4.6: Operators
( \newcommand{\kernel}{\mathrm{null}\,}\)
An operator, 0 (say), is a mathematical entity which transforms one function into another: i.e.,
O(f(x))→g(x)
For instance, x is an operator, since xf(x) is a different function to f(x) and is fully specified once f(x) is given. Furthermore, d/dx is also an operator, since df(x)/dx is a different function to f(x), and is fully specified once f(x) is given. Now,
xdfdx≠ddx(xf)
This can also be written
xddx≠ddxx
where the operators are assumed to act on everything to their right, and a final f(x) is understood [where f(x) is a general function]. The above expression illustrates an important point: i.e., in general, operators do not commute. Of course, some operators do commute: e.g.,
xx2=x2x
Finally, an operator, O, is termed linear if
O(cf(x))=cO(f(x))
where f is a general function, and a general complex number. All of the operators employed in quantum mechanics are linear.
Now, from Eqs. (158) and (174),
⟨x⟩=∫∞−∞ψ∗xψdx⟨p⟩=∫∞−∞ψ∗(−iℏ∂∂x)ψdx
These expressions suggest a number of things. First, classical dynamical variables, such as x and p, are represented in quantum mechanics by linear operators which act on the wavefunction. Second, displacement is represented by the algebraic operator x, and momentum by the differential operator −iℏ∂/∂x: i.e.,
p≡−iℏ∂∂x
Finally, the expectation value of some dynamical variable represented by the operator O(x) is simply
⟨O⟩=∫∞−∞ψ∗(x,t)O(x)ψ(x,t)dx
Clearly, if an operator is to represent a dynamical variable which has physical significance then its expectation value must be real. In other words, if the operator O represents a physical variable then we require that ⟨O⟩=⟨O⟩∗, or
∫∞−∞ψ∗(Oψ)dx=∫∞−∞(Oψ)∗ψdx
where O∗ is the complex conjugate of O. An operator which satisfies the above constraint is called an Hermitian operator. It is easily demonstrated that x and p are both Hermitian. The Hermitian conjugate, O†, of a general operator, O, is defined as follows:
∫∞−∞ψ∗(Oψ)dx=∫∞−∞(O†ψ)∗ψdx
The Hermitian conjugate of an Hermitian operator is the same as the operator itself: i.e., p†=p. For a non-Hermitian operator, O (say), it is easily demonstrated that (O†)†=O, and that the operator O+Ot is Hermitian. Finally, if A and B are two operators, then (AB)†=B†A†
Suppose that we wish to find the operator which corresponds to the classical dynamical variable xp. In classical mechanics, there is no difference between xp and px. However, in quantum mechanics, we have already seen that xp≠px. So, should be choose xp or px? Actually, neither of these combinations is Hermitian. However, (1/2)[xp+(xp)†] is Hermitian. Moreover, (1/2)[xp+(xp)†]=(1/2)(xp+p†x†)=(1/2)(xp+px), which neatly resolves our problem of which order to put x and p.
It is a reasonable guess that the operator corresponding to energy (which is called the Hamiltonian, and conventionally denoted H) takes the form
H≡p22m+V(x)
Note that H is Hermitian. Now, it follows from Eq. (191) that
H≡−ℏ22m∂2∂x2+V(x)
However, according to Schrödinger's equation, (137), we have
−ℏ22m∂2∂x2+V(x)=iℏ∂∂t
so
H≡iℏ∂∂t
Thus, the time-dependent Schrödinger equation can be written
ℏ∂ψ∂t=Hψ
Finally, if O(x,p,E) is a classical dynamical variable which is a function of displacement, momentum, and energy, then a reasonable guess for the corresponding operator in quantum mechanics is (1/2)[O(x,p,H)+O†(x,p,H)], where p=−iℏ∂/∂x, and H=iℏ∂/∂t.