请求讲解Srednicki《量子场论》中式（7.5）的推导过程

Let's walk through this derivation step by step, starting from the vacuum-to-vacuum path integral with a source (equation 7.3) you provided:

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

We'll stick with the simplification $m=1$, and remember the $i\epsilon$ terms are there to shift poles away from the real axis, ensuring the integral converges properly.

Step 1: Fourier Transform the Field and Source #

First, we expand the position field $q(t)$ and source $f(t)$ into their Fourier components (using the time-domain Fourier transform convention):
$$q(t) = \int_{-\infty}^{+\infty} \frac{dE}{2\pi} e^{-iEt} q(E), \quad f(t) = \int_{-\infty}^{+\infty} \frac{dE}{2\pi} e^{-iEt} f(E)$$
Since $q(t)$ is a real function, it satisfies the hermiticity condition $q(-E) = q^*(E)$.

Step 2: Rewrite the Lagrangian Integral in Fourier Space #

We substitute these transforms into each term of the integral inside the exponent:

1. Kinetic Term: $\int dt \dot{q}^2$ #

The time derivative of $q(t)$ is $\dot{q}(t) = \int \frac{dE}{2\pi} (-iE)e^{-iEt} q(E)$. Squaring this and integrating over time gives:
$$\int_{-\infty}^\infty dt \dot{q}^2(t) = \int \frac{dE dE'}{(2\pi)^2} (-iE)(-iE') q(E)q(E') \int dt e^{-i(E+E')t}$$
Using the delta function identity $\int dt e^{-i(E+E')t} = 2\pi \delta(E+E')$, this simplifies to:
$$\int dt \dot{q}^2(t) = -\int \frac{dE}{2\pi} E^2 q(E) q(-E)$$

2. Potential Term: $\int dt q^2$ #

Expanding $q^2(t)$ and integrating over time follows the same logic:
$$\int_{-\infty}^\infty dt q^2(t) = \int \frac{dE dE'}{(2\pi)^2} q(E)q(E') 2\pi \delta(E+E') = \int \frac{dE}{2\pi} q(E) q(-E)$$

3. Source Term: $\int dt fq$ #

Substituting the Fourier transforms for $f(t)$ and $q(t)$:
$$\int_{-\infty}^\infty dt f(t)q(t) = \int \frac{dE dE'}{(2\pi)^2} f(E)q(E') 2\pi \delta(E+E') = \int \frac{dE}{2\pi} f(-E) q(E)$$

Step 3: Combine Terms into a Quadratic Form #

Plug these results back into the exponent of equation (7.3):
$$
\begin{align*}
i \int dt \mathcal{L} &= i \left[ \frac{1}{2}(1+i\epsilon)\left(-\int \frac{dE}{2\pi} E^2 q(E)q(-E)\right) - \frac{1}{2}(1-i\epsilon)\omega^2 \int \frac{dE}{2\pi} q(E)q(-E) + \int \frac{dE}{2\pi} f(-E)q(E) \right] \
&= -\frac{i}{2} \int \frac{dE}{2\pi} \left[ (1+i\epsilon)E^2 + (1-i\epsilon)\omega^2 \right] q(E)q(-E) + i \int \frac{dE}{2\pi} f(-E)q(E)
\end{align*}
$$

This is a Gaussian integral in the functional space of $q(E)$, fitting the standard form:
$$\int Dq \exp\left( -\frac{1}{2} q \cdot M \cdot q + b \cdot q \right)$$
where $M(E,E') = i\left[(1+i\epsilon)E^2 + (1-i\epsilon)\omega^2\right] 2\pi \delta(E+E')$ and $b(E) = i f(-E)$.

Step 4: Evaluate the Gaussian Integral #

The standard result for a Gaussian functional integral is:
$$\int Dq \exp\left( -\frac{1}{2} q \cdot M \cdot q + b \cdot q \right) = \frac{\exp\left( \frac{1}{2} b \cdot M^{-1} \cdot b \right)}{\sqrt{\det M}}$$

1. Compute the Inverse Kernel $M^{-1}$ #

We solve for $M^{-1}$ using the definition $\int \frac{dE''}{2\pi} M(E,E'') M^{-1}(E'',E') = \delta(E-E')$. Substituting $M$:
$$i\left[(1+i\epsilon)(-E)^2 + (1-i\epsilon)\omega^2\right] M^{-1}(-E,E') = \delta(E-E')$$
Which simplifies to:
$$M^{-1}(E,E') = \frac{\delta(E+E')}{i\left[(1+i\epsilon)E^2 + (1-i\epsilon)\omega^2\right]}$$

2. Evaluate the Exponential Source Term #

Calculate $\frac{1}{2} b \cdot M^{-1} \cdot b$:
$$
\begin{align*}
\frac{1}{2} b \cdot M^{-1} \cdot b &= \frac{1}{2} \int \frac{dE dE'}{(2\pi)^2} b(E) M^{-1}(E,E') b(E') \
&= \frac{1}{2} \int \frac{dE dE'}{(2\pi)^2} (i f(-E)) \frac{\delta(E+E')}{i\left[(1+i\epsilon)E^2 + (1-i\epsilon)\omega^2\right]} (i f(-E')) \
&= \frac{i}{2} \int \frac{dE}{2\pi} \frac{f(-E)f(E)}{(1+i\epsilon)E^2 + (1-i\epsilon)\omega^2}
\end{align*}
$$

3. Evaluate the Functional Determinant #

For the diagonal kernel $M$, the functional determinant is a product over all Fourier modes:
$$\det M = \prod_E i\left[(1+i\epsilon)E^2 + (1-i\epsilon)\omega^2\right]$$
Taking the square root and reciprocal gives:
$$\frac{1}{\sqrt{\det M}} = \prod_E \frac{1}{\sqrt{i\left[(1+i\epsilon)E^2 + (1-i\epsilon)\omega^2\right]}}$$
This term corresponds to the vacuum-to-vacuum amplitude when there's no source ($f=0$).

Step 5: Combine Results to Get Equation (7.5) #

Putting everything together, we arrive at the full expression for $\langle 0 | 0 \rangle_f$:
$$
\langle 0 | 0 \rangle_f = \exp\left( \frac{i}{2} \int_{-\infty}^\infty \frac{dE}{2\pi} \frac{f(-E)f(E)}{(1+i\epsilon)E^2 - (1-i\epsilon)(-\omega^2)} \right) \times \prod_E \frac{1}{\sqrt{i\left[(1+i\epsilon)E^2 + (1-i\epsilon)\omega^2\right]}}
$$

In Srednicki's notation, the determinant term is often absorbed into the vacuum normalization (or rewritten using a logarithmic integral for simplicity), leading directly to the final form of equation (7.5).

内容的提问来源于stack exchange，提问作者Thomas Moore