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 Q̂f(x) remains in Hilbert space are allowed in the domain of Q̂.
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 Q̂ and Q̂⁺ must also be identical. Such operators are called self adjoint.
(a) Consider the momentum operator, p̂ = -iћd/dx, on the finite interval 0 ≤ x ≤ α. 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 p̂f (x) are in L² (0, a) ). Show that p̂ is hermitian: , with p̂+ = p̂. But is it self-adjoint? as long as f(0) = f(a) = 0, there is no restriction on g(0) or g(a) —the domain of is much larger than the domain of p̂.
(b) Suppose we extend the domain of p̂ to include all functions of the form f(a) = λf(0) , for some fixed complex number λ. What condition must we then impose on the domain of  p̂+ in order that  p̂ be hermitian? What value(s) of λ will render p̂ self adjoint? 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 self-adjoint momentum operator in this case?

Transcribed Image Text:

(g|pf) = (pg|f),