Operators

An operator, $O$ (say), is a mathematical entity which transforms one function into another: i.e.,

$$O(f(x))\rightarrow g(x).$$ (184)

For instance, $x$ is an operator, since $x f(x)$ 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,

$$x \frac{df}{dx} \neq \frac{d}{dx}\left(x f\right).$$ (185)

This can also be written

$$x \frac{d}{dx} \neq \frac{d}{dx} x,$$ (186)

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.,

$$x x^2 = x^2 x.$$ (187)

Finally, an operator, $O$, is termed linear if

$$O(c f(x)) =c O(f(x)),$$ (188)

where $f$ is a general function, and $c$ a general complex number. All of the operators employed in quantum mechanics are linear.

Now, from Eqs. (158) and (174),

$$\langle x\rangle = \int_{-\infty}^{\infty}\psi^\ast x \psi dx,$$ (189)

$$\langle p\rangle = \int_{-\infty}^{\infty}\psi^{\ast}\left(-{\rm i} \hbar \frac{\partial}{\partial x}\right)\psi dx.$$ (190)

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 $-{\rm i} \hbar \partial/\partial x$: i.e.,

$$p \equiv -{\rm i} \hbar \frac{\partial}{\partial x}.$$ (191)

Finally, the expectation value of some dynamical variable represented by the operator $O(x)$ is simply

$$\langle O \rangle = \int_{-\infty}^{\infty}\psi^\ast(x,t) O(x) \psi(x,t) dx.$$ (192)

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 $\langle O\rangle = \langle O \rangle^\ast$, or

$$\int_{-\infty}^{\infty} \psi^\ast (O \psi) dx = \int_{-\infty}^{\infty}(O \psi)^\ast \psi dx,$$ (193)

where $O^\ast$ 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, $O^\dag$, of a general operator, $O$, is defined as follows:

$$\int_{-\infty}^{\infty} \psi^{\ast} (O \psi) dx=\int_{-\infty}^\infty (O^\dag \psi)^\ast \psi dx.$$ (194)

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

Suppose that we wish to find the operator which corresponds to the classical dynamical variable $x p$. In classical mechanics, there is no difference between $x p$ and $p x$. However, in quantum mechanics, we have already seen that $x p\neq p x$. So, should be choose $x p$ or $p x$? Actually, neither of these combinations is Hermitian. However, $(1/2) [x p + (x p)^\dag ]$ is Hermitian. Moreover, $(1/2) [x p + (x p)^\dag ]=(1/2) (x p+p^\dag x^\dag )=(1/2) (x p+p x)$, 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

$$H \equiv \frac{p^2}{2 m} + V(x).$$ (195)

Note that $H$ is Hermitian. Now, it follows from Eq. (191) that

$$H \equiv -\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V(x).$$ (196)

However, according to Schrödinger's equation, (137), we have

$$-\frac{\hbar^2}{2 m} \frac{\partial^2}{\partial x^2} + V(x) = {\rm i} \hbar \frac{\partial}{\partial t},$$ (197)

so

$$H \equiv {\rm i} \hbar \frac{\partial}{\partial t}.$$ (198)

Thus, the time-dependent Schrödinger equation can be written

$${\rm i} \hbar \frac{\partial\psi}{\partial t} = H \psi.$$ (199)

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 $(1/2) [O(x,p,H)+ O^\dag (x,p,H)]$, where $p=-{\rm i} \hbar \partial/\partial x$, and $H={\rm i} \hbar \partial/\partial t$.