4.11: Continuous Eigenvalues
- Page ID
- 1168
In the previous two sections, it was tacitly assumed that we were dealing with operators possessing discrete eigenvalues and square-integrable eigenstates. Unfortunately, some operators--most notably, and --possess eigenvalues which lie in a continuous range and non-square-integrable eigenstates (in fact, these two properties go hand in hand). Let us, therefore, investigate the eigenstates and eigenvalues of the displacement and momentum operators.
Let \(\begin{equation}\psi_{x}\left(x, x^{\prime}\right)\end{equation}\) be the eigenstate of corresponding to the eigenvalue \(\begin{equation}x^{\prime}\end{equation}\). It follows that
\begin{equation}x \psi_{x}\left(x, x^{\prime}\right)=x^{\prime} \psi_{x}\left(x, x^{\prime}\right)\end{equation}
for all . Consider the Dirac delta-function \(\begin{equation}\delta\left(x-x^{\prime}\right)\end{equation}\). We can write
\begin{equation}x \delta\left(x-x^{\prime}\right)=x^{\prime} \delta\left(x-x^{\prime}\right)\end{equation}
since \(\begin{equation}\delta\left(x-x^{\prime}\right)\end{equation}\) is only non-zero infinitesimally close to \(\begin{equation}x=x^{\prime}\end{equation}\). Evidently \(\begin{equation}\psi_{x}\left(x, x^{\prime}\right)\end{equation}\) is proportional to \(\begin{equation}\delta\left(x-x^{\prime}\right)\end{equation}\). Let us make the constant of proportionality unity, so that
\begin{equation}\psi_{x}\left(x, x^{\prime}\right)=\delta\left(x-x^{\prime}\right)\end{equation}
Now, it is easily demonstrated that
\begin{equation}\int_{-\infty}^{\infty} \delta\left(x-x^{\prime}\right) \delta\left(x-x^{\prime \prime}\right) d x=\delta\left(x^{\prime}-x^{\prime \prime}\right)\end{equation}
Hence, \(\begin{equation}\psi_{x}\left(x, x^{\prime}\right)\end{equation}\) satisfies the orthonormality condition
\begin{equation}\int_{-\infty}^{\infty} \psi_{x}^{*}\left(x, x^{\prime}\right) \psi_{x}\left(x, x^{\prime \prime}\right) d x=\delta\left(x^{\prime}-x^{\prime \prime}\right)\end{equation}
This condition is analogous to the orthonormality condition (263) satisfied by square-integrable eigenstates. Now, by definition, \(\begin{equation}\delta\left(x-x^{\prime}\right)\end{equation}\) satisfies
\begin{equation}\int_{-\infty}^{\infty} f(x) \delta\left(x-x^{\prime}\right) d x=f\left(x^{\prime}\right)\end{equation}
where \(\begin{equation}f(x)\end{equation}\) is a general function. We can thus write
\begin{equation}\psi(x)=\int_{-\infty}^{\infty} c\left(x^{\prime}\right) \psi_{x}\left(x, x^{\prime}\right) d x^{\prime}\end{equation}
where \(\begin{equation}c\left(x^{\prime}\right)=\psi\left(x^{\prime}\right)\end{equation}\), or
\begin{equation}c\left(x^{\prime}\right)=\int_{-\infty}^{\infty} \psi_{x}^{*}\left(x, x^{\prime}\right) \psi(x) d x\end{equation}
In other words, we can expand a general wavefunction \(\begin{equation}\psi(x)\end{equation}\) as a linear combination of the eigenstates, \(\begin{equation}\psi_{x}\left(x, x^{\prime}\right)\end{equation}\) , of the displacement operator. Equations (283) and (284) are analogous to Eqs. (261) and (264), respectively, for square-integrable eigenstates. Finally, by analogy with the results in Sect. 4.9, the probability density of a measurement of yielding the value \(\begin{equation}x^{\prime} \text { is }\left|c\left(x^{\prime}\right)\right|^{2}\end{equation}\), which is equivalent to the standard result \(\begin{equation}\left|\psi\left(x^{\prime}\right)\right|^{2}\end{equation}\). Moreover, these probabilities are properly normalized provided \(\begin{equation}\psi(x)\end{equation}\) is properly normalized [cf., Eq. (265)]: i.e.,
\begin{equation}\int_{-\infty}^{\infty}\left|c\left(x^{\prime}\right)\right|^{2} d x^{\prime}=\int_{-\infty}^{\infty}\left|\psi\left(x^{\prime}\right)\right|^{2} d x^{\prime}=1\end{equation}
Finally, if a measurement of yields the value \(\begin{equation}x^{\prime}\end{equation}\) then the system is left in the corresponding displacement eigenstate, \(\begin{equation}\psi_{x}\left(x, x^{\prime}\right)\end{equation}\) immediately after the measurement: i.e., the wavefunction collapses to a ``spike-function'', \(\begin{equation}\delta\left(x-x^{\prime}\right)\end{equation}\), as discussed in Sect. 3.16.
Now, an eigenstate of the momentum operator \(\begin{equation}p \equiv-i \hbar \partial / \partial x\end{equation}\) corresponding to the eigenvalue \(\begin{equation}p^{\prime}\end{equation}\) satisfies
\begin{equation}-i \hbar \frac{\partial \psi_{p}\left(x, p^{\prime}\right)}{\partial x}=p^{\prime} \psi_{p}\left(x, p^{\prime}\right)\end{equation}
It is evident that
\begin{equation}\psi_{p}\left(x, p^{\prime}\right) \propto \mathrm{e}^{+i p^{\prime} x / \hbar}\end{equation}
Now, we require \(\begin{equation}\psi_{p}\left(x, p^{\prime}\right)\end{equation}\) to satisfy an analogous orthonormality condition to Eq. (281): i.e.,
\begin{equation}\int_{-\infty}^{\infty} \psi_{p}^{*}\left(x, p^{\prime}\right) \psi_{p}\left(x, p^{\prime \prime}\right) d x=\delta\left(p^{\prime}-p^{\prime \prime}\right)\end{equation}
Thus, it follows from Eq. (210) that the constant of proportionality in Eq. (287) should be \(\begin{equation}(2 \pi \hbar)^{-1 / 2}\end{equation}\): i.e.,
\begin{equation}\psi_{p}\left(x, p^{\prime}\right)=\frac{\mathrm{e}^{+i p^{\prime} x / \hbar}}{(2 \pi \hbar)^{1 / 2}}\end{equation}
Furthermore, according to Eqs. (202) and (203),
\begin{equation}\psi(x)=\int_{-\infty}^{\infty} c\left(p^{\prime}\right) \psi_{p}\left(x, p^{\prime}\right) d p^{\prime}\end{equation}
where \(\begin{equation}c\left(p^{\prime}\right)=\phi\left(p^{\prime}\right)\end{equation}\) [see Eq. (203)], or
\begin{equation}c\left(p^{\prime}\right)=\int_{-\infty}^{\infty} \psi_{p}^{*}\left(x, p^{\prime}\right) \psi(x) d x\end{equation}
In other words, we can expand a general wavefunction \(\begin{equation}\psi(x)\end{equation}\) as a linear combination of the eigenstates, \(\begin{equation}\psi_{p}\left(x, p^{\prime}\right)\end{equation}\), of the momentum operator. Equations (290) and (291) are again analogous to Eqs. (261) and (264), respectively, for square-integrable eigenstates. Likewise, the probability density of a measurement of \(\begin{equation}p \text { yielding the result } p^{\prime} \text { is }\left|c\left(p^{\prime}\right)\right|^{2}\end{equation}\), which is equivalent to the standard result \(\begin{equation}\left|\phi\left(p^{\prime}\right)\right|^{2}\end{equation}\). The probabilities are also properly normalized provided \(\begin{equation}\psi(x)\end{equation}\) is properly normalized [cf., Eq. (221)]: i.e.,
\begin{equation}\int_{-\infty}^{\infty}\left|c\left(p^{\prime}\right)\right|^{2} d p^{\prime}=\int_{-\infty}^{\infty}\left|\phi\left(p^{\prime}\right)\right|^{2} d p^{\prime}=\int_{-\infty}^{\infty}\left|\psi\left(x^{\prime}\right)\right|^{2} d x^{\prime}=1\end{equation}
Finally, if a mesurement of \(\begin{equation}p\end{equation}\) yields the value \(\begin{equation}p^{\prime}\end{equation}\) then the system is left in the corresponding momentum eigenstate, \(\begin{equation}\psi_{p}\left(x, p^{\prime}\right)\end{equation}\), immediately after the measurement.
Contributors
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)
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