}dt H_{1}\left(\frac{1}{i}\frac{\delta}{\delta h(t)},\frac{1}{i}\frac{\delta}{\delta f(t)}\right)\right]}\times\int\mathcal{D}q\mathcal{D}p\exp{\left[i\int_{-\infty}^{\infty}dt(p\dot{q}-H_{0}(p,q)+fq+hp)\right]}[/tex]

    How did we get the second line of the equation?

    PS -- f and h are differentiable functions of time, used just to pull down suitable powers of q and p.

    Thanks in advance.
     
    Last edited: Aug 19, 2009
  2. jcsd
  3. Aug 20, 2009 #2
    View

    [tex]
    \exp{\left[-i\int_{-\infty}^{\infty}dt H_{1}\left(\frac{1}{i}\frac{\delta}{\delta h(t)},\frac{1}{i}\frac{\delta}{\delta f(t)}\right)\right]}
    [/tex]

    as an operator that operates on the thing on the right. Assume you can differentiate under the integral sign. Once you do this you get the first line.
     
  4. Aug 20, 2009 #3
    Ok, but how and couldn't I have exchanged the roles of [itex]H_0[/itex] and [itex]H_1[/itex] here? Plus, I don't know the functional form of [itex]H_1[/itex] so how do I "differentiate"?
     
  5. Aug 20, 2009 #4
    As you know, we can divide
    [tex]= \int\mathcal{D}q\mathcal{D}p\exp{\left[i\int_{-\infty}^{\infty}dt(p\dot{q}-H_{0}(p,q)+fq+hp)\right]}[/tex]
    into different times. (Please check the origin of this equation).

    So we can exchange these [itex]H_0[/itex] and [itex]H_1[/itex] here.

    [tex] \exp{\left[-i\int_{-\infty}^{\infty}dt H_{1}\left(\frac{1}{i}\frac{\delta}{\delta h(t)},\frac{1}{i}\frac{\delta}{\delta f(t)}\right)\right]}[/tex]

    is also made from many different times.

    We must divide the the exponential functions into (1+dt H(t1)..) x (1+dt H(t2)...) x (1+dt H(t3)...) .....
    and differentiate at each different time.
    We can easily exchange the order of H1 and H0 at this form.

    (Sorry for simple explanation).
     
    Last edited: Aug 20, 2009
  6. Aug 20, 2009 #5
    You can exchange H_0 and H_1 if you want. But the reason you're even splitting the path integral into two parts that way is because

    [tex]
    = \int\mathcal{D}q\mathcal{D}p\exp{\left[i\int_{-\infty}^{\infty}dt(p\dot{q}-H_{0}(p,q)+fq+hp)\right]}
    [/tex]

    usually can be solved! So once you are able to solve this [in terms of h(t) and f(t)], you can get the total path integral you want through differentiation, which is much easier than taking the path integral of the entire thing.

    So really this is just the trick of differentiating under the integral sign.

    As for the form of H_1, it can be arbitrary. Take H_1=q^3. Then:

    [tex]
    = \exp{\left[-i\int_{-\infty}^{\infty}dt \left(\frac{1}{i}\frac{\delta}{\delta f(t)}\right)^3\right]}\times\int\mathcal{D}q\mathcal{D}p\exp{\left[i\int_{-\infty}^{\infty}dt(p\dot{q}-H_{0}(p,q)+fq+hp)\right]}=

    \int\mathcal{D}q\mathcal{D}p\exp{\left[i\int_{-\infty}^{\infty}dt(p\dot{q}-H_{0}(p,q)-q^3+fq+hp)\right]}
    [/tex]

    If you want to convince yourself that this is true, perhaps you should expand the term with [tex] \exp{\left[-i\int_{-\infty}^{\infty}dt \left(\frac{1}{i}\frac{\delta}{\delta f(t)}\right)^3\right]}[/tex] in power series and operate to the right, remembering that:

    [tex]\frac{\delta f(r)}{\delta f(t)}=\delta (r-t) [/tex]

    where the last delta on this line is the Dirac delta function.
     
  7. Aug 20, 2009 #6
    I know what you're thinking. You're thinking why not pull both H_0+H_1 out! Then the path integral that's left is merely exp(fq+hp) which is easy to do, and then you can get the complete path integral by differentiation!

    But the problem is that the path integral doesn't converge. Usually things that converge are e^(-x^2), and not e^(x). The (-) sign is very important for convergence, and something that I don't get is how an expression like [tex]\int e^{(E^2-\omega^2)x^2} dE[/tex] converges when you take the integral over x! It only converges if E^2-\omega^2<0 . However, this is all ignored by absorbing this into the normalization of the ground state, saying that if the source is equal to zero then the result should be 1, so that [tex]\int e^{(E^2-\omega^2)x^2} dE[/tex] integrates out to 1 (this comment comes after eqn. (7.9) of Srednicki).
     
  8. Aug 27, 2009 #7
    Hi RedX, thanks for the explanation and apologies for the late reply...I completely forgot about this thread somehow, and got a chance to revisit my QFT notes only this evening after exams of other courses :-p.

    Well, thats not quite what I had in mind. I just didn't see how H_1 was being involved in an operator acting on the expression containing H_0. I am going through your replies now, and will work them out for myself. I might bug you again with trivial seeming queries :-p
     
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