Continuous Eigenvalues

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, $x$ and $p$--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 $\psi_x(x,x')$ be the eigenstate of $x$ corresponding to the eigenvalue $x'$. It follows that

$$x \psi_x(x,x') = x' \psi_x(x,x')$$ (277)

for all $x$. Consider the Dirac delta-function $\delta(x-x')$. We can write

$$x \delta(x-x') = x' \delta(x-x'),$$ (278)

since $\delta(x-x')$ is only non-zero infinitesimally close to $x=x'$. Evidently, $\psi_x(x,x')$ is proportional to $\delta(x-x')$. Let us make the constant of proportionality unity, so that

$$\psi_x(x,x') = \delta(x-x').$$ (279)

Now, it is easily demonstrated that

$$\int_{-\infty}^{\infty} \delta(x-x') \delta(x-x'') dx = \delta(x'-x'').$$ (280)

Hence, $\psi_x(x,x')$ satisfies the orthonormality condition

$$\int_{-\infty}^\infty \psi_x^\ast(x,x') \psi_x(x,x'') dx = \delta(x'-x'').$$ (281)

This condition is analogous to the orthonormality condition (263) satisfied by square-integrable eigenstates. Now, by definition, $\delta(x-x')$ satisfies

$$\int_{-\infty}^\infty f(x) \delta(x-x') dx = f(x'),$$ (282)

where $f(x)$ is a general function. We can thus write

$$\psi(x) = \int_{-\infty}^\infty c(x') \psi_x(x,x') dx',$$ (283)

where $c(x')=\psi(x')$, or

$$c(x') = \int_{-\infty}^\infty \psi_x^\ast(x,x') \psi(x) dx.$$ (284)

In other words, we can expand a general wavefunction $\psi(x)$ as a linear combination of the eigenstates, $\psi_x(x,x')$, 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 $x$ yielding the value $x'$ is $\vert c(x')\vert^{ 2}$, which is equivalent to the standard result $\vert\psi(x')\vert^{ 2}$. Moreover, these probabilities are properly normalized provided $\psi(x)$ is properly normalized [cf., Eq. (265)]: i.e.,

$$\int_{-\infty}^\infty \vert c(x')\vert^{ 2} dx'= \int_{-\infty}^\infty \vert\psi(x')\vert^{ 2} dx' =1.$$ (285)

Finally, if a measurement of $x$ yields the value $x'$ then the system is left in the corresponding displacement eigenstate, $\psi_x(x,x')$, immediately after the measurement: i.e., the wavefunction collapses to a ``spike-function'', $\delta(x-x')$, as discussed in Sect. 3.16.

Now, an eigenstate of the momentum operator $p\equiv-{\rm i} \hbar \partial/\partial x$ corresponding to the eigenvalue $p'$ satisfies

$$-{\rm i} \hbar \frac{\partial \psi_p(x,p')}{\partial x} = p' \psi_p(x,p').$$ (286)

It is evident that

$$\psi_p(x,p') \propto {\rm e}^{+{\rm i} p' x/\hbar}.$$ (287)

Now, we require $\psi_p(x,p')$ to satisfy an analogous orthonormality condition to Eq. (281): i.e.,

$$\int_{-\infty}^\infty \psi_p^\ast(x,p') \psi_p(x,p'') dx = \delta(p'-p'').$$ (288)

Thus, it follows from Eq. (210) that the constant of proportionality in Eq. (287) should be $(2\pi \hbar)^{-1/2}$: i.e.,

$$\psi_p(x,p') =\frac{ {\rm e}^{+{\rm i} p' x/\hbar}}{(2\pi \hbar)^{1/2}}.$$ (289)

Furthermore, according to Eqs. (202) and (203),

$$\psi(x) = \int_{-\infty}^\infty c(p') \psi_p(x,p') dp',$$ (290)

where $c(p') = \phi(p')$ [see Eq. (203)], or

$$c(p') = \int_{-\infty}^\infty \psi_p^\ast(x,p') \psi(x) dx.$$ (291)

In other words, we can expand a general wavefunction $\psi(x)$ as a linear combination of the eigenstates, $\psi_p(x,p')$, 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 $p$ yielding the result $p'$ is $\vert c(p')\vert^{ 2}$, which is equivalent to the standard result $\vert\phi(p')\vert^{ 2}$. The probabilities are also properly normalized provided $\psi(x)$ is properly normalized [cf., Eq. (221)]: i.e.,

$$\int_{-\infty}^\infty \vert c(p')\vert^{ 2} dp'= \int_{-\i... ...,dp' = \int_{-\infty}^\infty \vert\psi(x')\vert^{ 2} dx' =1.$$ (292)

Finally, if a mesurement of $p$ yields the value $p'$ then the system is left in the corresponding momentum eigenstate, $\psi_p(x,p')$, immediately after the measurement.