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I am studying QFT from Srednicki's book. I would appreciate if anyone could offer how he derived Eq. 7.5 from the textbook. He is considering the quantum harmonic oscillator and writes:

$$\langle 0 | 0 \rangle_{f} = \int Dq \exp i \int_{-\infty}^{+\infty} dt \left[(1/2)(1+i \epsilon) \dot{q}^2 - (1/2)(1-i\epsilon) \omega^2 q^2 + fq\right],\tag{7.3}$$

where $m = 1$ for notational convenience. Then, he introduces Fourier-transformed variables:

$$q(t) = \int_{-\infty}^{+\infty} \frac{dE}{2\pi} e^{-iEt}\tilde{q}(E).\tag{7.4}$$

He then says, the expression in the square brackets becomes:

$$\frac{1}{2}\int_{-\infty}^{+\infty} \frac{dE}{2\pi}\frac{dE'}{2\pi} e^{-i(E + E')t }$$ $$\times \left[ \left(-(1+i\epsilon)EE' - (1-i\epsilon)\omega^2\right)\tilde{q}(E)\tilde{q}(E') + \tilde{f}(E)\tilde{q}(E') + \tilde{f}(E')\tilde{q}(E)\right].\tag{7.5}$$

However, he offers no derivation as to how he gets this. Perhaps, it is simple, but I cannot see it. Can someone fill in the details?

As an example, if I compute $\dot{q}(t)$, I get:

$\dot{q}(t) = \frac{-i}{2\pi}\int_{-\infty}^{\infty} Ee^{-iEt}\tilde{q}(E) dE$, but I don't see how squaring this gives the $\tilde{q}(E) \tilde{q}(E')$ terms.

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    $\begingroup$ Can you show some of your work? Directly plugging $q(t)$ into the first equation gets you most of the way there. $\endgroup$
    – knzhou
    Feb 3, 2018 at 23:38
  • $\begingroup$ I tried doing that, but, I don't understand how to evaluate $\dot{q}$, do you take the time derivative into the integral. Also, where does $\tilde{f}$ come from? $\endgroup$ Feb 3, 2018 at 23:39
  • $\begingroup$ Yes, the time derivative goes through the integral. To get the $\tilde{f}$, just take the $f(t)$ in the original equation and plug in the definition of the Fourier transform, just like you did for $q(t)$. $\endgroup$
    – knzhou
    Feb 3, 2018 at 23:40
  • $\begingroup$ Oh, okay, but why the double products: $EE'$, and the two $fq$'s? $\endgroup$ Feb 3, 2018 at 23:42
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    $\begingroup$ Take a simpler case: squaring $\int_0^1 x dx$. It's just $\int_0^1 x dx \int_0^1 y dy = \int dx dy \, x y$. Does that make sense to you? $\endgroup$
    – knzhou
    Feb 4, 2018 at 0:41

1 Answer 1

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Consider the Fourier transformed variables.

$$\tilde{q}(E) = \int dt ~ e^{i E t} q(t) $$ $$q(t) = \int \frac{dE}{2 \pi} ~ e^{-i E t} \tilde{q}(E) $$

Re-write the $dt$-integral in the $<0|0>$ expression using the above as

$$\frac{1}{2} (1 + i \epsilon) \dot{q}^2 - \frac{1}{2} (1 - i \epsilon) \omega^2 q^2 + f q = \int \int \frac{(1+ i \epsilon)}{2} \frac{d}{dt} ~ \Big(\frac{dE}{2 \pi} e^{- i E t} \tilde{q}(E)\Big) \frac{d}{dt}\Big( \frac{dE'}{2 \pi} e^{- i E' t} \tilde{q}(E') \Big) - \frac{(1- i \epsilon) \omega^2}{2} \frac{dE}{2 \pi} e^{-i E t} \tilde{q}(E) \frac{dE'}{2 \pi} e^{- i E t'} \tilde{q}(E') + \frac{dE}{2 \pi} \frac{d E'}{2 \pi} e^{- i (E + E')t} \tilde{q}(E) \tilde{f}(E')$$

Find the common factors, perform a bit of algebra to simplify it a bit and obtain.

$$\frac{1}{2} \int \int \frac{dE}{2 \pi} \frac{d E'}{2 \pi} e^{- i (E + E')t} \Bigg((-(1+i\epsilon)E E' - (1-i \epsilon) \omega^2) \tilde{q}(E) \tilde{q}(E') + \tilde{f}(E)\tilde{q}(E') + \tilde{q}(E) \tilde{f}(E') \Bigg)$$

Finally, calculate the action.

Cheers!!!

Edit: The integrals are of course from $-\infty$ to $\infty$

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  • $\begingroup$ Hi, thanks this is very helpful! However, I do not see where the $\tilde{f}\tilde{q}$, i.e., the third term in the brackets comes from. I only see the fourth term. Could you please clarify? Thanks. $\endgroup$ Feb 4, 2018 at 4:35
  • $\begingroup$ Hi Thomas. If you observe carefully the quantities you have in $ <0|0>_f$ and the expression you want to obtain you will see the following: from the first two terms you have a common factor of $1/2$, while the third term is just $f q$. You can write $f q = \frac{1}{2} f q + \frac{1}{2} f q$, then the factor of $1/2$ is everywhere and you pull it front and you have the third and fourth term. I hope this helps. Cheers!!! $\endgroup$
    – user172341
    Feb 4, 2018 at 12:20
  • $\begingroup$ I have a doubt in the previous answer by user172341. In 7.3, the whole integral is in exponential while in 7.5 the integral is not in exponential. How does that happen ? $\endgroup$
    – Aum Rawal
    Jan 6 at 10:27

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