Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 65 / 2

Because $A$ is a real symmetric positive-definite matrix, its inverse exists and all its eigenvalues are positive. Completing the square givesThe translation $v=u-A^{-1}b$ has unit Jacobian determinant. Choose an orthogonal matrix $O$ with $O^{T}AO=\mathrm{diag}(a_1,\dots ,a_n)$, where $a_i>0$. The change $v=Oz$ also has absolute Jacobian one. The remaining integral is the product of $n$ one-dimensional Gaussian integrals, each equal to $\sqrt{2\pi /a_i}$. Consequently the linear-source multivariate Gaussian integral isThe determinant square root is the positive one, as required by convergence of the original real integral.

A linear operator on finite-dimensional vectors acts by $(Au{)}_i={\sum }_j{A}_{ij}{u}_j$. For a field the discrete index becomes a continuous coordinate, and a suitable operator is represented by an integral kernel, possibly a distribution:The identity operator has kernel ${\delta }^{(d)}(x-y)$. If an inverse exists on the chosen domain and with the chosen boundary prescription, its kernel satisfieswith the analogous reversed identity. A differential operator is a distributional kernel: for example, $A(x,y)=(-{\square }_x-m^2){\delta }^{(d)}(x-y)$. Its inverse is a Green function. Boundary conditions, a causal prescription and the treatment of zero modes are part of the inverse's definition, not consequences of formal matrix notation. An operator with an untreated zero mode does not have an inverse on the full field space.

The functional derivative is defined through the first variation: for arbitrary smooth test variation $h$,In particular,For a symmetric kernel, differentiating $F[\varphi ]=\frac{1}{2}\int dxdy\varphi (x)A(x,y)\varphi (y)$ gives $\delta F/\delta \varphi =A\varphi$. The two terms from varying the two fields coincide; for a nonsymmetric kernel the symmetric part $(A+A^T)/2$ appears instead. Integration by parts is understood when $A$ is differential and the boundary variations vanish. A finite-volume, finite-mode regulator makes these statements ordinary matrix identities before a continuum limit.

Use the same metric signature as above and define the Minkowski generating functional byHere $\hslash =1$. Integration by parts writes the quadratic action as $S_0=\frac{1}{2}\varphi K\varphi$ with $K=-\square -m^2+i0$; the infinitesimal imaginary part damps the oscillatory Gaussian functional integral and selects the vacuum Feynman propagator. Abbreviating repeated spacetime integrations, completing the square givesTranslation of the regulated integration variables therefore yieldsThe source-independent normalization contains the regulated functional determinant and its oscillatory phase; it cancels in $Z_0[J]/Z_0[0]$. Define ${\Delta }_F=i{K}^{-1}$. The Gaussian evaluation of a free scalar generating functional is equivalently $Z_0[J]/Z_0[0]=\exp [-J{\Delta }_{F}J/2]$.

With Fourier convention $e^{-ip\cdot x}$, $-\square$ acts as $p^2$. The inverse-kernel equation $K{K}^{-1}=I$ then givesThis is the time-ordered free two-point correlation function: differentiating twice with respect to $J$ and multiplying by $1/i^2$ recovers ${\Delta }_F$. It obeys $(\square +m^2){\Delta }_F=-i{\delta }^{(d)}$ in the distributional limit. The $i0$ is essential: the algebraic denominator alone would not distinguish Feynman, advanced and retarded Green functions.

For a finite-dimensional integral, $u_i{e}^{b\cdot u}={\partial }_{b_i}{e}^{b\cdot u}$. Expanding a polynomial interaction $V(u)$ formally and interchanging each finite-order derivative with the regulated integral giveswhere $I_0$ is the free Gaussian integral. This is a perturbation-series identity; it does not assert convergence of the full expansion for an arbitrary potential.

For the Minkowski functional integral, each field insertion is generated byExpand $e^{i{S}_I[\varphi ]}$, replace its field factors by these functional derivatives, and take the derivatives outside the free integral. The result is polynomial interactions as source differential operators:If the field interaction term contains derivatives of the field, the corresponding spacetime derivatives act on the source-insertion operators as well. For the scalar polynomial interactions the operator in the exponent isThe free quadratic source exponential then supplies all Wick contractions, reproducing the Feynman rules. For normalized correlation functions, divide the resulting interacting functional by $Z[0]$; the interaction operator changes this denominator too, so it cannot simply be ignored.