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The black vegetable
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- TL;DR
- I am trying to calculate one loop contribution to the vacuum energy from a fermion.
Following the method by Peskin and Shroesder 11.4 Trying to calculate the vacuum energy of a fermion. If my method is correct so far the next step is to find gamma function , the formula I have for gamma fuctions doesn't match this equation. Can anyone help with the next step?
Starting with the Lagrangian $$ L=i \bar{\Psi} \partial / \Psi-m_{e} \bar{\Psi} \Psi-\lambda \Psi \bar{\Psi} \phi $$Expanding about the classical field
$$ \Psi_{c l}+\zeta \quad \bar{\Psi}=\bar{\Psi}_{c l}+\bar{\zeta} \quad \phi \rightarrow \phi_{c l}+\rho $$
The only terms quadratic with with ##\zeta\bar{\zeta}##
$$\bar{\zeta}i \gamma^{\mu} \partial_{\mu} \zeta-m_{e} \bar{\zeta} \zeta-\lambda \bar{\zeta} \zeta\left(\phi_{c l}+\rho\right)$$When comparing this to the formula for the effective action this coincides with
$$ \left[-\frac{\delta^{2} L_{1}}{\delta\bar{\Psi}(x) \delta\Psi(y)}\right]=i \gamma^{\mu} \partial_{\mu}-m_{e}-\lambda\left(\phi_{c l}+\rho\right)=i \gamma^{\mu} \partial_{\mu}-M_{e} $$
In Peskin and Schroder P374 they are doing this with a scalable field Lagrangian, where they get the Klein Gordon operator instead of the dirac operator. If I follow the method the next stage is to find the Gamma function for
$$\operatorname{Tr} \log \left(\gamma^{\mu} \partial_{\mu}+m\right)=\sum_{p} \log \left(\gamma^{\mu} p_{\mu}+m\right)$$
Where after a wicks rotation they get something similar to this but for a scaler field.
$$=V T \int \frac{d^{4} p}{(2 \pi)^{4}} \log \left(\gamma^{\mu} p_{\mu}+m\right)=V T \frac{\partial}{\partial a} \int \frac{d^{4} p}{(2 \pi)^{4}} \frac{1}{\left(\gamma^{\mu} p_{\mu}+m\right)^{a}}|_{a=0}$$
How do I find the gamma function for this, it doesn't fit my equation?
Many thanks for your time
Starting with the Lagrangian $$ L=i \bar{\Psi} \partial / \Psi-m_{e} \bar{\Psi} \Psi-\lambda \Psi \bar{\Psi} \phi $$Expanding about the classical field
$$ \Psi_{c l}+\zeta \quad \bar{\Psi}=\bar{\Psi}_{c l}+\bar{\zeta} \quad \phi \rightarrow \phi_{c l}+\rho $$
The only terms quadratic with with ##\zeta\bar{\zeta}##
$$\bar{\zeta}i \gamma^{\mu} \partial_{\mu} \zeta-m_{e} \bar{\zeta} \zeta-\lambda \bar{\zeta} \zeta\left(\phi_{c l}+\rho\right)$$When comparing this to the formula for the effective action this coincides with
$$ \left[-\frac{\delta^{2} L_{1}}{\delta\bar{\Psi}(x) \delta\Psi(y)}\right]=i \gamma^{\mu} \partial_{\mu}-m_{e}-\lambda\left(\phi_{c l}+\rho\right)=i \gamma^{\mu} \partial_{\mu}-M_{e} $$
In Peskin and Schroder P374 they are doing this with a scalable field Lagrangian, where they get the Klein Gordon operator instead of the dirac operator. If I follow the method the next stage is to find the Gamma function for
$$\operatorname{Tr} \log \left(\gamma^{\mu} \partial_{\mu}+m\right)=\sum_{p} \log \left(\gamma^{\mu} p_{\mu}+m\right)$$
Where after a wicks rotation they get something similar to this but for a scaler field.
$$=V T \int \frac{d^{4} p}{(2 \pi)^{4}} \log \left(\gamma^{\mu} p_{\mu}+m\right)=V T \frac{\partial}{\partial a} \int \frac{d^{4} p}{(2 \pi)^{4}} \frac{1}{\left(\gamma^{\mu} p_{\mu}+m\right)^{a}}|_{a=0}$$
How do I find the gamma function for this, it doesn't fit my equation?
Many thanks for your time