The normalization of wave functions of the continuous spectrum

The continuous spectrum of a quantum mechanical (QM) system contains important information on the system. However, the normalization of wave functions of the continuous spectrum is often difficult and therefore is often omitted from the books on QM. We discuss this problem and show how we can overcome this problem technically, with some examples.


Introduction
A physical system in quantum mechanics is characterized by its Hamiltonian, which is a self-adjoint operator in a Hilbert Space. Assume for now that the system is one dimensional. Then the Hilbert space are complex functions of a real variable ψ(x), such that The symbol * stands for complex conjugate. The functions ψ(x) are the states of the systems. If we define we say that we have normalized the wave function and we have Remark 1 In three dimensions, for central potentials, the differential equations for the radial part of the eigenfunction is identical to the equation for one dimension except by the centrifugal barrier. Also, the integrations in equations (1) and (3) are from 0 to ∞.
Let u E (x) be an eigenfunction of a Hamiltonian H, i.e., Sometimes, only some (or none) of the eigenfunctions are normalizable in the usual sense, the others (or all of them) are such that * Correspondence email address: coutinho@dim.fm.usp.br The values of E for such eigenfunctions form a continuum interval, as we shall see. The eigenfunctions u E (x) are said to be eigenfunctions of the continuous spectrum of the Hamiltonian operator.
A very important property of the Hamiltonian is that it has a complete set of eigenfunctions. They form a complete set of functions in the sense that any state of the system can be expanded in a series, integrals or a mixture of both series and integrals of these functions, that is, where and However, u Ei and u E (x) must be properly normalized. The summation over the bound states i (equation (7)) and the values of E in equation (8) depend on the system under study, as we shall see later.
The eigenfunctions on the second term of the equation (6) belong to the continuous spectrum and it is important that the eigenfunctions be normalized, in a special way, as we will explain below.

The location of the continuous spectrum and its normalization
Consider the problem of a particle in one dimension subjected to a potential that has the form shown in Figure 1. Almost all books on quantum mechanics show that for energies between E −1 and E 0 there exist eigenfunctions that are bound states, whose wave functions ψ Ei (x) are normalizable, that is, If we have two or more normalizable states with the same energy E i = E j we can arrange things so the In the above formula, δ ij is 1 if i = j and zero otherwise.
For energies greater than E 0 , the eigenfunctions are not normalizable as in equation (2) because these wave functions u E (x) are such that These eigenfunctions are said to be of the continuous spectrum of the Hamiltonian that in the above case extends from E 0 to ∞. Figure 2 shows an even more interesting situation. In this case, the potential is given by [1] In this case, the system have no normalizable eigenfunctions and the spectrum is continuous from −∞ < E < ∞.
For the eigenfunctions of the continuous spectrum to be useful, however, we have to "normalize" them in a special way. It is possible to proof [2, p. 99] that, when correctly normalized, any eigenfunction of the continuous spectrum, say u E (x), is such that where δ is the Dirac delta function. However, as stressed above, one has to correctly normalize the u E (r). This involves the difficult evaluation of divergent integrals to show that the resulting mathematical objects are δ functions [3, p. 237] [4,5], that is, that in fact they obey equation (13). The purpose of this article is to show ways of performing these difficult calculations. In the next section, we show three methods of normalizing the wave functions of the continuous spectrum.

Calculating divergent integrals
This is, in general, the most difficult method and should be used parsimoniously. It should be noted that divergent integrals sometimes produces distributions but they are not always the Dirac delta function distribution [6, p. 71], that is, they do not obey equation (13). We shall elaborate below about how to proof that a certain normalization results in a delta function.

Example 1) Normalize the following eigenfuntions of the continuous spectrum of the following problem [7]
Consider the Hamiltonian To normalize φ we have to use the relation (14) so that A = 2 π 1/2 .
To prove relation (14) we show that it obeys the following defining property of the Dirac Delta function, namely: This property is actually a rigorous definition of the Dirac delta function. If the reader feels uncomfortable with this definition, he/she should read Remark 2 for clarification.
To demonstrate (14) we calculate The details of this calculation are presented in Brownstein [7] and repeated here for completeness and also because there are a few misprints in his article Consider the first integral and change variable Then the first integral becomes We have used (see [8, p. 285 The second integral gives zero in the limit L → ∞, as can be easily verified. In fact, the second integral after the transformation that vanishes because the lower limit of integration approaches ∞ as L → ∞.

Example 2) Normalize the following eigenfunction of the continuous spectrum of the following problem
Consider the Hamiltonian − d 2 and hence The proof of this relation is identical to the proof of relation (16).

Example 3) Normalize the following eigenfuntions of the continuous spectrum
Consider the Hamiltonian − d 2 dx 2 defined for 0 < x < ∞ with boundary condition ψ (0) = αψ(0). The eigenfunctions of this problem are where k 2 = E.
To normalize this function we need the relations (14), (23) and the following relation To prove equation (25) we replace the upper limit of the integral by L, multiply by a function f (k 1 ) and integrate to get where we have used the relation (see [9, p. 310 After some more calculations we get the properly normalized eigenfunction which is We shall give another demonstration of this formula in the example 5.
We have only eigenfunctions of the continuous so that we can write equations (6) and (8) as and The correct normalization of a eigenfunction of the continuous spectrum of a problem such that V (x) → 0 as x → ∞ is obtained by matching the eigenfunction to equation (33) when x → ∞ [1].

Example 4) A square well
Consider the case when V (x) is given by V (x) = −V 0 , (V 0 > 0) for 0 < x < a and V (x) = 0 for x > a with boundary condition ψ(0) = 0 (an impenetrable barrier to the left).
The eigenfuction for E > 0 is given by and where K = (E + V 0 ) 1/2 and k = E 1/2 . The ratio B(k)/A(k) = C(k) can be determined by imposing continuity of the eigenfunction at x = a and by imposing continuity of the derivative of the eigenfunction at x = a.
To proof that this recipe produces the proper normalized eigenvalue we proceed as follows: Consider the eigenfunction at two values of the energy k 1 and k 2 . Then using the Schrodinger equation we get At x = 0 the left hand side of the previous equation gives zero and using the asymptotic value of the eigenfunctions, namely C(k 1 ) sin(k 1 x+δ 1 ) and C(k 2 ) sin(k 2 x+δ 2 ), we get Now, we want to show that the mathematical object To do this, we have to show that this object multiplied by any good function f (k 2 ) and integrated in R in this limit gives f (k 1 ).
The second part gives zero because the term between brackets oscillates violently as R → ∞.
The first part also goes to zero if k 1 = k 2 . However when k 1 → k 2 , δ 1 cancels with δ 2 and we are left with Therefore, in the limit as R → ∞, we get as we showed in equation (16) above. A different approach to this problem is shown in Remark 2.
Hence the proper normalization is obtained by making |C(k)| = (2/π) 1/2 as claimed. The eigenfunctions for E > 0 are then given by (sin δ cos kx + cos δ sin kx) where K = (E + V 0 ) 1/2 and k = E 1/2 . The above calculation was adapted from the book by Perelomov and Zel'dovich [10, p. 53-55]. It is however more convenient to use the following result, that can be found in [3, p. 247] and that is based in the following theorem: Theorem 1 (Friedman [3]) The eigenfunctions of the continuous spectrum u k (x) of the operator − d 2 dx 2 + V (x), where V (x) vanishes when x → ∞ and u k (x) satisfies some boundary condition at x = 0 are properly normalized if they are solutions of the equation and behave at infinity as Let's apply this theorem to the problem considered above.
Using equations (42) and (43) we can write and u k (x) = 1 2π which is normalized according to the above theorem.

The Titchmarsh-Weil m-coefficient (Everitt [11])
To normalize the eigenfunctions of the continuous spectrum of a problem that result from a potential like the one shown in Figure 2 the following method is easier. We mean easier in the sense that it does not require evaluation of divergent integrals but may be very laborious. We rewrite equation (6) as The function ρ(E) is called spectral density and is related to the m-coefficient, that we shall define later in section 3.3.1, by the formula: Recipe 1: where Im{m(E + i )} means imaginary part of m(E + i ).
In formula (48) the normalization of u E (x) is arbitrary but ρ(E) must be calculated and it contains all the information we need about the continuous spectrum of the problem.
The normalization of u E (x) is obtained from equation (48) as follows where we have replaced C(E) in equation (49) by its value given by equation (8). So The function is properly normalized. We shall illustrate now the method with examples. (An excellent example, the only one we could find in the literature dedicated to physics teaching, is given in reference [1].)

How to calculate the m-coefficient
We start with an operator given by its action (that is, what it does when acting on the function) and a boundary condition For notational simplicity, we shall consider only the case where the operator has only continuous spectrum.
Given a function f (x) we can write formula (48) as and its inverse where u E (x) are eigenfunctions of the operator (53), that is, The normalization of u E (x) is arbitrary, but ρ(E) must be calculated using equation (49). We now sketch how to calculate m(E).
First we calculate two eigenfunctions of the operator, u E (x) and γ E (x), with Im{E} > 0, such that and More generally we want two fundamental solutions of the equation. That is, u E (x) and γ E (x) must be linearly independent (Wronskian not zero) solutions of equation (57).
Recipe 2: The m-coefficient is determined uniquely by the condition that γ E (x) + m(E)u E (x) is square integrable, that is, belongs to L 2 (0.∞).

Example 5) As a first example of the procedure let's return to the Example 3 of section 3.1.
We want to find the normalization of the eigenfunctions of the operator with boundary condition u k (0) = αu k (0). It is important to remember that k 2 is complex with Im{k} > 0. Two fundamental solutions of equation (60) are and Note that the function (61) is an eigenvalue of equation (60) that satisfies the boundary condition u k (0) = αu k (0).
The m-coefficient can then be easily determined This is square integrable if To see this, note that Im{k} > 0, therefore Re{e −ikx } = e − Im{k}x but Re{e ikx } = e Im{k}x and so equation (64) has to be imposed. Re means real part.
From equation (64), we get So that using equation (49) We now return to equation (50) Remembering that E = k 2 so that dE = 2 k dk and using equation (66), we get (67) This is the same as equation (28).

Example 6) Normalize the functions of the Hamiltonian given by
where the potential V (x) is given by −F x, x > a and the boundary condition u E (0) = 0. This problem was completely solved in Dean and Fulling [1] or Titchmarsh [12, p. 92]. However here we describe only the solution for which V 0 = 0 and we follow Dean and Fulling [1].
First we put V 0 = 0 in equation (69) and change variables and Then equation with boundary condition ψ(0) = 0 becomes which is the Airy's equation [13]. The Airy's equation has two linearly independet solutions: Ai(z) and Bi(z).
We need to find combinations of Ai(z) and Bi(z) that satisfy (58) with α = 1 and (59). Write them as Now we invert (74) and (75) to write where Taking advantage of the fact that the Wronskian of the Airy function is W [Ai(z), Bi(z)] = π −1 , we calculate We now have two functions u E (x) and γ E (x) that satisfy the differential equation (72) and the boundary conditions (58) and (59).
We now must find a linear combination of these solutions that is square integrable. This linear combination found by asymptotic analysis (see Dean and Fulling [1]) is Readers unfamiliar with asymptotic analysis should consult Wasow [14], Olver [15, p.54] or Mazzitelli et al. [16].
Hence, by Recipe 2, and, by Recipe 1, (89) By replacing in equation (52) u E (x) given by equation (74) and ρ 1/2 (E) given by the square root of equation (89), we get which completes the solution.

Remark 2 Divergent integrals and the delta function
In section 3.1 we considered the following integral ∞ 0 sin(k 1 x) sin(k 2 x)dx (91) which in the usual sense is at least dubious. We want to show that it is in fact equal to π 2 δ(k 1 − k 2 ). The symbol δ(k 1 − k 2 ) is a distribution in this case the so-called Dirac Delta Function. A distribution is a linear functional that when acting on a good function produces a number. A simple example of a functional is one generated by a function F (x). In fact, given any good function f (x), this functional is given by A delta function is a functional that acts in a function as follows.
Note that there is no ordinary function that generates this functional. So, we want to show that the mathematical object ∞ 0 sin(k 1 x) sin(k 2 x)dx is a delta function. That is, we want to show that Another approach to obtain the δ function can be used, and this was probably noticed by attentive readers. In this approach, the δ function, δ(x), is defined as limits of sequences that approach a function that is 0 if x = 0 and ∞ if x = 0.
Note that, when k 1 → k 2 , equation (37) becomes and, therefore, taking the limit as R → ∞ we get from equation (97) Many other sequences that tend to a δ function can be found in [2, p.103] and the whole approach is explained in [17].