Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Physics LibreTexts

3.5: Operators

( \newcommand{\kernel}{\mathrm{null}\,}\)

An operator, O (say), is a mathematical entity that transforms one function into another: that is,

O(f(x))g(x).

For instance, x is an operator, because 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, because df(x)/dx is a different function to f(x), and is fully specified once f(x) is given. Now,

xdfdxddx(xf).

This can also be written xddxddxx,
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 previous expression illustrates an important point. Namely, in general, operators do not commute with one another. Of course, some operators do commute. For instance,

xx2=x2x.

Finally, an operator, O, is termed linear if O(cf(x))=cO(f(x)),
where f is a general function, and c a general complex number. All of the operators employed in quantum mechanics are linear.

Now, from Equations ([e3.22]) and ([e3.38]),

x=ψxψdx,p=ψ(ix)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 that act on the wavefunction. Second, displacement is represented by the algebraic operator x, and momentum by the differential operator i/x: that is, \[\label{e3.54} p \equiv -{\rm i}\,\hbar\,\frac{\partial}{\partial x}.\]

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 that 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 that satisfies the previous 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=(Oψ)ψdx.

The Hermitian conjugate of an Hermitian operator is the same as the operator itself: that is, p=p. For a non-Hermitian operator, O (say), it is easily demonstrated that (O)=O, and that the operator O+O is Hermitian. Finally, if A and B are two operators, then (AB)=BA.

Suppose that we wish to find the operator that 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 xppx. So, should we 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+px)=(1/2)(xp+px), which neatly resolves our problem of the order in which to place 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 Hp22m+V(x).

Note that H is Hermitian. Now, it follows from Equation ([e3.54]) that H22m2x2+V(x).
However, according to Schrödinger’s equation, ([e3.1]), we have 22m2x2+V(x)=it,
so Hit.
Thus, the time-dependent Schrödinger equation can be written iψt=Hψ.

Finally, if O(x,p,E) is a classical dynamical variable that 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=i/x, and H=i/t.

Contributors and Attributions

  • Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)


This page titled 3.5: Operators is shared under a not declared license and was authored, remixed, and/or curated by Richard Fitzpatrick.

  • Was this article helpful?

Support Center

How can we help?