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A book on Quantum Mechanics by Schwinger states, "A unitary operator can be considered to be a complex valued function of a Hermitian operator."

Please give a hint on how to prove this assertion.

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I think that Schwinger alludes to the Cayley transform. –  Mark Oct 19 '11 at 20:29
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2 Answers

If $U$ is unitary, then its spectrum is contained in the unit circle $\mathbb T$. A Borel logarithm on $\mathbb T$ is a Borel-measurable function $g:\mathbb T\to\mathbb C$ such that $\exp(g(z))=z$ for all $z\in\mathbb T$. Such functions exist: for example, $g(e^{i\theta})=i\theta$, $-\pi<\theta\leq\pi$. Then $g(U)$ can be defined with Borel function calculus, and $H=-ig(U)$ is a self-adjoint operator such that $U=\exp(iH)$ as in Savinov's answer.

If the spectrum of $U$ is not the entire circle, you can get away with much less than Borel function calculus. Multiplying by a unit length scalar if needed, WLOG assume $1$ is not in the spectrum of $U$. Then the inverse Cayley transform of $U$ is the self-adjoint operator $A=i(I+U)(I-U)^{-1}$, and $U$ is the Cayley transform of $A$, $U=(A-iI)(A+iI)^{-1}$, as in Mark's comment.

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Let $H$ is Hermitian operator, check that $U=\exp(iH)$ is unitary by definition. If you need $\exp(A)=\sum{\frac{A^k}{k!}}$

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But, that's the converse of what I want to prove. I thought about it but it's not leading to the representation of general unitary operator. –  Lakshya Bhardwaj Oct 19 '11 at 17:08
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