Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 44 / 1

For a unitary matrix, $U{U}^{†}=1$, so $(U^{†}MU{)}^r=U^{†}{M}^{r}U$. Cyclicity of the trace then gives $\mathrm{tr}(U^{†}{M}^{r}U)=\mathrm{tr}(M^r)$ for both powers entering the action. Thus the action is invariant under unitary conjugation.

The spectral theorem for normal operators diagonalizes a finite-dimensional Hermitian matrix by a unitary matrix. Applying the invariance to that diagonal form givesOnly the eigenvalues enter, not the choice of eigenvectors.

Use the normalized Gaussian integral with action $\mathrm{tr}(M^2)/2$. Its Wick contraction isThere is no factor $1/N$ here: this part uses $V$, whereas the following part uses $NV$. Expanding the normalized Hermitian matrix model integral to first order givesThe subtracted term removes Vacuum Feynman diagrams disconnected from the external pair. Since the one-point function vanishes by $M\mapsto -M$, the resulting two-point function is a connected correlation function.

Write the vertex as $M^a{}_b{M}^b{}_c{M}^c{}_d{M}^d{}_a$. Each external field must contract with a different vertex field, leaving the other two to form a tadpole diagram. Eight of the twelve connected pairings attach the external fields at adjacent cyclic positions. The remaining index loop gives $N{\delta }^i{}_l{\delta }^k{}_j$ in each case. The other four attach them at opposite positions and give ${\delta }^i{}_j{\delta }^k{}_l$, with no free index loop. ThereforeBoth index structures are required at finite $N$. For $N=1$ the answer is $1-3g+O(g^2)$, agreeing with the ordinary zero-dimensional quartic integral. This is a formal perturbative quantum field theory expansion; the real integral is convergent for $g\ge 0$.

The off-diagonal auxiliary integrations impose $M^i{}_j=0$ for $i\ne j$, by the Fourier representation of a Dirac delta function. After this constraint, $M=\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_N)$ andThe Grassmann Gaussian integral over each off-diagonal pair gives its coefficient. Consequently the ghost determinant isThe prime omits the diagonal directions, which were excluded from the outset. The constant sign depends on the Berezin integral ordering and can be absorbed in normalization. Thus the Vandermonde determinant squared is the eigenvalue measure factor in this matrix diagonalization ghost determinant.

Up to an eigenvalue-independent constant, the remaining integral is $\int {\prod }_{i}d{\lambda }_i{e}^{-S_{\mathrm{eig}}}$, whereThe determinant supplies logarithmic eigenvalue repulsion. An ordering restriction on the eigenvalues changes only a constant factorial; no remaining eigenvalue integral needs to be performed. The off-diagonal bosonic contours and the displayed normalization are understood in the usual Fourier-delta prescription.