4.6: Operators

An operator, \(\begin{equation}0\end{equation}\) (say), is a mathematical entity which transforms one function into another: i.e.,

\begin{equation}O(f(x)) \rightarrow g(x)\end{equation}

For instance, \(\begin{equation}x\end{equation}\) is an operator, since  \(\begin{equation}x f(x)\end{equation}\) is a different function to \(\begin{equation}f(x)\end{equation}\)  and is fully specified once \(\begin{equation}f(x)\end{equation}\) is given. Furthermore, \(\begin{equation}d / d x\end{equation}\) is also an operator, since  \(\begin{equation}d f(x) / d x\end{equation}\) is a different function to \(\begin{equation}f(x)\end{equation}\), and is fully specified once \(\begin{equation}f(x)\end{equation}\) is given. Now,

\begin{equation}x \frac{d f}{d x} \neq \frac{d}{d x}(x f)\end{equation}

 

This can also be written
\begin{equation}x \frac{d}{d x} \neq \frac{d}{d x} x\end{equation}

where the operators are assumed to act on everything to their right, and a final \(\begin{equation}f(x)\end{equation}\)  is understood [where \(\begin{equation}f(x)\end{equation}\) 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.,

\begin{equation}x x^{2}=x^{2} x\end{equation}

Finally, an operator, \(\begin{equation}O\end{equation}\), is termed linear if

\begin{equation}O(c f(x))=c O(f(x))\end{equation}

where \(\begin{equation}f\end{equation}\)  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),

\begin{equation}\begin{array}{l}
\langle x\rangle=\int_{-\infty}^{\infty} \psi^{*} x \psi d x \\
\langle p\rangle=\int_{-\infty}^{\infty} \psi^{*}\left(-i \hbar \frac{\partial}{\partial x}\right) \psi d x
\end{array}\end{equation}

These expressions suggest a number of things. First, classical dynamical variables, such as \(\begin{equation}x \text { and } p\end{equation}\),  are represented in quantum mechanics by linear operators which act on the wavefunction. Second, displacement is represented by the algebraic operator \(\begin{equation}x\end{equation}\), and momentum by the differential operator \(\begin{equation}-i \hbar \partial / \partial x\end{equation}\): i.e.,

\begin{equation}p \equiv-i \hbar \frac{\partial}{\partial x}\end{equation}

Finally, the expectation value of some dynamical variable represented by the operator \(\begin{equation}O(x)\end{equation}\) is simply

\begin{equation}\langle O\rangle=\int_{-\infty}^{\infty} \psi^{*}(x, t) O(x) \psi(x, t) d x\end{equation}

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 \(\begin{equation}\langle O\rangle=\langle O\rangle^{*}\end{equation}\), or 

\begin{equation}\int_{-\infty}^{\infty} \psi^{*}(O \psi) d x=\int_{-\infty}^{\infty}(O \psi)^{*} \psi d x\end{equation}

where \(\begin{equation}O^{*}\end{equation}\) is the complex conjugate of \(\begin{equation}O\end{equation}\). An operator which satisfies the above constraint is called an Hermitian operator. It is easily demonstrated that \(\begin{equation}x \text { and } p\end{equation}\) are both Hermitian. The Hermitian conjugate, \(\begin{equation}O^{\dagger}\end{equation}\), of a general operator, \(\begin{equation}O\end{equation}\), is defined as follows:

\begin{equation}\int_{-\infty}^{\infty} \psi^{*}(O \psi) d x=\int_{-\infty}^{\infty}\left(O^{\dagger} \psi\right)^{*} \psi d x\end{equation}

The Hermitian conjugate of an Hermitian operator is the same as the operator itself: i.e., \(\begin{equation}p^{\dagger}=p\end{equation}\). For a non-Hermitian operator, \(\begin{equation}O\end{equation}\) (say), it is easily demonstrated that \(\begin{equation}\left(O^{\dagger}\right)^{\dagger}=O\end{equation}\), and that the operator  \(\begin{equation}O+O^{t}\end{equation}\) is Hermitian. Finally, if  \(\begin{equation}A \text { and } B\end{equation}\) are two operators, then \(\begin{equation}(A B)^{\dagger}=B^{\dagger} A^{\dagger}\end{equation}\)

Suppose that we wish to find the operator which corresponds to the classical dynamical variable \(\begin{equation}x p\end{equation}\). In classical mechanics, there is no difference between \(\begin{equation}x p \text { and } p x\end{equation}\). However, in quantum mechanics, we have already seen that \(\begin{equation}x p \neq p x\end{equation}\). So, should be choose \(\begin{equation}x p\end{equation}\) or \(\begin{equation}p x\end{equation}\)? Actually, neither of these combinations is Hermitian. However, \(\begin{equation}(1 / 2)\left[x p+(x p)^{\dagger}\right]\end{equation}\) is Hermitian. Moreover, \(\begin{equation}(1 / 2)\left[x p+(x p)^{\dagger}\right]=(1 / 2)\left(x p+p^{\dagger} x^{\dagger}\right)=(1 / 2)(x p+p x)\end{equation}\), which neatly resolves our problem of which order to put \(\begin{equation}x\end{equation}\) and \(\begin{equation}p\end{equation}\).

It is a reasonable guess that the operator corresponding to energy (which is called the Hamiltonian, and conventionally denoted \(\begin{equation}H\end{equation}\)) takes the form

\begin{equation}H \equiv \frac{p^{2}}{2 m}+V(x)\end{equation}

Note that \(\begin{equation}H\end{equation}\) is Hermitian. Now, it follows from Eq. (191) that

\begin{equation}H \equiv-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)\end{equation}

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

\begin{equation}-\frac{\hbar^{2}}{2 m} \frac{\partial^{2}}{\partial x^{2}}+V(x)=\mathrm{i} \hbar \frac{\partial}{\partial t}\end{equation}

so

\begin{equation}H \equiv i \hbar \frac{\partial}{\partial t}\end{equation}

 

Thus, the time-dependent Schrödinger equation can be written
\begin{equation}\hbar \frac{\partial \psi}{\partial t}=H \psi\end{equation}

Finally, if  \(\begin{equation}O(x, p, E)\end{equation}\)  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  \(\begin{equation}(1 / 2)\left[O(x, p, H)+O^{\dagger}(x, p, H)\right], \text { where } p=-\mathrm{i} \hbar \partial / \partial x\end{equation}\), and \(\begin{equation}H=\mathrm{i} \hbar \partial / \partial t\end{equation}\).