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I would like to calculate $\langle\bar{\psi}\psi\rangle$ in free theory. I start from the following generating functional: $$Z[J]=\int\mathcal{D}[\bar{\psi},\,\psi]\exp\left(i\int d^dx\,[\bar{\psi}(i\gamma\partial-m)\psi+J\bar{\psi}\psi]\right)\tag{1}$$ and conclude that $$\langle\bar{\psi}\psi\rangle=\frac{1}{\mathcal{Z}}\left.\frac{\partial}{\partial J}Z[J]\right|_{J=0}=\frac{\partial}{\partial J}\left(\ln Z[J]\right).\tag{2}$$ Then, the path integral is gaussian, therefore $$Z[J]=\det(i\gamma\partial -m+J)\tag{3}$$ and $\ln\det=\mathrm{Tr}\ln$, where $\mathrm{Tr}$ is the trace over all indices. I don't understand how to deal with obtain functional determinant. If it has the form $\det(i\gamma\partial -m +J)/\det(i\gamma\partial-m)$, it will be more clear. May be I am wrong in my derivations?

I would like to calculate $\langle\bar{\psi}\psi\rangle$ in free theory. I start from the following generating functional: $$Z[J]=\int\mathcal{D}[\bar{\psi},\,\psi]\exp\left(i\int d^dx\,[\bar{\psi}(i\gamma\partial-m)\psi+J\bar{\psi}\psi]\right)\tag{1}$$ and conclude that $$\langle\bar{\psi}\psi\rangle=\frac{1}{\mathcal{Z}}\left.\frac{\partial}{\partial J}Z[J]\right|_{J=0}=\frac{\partial}{\partial J}\left(\ln Z[J]\right).\tag{2}$$ Then, the path integral is gaussian, therefore $$Z[J]=\det(i\gamma\partial -m+J)\tag{3}$$ and $\ln\det=\mathrm{Tr}\ln$, where $\mathrm{Tr}$ is the trace over all indices. I don't understand how to deal with obtain functional determinant. If it has the form $\det(i\gamma\partial -m +J)/\det(i\gamma\partial-m)$, it will be more clear. May be I am wrong in my derivations?

In fact, it is still unclear. I can rewrite obtained result as $$\frac{1}{2}\mathrm{Tr}\ln((i\gamma\partial-m+J)(-i\gamma\partial-m+J))=\frac{1}{2}\mathrm{Tr}\ln(\partial^2+m^2+2mJ+J^2),$$ and then try to factor out $\partial^2+m^2$, expand $\ln(1+\text{smth})$ and calculate $\mathrm{Tr}$ in momentum space. But it seems wrong.

I would like to calculate $\langle\bar{\psi}\psi\rangle$ in free theory. I start from the following generating functional: $$Z[J]=\int\mathcal{D}[\bar{\psi},\,\psi]\exp\left(i\int d^dx\,[\bar{\psi}(i\gamma\partial-m)\psi+J\bar{\psi}\psi]\right)\tag{1}$$ and conclude that $$\langle\bar{\psi}\psi\rangle=\frac{1}{\mathcal{Z}}\left.\frac{\partial}{\partial J}Z[J]\right|_{J=0}.\tag{2}$$ Then, the path integral is gaussian, therefore $$Z[J]=\det(i\gamma\partial -m+J)\tag{3}$$ and $\ln\det=\mathrm{Tr}\ln$, where $\mathrm{Tr}$ is the trace over all indices. I don't understand how to deal with obtain functional determinant. If it has the form $\det(i\gamma\partial -m +J)/\det(i\gamma\partial-m)$, it will be more clear. May be I am wrong in my derivations?

I would like to calculate $\langle\bar{\psi}\psi\rangle$ in free theory. I start from the following generating functional: $$Z[J]=\int\mathcal{D}[\bar{\psi},\,\psi]\exp\left(i\int d^dx\,[\bar{\psi}(i\gamma\partial-m)\psi+J\bar{\psi}\psi]\right)\tag{1}$$ and conclude that $$\langle\bar{\psi}\psi\rangle=\frac{1}{\mathcal{Z}}\left.\frac{\partial}{\partial J}Z[J]\right|_{J=0}=\frac{\partial}{\partial J}\left(\ln Z[J]\right).\tag{2}$$ Then, the path integral is gaussian, therefore $$Z[J]=\det(i\gamma\partial -m+J)\tag{3}$$ and $\ln\det=\mathrm{Tr}\ln$, where $\mathrm{Tr}$ is the trace over all indices. I don't understand how to deal with obtain functional determinant. If it has the form $\det(i\gamma\partial -m +J)/\det(i\gamma\partial-m)$, it will be more clear. May be I am wrong in my derivations?

I would like to calculate $\langle\bar{\psi}\psi\rangle$ in free theory. I start from the following generating functional: $$Z[J]=\int\mathcal{D}[\bar{\psi},\,\psi]\exp\left(i\int d^dx\,[\bar{\psi}(i\gamma\partial-m)\psi+J\bar{\psi}\psi]\right)\tag{1}$$ and conclude that $$\langle\bar{\psi}\psi\rangle=\left.\frac{\partial}{\partial J}Z[J]\right|_{J=0}.\tag{2}$$ Then, the path integral is gaussian, therefore $$Z[J]=\det(i\gamma\partial -m+J)\tag{3}$$ and $\ln\det=\mathrm{Tr}\ln$, where $\mathrm{Tr}$ is the trace over all indices. I don't understand how to deal with obtain functional determinant. If it has the form $\det(i\gamma\partial -m +J)/\det(i\gamma\partial-m)$, it will be more clear. May be I am wrong in my derivations?

I would like to calculate $\langle\bar{\psi}\psi\rangle$ in free theory. I start from the following generating functional: $$Z[J]=\int\mathcal{D}[\bar{\psi},\,\psi]\exp\left(i\int d^dx\,[\bar{\psi}(i\gamma\partial-m)\psi+J\bar{\psi}\psi]\right)\tag{1}$$ and conclude that $$\langle\bar{\psi}\psi\rangle=\frac{1}{\mathcal{Z}}\left.\frac{\partial}{\partial J}Z[J]\right|_{J=0}.\tag{2}$$ Then, the path integral is gaussian, therefore $$Z[J]=\det(i\gamma\partial -m+J)\tag{3}$$ and $\ln\det=\mathrm{Tr}\ln$, where $\mathrm{Tr}$ is the trace over all indices. I don't understand how to deal with obtain functional determinant. If it has the form $\det(i\gamma\partial -m +J)/\det(i\gamma\partial-m)$, it will be more clear. May be I am wrong in my derivations?