I am working the following exercise:

Calculate the generating functional $$Z[j]=\int \mathcal{D}\Phi \exp\left(\frac{i}{\hbar}S[\Phi,j]\right),\quad S[\Phi,j]=\int d^4x(\mathcal{L}(\Phi)+j\Phi),$$ $$\mathcal{L}=\frac{1}{2}\partial_\mu\Phi\partial^\mu\Phi-\frac{1}{2}m^2\Phi^2+V(\Phi)$$ using the method of stationary phase for fluctuations $\delta\Phi$ around the classical field configuration $\Phi_c$, for which $\delta S/\delta\Phi|_{\Phi_c}=0$. Show that this approximation is an asymptotic series in $\hbar$.

Now let me present you my calculations.

The method of stationary phase says that $$\int d^nx e^{i\lambda S(x)}\simeq \sqrt{\frac{2\pi i}{\lambda^n \det S^{\prime\prime}_0}}e^{i\lambda S_0},$$

where $S_0$ denotes the extremum.

Generalising this to functional integrals, this gives $$\int \mathcal{D}\Phi \exp\left(\frac{i}{\hbar}S[\Phi,j]\right)\simeq N(n,\hbar)\sqrt{\det S^{\prime\prime}_0}^{-1}e^{i\lambda S_0},$$ where $S^{\prime\prime}_0=\delta^2S/\delta\Phi^2|_{\Phi_c}=\left(V^{\prime\prime}(\Phi_c)-(\square_x+m^2)\right)\delta(x-y)$.

Now, I don't see how this gives us any asymptotic series in $\hbar$.

My first idea was to do the following instead. We substitute $\Phi=\Phi_c+\Delta$, which gives $\mathcal{D}\Phi=\mathcal{D}\Delta$. Then we insert the substitution into the action and obtain three contributions:

$$S[\Phi,j]=\int d^4x\left(\frac{1}{2}\partial_\mu\Phi_c\partial^\mu\Phi_c-\frac{1}{2}m^2\Phi^2_c+V(\Phi_c)+j\Phi_c\right)\\ +\int d^4x\left(V^\prime(\Phi_c)+j-(\square+m^2)\Phi_c\right)\Delta\\ +\int d^4x\left(\frac{1}{2}\partial_\mu\Delta\partial^\mu\Delta-\frac{1}{2}m^2\Delta^2\right).$$ The middle term vanishes, since $\delta S/\delta\Phi|_{\Phi_c}=0$ and thus $(\square+m^2)\Phi_c-V^\prime(\Phi_c)=j$, i.e. the classical field fulfills the EOM with source term.

This then gives us $$Z[j]=\int \mathcal{D}\Phi \exp\left(\frac{i}{\hbar}S[\Phi,j]\right)\\ =\exp\left(\frac{i}{\hbar}S[\Phi_c,j]\right)\int\mathcal{D}\Delta\exp\left[\frac{i}{\hbar}\int d^4x\left(\frac{1}{2}\partial_\mu\Delta\partial^\mu\Delta-\frac{1}{2}m^2\Delta^2\right)\right],$$ which can then be solved as a Gaussian integral:

$$Z[j]\simeq\exp\left(\frac{i}{\hbar}S[\Phi_c,j]\right)\times N(n,\hbar)\sqrt{\det(\square+m^2)}^{-1}.$$

But again this is not an asymptotic series in $\hbar$.

To me it seems, we are only ever going to get an expansion in terms of $\hbar$, if we expand out the exponential. But then we cannot use the method of stationary phase anymore. However, if we use the method of stationary phase, we will only ever get $\hbar$ to appear in the divergent prefactor, which is divided out by normalisation anyway.

How do I make this work?

  • $\begingroup$ Briefly, (i) define $\lambda\equiv 1/\hbar$ and (ii) divide the path integral definition with the normalization factor $N(n,\hbar)$, and (iii) rescale fluctuations $\Phi=\Phi_c+\sqrt{\hbar}\Delta$. $\endgroup$
    – Qmechanic
    Commented Jun 24, 2019 at 22:02
  • $\begingroup$ Sure, but that still gives us the classical result $Z[j]/Z[0]=\exp(i/\hbar S_0)$. Where are the correction terms? $\endgroup$ Commented Jun 24, 2019 at 22:07
  • $\begingroup$ It seems some parts of the interactions are missing. $\endgroup$
    – Qmechanic
    Commented Jun 24, 2019 at 22:27
  • $\begingroup$ No, the exercise is explicitly about the free theory without any interaction. Also, we are told to use fluctuations around the classical field configuration of the form $\Phi=\Phi_c+\Delta$. I would have expected to yield $\exp(i/\hbar S_0)$ as the classical result, multiplied by an asymptotic series in $\hbar$, which then give the quantum corrections. $\endgroup$ Commented Jun 24, 2019 at 22:35

2 Answers 2


We have

$$S[\Phi_c +\sqrt\hbar\Delta]=S[\Phi_c]+\sqrt\hbar\int dx\ \overbrace{\frac{\delta S}{\delta\Phi}[\Phi_c(x)]}^{=0}\Delta(x)+\frac{\hbar}{2}\iint dxdy\ \frac{\delta^2 S}{\delta \Phi(x)\delta\Phi(y)}\Delta(x)\Delta(y)+\ldots$$

So in particular

$$S[\Phi,j]=\int d^4x\left(\frac{1}{2}\partial_\mu\Phi_c\partial^\mu\Phi_c-\frac{1}{2}m^2\Phi^2_c+V(\Phi_c)+j\Phi_c\right)\\ +\sqrt\hbar\int d^4x\left(j-(\square+m^2)\Phi_c+\right)\Delta\\ +\frac{\hbar}{2}\int d^4x\left(\frac{1}{2}\partial_\mu\Delta\partial^\mu\Delta-\frac{1}{2}m^2\Delta^2\right)\\ +\sum_{n=1}^\infty\frac{\hbar^{n/2}}{n!}\int d^4x \iint\ldots \int d\xi_1d\xi_2\ldots d\xi_n \frac{\delta^n V(\Phi_c)}{\delta\Phi(\xi_1)\delta\Phi(\xi_2)\ldots\delta\Phi(\xi_n)}\Delta(\xi_1)\Delta(\xi_2)\ldots\Delta(\xi_n)$$

The $\hbar$ terms act like perturbations to a gaussian integral.

  • $\begingroup$ So basically, I expand $S$ with $\hbar$ as a small parameter and then I would have to calculate the additional terms in the stationary phase approximation, right? But I doubt it's even analytically possible to solve those kind of integrals. Is there any particular reason why you use $\sqrt{\hbar}$? $\endgroup$ Commented Jun 25, 2019 at 7:18
  • $\begingroup$ @ThomasWening I used $\sqrt\hbar$ so that the first quantum correction is proportional to $\hbar$. I don't know if that is customary as the texts I've seen use units with $\hbar=1$. I have not yet tried to do such integrals, but would it be possible to simply evaluate the expression in $n$ dimensions then take $n\to\infty?$ (possibly extending special functions to special functionals?) $\endgroup$
    – user195162
    Commented Jun 25, 2019 at 17:09


  1. Identify $\lambda\equiv 1/\hbar$,
  2. Divide the path integral measure ${\cal D}\Phi$ with the normalization factor $N(n,\hbar)$. More explicitly, this means replace ${\cal D}\Phi\to {\cal D}\frac{\Phi}{\sqrt{\hbar}}$.
  3. Rescale fluctuations $\Phi=\Phi_c+\sqrt{\hbar}\Delta$ with a factor $\sqrt{\hbar}$.

See e.g. this related Phys.SE post.


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