Diffusion equation Lagrangian: what is the conjugate field?

Question

Morse and Feshbach state on p. 313 without elaboration that the diffusion equation for temperature or concentration $\psi$ and its "conjugate" $\psi^*$ (quotation marks theirs) has the Lagrangian density:

$$L=-\nabla\psi\cdot\nabla\psi^* -\frac{1}{2}a^2(\psi^*\frac{\partial\psi}{\partial t}-\psi\frac{\partial\psi^*}{\partial t}). $$

I don't understand what the conjugate field, $\psi^*$, is. Since the classical (non-Schrödinger) field should be real, I suspect the conjugation symbol * refers to something other than complex conjugation. With a real field, $\psi^*=\psi$, and only $-\nabla\psi\cdot\nabla\psi$ remains, which would be the Lagrangian for the Laplace equation (steady state diffusion).

Answer 1

The conjugate field ψ∗ is but the complex conjugate of ψ, so an extra degree of freedom to expedite derivation of the diffusion equation, $$ \nabla^2 \psi = a^2 \partial_t \psi , $$ analogous to the Lagrangian of the free Schroedinger equation, real in that case--only.

• But, since this equation does not mix real and imaginary parts, take its imaginary part to be zero at the very end, and safely interpret ψ as a concentration, etc.

It is just computational expedience, namely extending to the complex plane, as, e.g. in electromagnetism, to avoid the quandary you observed if you take the imaginary part of the Lagrangian, (proportional to a²) to vanish.

Alternatively, integrating by parts in the action and discarding the surface terms nets a Lagrangian density $$ L= \psi^* (\nabla^2 -a^2 \partial_t )\psi, $$ so ψ∗ may be thought of as an extraneous Lagrange multiplier gimmick to brutally enforce the diffusion equation as is, and concentrate on its real solutions.

Note $\int dx \psi $ is a constant in time, as physically required for your diffusing quantity.

A central solution of this equation underlying its propagator is $$ \psi({\mathbf x},t)= \frac{a^3}{8 (\pi t )^{3/2}} ~ e^{-a^2 {\mathbf x}^2 /4t} $$ which starts out at t =0 as a Dirac $\delta ({\mathbf x})=\psi({\mathbf x},0)$. As a result, any initial concentration profile f can be written as a linear superposition of such δs, $$ \tilde \psi({\mathbf x},0)= \int d^3 y ~ f({\mathbf y} ) \delta ({\mathbf x} -{\mathbf y}) , $$ and propagated through each component thereof, by the above solution, $$ \tilde \psi({\mathbf x},t)= \frac{a^3}{8 (\pi t )^{3/2}} ~\int d^3 y ~ \tilde \psi({\mathbf y},0 ) ~ e^{-a^2 ({\mathbf x}-{\mathbf y})^2 /4t} . $$

Answer 2

This is actually a subtler question than it first seems. In fact, you absolutely cannot interpret $\psi^*$ as the complex conjugate of $\psi$, either in the Lagrangian or at the level of the equations of motion. Let's see why that is. As was said in Zachos' answer, the Lagrangian may be written as $$L = \psi^*(\nabla^2 - a^2 \partial_t)\psi.$$ Let us also integrate by parts on the Laplacian so that the Lagrangian only depends on the first partial derivatives of the field: $$L = -(\nabla_i \psi^*) (\nabla_i \psi) - a^2 \psi^* \partial_t \psi.$$

The Euler Lagrange equation for $\psi^*$ is then just that of a Lagrange multiplier, enforcing the equation of motion for $\psi$: $$\frac{\partial L}{\partial \psi^*} - \partial_\mu \frac{\partial L}{\partial (\partial_\mu \psi^*)} = 0 \implies (\nabla^2 - a^2 \partial_t)\psi = 0.$$ However, this is not the end of the story, because the Euler Lagrange equation must also be satisfied for $\psi$. This reads, $$\frac{\partial L}{\partial \psi} - \partial_\mu \frac{\partial L}{\partial (\partial_\mu \psi)} = 0 \implies (\nabla^2 + a^2 \partial_t)\psi^* = 0.$$ So, it is a bit of a cheat to call $\psi^*$ a Lagrange multiplier because it in fact must satisfy its own equation of motion for the action to be extremized. Now if you require that $\psi = \psi^*$, you get the system of equations $$\begin{split}(\nabla^2 + a^2 \partial_t)\psi = 0\\ (\nabla^2 - a^2 \partial_t)\psi=0\end{split}\implies \nabla^2\psi = 0 \;\;\;\text{and}\;\;\; \partial_t \psi = 0.$$ In fact, even just requiring $\psi^*$ to be the conjugate of $\psi$ (and allowing $\psi$ to be complex) leads to the same conclusion. So, if you want to write this Lagrangian down to obtain the heat equation, you must simply interpret $\psi^*$ as an unphysical auxiliary field with no relation to $\psi$.

You can actually play this trick to implement any equation of motion in a Lagrangian, i.e. include a Lagrange multiplier which merely sets the desired function $\psi$ to obey its equation of motion. However, this is not always physical, as occured in this case. In general, Lagrangians are not good at implementing parabolic differential equations, which the heat equation is. Physically, this is because parabolic equations do not have a conserved energy, a contradiction for Lagrangian dynamics because all time independent Lagrangians have a conserved energy. The energy is "saved" by the multiplier field, whose time reversed equation of motion has an unphysical energy which, when added to the "energy" of the heat equation field, produces a constant integral of motion.