Question 8.1: The QCD Vacuum Energy Density Derive (8.31) from (8.30)....
The QCD Vacuum Energy Density
Derive (8.31) from (8.30).
\begin{aligned} \varepsilon-\varepsilon(H=0)= & \underset{\eta, \delta \rightarrow 0}{\lim} \left[\frac{g H}{2 \pi^2} \frac{\mathrm{i}^{-1 / 2+\eta}}{\Gamma\left(-\frac{1}{2}+\eta\right)} \int\limits_0^{\infty} \mathrm{d} k \int\limits_0^{\infty} \mathrm{d} \tau \tau^{-3 / 2+\eta}\right. \\ & \times \mathrm{e}^{-\mathrm{i} \tau\left(k^2-\mathrm{i} \delta\right)}\left\{\sum\limits_{n=1}^{\infty} \mathrm{e}^{-\mathrm{i} \tau 2 g H\left(n-\frac{1}{2}\right)}+\sum\limits_{n=0}^{\infty} \mathrm{e}^{-\mathrm{i} \tau 2 g H\left(n+\frac{3}{2}\right)}\right\} \\ & -\frac{\mathrm{i}^{-\frac{1}{2}+\eta}}{\Gamma\left(-\frac{1}{2}+\eta\right)} \frac{1}{\pi^2} \int\limits_0^{\infty} \mathrm{d} k \int\limits_0^{\infty} \mathrm{d} \tau k^2 \tau^{-\frac{3}{2}+\eta} \mathrm{e}^{-\mathrm{i} \tau\left(k^2-\mathrm{i} \delta\right)} \\ & \left.+\frac{g H \mathrm{i}^{-\frac{1}{2}+\eta}}{2 \pi^2 \Gamma\left(-\frac{1}{2}+\eta\right)} \int\limits_0^{\infty} \mathrm{d} k \int\limits_0^{\infty} \mathrm{d} \tau \tau^{-\frac{3}{2}+\eta} \mathrm{e}^{-\mathrm{i} \tau\left(k^2-\mathrm{i} \delta-g H\right)}\right] . &(8.30) \end{aligned}
\varepsilon-\varepsilon(H=0)=\frac{11(g H)^2}{48 \pi^2} \ln \frac{g H}{\mu^2}-\mathrm{i} \frac{(g H)^2}{8 \pi} . (8.31)
The sums in (8.30) can be directly evaluated:
\begin{aligned} \sum\limits\limits\limits\limits\limits\limits\limits_{n=1}^{\infty} & \mathrm{e}^{-\mathrm{i} \tau 2 g H\left(n-\frac{1}{2}\right)}+\sum\limits\limits\limits\limits\limits\limits\limits_{n=0}^{\infty} \mathrm{e}^{-\mathrm{i} \tau 2 g H\left(n+\frac{3}{2}\right)} \\ & =\frac{\mathrm{e}^{\mathrm{i} \tau g H}}{1-\mathrm{e}^{-\mathrm{i} \tau 2 g H}}-\mathrm{e}^{\mathrm{i} \tau g H}+\frac{\mathrm{e}^{-3 \mathrm{i} \tau g H}}{1-\mathrm{e}^{-2 \mathrm{i} \tau g H}} \\ & =\frac{\mathrm{e}^{2 \mathrm{i} \tau g H}+\mathrm{e}^{-2 \mathrm{i} \tau g H}}{\mathrm{e}^{\mathrm{i} \tau g H}-\mathrm{e}^{-\mathrm{i} \tau g H}}-\mathrm{e}^{\mathrm{i} \tau g H}=\frac{\cos (2 g H \tau)}{\mathrm{i} \sin (g H \tau)}-\mathrm{e}^{\mathrm{i} \tau g H} . &(1) \end{aligned}
Also the k integrations can be directly performed:
\begin{aligned} & \int\limits_{0}^{\infty} \mathrm{d} k \mathrm{e}^{-\mathrm{i} \tau k^{2}}=\frac{1}{2} \sqrt{\frac{\pi}{\tau}} \mathrm{e}^{-\mathrm{i} \frac{\pi}{4}} \\ & \int\limits_{0}^{\infty} \mathrm{d} k k^{2} \mathrm{e}^{-\mathrm{i} \tau k^{2}}=\mathrm{i} \frac{\partial}{\partial \tau} \int\limits_{0}^{\infty} \mathrm{d} k \mathrm{e}^{-\mathrm{i} \tau k^{2}}=-\frac{\mathrm{i}}{4} \frac{\sqrt{\pi}}{\tau^{\frac{3}{2}}} \mathrm{e}^{-\mathrm{i} \frac{\pi}{4}} . & (2) \end{aligned}
Therefore we get
\begin{aligned} \varepsilon-\varepsilon(H=0) & \\ = & \underset{\eta, \delta \rightarrow 0}{\lim}\left[\frac{g H}{4 \pi^{\frac{3}{2}}} \frac{\mathrm{e}^{-\mathrm{i} \frac{\pi}{2}+\mathrm{i} \frac{\pi}{2} \eta}}{\Gamma\left(-\frac{1}{2}+\eta\right)} \int\limits_{0}^{\infty} \mathrm{d} \tau \tau^{-2+\eta} \mathrm{e}^{-\delta \tau}\left(\frac{\cos (2 g H \tau)}{\mathrm{i} \sin (g H \tau)}-\mathrm{e}^{\mathrm{i} \tau g H}\right)\right. \\ & +\frac{\mathrm{i}}{4} \frac{\mathrm{e}^{-\mathrm{i} \frac{\pi}{2}+\mathrm{i} \frac{\pi}{2} \eta}}{\pi^{\frac{3}{2}} \Gamma\left(-\frac{1}{2}+\eta\right)} \int\limits_{0}^{\infty} \mathrm{d} \tau \tau^{-3+\eta} \mathrm{e}^{-\delta \tau} \\ & \left.+\frac{g H}{4} \frac{\mathrm{e}^{-\mathrm{i} \frac{\pi}{2}+\mathrm{i} \frac{\pi}{2} \eta}}{\pi^{\frac{3}{2}} \Gamma\left(-\frac{1}{2}+\eta\right)} \int\limits_{0}^{\infty} \mathrm{d} \tau \tau^{-2+\eta} \mathrm{e}^{\mathrm{i} \tau g H-\delta \tau}\right] . & (3) \end{aligned}
Now we employ the relation
\cos (2 g H \tau)=1-2 \sin ^{2}(g H \tau), (4)
which yields
\begin{aligned} & \frac{\cos (2 g H \tau)}{\mathrm{i} \sin (g H \tau)}-\mathrm{e}^{\mathrm{i} \tau g H} \\ & \quad=\frac{1}{\mathrm{i} \sin (g H \tau)}-\frac{2}{\mathrm{i}} \frac{1}{2 \mathrm{i}}\left(\mathrm{e}^{\mathrm{i} \tau g H}-\mathrm{e}^{-\mathrm{i} \tau g H}\right)-\mathrm{e}^{\mathrm{i} \tau g H} \\ & \quad=\frac{1}{\mathrm{i} \sin (g H \tau)}-\mathrm{e}^{-\mathrm{i} \tau g H} \\ & \quad \rightarrow \frac{1}{\mathrm{i} \sin (g H \tau-\mathrm{i} \tilde{\eta})}-\mathrm{e}^{-\mathrm{i} \tau g H}. & (5) \end{aligned}
Here we have defined the singularities by inserting i \tilde{\eta}(\tilde{\eta}>0). The integral exists provided \eta>2. We calculate it in this region and determine its value for \eta \rightarrow 0 by analytic continuation. This very procedure has already been used in dimensional regularization. Identity (5) ensures that the integrand vanishes for \tau=R \mathrm{e}^{\mathrm{i} \phi},-\frac{\pi}{2}<\phi<0, R \rightarrow \infty. Therefore the integral surrounding the fourth quadrant vanishes:
\int\limits \mathrm{d} \tau \cdots=\int\limits_{0}^{\infty} \mathrm{d} \tau \cdots+\int\limits_{-\mathrm{i} \infty}^{0} \mathrm{~d} \tau \cdots=0, (6)
Since the integrand has no singularities, we substitute \tau=-\mathrm{i} s and evaluate the last integral with the help of (8.29):
\left(\omega^2-\mathrm{i} \delta\right)^{\nu-\mu}=\frac{\mathrm{i}^{-\nu+\eta}}{\Gamma(-\nu+\eta)} \int\limits_0^{\infty} \mathrm{d} \tau \tau^{-\nu+\eta-1} \mathrm{c}^{-\mathrm{i} \tau\left(\omega^2-\mathrm{i} \delta\right)} (8.29)
\begin{aligned} \varepsilon-\varepsilon(H=0) & =\underset{\eta, \delta \rightarrow 0}{\lim}\left[\frac{g H}{4 \pi^{\frac{3}{2}}} \frac{-\mathrm{i} \mathrm{e}^{\mathrm{i} \frac{\pi}{2} \eta}}{\Gamma\left(-\frac{1}{2}+\eta\right)}(-\mathrm{i})^{-1+\eta}\right. \\ & \times \int\limits_{0}^{\infty} \mathrm{d} s s^{-2+\eta} \mathrm{e}^{-\delta r}\left(\frac{1}{\sinh (g H s)}-\mathrm{e}^{-s g H}\right) \\ & +\frac{\mathrm{e}^{\mathrm{i} \frac{\pi}{2} \eta}(-\mathrm{i})^{-2+\eta}}{4 \pi^{\frac{3}{2}} \Gamma\left(-\frac{1}{2}+\eta\right)} \int\limits_{0}^{\infty} \mathrm{d} s s^{-3+\eta} \frac{g H}{4} \frac{\mathrm{e}^{-\mathrm{i} \frac{\pi}{2}+\mathrm{i} \frac{\pi}{2} \eta}}{\pi^{\frac{3}{2}} \Gamma\left(-\frac{1}{2}+\eta\right)} \\ & \left. \times \frac{(-1+\eta)}{\left.\mathrm{e}^{\mathrm{i} \frac{\pi}{2}(-1+\eta)}\right.} (-g H+\mathrm{i} \delta)^{1-\eta} \right] . & (7) \end{aligned}
Finally we substitute s=v / g H :
\begin{array}{l} \varepsilon-\varepsilon(H=0)=\underset{\eta \rightarrow 0}{\lim} \frac{(g H)^{2-\eta}}{4 \pi^{\frac{3}{2}}} \frac{1}{\Gamma\left(-\frac{1}{2}+\eta\right)} \\ \quad \times\left[\int\limits_{0}^{\infty} \mathrm{d} v v^{-2+\eta}\left(\frac{1}{\sinh (v)}-\mathrm{e}^{-v}-\frac{1}{v}\right)-(-1+\eta)(-)^{n}\right] . & (8) \end{array}
The integral of \mathrm{e}^{-v} leads to a gamma function:
\int\limits_{0}^{\infty} \mathrm{d} v v^{-2+\eta} \mathrm{e}^{-v}=\Gamma(-1+\eta) (9)
while the integral over 1 / \sinh v-1 / v can be split into a finite part and one that diverges in the limit \eta \rightarrow 0 :
\begin{aligned} \int\limits_{0}^{\infty} \mathrm{d} v & v^{-2+\eta}\left(\frac{1}{\sinh (v)}-\frac{1}{v}\right) \\ & =\int\limits_{0}^{\infty} \mathrm{d} v v^{-2+\eta}\left(\frac{1}{\sinh (v)}-\frac{1}{v}+\frac{1}{6} v\right)-\frac{1}{6} \int\limits_{0}^{\infty} \mathrm{d} v v^{-1+\eta} \\ & =C-\frac{1}{6} \frac{1}{\eta}, & (10)\\ \varepsilon & -\varepsilon(H=0)=\underset{\eta \rightarrow 0}{\lim} \frac{(g H)^{2-\eta}}{4 \pi^{\frac{3}{2}} \Gamma\left(-\frac{1}{2}+\eta\right)} \\ & \times\left\{C-\frac{1}{6} \frac{1}{\eta}-\Gamma\left[(-1+\eta)\left(1+(-)^{\eta}\right)\right]\right\} . & (11) \end{aligned}
With
\Gamma(-1+\eta)=\frac{1}{-1+\eta} \Gamma(\eta)=\frac{\Gamma(1+\eta)}{(-1+\eta) \eta} (12)
we are now able to calculate the limiting case \eta \rightarrow 0 :
\begin{aligned} (g H)^{2-\eta} & =(g H)^{2}(1-\eta \ln (g H)+\cdots), \\ \Gamma\left(-\frac{1}{2}+\eta\right) & =\Gamma\left(-\frac{1}{2}\right)\left[1+\eta \Psi\left(-\frac{1}{2}\right)\right]+\cdots \\ & =-2 \sqrt{\pi}\left[1+\eta \Psi\left(-\frac{1}{2}\right)\right]+\cdots, & (13) \end{aligned}
\begin{aligned} \Gamma(-1+\eta) & =\frac{-1-\eta}{\eta}[1+\eta \Psi(1)]+\cdots, \\ (-)^{\eta} & =1+\eta \mathrm{i} \pi \\ \Rightarrow \varepsilon-\varepsilon(H=0) & =\underset{\eta \rightarrow 0}{\lim}\left\{\frac{(g H)^{2}}{8 \pi^{2} \eta}+\frac{(g H)^{2}}{8 \pi^{2}}\left[\frac{11}{6} \ln (g H)+C^{\prime}-\mathrm{i} \pi\right]\right\} . & (14) \end{aligned}
Here all constants have been absorbed into C^{\prime}, which does not depend on H. The divergent first part can be renormalized. The renormalized, i.e., the physical, energy density is then
\varepsilon-\varepsilon(H=0)=\frac{11}{6} \frac{(g H)^{2}}{8 \pi^{2}} \ln (g H)+C^{\prime}-\mathrm{i} \frac{(g H)^{2}}{8 \pi} . (15)
In the transition to (8.28),
\begin{aligned} \varepsilon-\varepsilon(H & =0)=\underset{\eta, \delta \rightarrow 0}{\lim} \left[\frac { g H } { \pi } \int\limits\limits _ { 0 } ^ { \infty } \frac { \mathrm { d } k _ { 3 } } { 2 \pi } \left\{\sum\limits_{n=1}^{\infty}\left[2 g H\left(n-\frac{1}{2}\right)+k_3^2-\mathrm{i} \delta\right]^{1 / 2+\eta}\right.\right. \\ & \left.+\left[2 g H\left(n+\frac{3}{2}\right)+k_3^2-\mathrm{i} \delta\right]^{1 / 2-\eta}\right\} \\ & \left.-\frac{1}{\pi^2} \int\limits\limits_0^{\infty} \mathrm{d} k k^2\left(k^2-\mathrm{i} \delta\right)^{1 / 2-\eta}+\frac{g H}{2 \pi^2} \int\limits\limits_0^{\infty} \mathrm{d} k_3\left(k_3^2-g H-\mathrm{i} \delta\right)^{1 / 2-\eta}\right] \quad (8.28) \end{aligned}
however, we should have introduced a factor m^{2 \eta} in order to conserve the dimension ( m is supposed to be an energy). Then we obtain
\frac{11(g H)^{2}}{48 \pi^{2}}\left[\ln \left(\frac{g H}{m^{2}}\right)+\frac{6}{\pi} C^{\prime}\right]=\frac{11(g H)^{2}}{48 \pi^{2}} \ln \left(\frac{g H}{\mu^{2}}\right), (16)
where C^{\prime} has been absorbed by the definition of \mu
\begin{aligned} & \mu^{2}=m^{2} \mathrm{e}^{-6 C^{\prime} / 11} , & (17)\\ & \varepsilon-\varepsilon(H=0)=\frac{11(g H)^{2}}{48 \pi^{2}} \ln \left(\frac{g H}{\mu^{2}}\right)-\mathrm{i} \frac{(g H)^{2}}{8 \pi} . & (18) \end{aligned}
It should be noticed that the techniques used in this exercise to calculate (8.31) are practically the same as those introduced systematically in Sect. 4.3 in the context of dimensional regularization.