Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
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 3 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.)

share|improve this answer

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



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

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

share|improve this answer
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? –  Lakshya Bhardwaj Oct 19 '11 at 17:26
@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


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

share|improve this answer

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.