4
$\begingroup$

Going through my notes I stumbled upon something I can't wrap my head around.

I'd like to write the grand canonical partition function for a system of identical charged particles (charge $e$) interacting with external e.m. fields. The scalar potential enters the Hamiltonian in $$\hat H = \ldots + \int d\vec x \,\, e\phi(x,t) \, \psi^\dagger(\vec x) \psi(\vec x) \ $$ So, in coherent-states representation I'd write the partition function as $$ Z=Tr\, e^{-\beta( H-\mu N) }= \int \mathcal D (\psi^*,\psi)\, e^{-S}= \int \mathcal D (\psi^*,\psi) e^{\int dtd\vec x \,\,\psi^* (\partial_t -\mu +e\phi) \psi + \ldots } \qquad [1] $$

This is where my problem begins. Both my lectures notes and the books I've checked ( Nagaosa, QFT in cond. mat. phys., eq 5.1.38 ; Altland&Simons, cond. mat. field theory, eq. 6.25) are instead writing $i e \phi\, \psi^\dagger \psi $ in the action .
Altland&Simons motivates the extra $i$ saying that the partition function comes from the Wick-rotation of the "minkowsian" functional integral, through which the zeroth component of the gauge field aquires an $i$.

My problem with this reasoning is that one can obtain the functional integral [1] directly from the trace, simply inserting completeness relations for coherent states and going to the continuum limit, without adding any $i$s.

What am I missing?

$\endgroup$
  • $\begingroup$ Hint: you want $\hat H$ to be hermitian; but, due to Fermi statistics, $(\psi^\dagger\psi)^\dagger=-(\psi^\dagger\psi)$. $\endgroup$ – AccidentalFourierTransform Apr 28 '18 at 15:04
  • $\begingroup$ I think that you are missing is the fact that the action S appearing in the exponent is the Euclidean action, not the plane action one encounters in classical mechanics. The Euclidean action is precisely the plane action but after Wick rotation, i.e. in imaginary times. The time integral, thus, acquires an overall factor of i and the kinetic term changes sign. Eventually, the Euclidean action is (almost) the Hamiltonian of the system. $\endgroup$ – Panos C. Apr 28 '18 at 15:33
  • $\begingroup$ @PanosC. : I called action the exponent that appears in the functional integral [1], following the nomenclature of many books. It is understood that it is "euclidean" in this context, sorry for the confusion. I addressed in the question that one doesn't need a Wick rotation to obtain [1] (i refer to the Negele-Orland book eq 2.66 for a direct derivation, for example, or Altland-Simons eq 4.23 and following) $\endgroup$ – tbt Apr 28 '18 at 16:30
  • 1
    $\begingroup$ @AccidentalFourierTransform if that were the case, the particle number wouldn't be hermitian too, and we'd need an i also for the chemical potential, but it doesn't appear there... $\endgroup$ – tbt Apr 28 '18 at 16:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.