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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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One may roughly rephrase Schwinger's analogy as a Hermitian operator corresponds to an angle $\varphi\in\mathbb{R}$ in the same way as an unitary operator corresponds to a phase factor $e^{i\varphi}\in S^1$. –  Qmechanic Oct 19 '11 at 17:38
    
This seems pretty close to a pure math question to me... –  David Z Oct 19 '11 at 18:14

4 Answers 4

up vote 2 down vote accepted

Typically one is introduced to the spectral theorem for Hermitian operators. Recall: if $A$ is Hermitian then $$A = \sum_k a_k | k\rangle\langle k|,$$ where each $a_k$ is real and $\{| k \rangle\}$ is an orthonormal basis. If we have a function $f:\mathbb R\to \mathbb R$ (i.e. a real valued function), then we define (overloading the definition) $f:$Hermitian operators $\to$ Hermitian operators as $$f(A):=\sum_k f(a_k) | k\rangle\langle k|.$$ But we need not restrict ourselves to real valued functions. We could have a complex valued function $f:\mathbb R\to\mathbb C$. Now, however, the function defined on Hermitian operators will have a more general range, i.e. $f:$Hermitian operators $\to$ Linear operators. Consider the specific function $f(a)=e^{i a}$ applied to $A$. By definition $$f(A) = e^{iA} = \sum_k e^{i a_k} | k\rangle\langle k|,$$ which you can prove to yourself is unitary. It turns out that every unitary $U$ can be obtained by applying this function to a (non-unique) Hermitian operator (the canonical one being $-i\log U)$.

(Stone's theorem generalizes this a bit to parameterized groups of unitaries with the upshot that the Hermitian operator is uniquely determined.)

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http://en.wikipedia.org/wiki/Stone%27s_theorem_on_one-parameter_unitary_groups

Hopefully the hint of the name you need is enough. Look more widely for Stone's theorem than just Wikipedia.

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Define a unitary operator as one that preserves inner products, so $U$ is unitary iff

$$\langle U \Psi | U \Phi\rangle = \langle \Psi | \Phi \rangle$$

for all $\langle \Psi |$ and $|\Phi \rangle$.

Suppose $|\lambda\rangle$ is an eigenvector of $U$ with eigenvalue $\lambda$. Using the above, you can show that $\lambda^*\lambda = 1$, or $\lambda = e^{i\theta}$ for some real number $\theta$.

If we diagonalize $U$, it looks like

$$\left(\begin{array}{cccc} e^{i\lambda_1} & 0 & 0 & \ldots \\ 0 & e^{i\lambda_2} & 0 & \ldots \\ 0 & 0 & e^{i\lambda_3} & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{array}\right)$$

That's the same as

$$e^{i\mathbf{H}}$$

with

$$\mathbf{H} = \left(\begin{array}{cccc} \lambda_1 & 0 & 0 & \ldots \\ 0 & \lambda_2 & 0 & \ldots \\ 0 & 0 & \lambda_3 & \ldots \\ \vdots & \vdots & \vdots & \ddots \end{array}\right)$$

where $\mathbf{H}$ is a Hermitian matrix.

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So, you have proved that $DUD^{-1}=e^{iH}$ where D is the diagonalizing matrix. This implies $U=e^{iD^{-1}HD}$. Now, how do you justify that D is unitary because only then exponential has a Hermitian matrix? –  user4235 Oct 19 '11 at 17:26
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@lak Sure. We're really interested in a relationship between the operators, not the matrices, so just choose to work in the eigenbasis of $U$ and its matrix will be diagonal to begin with. However, to make this a proof would still require a lot of detail (proving that unitary operators have eigenbases, for example). I actually don't know all the details of such a proof - this was the heuristic at the top of my mind. –  Mark Eichenlaub Oct 19 '11 at 20:10
    
There are some exceptions to your general statement, but they don't matter for Schwinger's purposes (they can never be a time evolution operator, for example). –  joseph f. johnson Jan 2 '12 at 6:45

the answer has to do with a branchlet of mathematics called the Operational Calculus. Although it is maths, oddly enough it was first developed by engineers such as Heaviside. The broad idea is that just as for $z$ a complex number you can develop the whole theory of analytic functions and calculus (derivatives, integrals, etc) from power series such as $$1+z+\frac{z^2}2+\dots + \frac {z^n}{n!}+\dots,$$ and then study expressions even such as $\int \frac 1 {1-z} dz$, you can try to define functions of an operator such as $D$ the derivative operator (or even it's sort-of-inverse, an integration operator), and try to make sense of $\int \frac 1 {1-D} dD$. Even undergraduates can study the linear differential equation $(D^2 - 6D + I) \cdot f = \cos(x)$ by using a partial fractional expansion of $ I \over D^2-6D+I$ and applying it to $\cos$. (Here, $I$ means the identity operator.)

Thus, given any analytic function $f(z)$ of a complex variable, you could try to make sense of $f(H)$ where $H$ is an interesting operator. Stone's theorem studies even the family of operators $tH$ where $t$ is the time, and talks about the resulting family of operators $$e^{itH}.$$ Engineers still use this type of maths routinely, and its theoretical development by von Neumann and Gelfand and Dixmier are at the basis of many approaches to a kind of algebraic quantum theory.

The other answers posted here fall into this general category of giving meaning to applying an analytic function of a complex variable to an operator and getting another operator as the result. So, what Schwinger was getting at, is this:

If $U$ is a unitary opeartor, then there exists a complex-analytic function $f(z)$ and a Hermitian operator $H$ such that $$U=f(H).$$

(The proof is the Stone-von Neumann theorem. But Heaviside or Dirac would not have needed any proof....) I hope this puts the other detailed answers into context for you.

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