Why is the path integral action not real in most condensed matter QFT?

Question

The partition function in statistical physics $Z=Tr\exp(-\beta H)$, with the second quantized Hamiltonian like $$ H=\sum \epsilon_k c_k^\dagger c_k +...$$

can be represented by using the path-integral formalism as

$$ Z=\int D[c^*,c]exp[-S] $$

Here $$ S=\sum_k \int_0^\beta c_k^*(\tau)(\frac{\partial}{\partial \tau}+\epsilon_k)c_k(\tau) + ... $$ is the action functional of the Grassmann field (Fermion) of complex field (boson) $c_k(\tau)$.

But now you see that the partial derivative term is purely imaginary, since $$ \{\int c_k^*(\tau)\frac{\partial}{\partial \tau}c_k(\tau) \}^* =\int \{\frac{\partial}{\partial \tau}c_k(\tau) \}^* c_k(\tau)=-\int c_k^*(\tau)\frac{\partial}{\partial \tau}c_k(\tau)$$

My question is how and why dese the action become complex ?

Answer 1

I think the more accurate question is "why is this term in this imaginary-time path-integral imaginary?"

Actually, in path-integral formalism (imaginary time), there is no generic reason for an action to be real. In fact, you can same that's exactly one feature of quantum mechanics: the amplitude of a path can be complex. In real time path-integral, the action part is real is demanded by unitarity. However, some term can become imaginary after doing Wick rotation. The easiest example is probably the Berry phase term by external magnetic field for single-particle:

$$ \exp{\left( i \int \mathrm{d}t \, \left[ \frac{\mathrm{d}x}{\mathrm{d}t} \cdot A \right]\right)} ~~=~~\exp{\left[- \int \mathrm{d}\tau \, \left(-i\right) \frac{\mathrm{d}x}{\mathrm{d}\tau} \cdot A \right]} $$

The answer to your question (for simplicity, let me take single boson): $\int \mathrm{d}\tau \, b^{*} \frac{\mathrm{d}}{\mathrm{d} \tau} b$ is actually also a Berry phase term when one does coherent state path integral. These kinds of Berry phase term is not from some external gauge field, but from using some over-complete basis.

A heuristic way to see it: the term $\int \mathrm{d}\tau \, b^{*} \frac{\mathrm{d}}{\mathrm{d} \tau} b$ came from $\int \mathrm{d}t \, i b^{*} \frac{\mathrm{d}}{\mathrm{d} t} b$ before doing Wick rotation, which came from Legendre transition from Hamiltonian to Lagrangian( because $\left[i b^{\dagger},b\right]=i$ and can be viewed as the conjugate momentum of each others). The corresponding term in the single-particle path integral analogy is$$ \exp{\left(i \int \mathrm{d}t \, p \frac{\mathrm{d}q}{\mathrm{d} t} \right)} ~~=~~\exp{\left(- \int \mathrm{d}\tau \, \left[ -i p \frac{\mathrm{d}q}{\mathrm{d} \tau}\right]\right)} \,,$$ which is also imaginary.

The source of this term from canonical quantization ($Z=Tr(e^{-i H t })$) to path integral ($Z=\int Dq \, e^{i S[q] })$) is by pluging in coherent states $|\alpha \rangle=|x,p\rangle$ as the over-complete basis,$$ \int \mathrm{d}x \, \mathrm{d}p \, \left|x,p \right> \, \left< x, p \right| ~\propto~ I \,,$$and the non-orthogonality between $|x,p\rangle$ and $|x',p'\rangle$ gives $ \exp{\left(i \int \mathrm{d}t \, p \, \frac{\mathrm{d}q}{\mathrm{d} t}\right)}.$ Exactly the same thing can be done for the field theory you wrote.

Note that this derivation is different from the derivation in the elementary quantum mechanics textbook. It's a fun practice. In case you encounter difficulty, a random source I just found by google is a note by Fradkin.

Answer 2

Your formulation is based on the imaginary time path integral. The complex conjugation of $\frac{\partial}{\partial \tau}$, $$ \left(\frac{\partial}{\partial \tau}\right)^\star = -\frac{\partial}{\partial \tau} $$ gives you an extra minus sign. Therefore the action functional is Hermitian.