4 Evaluation of the vacuum energy and of its imaginary part

To compute the vacuum energy, apart from more standard contributions, we have to compute the following integral

$$I _{0} ( k , \bar{k} ; \kappa , m _{\mathrm{A}}) \equiv \in... ...\kappa \bar{k} ^{2}}{k ^{2} - m _{\mathrm{A}} ^{2}} \right) .$$ (26)

It can be evaluated in closed form and we first rewrite it as

$$I _{0} ( k , \bar{k} ; \kappa , m _{\mathrm{A}} ) = I ( k ,... ...{(2\pi)^4} \ln \left( k ^{2} - m _{\mathrm{A}} ^{2} \right) ,$$ (27)

so that we can separate the second common contribution, from the first one, i.e.

$$I ( k , \bar{k} ; \kappa , m _{\mathrm{A}} ) = \int \frac{... ...^{2} - m _{\mathrm{A}} ^{2} ) - \kappa \bar{k} ^{2} \right] .$$ (28)

Using (27) for the right hand side of (26) appearing in (24), we get for the free energy

$$W = \frac{V T}{2} I ( k , \bar{k} ; \kappa , m _{\mathrm{A}} ) + V T \int \frac{d^4k}{(2\pi)^4} \ln \left( k ^{2} \right).$$ (29)

The computation of $I ( k , \bar{k} ; \kappa , m _{\mathrm{A}} )$ is performed in appendix C. The infrared behavior is then extracted in appendix C.1. Ultraviolet divergences are regularized by a cut-off $\Lambda$ and the leading contributions to the free energy are computed in appendix C.2. Thus the final result for the vacuum energy density is obtained multiplying (61) by $1 / ( 4 \pi ^{2} )$ (because of (37)) and substituting for $I ( k , \bar{k} ; \kappa , m _{\mathrm{A}} )$ in (29), where the second contribution can also be exactly evaluated. The final result is

$$W = \frac{VT}{8 \pi ^{2}} \left [ I _{\Lambda} ^{(4)} \Lamb... ...} \Lambda ^{2} + I ^{(0)} + I ^{(0)} _{\ln \Lambda} \right ],$$ (30)

where the various contributions are defined at the end of appendix C. We will be especially interested in $V T I ^{(0)} / (8 \pi ^{2})$, which, according to equation (62), is

$$I ^{(0)} = \frac{VT}{8 \pi ^{2}} \biggl [ \frac{49 \kappa ^{2} + 132 \kappa ... ...appa ^{2} + 3 \kappa m _{\mathrm{A}} ^{2} + 3 m _{\mathrm{A}} ^{4}}{48} \ln 2 -$$

$$\hphantom{\frac{VT}{8 \pi ^{2}} \biggl [} - \frac{( \kappa + m _{... ...{A}} ^{2} ) ^{3}}{24 \kappa} \ln \left( m _{\mathrm{A}} ^{2} \right) \biggr]\,.$$ (31)

From this expression we see that the vacuum energy acquires an imaginary part in the case $\kappa + m _{\mathrm{A}} ^{2} < 0$, i.e. when the tachyonic modes are present. If we make use of the prescription $\kappa + m _{\mathrm{A}} ^{2}$ $\to$ $\kappa + m _{\mathrm{A}} ^{2} + \imath \epsilon$, the value of the imaginary part is

$$- \pi \frac{V T}{ 8 \pi ^{2}} \frac{\left( \kappa + m _{\mathrm{A}} ^{2} \right) ^{3}}{24 \kappa} \,.$$ (32)

Notice that, as opposed to the real part, the imaginary part of the vacuum energy is cut-off independent. This is so because only the infrared region of the integrand in (28) contributes to the imaginary part.

We also observe that all contributions in (30) are finite in the limit of vanishing external field, i.e. when $\kappa \to 0$. This is immediately evident for the terms $I _{\Lambda} ^{(4)}$, $I _{\ln \Lambda} ^{(4)}$, $I _{\Lambda} ^{(2)}$, $I _{\ln \Lambda} ^{(2)}$ and $I _{\ln \Lambda} ^{(0)}$, from their expressions at the end of appendix C. Concerning the contribution $I ^{(0)}$, the two divergent terms of opposite sign in the second line of (31) give a finite contribution in the limit, as shown in (48).

The cut-off dependent parts are regularization dependent, as it is known from general experience with the regularization of divergent integrals. In particular subleading divergences (i.e. the $\Lambda ^{2}$ terms in (30)) are highly dependent upon the regularization scheme and may vanish in certain of them. In the present problem, due to the complexity of the integrand in (28), other regularization schemes, like $\zeta$-function regularization or dimensional regularization, are difficult to implement.