676
15 The Principles of Quantum Mechanics. I. De Broglie Waves and the Schrödinger Equation
relation pick the other values, then Eq. (15.4-4) gives a solution to the Schrödinger
equation.
We must now apply the appropriate boundary conditions to our solution. The solution
must be finite all values of x, including very large values of |x|. If the series is permitted
to have infinitely many terms, it is found that the series becomes large very rapidly for
large values of |x| and the wave function fails to remain finite. 5 Therefore, the series
cannot have infinitely many terms. It must be a polynomial.
We cannot simply require all coefficients past a certain point in the series to vanish if
this violates the recursion relation. The function would then fail to satisfy the Hermite
equation. Let us assume that c v+2 is a vanishing coefficient and that c v does not
vanish. The numerator in the right-hand side of Eq. (15.4-6) must then vanish for
n v:
2av + a − b 2av + a − 2mE/ ¯
h
2
2a(v + 1/2) − 2mE/ ¯
h
2
0
(15.4-7)
We can solve this equation for the energy eigenvalues:
E E v
¯
h
2
2m
2a
v +
1
2
h
2π
k
m
v +
1
2
hν class
v +
1
2
(15.4-8)
where ν class is the frequency of the oscillator as predicted by classical mechanics,
given in Eq. (14.2-29), and where the quantum number ν equals 0, 1, 2, 3, and so on.
The energy is quantized and there is a zero-point energy:
E 0
1
2
hν class (zero-point energy)
(15.4-9)
The recursion relation relates the coefficients of even powers to each other, and the
coefficients of the odd powers to each other. Since there is only one recursion relation,
each polynomial can contain only even powers or only odd powers. If the power series
contained both even and odd powers, the recursion relation would cause only the even
powers to terminate or the odd powers to terminate, and the other powers would not
terminate, violating the finiteness condition. The series in our solution now represents
a polynomial that contains only even powers of x or only odd powers of x. These
polynomials are called Hermite polynomials.
In the case of the particle in a box, the quantization was produced by the condition
that the wave function must be continuous at the ends of the box. In the case of the
harmonic oscillator, the energy quantization is produced by the condition that the wave
function must be finite for all values of x, which causes the series to terminate.
E X A M P L E 15.5
If a harmonic oscillator has an electric charge, it can absorb or emit electromagnetic radiation.
a. Find a formula for the frequency of a photon with energy equal to the difference in energy
between the v 0 state and the v 1 state.
b. How does the frequency of the photon compare with the classical frequency of the oscillator? How do you interpret this comparison?
5 This assertion is not obvious. See I. N. Levine, Quantum Chemistry, 5th ed., Prentice-Hall, Englewood
Cliffs, NJ, 2000, pp. 67–68.
15 The Principles of Quantum Mechanics. I. De Broglie Waves and the Schrödinger Equation
relation pick the other values, then Eq. (15.4-4) gives a solution to the Schrödinger
equation.
We must now apply the appropriate boundary conditions to our solution. The solution
must be finite all values of x, including very large values of |x|. If the series is permitted
to have infinitely many terms, it is found that the series becomes large very rapidly for
large values of |x| and the wave function fails to remain finite. 5 Therefore, the series
cannot have infinitely many terms. It must be a polynomial.
We cannot simply require all coefficients past a certain point in the series to vanish if
this violates the recursion relation. The function would then fail to satisfy the Hermite
equation. Let us assume that c v+2 is a vanishing coefficient and that c v does not
vanish. The numerator in the right-hand side of Eq. (15.4-6) must then vanish for
n v:
2av + a − b 2av + a − 2mE/ ¯
h
2
2a(v + 1/2) − 2mE/ ¯
h
2
0
(15.4-7)
We can solve this equation for the energy eigenvalues:
E E v
¯
h
2
2m
2a
v +
1
2
h
2π
k
m
v +
1
2
hν class
v +
1
2
(15.4-8)
where ν class is the frequency of the oscillator as predicted by classical mechanics,
given in Eq. (14.2-29), and where the quantum number ν equals 0, 1, 2, 3, and so on.
The energy is quantized and there is a zero-point energy:
E 0
1
2
hν class (zero-point energy)
(15.4-9)
The recursion relation relates the coefficients of even powers to each other, and the
coefficients of the odd powers to each other. Since there is only one recursion relation,
each polynomial can contain only even powers or only odd powers. If the power series
contained both even and odd powers, the recursion relation would cause only the even
powers to terminate or the odd powers to terminate, and the other powers would not
terminate, violating the finiteness condition. The series in our solution now represents
a polynomial that contains only even powers of x or only odd powers of x. These
polynomials are called Hermite polynomials.
In the case of the particle in a box, the quantization was produced by the condition
that the wave function must be continuous at the ends of the box. In the case of the
harmonic oscillator, the energy quantization is produced by the condition that the wave
function must be finite for all values of x, which causes the series to terminate.
E X A M P L E 15.5
If a harmonic oscillator has an electric charge, it can absorb or emit electromagnetic radiation.
a. Find a formula for the frequency of a photon with energy equal to the difference in energy
between the v 0 state and the v 1 state.
b. How does the frequency of the photon compare with the classical frequency of the oscillator? How do you interpret this comparison?
5 This assertion is not obvious. See I. N. Levine, Quantum Chemistry, 5th ed., Prentice-Hall, Englewood
Cliffs, NJ, 2000, pp. 67–68.