 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-44/1/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-06

U U^{†} = 1

, so

(U^{†} M U)^{r} = U^{†} M^{r} U

. Cyclicity of the trace then gives

tr (U^{†} M^{r} U) = tr (M^{r})

for both powers entering the action. Thus the action is invariant under unitary conjugation.

V (M) = i = 1 \sum N (\frac{λ _{i}^{2}}{2} + \frac{g}{4} λ_{i}^{4}) .

Only the eigenvalues enter, not the choice of eigenvectors.

 New to topics? Read the docs here!