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Consider the partition function for a scalar field $\{\phi:\mathbb{R}_{\geq 0}\to\mathbb{R}\}$, $Z=\int D\phi D\lambda\exp(-S)$ with the action $$S=\underbrace{\int_0^\infty dx \frac{1}{2}(\partial_x\phi)^2}_{S_0}+\underbrace{\int_0^\infty dx \hspace{1mm}i\lambda(\partial_x\phi-c\phi)\delta(x)}_{S_\lambda},$$ where $\lambda$ is a Lagrange multiplier field to enforce the Robin boundary condition $\partial_x\phi(x=0)=c\phi(x=0)$. The second term in the action $$S_\lambda=\int_0^\infty dx \Big[ i\lambda(\partial_x\phi-c\phi)\delta(x)\Big]= i\lambda(\partial_x\phi-c\phi)\vert_{x=0},$$ generates the functional delta $$\int D\lambda \exp\Big(\int_0^\infty dx \hspace{1mm} i\lambda(\partial_x\phi-c\phi)\delta(x)\Big)=\int D\lambda\exp\Big( i\lambda(\partial_x\phi-c\phi)\vert_{x=0}\Big)=\delta((\partial_x\phi-c\phi)\vert_{x=0}).$$

Assuming that the fields and their derivatives (at least the first derivative) vanish at $x=\infty$, the saddle-point equations are obtained by enforcing $S[\phi+\delta\phi,\lambda+\delta\lambda]-S[\phi,\lambda]=0$: $$(-\partial_x\phi)\delta\phi\vert_{x=0}+i\lambda (\partial_x\delta\phi-c\delta\phi)\vert_{x=0}=0,$$ and $$\delta\lambda(\partial_x\phi-c\phi)\vert_{x=0}=0.$$ Now, how is one supposed to deal with the $\partial_x\delta\phi $ term, over which we do not have any assumptions? $\delta\phi$ are the off-shell fluctuations for which, from what I have seen earlier, we just assume $\delta \phi(x=0,\infty)=0$, with no assumptions for $\partial_x\delta\phi$. In this sense, (only) the on-shell solution of the equations of motion satisfies the required boundary condition at $x=0$, and also possibly any correlators, i.e. $(\partial_x-c)\langle\phi(x)\phi(x')\rangle\vert_{x=0}$, but that's for another day.

P.S.: Also, another method would be to not use the stationary phase value for $\lambda$ and later integrate over it after finding the required correlators, i.e. $$\langle\phi(x)\phi(x')\rangle=\frac{\int D\lambda \phi(x)\phi(x')\exp(-S)}{\int D\lambda \exp(-S)},$$ but at the moment I am just concerned with the method chalked out above.

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2 Answers 2

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Comments to the post (v5):

  1. First of all: an integral where the support of the Dirac delta distribution coincides with one of the integration limits is ill-defined.

    However, in OP's case this can be avoided altogether. Just add the term $i\lambda(\phi^{\prime}(0)-c\phi(0))$ to the action $S$ instead.

  2. For consistency, we need 2 boundary conditions (BCs): 1 initial BC at $x_i=0$ and 1 final BC at $x_f=0$.

    Moreover, for the functional/variational derivative to be well-defined, the BCs should be either essential or natural. The Robin BC is neither.

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  • $\begingroup$ Thanks for the answer. So it seems one can either have a pure Dirichlet or a pure Neumann on any connected boundary domain. As such, forcibly introducing the Lagrange multiplier still yields the term (Dirichlet at $x=0$) $i\lambda\delta\phi(0)$ or (Neumann at $x=0$) $i\lambda \partial_x\delta\phi(0)$, which still leaves open my confusion regarding $\partial_x\delta\phi(0)$. Should the BCs for the off-shell variations be inherited from the on-shell ones? In this regard: iopscience.iop.org/article/10.1070/PU1993v036n09ABEH002311/meta see Eq. (4) and the text below it. $\endgroup$ Commented Oct 10, 2020 at 17:15
  • $\begingroup$ ... seem to be the natural BCs. However, they don't have an explicit derivative term at the boundary and hence only the variation of the boundary field, and not its derivative, enters the picture. $\endgroup$ Commented Oct 10, 2020 at 20:27
  • $\begingroup$ The system described in the Ginzburg article seems different from OP's action. $\endgroup$
    – Qmechanic
    Commented Oct 11, 2020 at 17:30
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Correct me if I’m wrong. You try 2 functions $\phi$ and $\phi_2=\phi+\delta\phi$ which both must satisfy boundary condition. Since boundary condition is linear, logically $\delta\phi$ satisfies it too.

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  • $\begingroup$ I get your point, but I was trying to "naturally" end up with the constraint while beginning with unconstrained fields, by using the Lagrange multipliers. As such, I was not starting with an assumption on $\phi$ and $\phi_2=\phi+\delta\phi$. But then, I suppose from $\delta\lambda(\partial_x\phi-c\phi)\vert_{x=0}=0$, one derives $(\partial_x\phi-c\phi)\vert_{x=0}=0$, which is further used to conclude $(\partial_x\delta\phi-c\delta\phi)\vert_{x=0}=0$ as you said. I was wondering, in the spirit mentioned above, is it fine to limit the behaviour of the variations to mimic that of the $\phi$. $\endgroup$ Commented Oct 11, 2020 at 18:49
  • $\begingroup$ ... Check out arxiv.org/pdf/cond-mat/0110532.pdf, I am looking at the saddle point of such a system. Here, in Eq. (4) the free Green's function corresponds to that of an unbounded space. In that spirit, maybe it's not correct to keep the first boundary term in $(-\partial_x\phi)\delta\phi\vert_{x=0}+i\lambda (\partial_x\delta\phi-c\delta\phi)\vert_{x=0}=0$ as that assumes a bounded system to start with. $\endgroup$ Commented Oct 11, 2020 at 19:28

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