Verify the matrix exponential $e^{i\hat{H}t/\hbar}$ is unitary

Question

What could be an example of an anti-hermitian matrix $i\hat{H}t$ , which satisfies the matrix exponential $$e^{i\hat{H}t/\hbar}$$ being a unitary matrix?

Answer 1

For any skew hermitian matrix $S$,

$S^\dagger = -S, \tag 1$

we have

$(e^S)^\dagger = e^{S^\dagger} = e^{-S}, \tag 2$

as follows by applying the adjoint operation ${}^\dagger$ term-by-term to the power series

$e^S = \displaystyle \sum_0^\infty \dfrac{S^n}{n!}, \tag 3$

viz

$(e^S)^\dagger = \displaystyle \sum_0^\infty \dfrac{(S^n)^\dagger}{n!} = \sum_0^\infty \dfrac{(S^\dagger)^n}{n!} = e^{S^\dagger}; \tag 4$

therefore, from (2),

$(e^S)^\dagger e^S = e^S (e^S)^\dagger = I, \tag 5$

that is, $e^S$ is unitary.

Now for hermitian $H$,

$H^\dagger = H, \tag 6$

it follows that

$(iHt/\hbar)^\dagger = -iH^\dagger t /\hbar = -iHt/\hbar, \tag 7$

i.e., $iHt/\hbar$ is skew-hermitian; we conclude that $e^{iHt/\hbar}$ is unitary.

It is then a simple matter to present a large host of examples illustrating these arguments; we merely need pick any hermitian $H$ and form $e^{iHt/\hbar}$, for instance, with

$H = \begin{bmatrix} a & bi \\ -bi & a \end{bmatrix} = H^\dagger, \; a, b \in \Bbb R, \tag 8$

$iHt/\hbar = \begin{bmatrix} iat/\hbar & -bt/\hbar \\ bt/\hbar & iat/\hbar \end{bmatrix} = (-iHt/\hbar)^\dagger \tag 9$

is skew-, and so

$e^{iHt/\hbar}= \exp \left ( \begin{bmatrix} iat/\hbar & -bt/\hbar \\ bt/\hbar & iat/\hbar \end{bmatrix} \right) \tag{10}$

is unitary.