An operator is defined not just by its action (what it does to the vector it is applied to) but its domain (the set of vectors on which it acts). In a finite-dimensional vector space the domain is the entire space, and we don’t need to worry about it. But for most operators in Hilbert space the

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An operator is defined not just by its action (what it does to the vector it is applied to) but its domain (the set of vectors on which it acts). In a finite-dimensional vector space the domain is the entire space, and we don’t need to worry about it. But for most operators in Hilbert space the domain is restricted. In particular, only functions such that [latex] \hat{Q} f(x) [/latex] remains in Hilbert space are allowed in the domain of [latex] \hat{Q}[/latex]. (As you found in Problem 3.2, the derivative operator can knock a function out of L².)

A hermitian operator is one whose action is the same as that of its adjoint (Problem 3.5). But what is required to represent observables is actually something more: the domains of [latex] \hat{Q} ext { and } \hat{Q}^{\dagger} [/latex] must also be identical. Such operators are called self-adjoint.

(a) Consider the momentum operator, [latex] \hat{p}=-i \hbar d / d x [/latex], on the finite interval 0 ≤ x ≤ a . With the infinite square well in mind, we might define its domain as the set of functions f(x) such that f(0) = f(a) = 0 (it goes without saying that f(x) and [latex] \hat{p} f(x)
[/latex] are in L²(0,a)) . Show that is hermitian: [latex] \langle g \mid \hat{p} f angle=\left\langle\hat{p}^{\dagger} g \mid f ight angle [/latex], with [latex] \hat{p}^{\dagger}=\hat{p} [/latex] . But is it self-adjoint?
Hint: as long as f(0) = f(a) = 0, there is no restriction on g(0) or g(a) - the domain of [latex] \hat{p}^{\dagger} [/latex] is much larger than the domain of [latex] \hat{p} [/latex].

(b) Suppose we extend the domain of [latex] \hat{p} [/latex] to include all functions of the form [latex] f(a)=\lambda f(0) [/latex], for some fixed complex number λ . What condition must we then impose on the domain of [latex] \hat{p}^{\dagger} [/latex] in order that [latex] \hat{p} [/latex] be hermitian? What value(s) of will render [latex] \hat{p} [/latex] self-adjoint? Comment: Technically, then, there is no momentum operator on the finite interval—or rather, there are infinitely many, and no way to decide which of them is “correct.” (In Problem 3.34 we avoided the issue by working on the infinite interval.)

(c) What about the semi-infinite interval, 0 ≤ x ≤ ∞? Is there a selfadjoint momentum operator in this case?.

Accepted Answer

(a) Integrate by parts:

[latex] \langle g \mid \hat{p} f angle=\int_{0}^{a} g^{*}\left(-i \hbar \frac{d f}{d x} ight) d x=-\left.i \hbar g^{*} f ight|_{0} ^{a}+i \hbar \int_{0}^{a} \frac{d g^{*}}{d x} f d x=\int_{0}^{a}\left(-i \hbar \frac{d g}{d x} ight)^{*} f d x=\langle\hat{p} g \mid f angle [/latex].

The boundary term drops out because f(0) = f(a) = 0, regardless of what g(x) does at the end points. No.

It's not self-adjoint, because the domain of [latex] \hat{p}^{\dagger} [/latex] (the set of allowable g's) is not the same as that of [latex] \hat{p} [/latex] (the allowed f's).

(b) For the boundary term to vanish we need

[latex] \left.g^{*} f ight|_{0} ^{a}=g^{*}(a) f(a)-g^{*}(0) f(0)=g^{*}(a) \lambda f(0)-g^{*}(0) f(0)=\left[\lambda g^{*}(a)-g^{*}(0) ight] f(0)=0 [/latex].

But f(0) is not necessarily zero-else we are back to part (a)|so we need

[latex] \lambda g^{*}(a)=g^{*}(0) [/latex], or [latex] g(a)=\frac{1}{\lambda^{*}} g(0) [/latex].

The two domains will be identical if [latex] 1 / \lambda^{*}=\lambda, ext { or }|\lambda|^{2}=1 [/latex] which is to say, [latex] \lambda=e^{i \phi} [/latex] for some real number Φ.

(c) For f(x) and g(x) to be in [latex] L^{2}(0, \infty), f(\infty)=g(\infty)=0 [/latex], so to kill the boundary term we need [latex] g^{*}(0) f(0)=0 [/latex] , Either f(0) = 0, in which case g(0) can be anything, or else g(0) = 0, in which case f(0) can be anything; there is no way to make the two domains equal. No, there is no self-adjoint momentum operator on the semi-infinite interval.

Question 3.48: An operator is defined not just by its action (what it does ...

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