Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 60 / 3 / a

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- Solution a

Solution

Since

A

is a bounded operator, its exponential power series converges in operator norm. The identity

A^{2} = I

gives

A^{2 k} = I

and

A^{2 k + 1} = A

. Separating the even and odd powers therefore gives

e^{- i θ A} = \sum_{k = 0}^{\infty} \frac{( - 1 ) ^{k} θ ^{2 k}}{( 2 k )!} I - i \sum_{k = 0}^{\infty} \frac{( - 1 ) ^{k} θ ^{2 k + 1}}{( 2 k + 1 )!} A = cos θ I - i sin θ A .

The algebraic identity does not require

A

to be Hermitian. Hermiticity is additionally needed for this exponential to be a unitary operator for real

θ

.

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