dk π ‚k

The density of allowed k values is therefore

1 ‚δ

ρ(k) = R+ . (4.140)

π ‚k

For our delta-shell example, a plot of ρ(k) looks like

4.3. COMPLETENESS OF EIGENFUNCTIONS 111

ka

ρ

3π

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¡¢¡¢¡¢¡¢

¢¡¡ ¡ ¡

2π ¡ ¢ ¡ ¡ ¡ ¢ ¡

¢¡¢¡¢¡¢¡¢¡

¡ ¡ ¡ ¡ ¡

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(R’a) π π

π 2π 3π ka r

The density of states for a system with resonances. The extended states are

so close in energy that we need an optical aid to resolve individual levels.

The almost-bound resonance levels have to squeeze in between them.

which is understood as the resonant bound states at ka = nπ superposed

on the background continuum density of states appropriate to a large box of

length (R ’ a). Each “spike” contains one extra state, so the average density

of states is that of a box of length R. We see that changing the potential

does not create or destroy eigenstates, it just moves them around.

The spike is not exactly a delta function because of level repulsion between

nearly degenerate eigenstates. The interloper elbows the nearby levels out of

the way, and all the neighbours have to make do with a bit less room. The

stronger the coupling between the states on either side of the delta-shell, the

stronger is the inter-level repulsion, and the broader the resonance spike.

Normalization Factor

We now evaluate

R

dr|ψk |2 = Nk ,

’2

(4.141)

0

so as to ¬nd the the normalized wavefunctions

χk = Nk ψk . (4.142)

Let ψk (r) be a solution of

d2

Hψ = ’ 2 + V (r) ψ = k 2 ψ (4.143)

dr

112 CHAPTER 4. LINEAR DIFFERENTIAL OPERATORS

satisfying the boundary condition ψk (0) = 0, but not necessarily the bound-

ary condition at r = R. Such a solution exists for any k. We scale ψk

by requiring that ψk (r) = sin(kr + δ) for r > R0 . We now use Lagrange™s

identity to write

R R

2

2

(k ’ k ) dr {(Hψk )ψk ’ ψk (Hψk )}

dr ψk ψk =

0 0

R

= [ψk ψk ’ ψk ψk ]0

= sin(kR + δ)k cos(k R + δ)

’k cos(kR + δ) sin(k R + δ). (4.144)

Here, we have used ψk,k (0) = 0, so the integrated out part vanishes at the

lower limit, and have used the explicit form of ψk,k at the upper limit.

Now di¬erentiate with respect to k, and then set k = k . We ¬nd

1 ‚δ

R

dr(ψk )2 = ’ sin 2(kR + δ) + k R +

2k . (4.145)

2 ‚k

0

In other words,

1 ‚δ 1

R

dr(ψk )2 = ’

R+ sin 2(kR + δ) . (4.146)

2 ‚k 4k

0

At this point, we impose the boundary condition at r = R. We therefore

have kR + δ = nπ and the last term on the right hand side vanishes. The

¬nal result for the normalization integral is therefore

1 ‚δ

R

dr|ψk |2 = R+ . (4.147)

2 ‚k

0

Observe that the same expression occurs in both the density of states and

the normalization integral.

The sum over the continuous spectrum in the completeness integral is

therefore

dn 2

∞ ∞

2

dk Nk ψk (r)ψk (r ) = dk ψk (r)ψk (r ). (4.148)

dk π

0 0

Both the density of states and the normalization factor have disappeared

from the end result. This is a general feature of scattering problems: The

4.3. COMPLETENESS OF EIGENFUNCTIONS 113

completeness relation must give a delta function when evaluated far from

the scatterer where the wavefunctions look like those of a free particle. So,

provided we normalize ψk so that it reduces to a free particle wavefunction

at large distance, the measure in the integral over k must also be the same

as for the free particle.

Including any bound states in the discrete spectrum, the full statement

of completeness is therefore

2 ∞

dk ψk (r) ψk (r ) = δ(r ’ r ).

ψn (r)ψn (r ) + (4.149)

π 0

bound states

Example: We will exhibit a completeness relation for a problem on the entire