heat kernel method

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Motivation

We wish to find the inverse of the differential operator D_x\,, i.e., D_x D^{-1}(x,y) = \delta(x,y)\,. Formally,

D^{-1}\, = i\int_0^\infty\!ds\,e^{-s\, i D}\,,
= i\int_0^\infty\!ds\, K(D,s)\,,

where K(D,s)\, is called the Heat Kernel and satisfies the Schrödinger equation

i \frac{\partial}{\partial s} K(D,s) = D K(D,s)\,,

and the initial condition K(D,0) = 1\,. Abstractly, this can be thought of as the wave-function of a particle with spacetime coordinates x,s\, whose time evolution is described by a Hamiltonian H = -D\, with the initial condition \left\langle x,0 \vert y,0\right\rangle = \delta(x,y)\,. Then K(D,s) = \left\langle x,s \vert y,0\right\rangle\,.

K(D,s) = e^{-i D_x}\delta(x,y)\,.

Covariance

Let D_x\, operate on a Hilbert space of functions \psi(x)\,, and consider a local gauge transformation \psi(x) \to e^{-\Lambda(x)}\psi(x)\, under which D_x\, transforms covariantly, i.e., D_x \to e^{-\Lambda(x)}D_x e^{\Lambda(x)}\,. The Schrödinger operator transforms as

\left(i\frac{\partial}{\partial s} - D_x \right) \to e^{-\Lambda(x)}\left(i\frac{\partial}{\partial s} - D_x \right)e^{\Lambda(x)}\,,

so that at the very least K(D, s)(x,y) \to e^{-\Lambda(x)} K(D,s)(x,y)\,. Actually, the initial condition is symmetric under the exchange of x\, and y\, in that K(D,0)(x,y) = K(D,0)^{\dagger}(y,x)\,, so it is necessary that

K(D,0)(x,y) \to e^{-\Lambda(x)}K(D,0)e^{\Lambda(y)}\,,

so that

K(D,s)(x,y) = e^{-i D_x} K(D,0)(x,y) \to e^{-\Lambda(x)}K(D,s)e^{\Lambda(y)}\,.

Regularization

Suppose we have some differential operator D_x + m^2\,, and (D_x + m^2) G(x,y) = \delta(x-y)\,. Write

G(x,y) = i \int_0^\infty\!ds\left\langle x,s \vert y,0\right\rangle e^{-i m^2 s}\,.

Then,

D_x G(x,y)\, = \delta(x-y) - m^2 G(x,y)\,,
= \delta(x-y) + \int_0^\infty\!ds \left\langle x,s \vert y,0\right\rangle \frac{\partial}{\partial s}e^{-i m^2 s}\,,
= \delta(x-y) + \left[\left\langle x,s \vert x,0\right\rangle e^{-i m^2 s} \right]_0^\infty - \int_0^\infty\!ds\,e^{-i m^2 s} \frac{\partial}{\partial s} \left\langle x,s \vert y,0\right\rangle \,,
= \delta(x-y) - \left\langle x,0 \vert y,0\right\rangle - \int_0^\infty\!ds\,e^{-i m^2 s} \frac{\partial}{\partial s} \left\langle x,s \vert y,0\right\rangle \,
= i \int_0^\infty\!ds\, e^{-i m^2 s} D_x\left\langle x,s \vert y,0\right\rangle\,.

So

i\frac{\partial}{\partial s} \left\langle x,s \vert y,0\right\rangle = D_x \left\langle x,s \vert y,0\right\rangle\,, \left\langle x,0 \vert y,0\right\rangle = \delta(x-y)\,,

which is essentially the Schrödinger equation. The solution is

\left\langle x,s \vert y,0\right\rangle = e^{-i s D_x} \left\langle x,0 \vert y,0\right\rangle\,.
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