Introduction to Quantum Mechanics · Problem 3.48

Formalism — David J. Griffiths

Physics — Quantum Mechanics · self_adjoint_operators

Problem

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 Q^f(x)\hat{Q}f(x) remains in Hilbert space are allowed in the domain of Q^\hat{Q}. (As you found in Problem 3.2, the derivative operator can knock a function out of L2L^2.) 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 Q^\hat{Q} and Q^†\hat{Q}^\dagger must also be identical. Such operators are called self-adjoint. (a) Consider the momentum operator, p^=−iℏd/dx\hat{p} = -i\hbar d/dx, on the finite interval 0≤x≤a0 \le x \le a. With the infinite square well in mind, we might define its domain as the set of functions f(x)f(x) such that f(0)=f(a)=0f(0) = f(a) = 0 (it goes without saying that f(x)f(x) and p^f(x)\hat{p}f(x) are in L2(0,a)L^2(0,a)). Show that p^\hat{p} is hermitian: ⟨g∣p^f⟩=⟨p^†g∣f⟩\langle g | \hat{p}f \rangle = \langle \hat{p}^\dagger g | f \rangle, with p^†=p^\hat{p}^\dagger = \hat{p}. But is it self-adjoint? *Hint:* as long as f(0)=f(a)=0f(0) = f(a) = 0, there is no restriction on g(0)g(0) or g(a)g(a)—the domain of p^†\hat{p}^\dagger is much larger than the domain of p^\hat{p}. (b) Suppose we *extend* the domain of p^\hat{p} to include all functions of the form f(a)=λf(0)f(a) = \lambda f(0), for some fixed complex number λ\lambda. What condition must we then impose on the domain of p^†\hat{p}^\dagger in order that p^\hat{p} be hermitian? What value(s) of λ\lambda will render p^\hat{p} 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<∞0 \le x < \infty? Is there a self-adjoint momentum operator in this case?
Intuitive explanation

The key insight is that hermiticity requires matching actions, but self-adjointness requires matching both actions and domains—a much stricter condition. On a finite interval, boundary conditions that make p^\hat{p} hermitian (like f(0)=f(a)=0f(0) = f(a) = 0) don't automatically restrict p^†\hat{p}^\dagger to the same domain, so the operator isn't self-adjoint. To represent observables faithfully, you must ensure that if gg is in the domain of p^†\hat{p}^\dagger, then p^†g\hat{p}^\dagger g stays in L2L^2 under the same boundary constraints as p^\hat{p}—a constraint that becomes impossible to satisfy consistently on finite intervals without losing physical generality.

Intuitive explanation

The key insight is that hermiticity requires matching actions, but self-adjointness requires matching both actions and domains—a much stricter condition. On a finite interval, boundary conditions that make p^\hat{p} hermitian (like f(0)=f(a)=0f(0) = f(a) = 0) don't automatically restrict p^†\hat{p}^\dagger to the same domain, so the operator isn't self-adjoint. To represent observables faithfully, you must ensure that if gg is in the domain of p^†\hat{p}^\dagger, then p^†g\hat{p}^\dagger g stays in L2L^2 under the same boundary constraints as p^\hat{p}—a constraint that becomes impossible to satisfy consistently on finite intervals without losing physical generality.

Related theory

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