Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 58 / 4 / b / i

Table of contents
- Solution i

Solution

For a unitary operator

W

,

W W^{†} = I

. Thus for every nonnegative integer

j

,

(W^{†} H W)^{j} = W^{†} H^{j} W .

Insert this into the convergent matrix exponential series in the finite-dimensional qubit setting:

W^{†} e^{i H} W = \sum_{j = 0}^{\infty} \frac{i ^{j}}{j !} W^{†} H^{j} W = \sum_{j = 0}^{\infty} \frac{i ^{j}}{j !} (W^{†} H W)^{j} .

Consequently unitary conjugation commutes with the exponential:

W^{†} e^{i H} W = e^{i W^{†} H W} .

For an unbounded self-adjoint Hamiltonian, the same identity follows from the spectral theorem for normal operators, with the domain transported by

W

; no unbounded power-series manipulation is needed.

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