Lagrange multiplier for Robin boundary condition in variational minimisation

Consider the partition function for a scalar field ${phi:mathbb{R}_{geq 0}tomathbb{R}}$, $Z=int Dphi Dlambdaexp(-S)$ with the action $$S=underbrace{int_0^infty dx frac{1}{2}(partial_xphi)^2}_{S_0}+underbrace{int_0^infty dx hspace{1mm}ilambda(partial_xphi-cphi)delta(x)}_{S_lambda},$$
where $lambda$ is a Lagrange multiplier field to enforce the Robin boundary condition $partial_xphi(x=0)=cphi(x=0)$. The second term in the action $$S_lambda=int_0^infty dx Big[ ilambda(partial_xphi-cphi)delta(x)Big]= ilambda(partial_xphi-cphi)vert_{x=0},$$
generates the functional delta $$int Dlambda expBig(int_0^infty dx hspace{1mm} ilambda(partial_xphi-cphi)delta(x)Big)=int DlambdaexpBig( ilambda(partial_xphi-cphi)vert_{x=0}Big)=delta((partial_xphi-cphi)vert_{x=0}).$$

Assuming that the fields and their derivatives (at least the first derivative) vanish at $x=infty$, the saddle-point equations are obtained by enforcing $S[phi+deltaphi,lambda+deltalambda]-S[phi,lambda]=0$:
$$(-partial_xphi)deltaphivert_{x=0}+ilambda (partial_xdeltaphi-cdeltaphi)vert_{x=0}=0,$$
Now, how is one supposed to deal with the $partial_xdeltaphi $ term, over which we do not have any assumptions? $deltaphi$ 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 $partial_xdeltaphi$. 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. $(partial_x-c)langlephi(x)phi(x’)ranglevert_{x=0}$, 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. $$langlephi(x)phi(x’)rangle=frac{int Dlambda phi(x)phi(x’)exp(-S)}{int Dlambda exp(-S)},$$ but at the moment I am just concerned with the method chalked out above.

Physics Asked by evening silver fox on December 29, 2020

2 Answers

2 Answers

Correct me if I’m wrong. You try 2 functions $phi$ and $phi_2=phi+deltaphi$ which both must satisfy boundary condition. Since boundary condition is linear, logically $deltaphi$ satisfies it too.

Answered by Alex on December 29, 2020

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 $ilambda(phi^{prime}(0)-cphi(0))$ 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.

Answered by Qmechanic on December 29, 2020

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