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##  Appendix A A Calculation Without Auxiliary Fields

Calculating the effects of interactions on anomalous relation through an
auxiliary field is insightful on its own, but as additional clarification we
present a slightly different calculation here, which does not rely on
auxiliary fields. We begin by dividing the spinor fields into left and right
moving parts,

$\begin{split}&\psi_{R}=\frac{1+\gamma_{5}}{2}\psi,\ \ \ \
\bar{\psi}_{R}=\bar{\psi}\frac{1-\gamma_{5}}{2},\\\
&\psi_{L}=\frac{1-\gamma_{5}}{2}\psi,\ \ \ \
\bar{\psi}_{L}=\bar{\psi}\frac{1+\gamma_{5}}{2}.\end{split}$ (134)

This way both the electrical current and the chiral current will also be
divided into $L$ and $R$ parts. Therefore, the current-current interaction
(which is nothing but a density-density interaction made Lorentz invariant) is
given by

$-\frac{\lambda^{2}}{2}\bar{\psi}\gamma^{\mu}\psi\bar{\psi}\gamma_{\mu}\psi=-\frac{\lambda^{2}}{2}\left(j_{R}^{\mu}j^{R}_{\mu}+j_{L}^{\mu}j^{L}_{\mu}+2j_{R}^{\mu}j^{L}_{\mu}\right)\,,$
(135)

where $j^{\mu}_{R,L}\equiv\bar{\psi}_{R,L}\gamma^{\mu}\psi_{R,L}$ and we used
the fact that paired Grassmann numbers are commutative. The quantum system is
therefore described by the following path-integral,

$I=\int\mathcal{D}\bar{\psi}_{R}\mathcal{D}\psi_{R}\mathcal{D}\bar{\psi}_{L}\mathcal{D}\psi_{L}e^{iS[\bar{\psi}_{R},\psi_{R},\bar{\psi}_{L},\psi_{L},A_{\mu}]}\,.$
(136)

The action and the path-integral are both symmetric under $R\leftrightarrow L$
transformation and the action $S$ for the (1+1)-dimensional case is given by

$\displaystyle S=\int d^{2}x\bigg{[}$
$\displaystyle\bar{\psi}_{R}i\gamma^{\mu}D_{\mu}\psi_{R}+\bar{\psi}_{L}i\gamma^{\mu}D_{\mu}\psi_{L}-\frac{\lambda^{2}}{2}\left(j_{R}^{\mu}j^{R}_{\mu}+j_{L}^{\mu}j^{L}_{\mu}+2j_{R}^{\mu}j^{L}_{\mu}\right)\bigg{]}\,,$
(137)

with $D_{\mu}\equiv\partial_{\mu}-ieA_{\mu}$.

By an infinitesimal chiral rotation in either right or left moving pairs we
can obtain corresponding anomalous relations. Since we have divided our
original fields into two distinct $R$ and $L$ parts, the anomalous term, which
was originally equal to the difference of right and left moving zero modes
$n_{+}-n_{-}$, gets divided by a factor of two; so that in the absence of
interactions the anomalous relation becomes

$\partial_{\mu}{j_{5}^{\mu}}_{R,L}=\frac{e}{2\pi}\epsilon^{\mu\nu}\partial_{\mu}A_{\nu}$
(138)

instead of

$\partial_{\mu}{j_{5}^{\mu}}=\partial_{\mu}{j_{5}^{\mu}}_{R}+\partial_{\mu}{j_{5}^{\mu}}_{L}=\frac{e}{\pi}\epsilon^{\mu\nu}\partial_{\mu}A_{\nu}\,.$
(139)

Since $\psi_{R}$ pairs are independent from $\psi_{L}$ pairs, each time we
chirally rotate one pair we can treat the other as constant fields and its
corresponding current a constant four-vector. For example, let us rotate
$\psi_{R}$ to $e^{i\alpha\gamma_{5}}\psi_{R}$ while we have prepared the
action in the following form,

$\begin{split}S=\int\\!d^{2}x\bigg{[}\bar{\psi}_{R}i\gamma^{\mu}\left(D_{\mu}+i\lambda^{2}j^{L}_{\mu}\right)\psi_{R}&+\mathcal{L}[\psi_{L},\bar{\psi}_{L}]\bigg{]}\,.\end{split}$
(140)

We see that $j^{L}_{\mu}$ is now acting as an external field for right moving
fermions. (And it does so through its field strength
$F^{j_{L}}_{\mu\nu}\equiv\partial_{\mu}j^{L}_{\nu}-\partial_{\nu}j^{L}_{\mu}$.)

Now let us assume that in the presence of interactions the total anomalous
term, i.e. the right hand side of equation (139), will be modified by a
constant factor $\kappa(\lambda^{2})$ where $\lambda^{2}/2$ is the interaction
coupling:

$\partial_{\mu}{j_{5}^{\mu}}=\kappa(\lambda^{2})\frac{e}{\pi}\epsilon^{\mu\nu}\partial_{\mu}A_{\nu}\,.$
(141)

For the right and left moving anomalous relations the interaction term is
divided into these two parts as well. Note that before, the whole field,
containing both the right and left moving modes, was interacting with itself.
But now only half of that self-interaction appears as a self-interaction for
each left and right moving field, while the other half appears as the
interaction between one field and the other. Therefore, the corresponding
coefficient for their anomalous term will be $\kappa(\lambda^{2}/2)$ instead
of $\kappa(\lambda^{2})$. All in all, the anomalous identity would then be

$\partial_{\mu}{j_{5}^{\mu}}_{R}=\kappa\left(\frac{\lambda^{2}}{2}\right)\left[\frac{e}{2\pi}\epsilon^{\mu\nu}\partial_{\mu}A_{\nu}-\frac{\lambda^{2}}{2\pi}\epsilon^{\mu\nu}\partial_{\mu}j^{L}_{\nu}\right]\,.$
(142)

Since the transformation $R\leftrightarrow L$ is a symmetry of the path-
integral, we can write down the same relation for left moving part of the
chiral current. One can then sum up these relations to obtain the
corresponding anomalous identity for the total chiral current:

$\partial_{\mu}j^{\mu}_{5}=\kappa\left(\frac{\lambda^{2}}{2}\right)\left[\frac{e}{\pi}\epsilon^{\mu\nu}\partial_{\mu}A_{\nu}-\frac{\lambda^{2}}{2\pi}\partial_{\mu}j^{\mu}_{5}\right]\,,$
(143)

where we have used the relation
$\epsilon^{\mu\nu}\gamma_{\nu}=\gamma^{\mu}\gamma_{5}$ to change $j^{\mu}$ to
$j_{5}^{\mu}$. Algebraic rearrangements then give,

$\partial_{\mu}j^{\mu}_{5}=\frac{\kappa(\lambda^{2}/2)}{1+\frac{\lambda^{2}}{2\pi}\kappa(\lambda^{2}/2)}\frac{e}{\pi}\epsilon^{\mu\nu}\partial_{\mu}A_{\nu}\,.$
(144)

Looking at equation (141) prompts a consistency equation:

$\kappa(\lambda^{2})=\frac{\kappa(\lambda^{2}/2)}{1+\frac{\lambda^{2}}{2\pi}\kappa(\lambda^{2}/2)}\,.$
(145)

In terms of $\kappa(\lambda^{2})^{-1}$ the form of the above equation becomes
simpler,

$\kappa\left(\lambda^{2}\right)^{-1}=\kappa\left(\lambda^{2}/2\right)^{-1}+\frac{\lambda^{2}}{2\pi}\,.$
(146)

Knowing that $\kappa(0)$ is equal to one, this equation has a unique solution:
$\kappa(\lambda^{2})^{-1}=1+\lambda^{2}/\pi$. By this we eventually arrive at
the final stage of equation (141),

$\partial_{\mu}{j_{5}^{\mu}}=\frac{1}{1+\lambda^{2}/\pi}\frac{e}{\pi}\epsilon^{\mu\nu}\partial_{\mu}A_{\nu}\,.$
(147)

##  Appendix B 1+1 Dimensions Remaining Remarks

Here we answer two simple questions that are prone to be asked regarding the
decoupling procedure employed in our calculation, and another possible four
fermionic interaction.

There are different ways we can decouple the current-current term via a
Hubbard-Stratonovich field $a_{\mu}$. One might then tend to ask whether these
different ways will lead to different answers or not; if they do the result
would have been unphysical, but this is not the case. We can verify this
through the simple example below which is generalizable to more complicated
cases that appears in later sections when we discuss the four dimensional
case. Consider the action,

$S=\int\\!d^{2}x\left[\bar{\psi}i\gamma^{\mu}(\partial_{\mu}-ieA_{\mu}-i\text{L}a_{\mu})\psi+\frac{\Gamma}{2}a_{\mu}a^{\mu}\right].$
(148)

On-shell value of $a^{\mu}$ is $-(\text{L}/\Gamma)\bar{\psi}\gamma^{\mu}\psi$.
After a shift in the auxiliary field by its on-shell value and integrating the
shifted auxiliary field out, the action turns out as below,

$S=\int\\!d^{2}x\left[\bar{\psi}i\gamma^{\mu}(\partial_{\mu}-ieA_{\mu})\psi-\frac{\text{L}^{2}}{2\Gamma}\bar{\psi}\gamma^{\mu}\psi\bar{\psi}\gamma_{\mu}\psi\right].$
(149)

We set $\text{L}^{2}/\Gamma\equiv\lambda^{2}$ to make the action look like
(28). On the other hand, the anomalous identity reads,

$\partial_{\mu}j^{\mu}_{5}=\frac{e}{\pi}\epsilon^{\mu\nu}\partial_{\mu}A_{\nu}+\frac{\text{L}}{\pi}\epsilon^{\mu\nu}\partial_{\mu}a_{\nu}\,,$
(150)

which after the shift and integration of the auxiliary field becomes,

$\partial_{\mu}j^{\mu}_{5}=\frac{e}{\pi}\epsilon^{\mu\nu}\partial_{\mu}A_{\nu}-\frac{\text{L}^{2}}{\Gamma\pi}\epsilon^{\mu\nu}\partial_{\mu}a_{\nu},$
(151)

which gives back

$\partial_{\mu}j^{\mu}_{5}=\frac{1}{1+\lambda^{2}/\pi}\frac{e}{\pi}\epsilon^{\mu\nu}\partial_{\mu}A_{\nu}\,,$
(152)

regardless of how the decoupling of the interaction term is conducted.

Another question is regarding other possible interaction terms. One candidate
is $j^{\mu}_{5}j_{\mu}^{5}$; but in two spacetime dimensions this interaction
is no different than $j^{\mu}j_{\mu}$. One can also consider interaction terms
that do not respect the chiral symmetry such as $(\bar{\psi}\psi)^{2}$. The
fact that it manifestly breaks the chiral symmetry does not readily mean that
it will not alter the non-trivial Jacobian of the transformation.

The path-integral under consideration would be

$I=\int\mathcal{D}\bar{\psi}\mathcal{D}\psi\exp\int
d^{2}x\left[\bar{\psi}i\gamma^{\mu}D_{\mu}\psi+\Lambda^{2}(\bar{\psi}\psi)^{2}\right],$
(153)

which by using an auxiliary field $\phi$ can also be written in the following
form,

$\displaystyle\int\mathcal{D}\bar{\psi}\mathcal{D}\psi\mathcal{D}\phi\exp\int\\!d^{2}x\left[\bar{\psi}i\gamma^{\mu}D_{\mu}\psi+2\Lambda\phi\bar{\psi}\psi-\phi^{2}\right]$
$\displaystyle=\int\mathcal{D}\bar{\psi}\mathcal{D}\psi\mathcal{D}\phi\exp\int\\!d^{2}x\left[\bar{\psi}i(\not{D}-2i\Lambda\phi)\psi-\phi^{2}\right].$
(154)

Either by looking at the first line in the above and see that all we have done
is introducing a mass term, or by looking at the second line and following
strict calculations of anomaly, we can see that the Jacobian of chiral
transformation $\psi\rightarrow e^{i\alpha\gamma_{5}}\psi$ is not modified by
this term. Nevertheless, being a mass term, it does not remain invariant under
a chiral transformation, and contributes to the non-conservation as below:

$\langle\partial_{\mu}j_{5}^{\mu}\rangle_{\phi}=\mathcal{A}_{5}+\langle
4i\Lambda\phi\bar{\psi}\gamma_{5}\psi\rangle_{\phi}\,,$ (155)

where $\mathcal{A}_{5}$ is the usual anomalous term coming from the Jacobian
and there is a remaining integration over the $\phi$ field. The end result
after the integration is

$\partial_{\mu}j_{5}^{\mu}=\mathcal{A}_{5}+4i\Lambda^{2}\bar{\psi}\psi\bar{\psi}\gamma_{5}\psi\,,$
(156)

which agrees with a Ward identity derived directly from equation (153).

##  Appendix C Path-Integral Self Regularization

Unlike regular path-integration, fermionic path-integrals are defined by left
differentiation of Grassmann numbers and they need regularization. For
example, path integration over fermionic degrees of freedom of a free massive
Dirac fermion theory yields the determinant of the Dirac operator
$\det(i\not{D}+m)=\prod l_{n}$ only if the product is well-defined which is
the case when $\sum_{n}^{\infty}|l_{n}-1|$ converges InfiniteProduct . But
eigenvalues, $l_{n}$, of the Dirac operator are not even bounded. Thus the
well-defined path-integral must carry a type of regularization with itself.
This is where action functional intrudes into measure’s business. The path-
integral (or the partition function) has two elements; the measure and the
action; the regularization is not present in the latter. But how the measure
is regularized must only be determined by the action, since the partition
function is to be a self-sufficient object and should be indifferent about the
backstory of why it is written. Consequently, the regularization, being
determined only by the action, will share the symmetries of the action; even
if the measure does not respect them all.

For the sake of clarification let us investigate an instance of a
regularization which solely relies on the path-integral itself. Similar to (6)
we first expand the fermionic fields in the basis of a generic hermitian
operator. Then the action will be a function of Grassmann numbers
$\bar{b}_{n}$ and $a_{n}$.

$\int\mathcal{D}\bar{\psi}\mathcal{D}\psi
e^{S[\bar{\psi},\psi]}\rightarrow\int\prod_{n}d\bar{b}_{n}da_{n}\mathcal{F}\left[\\{\bar{b}_{n}\\},\\{a_{n}\\}\right]\,,$
(157)

which is not yet regularized. $\mathcal{F}$ is generally written as,

$\mathcal{F}=\sum_{k=0}^{\infty}\sum_{i_{1},\cdots,i_{2k}}c^{i_{1}\cdots
i_{k}i_{k+1}\cdots
i_{2k}}\bar{b}_{i_{1}}\cdots\bar{b}_{i_{k}}a_{i_{k+1}}\cdots a_{i_{2k}}\,,$
(158)

with $c$ being complex completely anti-symmetric tensors. We can then define a
complex number $\ell_{m}$ as follows,

$\ell_{m}\equiv\frac{\int\prod
d\bar{b}_{n}da_{n}\mathcal{F}\left[\\{\bar{b}_{n}\\},\\{a_{n}\\}\right]}{\int\prod
d\bar{b}_{n}da_{n}\bar{b}_{m}a_{m}\mathcal{F}\left[\\{\bar{b}_{n}\\},\\{a_{n}\\}\right]}\,,$
(159)

given directly by the path-integral and not some specific building block of
it. The reason for this definition becomes clearer when when we look at a
specific class of $\mathcal{F}$ given as

$\mathcal{F}_{D}=\prod_{n}\left(1+l_{n}\bar{b}_{n}a_{n}\right)=\exp\left\\{\sum_{n}l_{n}\bar{b}_{n}a_{n}\right\\}\,.$
(160)

In this situation we say $\mathcal{F}_{D}$ is formally diagonalized in terms
of Grassmann numbers $\\{\bar{b}_{n}\\}$ and $\\{a_{n}\\}$. Then $\ell_{m}$
defined as (159) is equal to $l_{m}$ from the above. Moreover, a formally
diagonalized $\mathcal{F}_{D}$ allows the path-integral to be completely
factorized as $\int\prod d\bar{b}_{n}da_{n}\exp\\{l_{n}\bar{b}_{n}a_{n}\\}$.
Having the path-integral written in this way makes it easy to set a cut-off in
$l_{n}$ for the modes of integration, leading to a natural regularization. For
general $\mathcal{F}$ which is not formally diagonalized $\ell_{n}$ plays the
role of $l_{n}$ and can be used for the regularization process instead.

We can then form the product

$\mathcal{I}=\prod_{n}f_{M}(\ell_{n})\,,$

as a regularized value for the path-integral, with $f_{M}(x)$ being a function
that rapidly declines to unity for $x^{2}>M^{2}$ and $f_{M}(x)=x$ for
$x^{2}<M^{2}$.

In other words, we can simply truncate the path-integration by limiting it to
only those modes whose corresponding $\ell^{2}_{n}$ is less than a certain
value $M^{2}$, which can also be done smoothly by weighting each mode of
integration $\int d\bar{b}_{n}da_{n}$ by a coefficient $f_{n}$ which is equal
to one for $\ell^{2}_{n}<M^{2}$, and smoothly but rapidly approaches to
$\ell_{n}^{-1}$ for $\ell^{2}_{n}>M^{2}$. We thus obtain a well-defined
fermionic path-integral.

One might then wonder if this approach include any other already known
regularizations; the answer is yes. This of course includes the regularization
used in Fujikawa’s approach, as we will see later, but is more general than
that. To see this, consider the case where we have chosen $f_{M}(x)$ to be
$(1+iM/x)^{-1}$:

$\displaystyle f_{M}(\ell_{m})^{-1}$
$\displaystyle=\frac{\ell_{m}+iM}{\ell_{m}}=1+\frac{iM}{\ell_{m}}$
$\displaystyle=\frac{\int\prod
d\bar{b}_{n}da_{n}\mathcal{F}_{\mathbf{n}}}{\int\prod
d\bar{b}_{n}da_{n}\mathcal{F}_{\mathbf{n}}}+\frac{\int\prod
d\bar{b}_{n}da_{n}\left(iM\bar{b}_{m}a_{m}\right)\mathcal{F}_{\mathbf{n}}}{\int\prod
d\bar{b}_{n}da_{n}\mathcal{F}_{\mathbf{n}}}$ $\displaystyle=\frac{\int\prod
d\bar{b}_{n}da_{n}\exp\left(iM\bar{b}_{m}a_{m}\right)\mathcal{F}_{\mathbf{n}}}{\int\prod
d\bar{b}_{n}da_{n}\mathcal{F}_{\mathbf{n}}}\,,$ (161)

where now the subscript $\mathbf{n}$ in $\mathcal{F}_{\mathbf{n}}$ designates
the dependence of $\mathcal{F}$ on $\\{\bar{b}_{n}\\}$ and $\\{a_{n}\\}$,
shortening the previous notation. Therefore, we can define the product
$\mathcal{I}$ as

$\displaystyle\prod_{m}\frac{\ell_{m}}{\ell_{m}+iM}$
$\displaystyle=\frac{\int\left[\prod_{m}\left(\prod_{n}d\bar{b}_{n}^{m}da_{n}^{m}\right)\mathcal{F}_{\mathbf{n}}^{m}\right]}{\int\left[\prod_{m}\left(\prod_{n}d\bar{b}_{n}^{m}da_{n}^{m}\right)e^{iM\bar{b}_{m}^{\tilde{m}}a_{m}^{\tilde{m}}}\mathcal{F}_{\mathbf{n}}^{m}\right]}$
$\displaystyle=\\!\frac{\int\left[\prod_{m}\left(\prod_{n}d\bar{b}_{n}^{m}da_{n}^{m}\right)\mathcal{F}_{\mathbf{n}}^{m}\right]}{\int\left[\prod_{m}\left(\prod_{n}d\bar{b}_{n}^{m}da_{n}^{m}\right)\mathcal{F}_{\mathbf{n}}^{m}\right]\exp\\{iM\sum_{m}\bar{b}_{m}^{\tilde{m}}a_{m}^{\tilde{m}}\\}}.$
(162)

Here each $(\\{\bar{b}_{n}\\}^{m},\\{a_{n}\\}^{m})$ is a distinct copy of
$(\\{\bar{b}_{n}\\},\\{a_{n}\\})$ for distinct values of $m$, each of which
span a similar space of Grassmann numbers with the same dimension. The number
of copies (domain of $m$) is equal to the dimensionality of each copy (domain
of $n$). Also, $\tilde{m}(m)$ is any permutation of $m$.

The above yields a regularized path-integral without any external information
which should be regarded irrelevant to the path-integral as a mathematical
object. The term inside the exponential in the last line, is a sum over all
possible values of $m$, hence $m$ is a dummy index and can be substituted with
any other letter. On the other hand, the form of
$\mathcal{F}^{m}_{\mathbf{n}}$ is the same for all $m$. Therefore, considering
the Grassmann algebra, one can separate the subspace spanned by
$\left(\\{\bar{b}^{\tilde{m}}_{m}\\},\\{a^{\tilde{m}}_{m}\\}\right)$ from the
rest of Grassmann space, taking a quotient of
$\prod_{m}\mathcal{F}^{m}_{\mathbf{n}}$ without changing the ultimate value of
the product, and rewrite it as,

$\displaystyle\mathcal{I}=\frac{\mathcal{Q}\times\int\left(\prod_{n}d\bar{b}^{\tilde{n}}_{n}da^{\tilde{n}}_{n}\right)\mathcal{F}_{\mathbf{n}}^{\tilde{n}}}{\mathcal{Q}\times\int\left(\prod_{n}d\bar{b}_{n}^{\tilde{n}}da_{n}^{\tilde{n}}\right)\mathcal{F}_{\mathbf{n}}^{\tilde{n}}\exp\\{iM\sum_{n}\bar{b}_{n}^{\tilde{n}}a_{n}^{\tilde{n}}\\}}\,,$
(163)

where the quotient $\mathcal{Q}$ is what remains from both numerator and
denominator when other terms are extracted out, and
$\mathcal{F}^{\tilde{n}}_{\mathbf{n}}\equiv\mathcal{F}\left[\\{\bar{b}^{\tilde{n}}_{n}\\},\\{a^{\tilde{n}}_{n}\\}\right]$.
After canceling $\mathcal{Q}$ the nominator will be the original path-integral
we began with while the denominator is the same path-integral with a mass term
$i\int d^{2}xM\bar{\psi}\psi$ introduced in its action. Therefore, the product
$\mathcal{I}$ can also be written as one path-integral,

$\mathcal{I}=\int\mathcal{D}\bar{\psi}\psi\mathcal{D}\bar{\varphi}\varphi\mbox{\large{
$e^{S[\bar{\psi},\psi]+S[\bar{\varphi},\varphi]+i\int
d^{2}xM\bar{\varphi}\varphi}$ }}\,,$ (164)

which is the famous Pauli-Villars regularization. Notice that
$S[\bar{\varphi},\varphi]$ has the similar form as $S[\bar{\psi},\psi]$ but
with spinor fields substituted by bosonic fields $\bar{\varphi}(x)$ and
$\varphi(x)$. Also recall that Grassmann integral is defined by left
derivative and therefore appears inversely to what a bosonic integral would.

Thus the regularized fermionic path-integral introduced here is a general case
that includes, but is not limited to, the Fujikawa and Pauli-Villars
regularizations.

As an interesting remark, let us now look at a global chiral rotation of both
fermionic (5) and bosonic fields,

$\displaystyle\bar{\varphi}\rightarrow\bar{\varphi}e^{i\alpha\gamma_{5}},\ \
\varphi\rightarrow e^{i\alpha\gamma_{5}}\varphi\,,$ (165)

for an angle $\alpha=\frac{\pi}{2}$. Jacobian of transformation for fermionic
fields is inversely equal to that of bosonic fields, therefore, the whole
measure $\mathcal{D}\bar{\psi}\psi\mathcal{D}\bar{\varphi}\varphi$ remains
preserved. But the same is not true for the action. The effect of this
transformation on the action is that the mass term obtains a minus sign. Thus,

$\displaystyle-\delta\ln\mathcal{I}$
$\displaystyle=\sum_{n}\ln\frac{\ell_{n}}{\ell_{n}+iM}-\sum_{n}\ln\frac{\ell_{n}}{\ell_{n}-iM}=2i\operatorname{Im}\mathcal{I}$
$\displaystyle=\sum_{n}\ln\frac{\ell_{n}-iM}{\ell_{n}+iM}\approx-i\pi\left(\sum_{\ell_{n}>0}1-\sum_{\ell_{n}<0}1\right)\,,$
(166)

for the large $M$ limit. But if there were no bosonic fields, the Jacobian of
fermionic measure (17), would have contributed in the exact same way according
to (15). Hence, $-\frac{2}{\pi}\operatorname{Im}\mathcal{I}$ is the
corresponding path-integral version of the index (15). Notice that it is
$\ell_{n}$, not eigenvalues of an externally given operator, that appears in
the index theorem, and thus in the anomalous term as well.

We saw that our regularization indeed comes down to well-known forms at
special cases, however, $\ell_{n}$, when isolated, is only well-defined for
certain bases, which our generic basis is assumed to be one of them. From a
different perspective, therefore, the question of finding a basis that makes
definition (159) realizable, leads to the set of path-integral elected bases.
We could have started by this question instead, namely, what basis is
preferred by the path-integral. As an answer we can ask for a basis
$\\{\bar{b}_{n},a_{n}\\}$ that turns the unregularized fermionic path-integral
into products of separate Grassmann integrations,

$\int\mathcal{D}\bar{\psi}\mathcal{D}\psi
e^{S[\bar{\psi},\psi]}\rightarrow\prod_{n}\left(\int
d\bar{b}_{n}da_{n}F_{n}\left[\bar{b}_{n},a_{n}\right]\right)\,,$ (167)

with $\mathcal{F}=\mathcal{F}_{D}\equiv\prod_{n}F_{n}$ so that (167) is a mere
rearrangement. Since each term after integration will be a complex number,
there exists a natural regularization:

$\int\mathcal{D}\bar{\psi}\mathcal{D}\psi
e^{S[\bar{\psi},\psi]}\equiv\prod_{n}f_{M}\left(\int
d\bar{b}_{n}da_{n}F_{n}\left[\bar{b}_{n},a_{n}\right]\right)\,,$ (168)

with $f_{M}$ defined as before. Again, this leads to a well-defined path-
integral.

Finding such a basis is subject to solving the following “eigenvalue”
equation,

$\lim_{K\to\infty}\left(\int
d\bar{b}_{n}da_{n}\left[\mathcal{F}_{\Lambda_{K}}\right]-\ell_{n}\mathcal{F}_{\Lambda_{K-1}}\right)=0\,,$
(169)

where $\mathcal{F}_{\Lambda_{K}}$ is defined as (158) when variable $k$ spans
from $0$ to $K$. $\Lambda_{K}$ is the space spanned by $k$ pairs of Grassmann
variables, and $\Lambda_{K-1}$ is the subspace of $\Lambda_{K}$ with
$\bar{b}_{n}$ and $a_{n}$ excluded. At certain times, the above equation
reduces to

$(c_{n}+\bar{b}_{n}a_{n})\int
d\bar{b}_{n}da_{n}\left[\mathcal{F}\right]=\ell_{n}\mathcal{F}\,.$ (170)

It is then easy to see that we can equivalently write down the following path-
integral

$\int\mathcal{D}\bar{\psi}\mathcal{D}\psi
e^{S}\rightarrow\int\prod_{n}d\bar{b}_{n}da_{n}\exp\left\\{\sum_{n}\ell_{n}\bar{b}_{n}a_{n}\right\\}\,,$
(171)

where we can say that the action is formally diagonalized in the chosen basis.

As a conclusion we can say that action determines what bases are preferable
for the path-integral to be regularized with respect to. This means that the
measure, which is regularized in the same manner, is affected by the action,
or the weight, of the fermionic path-integral. Therefore, adding interaction
terms to the action does not necessarily leave the measure, and the anomaly
that comes from it, untouched. At last, we saw that a way to correctly see how
the measure is affected is to move to a basis that formally diagonalizes the
action.