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Physics LibreTexts

3.5: Operators

  • Page ID
    15740
  • 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)

      \( \newcommand {\ltapp} {\stackrel {_{\normalsize<}}{_{\normalsize \sim}}}\) \(\newcommand {\gtapp} {\stackrel {_{\normalsize>}}{_{\normalsize \sim}}}\) \(\newcommand {\btau}{\mbox{\boldmath$\tau$}}\) \(\newcommand {\bmu}{\mbox{\boldmath$\mu$}}\) \(\newcommand {\bsigma}{\mbox{\boldmath$\sigma$}}\) \(\newcommand {\bOmega}{\mbox{\boldmath$\Omega$}}\) \(\newcommand {\bomega}{\mbox{\boldmath$\omega$}}\) \(\newcommand {\bepsilon}{\mbox{\boldmath$\epsilon$}}\)