In "Schrödinger Operator: Heat Kernel and Its Applications", Feng computes the heat kernels associated to Schrödinger operators with at most quadratic potentials.

I am trying to see how these work in one variable. So consider his formula for the heat kernel $K(x,y,t)$ associated to $$ L = -\Delta + (ax^2 + bx) \qquad \text{ where } a > 0$$ He gives, to the best of my understanding,

$$K(x,y,t) = \left(\frac{\sqrt{a}}{2\pi}\right)^{1/2} \left(\frac{1}{\sinh 2\sqrt{a} t}\right)^{1/2} e^{\frac{b^{2}}{4a}t} \times \exp\left\{-\frac{b^{2}}{8a^{3/2}}\coth 2\sqrt{a}t\right\} \\ \times \exp\left\{ -\sqrt{a}\left(\frac{1}{2}\coth 2\sqrt{a}t(x^{2} + y^{2}) - \frac{xy}{\sinh 2\sqrt{a}t}\right)\right\} \\ \times \exp\left\{-\frac{b}{2\sqrt{a}}\left(x\coth 2\sqrt{a}t - \frac{y}{\sinh 2\sqrt{a}t}\right)\right\}$$

So for example, setting $a=1$ and $b = 0$ we recover the Mehler kernel for the harmonic oscillator.

But I am very confused about what happens when $b \neq 0$. If $x = y$ then the last two exponentials are independent of $t$, but $K(x,x,t)$ is not singular as $t \to 0$ because the term

$$\exp\left\{-\frac{b^{2}}{8a^{3/2}}\coth 2\sqrt{a}t\right\}$$

decays faster than the blowup from the $(\sinh 2\sqrt{a}t)^{-1/2}$ term. So how can this actually be the correct formula for the kernel? Or what is going on?

`$\partial_{t} + L_{0}$`

where`$L_{0} = -\Delta + ax^{2}$`

? I don't know if that works. For the record, this should be a very special case Feng's Theorem 4.2 on page 20 of the paper. But I can't find any sources to check against. – Michael Tinker May 14 '13 at 15:54