0
$\begingroup$

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)?

$\endgroup$
5
  • 1
    $\begingroup$ I don't understand what this question is trying to do. If you want $\chi$ to be complex, then suddenly $A$ is complex, too (since otherwise the sum $A+\mathrm{d}\chi$ does not make sense), and there's various other things you then have to clarify in order to make this into a consistent physical theory. Just assuming $\chi$ is complex doesn't make a consistent theory - yet this question is so specifically about a gauge transformation as if it was unambiguous how to setup this new theory. What problem exactly are you trying to solve by assuming $\chi$ is complex? $\endgroup$
    – ACuriousMind
    Sep 16, 2023 at 14:05
  • $\begingroup$ @ACuriousMind, I'll explain in the case of $\chi$ being imaginary. There, $A$ should be imaginary as well. In the usual gauge transformation, where $\chi, A$ are real, $\Psi, \Psi^*$ transform in a way as to keep $\rho=\Psi^* \Psi$ invariant. If we use the same rule for the case of imaginary $\chi$, $\rho$ is not. The gauge transformation should change to the one mentioned in my second to last paragraph. I guess I want to know whether my reasoning is correct and what is the general reasoning behind gauge transformations of $SU(N)$ gauge theories when the parameters are complex. $\endgroup$ Sep 16, 2023 at 14:14
  • 1
    $\begingroup$ And my point is: Why do you think you can just assume a complex $A$ (note that the Hamiltonian is no longer real even classically!)? Why would the parameters of an SU(N) theory be complex? When we expand $A = A^a T^a$ the $T^a$ are always generators of the Lie algebra as a real Lie algebra, even if the algebra carries a complex structure. You simply have underspecified the theory you're talking about. I could just interpret a complex $A$ as two real $A_i$ $A_1 + iA_2$ transforming via $A_i\mapsto A_i + \mathrm{d}\chi_i$ and then state both these act via $\mathrm{e}^{iA_i}$ on wavefunctions $\endgroup$
    – ACuriousMind
    Sep 16, 2023 at 14:20
  • $\begingroup$ i.e. I think this question is an XY problem: You have some underlying problem and you've convinced yourself that somehow a "complex $\chi$" is a solution to that problem, but the misunderstanding lies deeper in that underlying part that you've omitted from the question $\endgroup$
    – ACuriousMind
    Sep 16, 2023 at 14:21
  • $\begingroup$ @ACuriousMind I think you misunderstood me. First, complex $A$ can be found in systems with dissipation, such as open systems (see the field of non-Hermitian topological matter). See an old model by Hatano and Nelson here: arxiv.org/abs/cond-mat/9705290 Second, $A$ is indeed of the form $Re(A)+iIm(A)$. Your interpretation is correct. The point where I think you misundestood is in that I used "complex $\chi$ as a solution". I have note. The problem is one of complex $A$ and I'm trying to figure out its gauge transformation that leaves the theory invariant. $\endgroup$ Sep 16, 2023 at 14:29

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.