Gauge transformation with complex parameter in quantum mechanics with minimal substitution

Question

This follows from the question Can an Electromagnetic Gauge Transformation be Imaginary?

It's about the Hamiltonian $$H=\frac{(p-A)^2}{2m}$$ in units where $c=e=\hbar=1$. The question regarded a gauge transformation $$\Psi\rightarrow e^{i\chi}$$ where $\chi$ is defined through $$A\rightarrow A-\nabla \chi$$ with $\chi$ being complex. Obviously, $A$ is also complex in this case. A physical motivation for this is to study systems with dissipation. For example, see this paper by Nelson and Hatano: https://arxiv.org/abs/cond-mat/9705290

As pointed out in the answer https://physics.stackexchange.com/a/21608/75628, when $\chi$ is complex, then the gauge group is no longer $U(1)$. Such gauge transformations might be relevant to the field of non-Hermitian matter (or associated continuum theories).

My question is what's the correct gauge transformation in this case? More generally, what's the criterion for finding what the gauge transformation is.

Upon thinking about it, my criterion has been that observable quantities such as $\Psi^* \Psi$ should remain invariant under the gauge transformation. While for real $\chi$ this quantity is invariant if $\Psi^*\rightarrow (e^{i\chi})^\dagger$ ($\dagger$ is put here in order to generalize more easily to $SU(N)$ gauge groups), for complex $\chi$ it is invariant when $\Psi^*\rightarrow (e^{i\chi})^{-1}$ so that the real part of $\chi$ would correspond to the previous $\dagger$ and the imaginary part of it would correspond to a real exponential which would be the inverse of the real exponential coming from the gauge transformation of $\Psi$.

Now, while I'm no expert in group theory in any way, I can see that if the above reasoning is correct, the gauge group is now not compact (as is the case with $SU(N)$). So, is the above reasoning correct? If it is, how do we reconcile $\Psi^*\rightarrow (e^{i\chi})^{-1}$ along with $(\Psi)^* \rightarrow (e^{i\chi})^*=e^{-i\chi^*}$? In general, how do we find the local symmetry transformation of such a theory? What changes if the parameters of an $SU(N)$ group are complex (with the standard convention in physics being to define generators so that the parameters are real in the standard case)?

Answer 1

I'll try to answer a portion of my question. It's incomplete, but maybe somebody else can try to contribute to this community's knowledge by completing it.

Upon further researching the topic, I've found that in the case where imaginary fields are included, the energy might become complex, so we might have to use the so-called biorthogonal quantum mechanics, in which we use left and right Hamiltonian eigenstates, $| \Psi_L\rangle, |\Psi_R \rangle$ defined via $$H|\Psi_{nR}\rangle=E_n\Psi_{nR}\rangle, \ \ H|\Psi_{nL}\rangle=E_n^*\Psi_{nL}\rangle$$ for energy eigenstates. Following the paper by Shen, Zhen, Fu named "Topological Band theory for non-Hermitian Hamiltonians" (https://arxiv.org/pdf/1706.07435.pdf), we see in section II of the supplamental material, the gauge transformation $$|\Psi_R\rangle \rightarrow r(k)e^{if(k)}|\Psi_R\rangle$$ $$|\Psi_L\rangle \rightarrow \frac{1}{r(k)}e^{-if(k)}|\Psi_L\rangle$$ Here, $r(k), f(k)$ are real and continuous (and taken to be periodic in this paper). It is my understanding that within the framework of biorthogonal quantum mechanics, $\langle\Psi_L|$ is what replaces $|\Psi\rangle$ (so $\Psi^*(x)$ in position representation) in conventional quantum mechanics. For example, for operator $\mathcal{O}$, $\langle \mathcal{O} \rangle=\langle \Psi_L|\mathcal{O}|\Psi_R\rangle=\int d^Dr \Psi_L^*(r)\ \mathcal{O}\ \Psi_R(r)$. Hence the above gauge transformation for left and right eigenstates do indeed correspond to what I suspected to be the gauge transformation $\Psi^*\rightarrow (e^{i\chi})^{-1}$ in the context of conventional quantum mechanics.

Hence, I believe my reasoning is correct. The gauge group is no longer $U(1)$ but the group of all complex numbers with non-zero and non-singular modulus. The questions I put in bold in my original question still remain. Any corrections will be appreciated.