Operators

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

$$O(f(x))\rightarrow g(x).$$ (166)

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,

$$x\,\frac{df}{dx} \neq \frac{d}{dx}\left(x\,f\right).$$ (167)

This can also be written

$$x\,\frac{d}{dx} \neq \frac{d}{dx}\,x,$$ (168)

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.,

$$x\,x^2 = x^2\,x.$$ (169)

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

$$O(c\,f(x)) =c\,O(f(x)),$$ (170)

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. (140) and (156):

$$\langle x\rangle = \int_{-\infty}^{\infty}\psi^\ast\,x\,\psi\,dx,$$ (171)

$$\langle p\rangle = \int_{-\infty}^{\infty}\psi^{\ast}\left(-{\rm i}\,\hbar\, \frac{\partial}{\partial x}\right)\psi\,dx.$$ (172)

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 wave-function. Second, displacement is represented by the algebraic operator $x$, and momentum by the differential operator $-{\rm i}\,\hbar\,\partial/\partial x$: i.e.,

$$p \equiv -{\rm i}\,\hbar\,\frac{\partial}{\partial x}.$$ (173)

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

$$\langle O \rangle = \int_{-\infty}^{\infty}\psi^\ast(x,t)\,O(x)\,\psi(x,t)\,dx.$$ (174)

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

$$\int_{-\infty}^{\infty} \psi^\ast\,(O\,\psi)\,dx = \int_{-\infty}^{\infty}(O\,\psi)^\ast\,\psi\,dx,$$ (175)

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:

$$\int_{-\infty}^{\infty} \psi^{\ast} \,(O\,\psi)\,dx=\int_{-\infty}^\infty (O^\dag\,\psi)^\ast\,\psi\,dx.$$ (176)

The Hermitian conjugate of an Hermitian operator is obviously 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

$$H \equiv \frac{p^2}{2\,m} + V(x).$$ (177)

Note that $H$ is Hermitian. Now, it follows from Eq. (173) that

$$H \equiv -\frac{\hbar^2}{2\,m}\,\frac{\partial^2}{\partial x^2} + V(x).$$ (178)

However, according to Schrödinger's equation, (119), we have

$$-\frac{\hbar^2}{2\,m}\,\frac{\partial^2}{\partial x^2} + V(x) = {\rm i}\,\hbar\,\frac{\partial}{\partial t}.$$ (179)

Hence,

$$H \equiv {\rm i}\,\hbar\,\frac{\partial}{\partial t}.$$ (180)

Thus, the time-dependent Schrödinger equation can be written

$${\rm i}\,\hbar\,\frac{\partial\psi}{\partial t} = H\,\psi.$$ (181)

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 $H={\rm i}\,\hbar\,\partial/\partial t$.