# A heat kernel for Schrödinger operator with low-order terms

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?

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Shouldn't I be able to simplify the exponentials in terms of the shifted variables $\tilde x=x+b/2a$, $\tilde y=y+b/2a$? That is the origin of the $\exp(b^2t/4a)$, which reflects the shift of the quadratic potential. The terms you have don't seem to simplify nicely, however. – Austen May 14 '13 at 9:41
Do you mean we should be able to write $\partial_{t} + L$ as the conjugation (with appropriate exponential $\exp\{f(x,t;b)\}$) of the operator $\partial_{t} + L_{0}$ 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

I think your formula is not correct. The right kernel is invariant by interchanging $x$ with $y$. This symmetry must be preserved.
Then, note that $$L=-\Delta+ax^2+bx=-\Delta+a\left(x+\frac{b}{2\sqrt{a}}\right)^2-\frac{b^2}{4a}$$ and this operator is invariant for translations. This means that the kernel for $b\ne 0$ can be obtained from the kernel for $b=0$ by translation of a shift $\frac{b}{2\sqrt{a}}$ and multiplying for the overall factor $e^{\frac{b^2}{4a}t}$.
Once this is observed, we can consider the kernel for $b=0$, just the argument of the exponential, to be $$S_0(x,y;t)=\sqrt{a}\left[(x^2+y^2)\coth 2\sqrt{a}t-\frac{xy}{2\sinh 2\sqrt{a}t}\right]$$ and apply the above translation to get $$S(x,y;t)=\sqrt{a}\left[\left(x^2+y^2+(x+y)\frac{b}{\sqrt{a}}+\frac{b^2}{4a}\right)\coth 2\sqrt{a}t\right.$$ $$\left.-\frac{1}{2\sinh 2\sqrt{a}t}\left(xy+\frac{b}{2\sqrt{a}}(x+y)+\frac{b^2}{4a}\right)\right]$$ and this gives the right kernel $$K(x,y;t)=\left(\frac{\sqrt{a}}{2\pi\sinh 2\sqrt{a}t}\right)^\frac{1}{2}e^{\frac{b^2}{4a}t}e^{-S(x,y;t)}.$$
Thank you Jon---I eventually realized I needed a shift by integrating in $y$ and determining where the kernel became singular as $t \to 0$, but your answer is much clearer. Can you recommend a good introduction to heat kernels and Schrodinger operators? – Michael Tinker May 25 '13 at 20:45