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I'm reading the article https://arxiv.org/abs/hep-th/9705200 and part of it has left me very confused. In order to speak about their equation (1) the authors make the following statement: \begin{align*} K(t) &= \lim_{x' \to x} \int \dfrac{d^4 k}{(2\pi)^4} e^{i k \cdot (x - x')} \left( e^{-i k \cdot (x - x')} e^{t D_a D_a} e^{i k \cdot (x - x')}\right) \\ &= \int \dfrac{d^4 k}{(2\pi)^4} e^{tX_a X_a} \end{align*} with $X_a = D_a + i k_a$, indices ($a$) refer to Euclidean vectors, and $D_a$ can be an Abelian covariant derivative or not, we can choose.

Could anyone give me a tip on how to show this passage?

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Suppose we wish to compute the quantum mechanics matrix element $$ \langle{x}|{e^{-tH(\hat p,\hat x)}}|{y}\rangle, \quad [\hat x, \hat p]=i. $$ We use $$ \langle{x}|{\hat p}| {\psi}\rangle= -i\partial_x \langle{x}|{\psi}\rangle, \quad \langle {x}|{\hat x}| {\psi}\rangle= x \langle {x}|{\psi}\rangle, $$ to proceed as follows $$ \langle {x}|{e^{-tH(\hat p,\hat x)}}|{\psi}\rangle= e^{-tH(-i\partial_x , x)}\langle{x}|{\psi}\rangle,\nonumber\\ =\int \frac {dk}{2\pi}e^{-tH(-i\partial_x , x)} \langle{x}|{k}\rangle \langle {k}|{\psi}\rangle ,\nonumber\\ = \int \frac {dk}{2\pi}e^{-tH(-i\partial_x , x)} e^{ikx} \langle{k}|{\psi}\rangle ,\nonumber\\ = \int \frac {dk}{2\pi} e^{ikx}e^{-tH(-i\partial_x+k , x)} \langle{k}|{\psi}\rangle ,\nonumber\\ = \int \frac {dk}{2\pi} e^{ikx}\langle{k}|{\psi}\rangle e^{-tH(-i\partial_x+k , x)} 1. \nonumber $$ Now set $|{\psi}\rangle =|{y}\rangle $ and recall that $\langle {k}|{y}\rangle = e^{-iky}$ to get $$ \langle {x}|{e^{-tH(\hat p,\hat x)}}|{y}\rangle = \int \frac {dk}{2\pi} e^{ik(x-y)}e^{-tH(-i\partial_x+k , x)}1 $$ where the $\partial_x$ derivatives act on everything to their right until they reach $\partial_x 1=0$.

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  • $\begingroup$ I am sorry, but I cannot see how this is conected with my question. $\endgroup$ Commented May 26 at 20:15
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    $\begingroup$ My Fujikawa heat kernel calculation recipe exactly the same as what you want surely? My $H$ is you $D_aD_a$. You have 4 dimensions but the way the derivative opertors work is identical. Of course I use "quantum mechanics" as a convenient name for the plane wave modes and their completeness relations,. and have $-i\partial +k$ rather than $D+ik$, but these are trivialities. You also set $x=y$ at the end, but sometimes the $x\ne y$ case is useful. $\endgroup$
    – mike stone
    Commented May 26 at 20:35

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