Continuous Spectra

Up to now, we have studiously avoided dealing with observables possessing eigenvalues that lie in a continuous range, rather than having discrete values. The reason for this is that continuous eigenvalues imply a ket space of non-denumerably infinite dimension. Unfortunately, continuous eigenvalues are unavoidable in quantum mechanics. In fact, the most important observables of all--namely position and momentum--generally have continuous eigenvalues. (See the following chapter.) Fortunately, many of the results that we obtained previously for a finite-dimensional ket space with discrete eigenvalues can be generalized to ket spaces of non-denumerably infinite dimensions.

Suppose that $\xi$ is an observable with continuous eigenvalues. We can still write the eigenvalue equation as

$$\xi \,\vert\xi'\rangle =\xi'\, \vert\xi'\rangle.$$ (1.87)

But, $\xi'$ now takes a continuous range of values. Let us assume, for the sake of simplicity, that $\xi'$ can take any value. The orthonormality condition (1.53) generalizes to

$$\langle \xi'\vert\xi''\rangle = \delta(\xi'-\xi''),$$ (1.88)

where $\delta(x)$ denotes the famous Dirac delta function [32,92], and satisfies

$$\delta(x\neq 0) =0,$$ (1.89)

$$\int_{-\infty}^\infty dx\,\delta(x) =1,$$ (1.90)

$$\int_{-\infty}^\infty dx\,f(x)\,\delta(x-x') =f(x')$$ (1.91)

for any function, $f(x)$ , that is well behaved at $x=x'$ . The Dirac delta function is a generalized function [104] that can be realized in many equivalent limiting forms. For instance,

$$\delta(x) = \lim_{\eta\rightarrow 0} \frac{1}{\pi}\,\frac{\eta}{x^{\,2}+\eta^{\,2}},$$ (1.92)

$$\delta(x) =\lim_{\eta\rightarrow 0} \frac{1}{\pi}\,\frac{\sin(x/\eta)}{x}.$$ (1.93)

Note from Equations (1.87) and (1.88) that there are a non-denumerably infinite number of mutually orthogonal eigenstates of $\xi$ . Hence, the dimensionality of ket space is non-denumerably infinite. Furthermore, eigenstates corresponding to a continuous range of eigenvalues cannot be normalized such that they have unit norms. In fact, it is clear from Equation (1.88), together with the well-known fact that $\delta(0)\rightarrow \infty$ , that these eigenstates have infinite norms. In other words, they are infinitely long. This is the major difference between eigenstates in a finite-dimensional and an infinite-dimensional ket space. The extremely useful relation (1.57) generalizes to

$$\int d\xi' \,\vert\xi'\rangle\langle \xi'\vert = 1.$$ (1.94)

(See Exercise 18.) Note that a summation over discrete eigenvalues goes over into an integral over a continuous range of eigenvalues. The eigenstates $\vert\xi'\rangle$ must form a complete set if $\xi$ is to be an observable. It follows that any general ket can be expanded in terms of the $\vert\xi'\rangle$ . In fact, the expansions (1.54)-(1.56) generalize to give

$$\vert A\rangle = \int d\xi'\,\vert\xi'\rangle\langle \xi'\vert A\rangle,$$ (1.95)

$$\langle A\vert = \int d\xi'\,\langle A\vert\xi'\rangle \langle \xi'\vert,$$ (1.96)

$$\langle A\vert A\rangle = \int d\xi'\,\langle A\vert\xi'\rangle\langle \xi'\vert A\rangle = \int d\xi' \,\vert\langle A\vert\xi'\rangle\vert^{\,2},$$ (1.97)

respectively. These results also follow simply from Equation (1.94). We have seen that it is not possible to normalize the eigenstates $\vert\xi'\rangle$ such that they have unit norms. Fortunately, this convenient normalization is still possible for a general state ket. In fact, according to Equation (1.97), the normalization condition can be written

$$\langle A\vert A\rangle =\int d\xi' \,\vert\langle A\vert\xi'\rangle\vert^{\,2} = 1.$$ (1.98)

We have now studied observables whose eigenvalues take a discrete number of values, as well as those whose eigenvalues take a continuous range of values. There are a number of other cases that we could look at. For instance, observables whose eigenvalues can take a (finite) continuous range of values, plus a set of discrete values. Such cases can be dealt with using a fairly straightforward generalization of the previous analysis [32].