2.1 Quantum Mechanics and Bound States
49
To get the correct eigenfunctions, we need to make a linear combination of odd and
even terms
η n (ξ) = aη 2n (ξ) + bη 2n+1 (ξ)
(2.37)
where a
2
+ b
2
= 1 and the radial wave function must be zero at r = 0 with
aη 2n (−r e ) + bη 2n+1 (−r e ) = 0 giving the second condition to determine the coefficients a and b. In fact, if r e = 0 then a = 0 and b = 1. The linear combination of
the even and odd eigenfunctions is, therefore, as follows:
η n (ξ) = A n N n H n (ξ)e
−ξ
2 /2
+ B n N n+1 H n+1 (n + 1)(ξ)e
−ξ
2 /2
(2.38)
giving
E n = A n ω [A n (n + 3/2) + B n (n + 5/2)]
(2.39)
where A n and B n are chosen to make η n (ξ r=r e ) = 0 and normalized. As an illustration,
let r e = 0 then we have half of the HO with the condition that at r = 0 the wave
function is zero for odd parity. This only occurs for odd values of n and therefore
η n (ξ) = (2n + 1)N 2n+1 H 2n+1 (ξ)e
−ξ
2 /2
(2.40)
giving energy eigenvalues, E n = ω(2n + 5/2). If one includes angular momentum,
the exact eigenenergies are given by the expression E n = ω(2k + l + 3/2) with l
being the angular momentum quantum number. This is equivalent to Eq. 2.36 with
a replacement of l by an odd integer. For an even and an odd value of l, one has
respectively an even and an odd parity defined like p = (−1)
l .
One should emphasize here again that in the above description of the HO we used
the variable ξ with range (−∞, ∞) with the potential being symmetric about ξ = 0.
2.2 Quantum Elastic Scattering
2.2.1 The Coulomb Potentials and the Hydrogen Atom
The Coulomb potential:
V (r) = ±q/r q = positive constant
is even more emblematic and important. For example, the solution for the attractive
V (r) = −q/r = −Ze
2
/r potential of Z protons and one electron can be obtained in
49
To get the correct eigenfunctions, we need to make a linear combination of odd and
even terms
η n (ξ) = aη 2n (ξ) + bη 2n+1 (ξ)
(2.37)
where a
2
+ b
2
= 1 and the radial wave function must be zero at r = 0 with
aη 2n (−r e ) + bη 2n+1 (−r e ) = 0 giving the second condition to determine the coefficients a and b. In fact, if r e = 0 then a = 0 and b = 1. The linear combination of
the even and odd eigenfunctions is, therefore, as follows:
η n (ξ) = A n N n H n (ξ)e
−ξ
2 /2
+ B n N n+1 H n+1 (n + 1)(ξ)e
−ξ
2 /2
(2.38)
giving
E n = A n ω [A n (n + 3/2) + B n (n + 5/2)]
(2.39)
where A n and B n are chosen to make η n (ξ r=r e ) = 0 and normalized. As an illustration,
let r e = 0 then we have half of the HO with the condition that at r = 0 the wave
function is zero for odd parity. This only occurs for odd values of n and therefore
η n (ξ) = (2n + 1)N 2n+1 H 2n+1 (ξ)e
−ξ
2 /2
(2.40)
giving energy eigenvalues, E n = ω(2n + 5/2). If one includes angular momentum,
the exact eigenenergies are given by the expression E n = ω(2k + l + 3/2) with l
being the angular momentum quantum number. This is equivalent to Eq. 2.36 with
a replacement of l by an odd integer. For an even and an odd value of l, one has
respectively an even and an odd parity defined like p = (−1)
l .
One should emphasize here again that in the above description of the HO we used
the variable ξ with range (−∞, ∞) with the potential being symmetric about ξ = 0.
2.2 Quantum Elastic Scattering
2.2.1 The Coulomb Potentials and the Hydrogen Atom
The Coulomb potential:
V (r) = ±q/r q = positive constant
is even more emblematic and important. For example, the solution for the attractive
V (r) = −q/r = −Ze
2
/r potential of Z protons and one electron can be obtained in