Stone's Theorem

Exponential of Operators

Exponential of a Self-adjoint Operator

Suppose is a self-adjoint operator on a Hilbert space . If is bounded, we can simply define the exponential of as If is unbounded, we can define the exponential of using the spectral theorem,

Proof We shall check it is well-defined when is bounded. Suppose , define as . Then for any , we have because the exponential series (Cauchy) converges in (ref. theorem). Therefore, is also Cauchy, thus converges in the Banach space .

Proposition

Let be a self-adjoint operator on a Hilbert space . Define , then

• For each , is unitary and ;
• If , then as , i.e. is strongly continuous;
• For , as ;
• If exists, then .

Proof

1. Consider the functional calculus of . Let be the map . Then since is a -homomorphism, we have Moreover,

The Stone’s Theorem

Conversely, given a group of unitary operators, can it be represented as the exponential? Stone’s theorem answers this question:

Stone's Theorem

Let be a strongly continuous one-parameter unitary group on a Hilbert space . Then, there is a self-adjoint operator on so that for all . In this case, we call the infinitesimal generator of the unitary group .

Proof For any , and , we define as a vector-valued integral, taking values in , because is strongly continuous. Define the following subset of : Now we claim that is dense in . In fact, let be the approximate identity, then , and we have Notice that by strong continuity of , we have so as . Hence, any has that can be arbitrarily close to , that is is dense in . Now we define by , and claim that its closure is our desired operator. We first show it is symmetric. Observe that where the last equation holds because the derivative of compactly supported smooth is bounded, so the bounded convergence theorem applies. Thus . Similarly, we can also compute and show that . For any , we have Therefore, is symmetric. Moreover, satisfies . We can see this by first identifying (so ) where . Then similarly compute: Now we show that it is essentially self-adjoint by showing that , so has a unique self-adjoint extension. Let such that . For any , consider the complex function . Its derivative is: The solution to this differential equation is . Note that since is unitary, is bounded (by the Cauchy-Schwarz Inequality): So, we must have for all . This can only be true if . Since this holds for all and is dense in the Hilbert space , we must have (ref. corollary). A similar argument shows that if , then . Thus, . Therefore, the symmetric operator is essentially self-adjoint. It admits a unique self-adjoint extension . It remains to show that . For any , let us define a function : Note that . Now, we compute the derivative of . Since , it follows that and . On the domain , the operator and its extension coincide. Hence, Now, let’s look at the derivative of the squared norm of : This implies that is a constant, for all . Thus, for all . This means: for all and . Since is dense in , we have for all .

e.g. Suppose we have a family of transition operators on , defined by . Then we can apply Stone’s theorem to find the infinitesimal generator: so the infinitesimal generator of the unitary group .