# I Determinant problem in an article about QCD phase diagram

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1. Jul 17, 2017

### Ken Gallock

Hi.
I want to derive eq(20), but I don't know how.
Does anyone know how to derive this?

Last edited by a moderator: Jul 20, 2017
2. Jul 18, 2017

### vanhees71

It just uses that the determinant of a matrix is given by the product of its eigenvalues. The eigenvalues of $D(0)$ come in pairs, $\gamma_i$ and $\gamma_i^*$. The Mass matrix is proportional to the unit matrix and thus the eigenvalues are $\gamma_i+m_q$ and $\gamma_i^*+m_q$. Do you get
$$\det (D(0)+m_q)=\prod_i (\gamma_i+m_q)(\gamma_i^*+m_q)$$
which is Eq. (20) in the paper.

3. Jul 18, 2017

### Ken Gallock

Thanks.
If there is no mass, ($m_q=0$), then will it be like this?:
$$\det D(0)=\prod_i \gamma_i \gamma_i^*.$$
I'm not familiar with 'a pair ($\gamma_i, \gamma_i^*$)' part. Why do we have to think about pair of eigenvalues?

4. Jul 18, 2017

### vanhees71

The point is to show that without baryo-chemical potential you have always pairs of conjugate omplex eigenvalues and that's why in this case the fermion determinant is positive. For finite $\mu_{\text{B}}$ it's not longer real (except for imaginary chemical potential). That's why you cannot use Lattice QCD so easily to evaluate the QCD phase diagram at finite $\mu_{\text{B}}$. Ways out of this trouble is subject of vigorous ungoing research in the nuclear-physics/finite-temperature-lattice community.

5. Jul 20, 2017

### Ken Gallock

Thanks!
Problem solved.