2.5 Basic Properties of Bound States
43
Exercise XI
Show that for the Hermitian Hamiltonian (2.5), k 2 has to be real if u(k, r) is
a bound state.
It is sufficient to study the asymptotic behavior of wave functions at infinity to
be able to know whether a one-body system can sustain a zero-energy bound state
or not. In the neutron case, the solutions of Eq. (2.5) for k = 0 read (see Eqs. (2.53)
and (2.54)):
u(k, r) = Ar
+ Br
− , r ≥ R .
(2.128)
If = 0, then Eq. (2.128) never vanishes for r → +∞, so that no bound state of
zero energy can exist therein. Conversely, if A = 0 in (2.128) when > 0, then
u(k, r) is the square-integrable wave function of a bound state.
For proton states, solutions of Eq. (2.5) when k = 0 are provided by Eqs. (2.56)
and (2.57)):
u(k, r) = A r
1/4 (2
√
v c r) i 2 (2
√
v c r)
+B r
1/4 (2
√
v c r) k 2 (2
√
v c r) , r ≥ R .
(2.129)
In the above equation, (2
√
v c r) k 2 (2
√
v c r) ∼ exp(−2
√
v c r) up to an
unimportant constant when r → +∞ ∀ ≥ 0. Therefore, u(k, r) is a bound state
if A = 0. Charged particles can then bear zero-energy bound states in all partial
waves.
2.5.1 Number of Bound States of a One-Body Hamiltonian
One will show now that if v c ≥ 0 u(0, r) is bound, then the number of bound
states is finite and equals to the number of zeros of the u(0, r), including the zero at
infinity. u(0, r) is either a zero-energy bound state (see Eqs. (2.128) and (2.129)) or
|u(0, r)| → +∞ for r → +∞, due to the properties of Coulomb wave functions
for k → 0. Thus, u(0, r) cannot oscillate along the real r-axis in the asymptotic
region, so that its number of zeros is necessarily finite.
To show that the number of bound states is finite, one needs to consider the
partial overlaps of two different states u(k a , r) and u(k b , r), not necessarily bound,
but with k 2
a < k 2
b ≤ 0. Note that no complex conjugate of wave functions will enter
radial overlaps because u(k a , r) and u(k b , r) wave functions can be chosen to bear
real values. A straightforward manipulation of Eq. (2.2) provides with the following
equation:
r 2
r 1
u
(k b , r)u(k a , r) − u
(k a , r)u(k b , r)
dr
= (k
2
a − k
2
b )
r 2
r 1
u(k a , r)u(k b , r) dr
43
Exercise XI
Show that for the Hermitian Hamiltonian (2.5), k 2 has to be real if u(k, r) is
a bound state.
It is sufficient to study the asymptotic behavior of wave functions at infinity to
be able to know whether a one-body system can sustain a zero-energy bound state
or not. In the neutron case, the solutions of Eq. (2.5) for k = 0 read (see Eqs. (2.53)
and (2.54)):
u(k, r) = Ar
+ Br
− , r ≥ R .
(2.128)
If = 0, then Eq. (2.128) never vanishes for r → +∞, so that no bound state of
zero energy can exist therein. Conversely, if A = 0 in (2.128) when > 0, then
u(k, r) is the square-integrable wave function of a bound state.
For proton states, solutions of Eq. (2.5) when k = 0 are provided by Eqs. (2.56)
and (2.57)):
u(k, r) = A r
1/4 (2
√
v c r) i 2 (2
√
v c r)
+B r
1/4 (2
√
v c r) k 2 (2
√
v c r) , r ≥ R .
(2.129)
In the above equation, (2
√
v c r) k 2 (2
√
v c r) ∼ exp(−2
√
v c r) up to an
unimportant constant when r → +∞ ∀ ≥ 0. Therefore, u(k, r) is a bound state
if A = 0. Charged particles can then bear zero-energy bound states in all partial
waves.
2.5.1 Number of Bound States of a One-Body Hamiltonian
One will show now that if v c ≥ 0 u(0, r) is bound, then the number of bound
states is finite and equals to the number of zeros of the u(0, r), including the zero at
infinity. u(0, r) is either a zero-energy bound state (see Eqs. (2.128) and (2.129)) or
|u(0, r)| → +∞ for r → +∞, due to the properties of Coulomb wave functions
for k → 0. Thus, u(0, r) cannot oscillate along the real r-axis in the asymptotic
region, so that its number of zeros is necessarily finite.
To show that the number of bound states is finite, one needs to consider the
partial overlaps of two different states u(k a , r) and u(k b , r), not necessarily bound,
but with k 2
a < k 2
b ≤ 0. Note that no complex conjugate of wave functions will enter
radial overlaps because u(k a , r) and u(k b , r) wave functions can be chosen to bear
real values. A straightforward manipulation of Eq. (2.2) provides with the following
equation:
r 2
r 1
u
(k b , r)u(k a , r) − u
(k a , r)u(k b , r)
dr
= (k
2
a − k
2
b )
r 2
r 1
u(k a , r)u(k b , r) dr