Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 40 / 3

A Grassmann variable is an odd generator of a Grassmann algebra: ${\eta }_i{\eta }_j=-{\eta }_j{\eta }_i$, so ${\eta }_i^2=0$. For one generator, any function is $f(\eta )=a+b\eta$. The Berezin integral is the linear operationThus integration extracts a coefficient, rather than assigning a length or volume. For many generators it extracts the coefficient of the highest-degree monomial with the sign fixed by the order of the measure. Odd coefficients and Grassmann derivatives must retain their order; exchanging two odd objects changes the sign.

This operation is invariant under odd translations, because a translation only changes terms of lower degree. For an invertible ordinary matrix $S$ and $\eta =S\xi$, the Grassmann change-of-variables formula isThe inverse Jacobian determinant, rather than the ordinary commuting-variable Jacobian, compensates for the factor $\det S$ multiplying the top monomial. Integration agrees with the appropriate ordered Grassmann derivatives, but the orientation must be specified when combining barred and unbarred variables.

Choose the orientation of the Berezin integral so thatHere the product has increasing $i$; this explicitly fixes the otherwise convention-dependent overall sign in the compact measure notation.

Since $A$ is a diagonalizable matrix, write $A=S\Lambda S^{-1}$ with $\Lambda =\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$. Make the independent changes $\eta =S\xi$ and $\bar{\eta }=\bar{\xi }{S}^{-1}$. The Grassmann change-of-variables formula gives the two factors $(\det S{)}^{-1}$ and $\det S$, so the complete measure is unchanged. The exponent becomes ${\sum }_i{\lambda }_i{\bar{\xi }}_i{\xi }_i$. Each summand is even and squares to zero, and the different even summands commute. Consequently,Only the term containing every generator survives the Berezin integral. Hence the Grassmann Gaussian integral isZero eigenvalues give zero on both sides, so invertibility of $A$ is unnecessary. In fact the identity extends to all ordinary matrices: the top-degree coefficient of the exponential is the alternating determinant expansion. The assumption that $A$ is a diagonalizable matrix makes the proof especially transparent.

The sources are independent odd Grassmann variables and anticommute with the Dirac field. Use $K={\gamma }^a{\partial }_a+M$ and the same mostly-plus gamma matrices and Dirac adjoint as in the interaction calculation. The inverse $K_F^{-1}$ is selected with vacuum Feynman i-epsilon prescription boundary conditions. With integral kernels and spinor contractions understood, the exponent can be completed to a square:Translations preserve the Berezin integral, so the Gaussian generating functional for a Dirac field isThe positions of the sources matter. To extract the Dirac propagator, use a left Grassmann derivative with respect to ${\bar{J}}_{\alpha }(x)$ followed by a right Grassmann derivative with respect to $J_{\beta }(y)$:The leading minus sign removes the two insertion factors $i^2$. It can also be checked by differentiating the quadratic source exponential: its ordered second derivative is $-iZ[0,0]K_F^{-1}$.

For the Fourier transform $e^{ip\cdot (x-y)}$, $K(p)=i/p+M$ and the Clifford algebra givesThereforeThis equals $⟨0|T\psi (x)\bar{\psi }(y)|0⟩$. The fermionic time ordering is explicitlywith the minus sign supplied by exchanging odd fields. The poles put positive energy forward in time and negative energy backward, which is the antiparticle contribution. Finally, $K_x{S}_F(x,y)=iI{\delta }^{(4)}(x-y)$ checks the numerator and overall sign. All formulas are distributional limits with the indicated boundary prescription.

Apply the Grassmann Gaussian integral to a regulated finite collection of field components. Up to a field-independent measure normalization and phase,In the continuum this is a formal functional determinant, including spinor and spacetime indices. Its meaningful definition requires a regulator and boundary conditions. A complex Dirac field supplies a determinant, not the inverse square root obtained for a real commuting field.

The comparison is clean after Wick rotation. A free real scalar field with positive Euclidean operator $P_E=-{\partial }_E^2+m^2$ hasTaking a logarithm gives $-\frac{1}{2}\mathrm{Tr}\log P_E$ for the scalar and $+\mathrm{Tr}\log K_E$ for the Dirac field. The opposite statistics sign is the determinant counterpart of the minus sign for a closed fermion loop. The magnitude also differs because a Dirac field has several independent spin and antiparticle degrees of freedom.

The vacuum energy sign of a fermionic oscillator makes the comparison explicit. A real scalar mode contributes $+E/2$; each independent fermionic oscillator contributes $-E/2$. There is one oscillator per scalar momentum, but a massive Dirac field has two particle and two antiparticle oscillators. Thus the vacuum energy densities are formallyEquivalently, the large Euclidean-time vacuum functional behaves as $\log Z_E\sim -\beta V{\rho }_{\mathrm{vac}}$. Both displayed vacuum energies are ultraviolet divergent; a common regularization in quantum field theory and the appropriate renormalization are needed before comparing them. The zero-point vacuum energies of bosons and fermions have opposite signs, but do not cancel without matching masses and degrees of freedom. Normal ordering removes an additive vacuum constant in nongravitating flat-space theory. Coupling to the metric in general relativity makes that constant contribute to the cosmological constant, so it cannot simply be discarded without a renormalization condition.