Past exam of the theoretical physics II course of the University of Cambridge / 2012 / Paper 1 / 1

For a differentiable family of invertible linear operators, differentiate the identity without changing the factor order:Multiplication on the right by $A^{-1}$ gives the derivative of matrix inversion,Commuting $A^{-1}$ past the derivative would generally be invalid.

Take an orthonormal basis $\{|j⟩\}$ and insert its completeness relation:No normalization or orthogonality of $u,v$ is required. This is the trace of a rank-one operator; in a Hilbert space the sum converges by the Cauchy-Schwarz inequality.

In a finite-dimensional basis, interchange the two finite sums:This cyclic property of the trace does not assert $AB=BA$ or permit arbitrary reordering of three factors. For infinite-dimensional linear operators, the traces and rearrangements need hypotheses, for example one trace-class operator and one bounded operator; an arbitrary pair of unbounded linear operators is not covered.

For real $ϵ$, unitarity to first order givesHence the generator of an infinitesimal unitary transformation obeys $F=F^{†}$: it is Hermitian. More precisely, $U(ϵ)=I+iϵF+O({ϵ}^2)$ is a differentiable unitary family at the identity. The omitted second-order terms matter if $ϵ$ is finite.

Set $C(\beta )=[A,e^{-\beta B}]$. The product rule givesMultiplying by the integrating factor $e^{\beta B}$ and integrating yields the commutator of an operator with an exponential:Equivalently, differentiate $e^{\beta B}A{e}^{-\beta B}-A$ directly; its derivative is the integrand and it vanishes at zero. The argument is exact for finite matrices, and for bounded operators with their norm-convergent exponentials; unbounded linear operators require an appropriate common domain.