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?
 A: 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
$$
   K(x,y;t)=\left(\frac{\sqrt{a}}{2\pi\sinh 2\sqrt{a}t}\right)^\frac{1}{2}e^{\frac{b^2}{4a}t}e^{-S(x,y;t)}.
$$
