$$\require{cancel}$$

# 3.5: Operators

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

$O(f(x))\rightarrow g(x).$ For instance, $$x$$ is an operator, because $$x\,f(x)$$ is a different function to $$f(x)$$, and is fully specified once $$f(x)$$ is given. Furthermore, $$d/dx$$ is also an operator, because $$df(x)/dx$$ is a different function to $$f(x)$$, and is fully specified once $$f(x)$$ is given. Now,

$x\,\frac{df}{dx} \neq \frac{d}{dx}\left(x\,f\right).$ This can also be written $x\,\frac{d}{dx} \neq \frac{d}{dx}\,x,$ 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,

$x\,x^{\,2} = x^{\,2}\,x.$ Finally, an operator, $$O$$, is termed linear if $O(c\,f(x)) =c\,O(f(x)),$ 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]),

\begin{aligned} \langle x\rangle &=\int_{-\infty}^{\infty}\psi^\ast\,x\,\psi\,dx,\\[0.5ex] \langle p\rangle &= \int_{-\infty}^{\infty}\psi^{\ast}\left(-{\rm i}\,\hbar\, \frac{\partial}{\partial x}\right)\psi\,dx.\end{aligned}

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 $$-{\rm i}\,\hbar\,\partial/\partial x$$: 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 $\label{e3.55} \langle O \rangle = \int_{-\infty}^{\infty}\psi^\ast(x,t)\,O(x)\,\psi(x,t)\,dx.$

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 $$\langle O\rangle = \langle O \rangle^\ast$$, or $\label{e3.55a} \int_{-\infty}^{\infty} \psi^\ast\,(O\,\psi)\,dx = \int_{-\infty}^{\infty}(O\,\psi)^\ast\,\psi\,dx,$

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

$\label{e5.48} \int_{-\infty}^{\infty} \psi^{\ast} \,(O\,\psi)\,dx=\int_{-\infty}^\infty (O^\dagger\,\psi)^\ast\,\psi\,dx.$ The Hermitian conjugate of an Hermitian operator is the same as the operator itself: that is, $$p^\dagger = p$$. For a non-Hermitian operator, $$O$$ (say), it is easily demonstrated that $$(O^\dagger)^\dagger=O$$, and that the operator $$O+O^\dagger$$ is Hermitian. Finally, if $$A$$ and $$B$$ are two operators, then $$(A\,B)^\dagger = B^\dagger\,A^\dagger$$.

Suppose that we wish to find the operator that 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 we choose $$x\,p$$ or $$p\,x$$? Actually, neither of these combinations is Hermitian. However, $$(1/2)\,[x\,p + (x\,p)^\dagger]$$ is Hermitian. Moreover, $$(1/2)\,[x\,p + (x\,p)^\dagger]=(1/2)\,(x\,p+p^\dagger\,x^\dagger)=(1/2)\,(x\,p+p\,x)$$, 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 $H \equiv \frac{p^{\,2}}{2\,m} + V(x).$ Note that $$H$$ is Hermitian. Now, it follows from Equation ([e3.54]) that $H \equiv -\frac{\hbar^{\,2}}{2\,m}\,\frac{\partial^{\,2}}{\partial x^{\,2}} + V(x).$ However, according to Schrödinger’s equation, ([e3.1]), we have $-\frac{\hbar^{\,2}}{2\,m}\,\frac{\partial^{\,2}}{\partial x^{\,2}} + V(x) = {\rm i}\,\hbar\,\frac{\partial}{\partial t},$ so $H \equiv {\rm i}\,\hbar\,\frac{\partial}{\partial t}.$ Thus, the time-dependent Schrödinger equation can be written $\label{etimed} {\rm i}\,\hbar\,\frac{\partial\psi}{\partial t} = H\,\psi.$

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

# Contributors

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

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