Question

An anti-Hermitian (or skew-Hermitian) operator is equal to minus its Hermitian conjugate:

${\stackrel{\mathbf{⏜}}{\mathbf{Q}}}^{\mathbf{t}}\mathbf{=}\mathbf{-}\stackrel{\mathbf{⏜}}{\mathbf{Q}}$

(a) Show that the expectation value of an anti-Hermitian operator is imaginary. (b) Show that the commutator of two Hermitian operators is anti-Hermitian. How about the commutator of two anti-Hermitian operators?

Short answer

(a) The expectation value of an anti-Hermitian operator is imaginary$<\mathrm{Q}>=-<\mathrm{Q}{>}^{*}$

(b) The commutator of two Hermitian operators is anti-Hermitian$[\stackrel{⏜}{Q},\stackrel{⏜}{R}{]}^{†}=-[\stackrel{⏜}{Q},\stackrel{⏜}{R}].$.

A Hermitian operator would be the result of two anti-Hermitian operators.

Step-by-step solution

The expectation value of an anti-Hermitian operator is imaginary.

(a)

A Hermitian operator is equal to its Hermitian conjugate, that is:

${\stackrel{⏜}{Q}}^{†}=\stackrel{⏜}{Q}$

which has the result that for inner products

The expectation value of the Hermitian operator is:

…… (1)

An anti-Hermitian operator is equal to the negative of its Hermitian conjugate, that is:

Show that the commutator of two Hermitian operators is anti-Hermitian.

(b)

Consider two Hermitian operators $\stackrel{⏜}{Q}$and $\stackrel{⏜}{R}$, their commutator is:

role="math" localid="1656321298089" $$\begin{gathered} \left[\stackrel{⏜}{Q},\stackrel{⏜}{R}\right]=\stackrel{⏜}{Q},\stackrel{⏜}{R}-\stackrel{⏜}{R}\stackrel{⏜}{Q} \\ \mathrm{The}\mathrm{conjugate}\mathrm{transpose}\mathrm{of}\mathrm{this}\mathrm{commutator}\mathrm{is}: \\ \left[\stackrel{⏜}{\mathrm{Q}},\stackrel{⏜}{\mathrm{R}}\right]={\left(\stackrel{⏜}{\mathrm{Q}},\stackrel{⏜}{R}\right)}^{†}={\stackrel{⏜}{R}}^{†}{\stackrel{⏜}{Q}}^{†},so: \\ {\left[\stackrel{⏜}{\mathrm{Q}},\stackrel{⏜}{R}\right]}^{†}={\stackrel{⏜}{R}}^{†}{\stackrel{⏜}{Q}}^{†}-{\stackrel{⏜}{Q}}^{†}{\stackrel{⏜}{R}}^{†} \\ \mathrm{For}\mathrm{a}\mathrm{Hermitian}\mathrm{operator}{\stackrel{⏜}{\mathrm{R}}}^{†}=\stackrel{⏜}{\mathrm{R}}\mathrm{and}{\stackrel{⏜}{\mathrm{Q}}}^{†}=\stackrel{⏜}{\mathrm{Q}},\mathrm{thus}: \\ {\left[\stackrel{⏜}{\mathrm{Q}},\stackrel{⏜}{R}\right]}^{†}=\stackrel{⏜}{R},\stackrel{⏜}{Q}-\stackrel{⏜}{\mathrm{Q}},\stackrel{⏜}{R} \\ =\left[{\stackrel{⏜}{R}}^{†}{\stackrel{⏜}{Q}}^{†}\right] \\ =-\left[\stackrel{⏜}{\mathrm{Q}},\stackrel{⏜}{R}\right] \\ \mathrm{Note}\mathrm{that}\mathrm{equation}\left(3\right)\mathrm{is}\mathrm{used},\mathrm{in}\mathrm{the}\mathrm{last}\mathrm{two}\mathrm{lines}.{\left[\stackrel{⏜}{\mathrm{Q}},\stackrel{⏜}{\mathrm{R}}\right]}^{†}=-\left[\stackrel{⏜}{\mathrm{Q}},\stackrel{⏜}{\mathrm{R}}\right]. \\ \mathrm{A}\mathrm{Hermitian}\mathrm{operator}\mathrm{would}\mathrm{be}\mathrm{the}\mathrm{result}\mathrm{of}\mathrm{two}\mathrm{anti}-\mathrm{Hermitian}\mathrm{operators}. \end{gathered}$$