 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/5/b/solution

A^{†} = - A

, then

H = - i A

H^{†} = i A^{†} = - i A = H

Q

with

H = Q diag (θ_{1}, \dots, θ_{n}) Q^{†}, θ_{j} \in R .

Thus

A = Q diag (i θ_{j}) Q^{†}

e^{A} = QC Q^{†}, C = diag (e^{i θ_{1}}, \dots, e^{i θ_{n}}) .

Each diagonal entry has modulus one, so

C^{†} C = I

. Using

Q^{†} Q = Q Q^{†} = I

(e^{A})^{†} e^{A} = Q C^{†} Q^{†} QC Q^{†} = I .

Hence the exponential of a skew-Hermitian matrix is unitary.

 New to topics? Read the docs here!