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Let $D$ be the Dirac operator, $O_N:=e^{-(D/N)^2}$ for $N\in\mathbf{N}$ and $\{\psi_n\}$ a complete set of eigenfunctions of $D$.

On page $69$ and $78$ of Path Integrals and Quantum Anomalies and in the paper Path Integral for Gauge Theories with Fermions, Fujikawa uses the equation \begin{equation}\tag{1} (\mathrm{tr}\,O_N)(x):=\sum_{n=1}^{\infty}\big\langle\psi_n(x)|(O_N\psi_n)(x)\big\rangle=\mathrm{tr}\int\frac{\mathrm{d}k}{(2\pi)^4}e^{ikx}O_Ne^{-ikx} \end{equation} to derive \begin{equation} \lim_{N\to\infty}(\mathrm{tr}\,O_N)(x)=-\frac{1}{8\pi^2}\mathrm{tr}(F_{ij}F_{kl})(x)\epsilon^{ijkl}. \end{equation} Unfortunately, I don't understand how the right hand side of $(1)$ is defined - however, he uses the relation \begin{equation} \nabla_\mu\mathrm{e}^{ikx}=\mathrm{e}^{ikx}(\nabla_\mu+ik_\mu) \end{equation} later on - since $\nabla_\mu(u\cdot\psi)=u\cdot(\nabla_\mu+ik_\mu)\psi$ if $u(x):=e^{ikx}$, this suggests that $Ae^{-ikx}$ is actually the operator $\psi\mapsto A(u\cdot\psi)$.

He also claims that $(1)$ is a "unitary transformation" from the basis $\{\psi_n\}$ to plane waves $\{e^{ikx}\}$, but this doesn't make sense to me - I think $\psi_n(x)\in\mathbf{C}^N\otimes \mathfrak{g}$, whereas $e^{ikx}\in\mathbf{C}$.

If you've seen similar equations somewhere else, please also let me know.

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  • $\begingroup$ Would choosing $e^{-ikx}$ as a complete set of states help? Or is your question about something else? $\endgroup$
    – nwolijin
    Mar 29 at 10:19
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You are probably confused because of the muddled notation such as $|\psi(x)\rangle$. Indeed I remember being confused when I first read Fujikawa's account. Let's do it with proper Dirac notation:

Firstly the LHS $$ \sum_n \langle \psi_n(x)|O(\psi_n(x)\rangle $$ is to be interpreteted as $$ \sum_n\langle \psi_n|x\rangle \langle x|\hat O|\psi_n\rangle $$ where the $|\psi_n\rangle$ are any complete set of states and the $\langle x|\psi_n\rangle\equiv \psi_n(x)$ the corresponding wavefunctions. This is the usual definition of the functional trace ${\rm Tr} \{\hat O\}$ for a trace-class operator on $L^2[\mathbb R]^n\otimes {\mathcal V}$ where ${\mathcal V}$ in the space of internal degrees of freedom such as group representation or spinor indices. As the set is complete $$ \sum_n |\psi_n\rangle \langle \psi_n|= {\rm Identity} $$ we could also use as a complete set the position eigenstates $|x\rangle$ write the trace as $$ {\rm Tr} \{\hat O\} = {\rm tr} \int d^nx \langle x|\hat O|x\rangle $$ where the trace with the lower case "t" is the trace only over the ${\mathcal V}$ internal indices, which are usually hidden. In more detail we would include those, so that the wavefinctions would be $$ \psi_{n,i}(x)= \langle x,i|\psi\rangle $$ where $i$ labels a basis for ${\mathcal V}$ and $$ {\rm Tr}\{\hat O\}= \sum_i \int d^nx \langle i,x|\hat O|i,x\rangle $$ For the Dirac operator acting on sections of a gauge bundle the internal indices are the spin labels $\alpha$ on which the gamma matrices act, and labels "$i$" on which the group representation matrices $\lambda_a$ in the bundle-connection gauge field $$ A_\mu= \lambda_a A^{(a)}_\mu $$ act.

We can of course use any complete set of eigenfunctions for the spatial part of the trace. For example $\langle x|k\rangle = e^{ikx}$.

As an illustration fo what Fujikawa doe with this, 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 $$ \langle{x}|{\hat p} |{\psi}\rangle = -i\partial_x \langle {x}|{\psi}\rangle, \quad \langle{x}|{\hat x}| {\psi}\rangle = x \langle{x}|{\psi}\rangle , $$ and $$ \langle x|x'\rangle = \delta(x-x'), \quad \langle k|k'\rangle = 2\pi \delta(k-k'), \quad \langle x|k\rangle=e^{ikx}, $$ to proceed as follows $$ \langle {x}|{e^{-tH(\hat p,\hat x)}}|{\psi}\rangle = e^{-tH(-i\partial_x , x)}\langle{x}|{\psi}\rangle,\\ =\int \frac {dk}{2\pi}e^{-tH(-i\partial_x , x)} \langle {x}|{k}\rangle \langle {k}|{\psi}\rangle,\\ = \int \frac {dk}{2\pi}e^{-tH(-i\partial_x , x)} e^{ikx} \langle {k}|{\psi}\rangle,\\ = \int \frac {dk}{2\pi} e^{ikx}e^{-tH(-i\partial_x+k , x)} \langle {k}|{\psi}\rangle,\\ = \int \frac {dk}{2\pi} e^{ikx}\langle{k}|{\psi}\rangle e^{-tH(-i\partial_x+k , x)} 1. $$ Now set $|{\psi}\rangle =|{y}\rangle $ so $\langle k|\psi\rangle\to \langle {k}|{y}\rangle = e^{-iky}$ 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, when we expand out the exponential, the $\partial_x$ derivatives act on everything to their right until they reach $\partial_x 1=0$.

To take the trace we set $x=y$ and integrate $$ {\rm Tr}\{e^{-tH}\} =\sum_n e^{-t\lambda_n}= \int dx\left\{ \int \frac {dk}{2\pi} e^{-tH(-i\partial_x+k , x)}1\right\} $$ where the $\lambda_n$ are the eigenvalues of $H$.

To answer your question about $\langle x|\hat O| x\rangle$: If $\hat O$ has eigenvectors $|\psi_n\rangle$ then inserting two complete sets of states, we have $$ \langle x|\hat O| x\rangle=\sum_{m,n} \langle x|\psi_n\rangle \langle \psi_n|\hat O|\psi_m\rangle\langle \psi_m|x\rangle\\ = \sum_{m,n} \lambda_m\langle x|\psi_n\rangle \langle \psi_n|\psi_m\rangle\langle \psi_m|x\rangle = \sum_{m,n} \delta_{mn}\lambda_m\langle x|\psi_n\rangle \langle \psi_m|x\rangle\\ = \sum_m \lambda_m \psi_m^*(x) \psi_m(x). $$ This assumes that the sum converges. Sometimes it does not, but the reason for Fujikawa using heat kernels is that $\sum e^{-t\lambda_n}$ is very nicely behaved when the $\lambda_n$ are positive.

For example $$ \langle x |e^{-t(-\partial_x^2)}|y \rangle = \frac 1{\sqrt{4\pi t}} \exp\{-(x-y)^2/4t\} $$ so $$ \langle x |e^{-t(-\partial_x^2)}|x \rangle= \frac 1{\sqrt{4\pi t}} $$ which can also be obtained from the eigenvector $\langle x|k\rangle= e^{ikx}$ and eigenvalues $k^2$ of $-\partial_x^2$ as $$ \int \frac {dk}{2\pi} e^{-tk^2}=\frac 1{\sqrt{4\pi t}} $$

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  • $\begingroup$ Sorry, but I don't see the correlation between the equation from the title and the equation you derived (I'm not saying that there isn't one). $\endgroup$
    – Filippo
    Mar 29 at 14:05
  • $\begingroup$ But I'd like to explain the reason for my notation: The $\psi_n$ are vector valued functions (I think the codomain is $\mathbf{C}^N\otimes\mathfrak{g}$) and the codomain is equipped with an inner product. Thus, the trace from the title is a function, defined pointwise. In his book, Fujikawa defines $\mathrm{Tr}\,O:=\int(\mathrm{tr}\,O)(x)\,dx$. $\endgroup$
    – Filippo
    Mar 29 at 14:10
  • $\begingroup$ I was under the impression that I had done what what was puzzling you. I'll add to my answer to include what the notation means. $\endgroup$
    – mike stone
    Mar 29 at 14:19
  • $\begingroup$ Thank you! I honestly think that the dirac notation is preventing me from understanding the calcukation and the result. $\endgroup$
    – Filippo
    Mar 29 at 14:31
  • $\begingroup$ Thank you very much for the edit, I like the direction this is going! In his paper from 1986 (Simple evaluation of chiral anomalies in the path-integral approach), Fujikawa wrote an equation that looks a bit similar: $\sum_{n=1}^{\infty}\psi_n(x)^\dagger\gamma_5e^{-\beta H}\psi_n(x)=\mathrm{Tr}\langle x|\gamma_5e^{-\beta H}|x\rangle=\mathrm{Tr}\int\frac{d^4k}{(2\pi)^4}e^{-ikx}\gamma_5e^{-\beta H}e^{ikx}$. $\endgroup$
    – Filippo
    Mar 29 at 15:47
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Think of the $\psi$'s as being a some spinor $u\in\mathbb{C}^4$ of functions $f:\mathbb{R}^4\rightarrow \mathbb{C}$, which is again the space where the Lie algebra acts, not the Lie algebra itself. So think that $$\psi(x)\propto e^{i k x} u_s $$

Now notice equation (1) still has a trace, which is the trace over possible remaining index $s$, in this case the spinor indices. While the functional part is taken care by the plane waves and as you might know computing an inner product can be done with the Fourier transform, namely momentum space instead of configuration space (spacetime points).

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Let's assume $O$ acts on complex valued functions. We can improve the notation by setting $\langle x|\psi\rangle:=\psi(x)$, $\langle\psi|x\rangle:=\overline{\psi(x)}$ (complex conjugate) and $|n\rangle:=\psi_n$. We can obtain the result from the title by introducing the plane wave basis defined by $\langle x|k\rangle=\mathrm{e}^{\mathrm{i}kx}$ and using the Fourier inversion theorem: \begin{align} \sum_n \langle \psi_n(x)|(O\psi_n)(x)\rangle=\sum_n \overline{\psi_n(x)}(O\psi_n)(x)=\sum_n\langle n|x\rangle \langle x|O|n\rangle=\sum_n\int\frac{\mathrm{d}k}{(2\pi)^d}\langle n|k\rangle\langle k|x\rangle\langle x|O|n\rangle\\ =\int\frac{\mathrm{d}k}{(2\pi)^d}\langle k|x\rangle\langle x|O\sum_n\langle n|k\rangle|n\rangle=\int\frac{\mathrm{d}k}{(2\pi)^d}\langle k|x\rangle\langle x|O|k\rangle=\int\frac{\mathrm{d}k}{(2\pi)^d}\mathrm{e}^{-\mathrm{i}kx}\langle x|O|k\rangle \end{align} Of course, writing $\langle x|O|k\rangle=O\langle x|k\rangle$ is simply wrong.

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