# Euclidean correlators for finite chemical potential

1. Feb 23, 2016

### Einj

Hello everyone,
my question is about Euclidean correlators (say a 2-pt function to be specific) in presence of non-zero chemical potential.

The question in particular is: is it still true that the Minkowski time ordered 2-pt function can be simply obtained from the Euclidean one by analytic continuation? Is this property spoiled by the presence of a chemical potential?

My confusion is mostly due to the fact that, if I'm not mistaken, a real chemical potential in Minkowski should be analytically continued to an imaginary one in Euclidean signature and I don't know if this is a problem or not.

Thanks!

2. Feb 23, 2016

### A. Neumaier

It can be a problem. Numerical stability is lost in naive lattice simulations.

3. Feb 23, 2016

### Einj

Thanks a lot for the quick answer! My question however, goes beyond lattice simulations. I don't have any lattice and everything is continuous. The questions is: is it still true that:

$$\langle O_1O_2\rangle (\tilde\omega,\vec k)\longrightarrow\langle T(O_1O_2)\rangle(\omega,\vec k)$$
when $\tilde\omega\to-i\omega$. This is all in the continuous limit. No lattice nor simulations of any sort.

Thanks!

4. Feb 23, 2016

### A. Neumaier

I can't tell off-hand but you'd probably be able to work it out yourself by writing both sides as path integrals and perform the analytic continuation explicitly.

5. Feb 24, 2016

### Einj

Thanks for your reply. I indeed checked that for an action of the kind:
$$S=-\int d^4x\left(-(\partial_t+i\mu)\Phi^*(\partial_t-i\mu)\Phi +\vec\nabla\Phi^*\cdot\vec\nabla\Phi+m^2|\Phi|^2\right)$$
the 2-pt function in the Euculidean and Minkowskian case are related by an analytic continuation.

I guess at this point my question is more general: does the fact that Euclidean correlators can be obtained from the Minkowskian one with an analytic continuation depend on what kinds of boundary conditions we are imposing on the field?
In particular, is it affected by the requirement $A_t(r=\infty)=\mu$?

Thanks again!

P.S. My question clearly has AdS/CFT in mind.

6. Feb 24, 2016

### A. Neumaier

I think it shoulldn't make a difference but I am not an expert on this.