In reading 14.4 of Gregory Moore's notes on abstract group theory, I was left with some questions on the computation he did of the path integral that may be general features.

Let consider a spacetime $M=\Sigma\times[t_0,t_f]$ on which we have a space of fields $C^\infty(M)$. Let $\mathcal E_1$ be the set of fields $\phi\in C^\infty(M)$ such that $\phi|_{\Sigma\times\{t_0\}}=\phi_0$ and $\phi|_{\Sigma\times\{t_f\}}=\phi_f$ for some fixed $\phi_0,\phi_f\in C^\infty(\Sigma)$. The technique employed in the notes above to compute $$\int_{\mathcal E_1}\mathcal{D}\phi\, e^{-\frac{1}{\hbar}S(\phi)},$$ is to first find a solution of the classical equations of motion $\phi_c\in\mathcal E_1$ and then reduce this to an integral $$\propto\int_{\mathcal E_2}\mathcal{D}\phi_q\, e^{-\frac{1}{\hbar}\tilde{S}(\phi_q)},$$ where $\mathcal{E}_2$ is the same as $\mathcal{E}_1$ except that $\phi_0=\phi_f=0$. He the proceeds to compute the integral of $\mathcal{E}_2$ using Gaussian integration. However, the original integral was also Gaussian. Why can't we compute the integral over $\mathcal{E}_1$ using Gaussian integration?

Of course, if one where to compute this integral through Gaussian integration the obvious problem appears of how to incorporate the boundary conditions. But that is at the root of my problem. Namely, what is special about the boundary conditions in $\mathcal{E}_2$ vs. $\mathcal{E}_1$? In usual Gaussian integration the integration of each variable $dx^i$ is on the range $-\infty$ to $\infty$. If one thinks naively of the measure as $\mathcal{D}\phi=\prod_{x\in M}d\phi(x)$, each variable of integration $\phi(x)$ is still being integrated in the range $-\infty$ to $\infty$ except for the ones at the boundaries.

I posted a similar question yesterday but, after a suggestion by QMechanic, I decided to delete it to focus on just this point. In that question I asked why it was that $S(\phi_c+\phi_q)=S(\phi_c)+S(\phi_q)$ when $\phi_c$ is a solution of the classical eoms. This turns out to be true whenever the theory is free (quadratic) since the second derivative of the action is independent of the fields.

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    $\begingroup$ This was a great idea! Doing this definitely clarified what is going on. I will try to post an answer with the details. $\endgroup$ Sep 6 '20 at 12:20

In order to solve this question with a finite-dimensional toy model, let us imagine that we have discretized spacetime so that it contains $N$ points. Then field configurations correspond to $\mathbb{R}^N$, for which we will use coordinates $x^\mu$, $\mu\in\{1,\dots,N\}$. We will assume the action is quadratic $$S(x)=\frac{1}{2}x^\mu A_{\mu\nu}x^\nu+b_\mu x^\mu+c.$$ A boundary condition fixes the value of these vectors on say $M$ of these $N$ points. After relabelling, one can then assume that the boundary condition is fixed by a vector $a\in\mathbb{R}^M$. The path integral with this boundary condition has a domain $\{a\}\times\mathbb{R}^{N-M}$, where we will use $\tilde{x}^r$, $r\in\{M+1,\dots,N\}$, as the standard coordinates. In this space the action restricts to $$\tilde{S}(\tilde{x})=\frac{1}{2}\tilde{x}^r \tilde{A}_{rs}\tilde{x}^s+\tilde{b}_r\tilde{x}^r+\tilde{c},$$ where $\tilde{A}_{rs}:=A_{rs}$, $\tilde{b}_r=b_r+A_{ri}a^i$, and $\tilde{c}=c+b_ia^i+\frac{1}{2}a^iA_{ij}a^j$. In here $i,j\in\{1,\dots,M\}$. This is again quadratic and the path integral can be solved as long as $\tilde{A}$ is positive definite. In this case, let $\tilde{A}^{rs}$ be its inverse. Then we have $$\int_{\{a\}\times\mathbb{R}^{N-M}}\text{d}^{N-M}\tilde{x}\,e^{-\tilde{S}(\tilde{x})}=\det\left(\frac{\tilde{A}}{2\pi}\right)^{-1/2}\exp\left(\frac{1}{2}\tilde{b}_r\tilde{A}^{rs}\tilde{b}^s\right).$$

Now, let us consider the semiclassical approximation (which is not an approximation in the quadratic case) approach. This starts by finding a solution of the equations of motion $$A_{\mu\nu}x^\nu+b_\mu=0,$$ satisfying the boundary conditions $x^i=a^i$. The existence of this solution already restricts the possible choices of boundary conditions, much like the hyperbolic or elliptic nature of the equations of motion does in the infinite dimensional case. This is most explicit in terms of the coordinates $\tilde{x}$, where the equations of motion and the boundary conditions reduce to $$A_{\mu r}\tilde{x}^r=-b_\mu-A_{\mu i}a^i.$$ Thus, admissible boundary conditions are those for which the vector $(b_\mu+A_{\mu i}a^i)_\mu$ is in the span of the vectors $(A_{\mu,M+1})_\mu,\dots,(A_{\mu,N})_\mu$.

Once we have a solution $x_{c}$, we proceed to do the change of coordinates $\tilde{x}^r\mapsto x_q^r:=\tilde{x}^r-x_c^r$. This is a translation, so that it does not introduce a Jacobian factor in the integral above. The new region of integration is now $\{0\}\times\mathbb{R}^{N-M}\subseteq\mathbb{R}^N$ and the action in the new coordinates is $$\tilde{S}(x_c+x_q)=\frac{1}{2}x_q^r\tilde{A}_{rs}x_q^s+b_r x_q^r+x_c^\mu A_{\mu r}x_q^r+c+b_\mu x_c^\mu+\frac{1}{2}x_c^\mu A_{\mu\nu}x_c^\nu=\frac{1}{2}x_q^r\tilde{A}_{rs}x_q^s+S(x_c).$$ Notice that the terms linear in $x_q$ vanish due to the equations of motion. Then, in the new coordinates the integral above can be computed as $$\int_{\{a\}\times\mathbb{R}^{N-M}}\text{d}^{N-M}\tilde{x}\,e^{-\tilde{S}(\tilde{x})}=\int_{\{0\}\times\mathbb{R}^{N-M}}\text{d}^{N-M}x_q\,e^{-\tilde{S}(x_c+x_q)}=\det\left(\frac{\tilde{A}}{2\pi}\right)^{-1/2}e^{-S(x_c)}.$$

The upshot of this is:

  1. You can in principle use Gaussian integration to compute the integral with non-trivial boundary conditions without using the semiclassical approximation. In doing there is no need to find classical solutions to the equations of motion and a higher variety of boundary conditions can be explored. On the other hand, one needs to compute the terms $\tilde{b}$ and $\tilde{c}$, which may prove to be difficult.
  2. The semiclassical approximation hides away the terms $\tilde{b}$ and $\tilde{c}$ into the action of the classical solution. On the other hand, one needs to solve the classical equations of motion. This restricts the possible boundary conditions (which may, in any case, be all of the physically interesting ones).
  3. As a final remark, this toy model shows explicitly that the determinant appearing in these path integrals is not the determinant of the original quadratic form $A$ but rather of its restriction $\tilde{A}$. This is related to the fact that in examples of computations with the semiclassical approximation a primed determinant $\det '$ appears. This determinant throws away the zero modes of the quadratic form, which I found confusing since the original problem may not have gauge symmetry.

In the future I will try to complement this answer with an infinite dimensional example which showcases the discussion above.


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