Section 3-5
Section 3-5 Quantum-Mechanical Average Value of the Potential Energy
3-4.J Summary of Solution of Harmonic-OscillatorSchr ¨odinger Equation
The detailed solution is so long that the reader may have lost the broad outline.
The basic steps were:
1. Determine the asymptotic behavior of the Schr¨odinger equation and of ψ. This produces a gaussian factor exp(−y2/2) times a function of y, f (y).
2. Obtain a differential equation for the rest of the wavefunction, f (y).
3. Represent f (y) as a power series in y, and find a recursion relation for the coefficients in the series. The symmetries of the wavefunctions are linked to the symmetries of the series.
4. Force the power series to be finite (i.e., polynomials) so as not to spoil the asymptotic behavior of the wavefunctions. This leads to a relation between α and β that produces uniformly spaced, quantized energy levels.
5. Recognize the polynomials as being Hermite polynomials, and utilize some of the known properties of these functions to establish orthogonality and normalization constants for the wavefunctions.
EXAMPLE 3-3 Which of the following expressions are, by inspection, unacceptable eigenfunctions for the Schr¨odinger equation for the one-dimensional harmonic oscillator?
a) (64y6 − 480y4 + 720y2 − 120) exp(y2/2) b) (64y6 − 480y5 + 720y3 − 120) exp(−y2/2) c) (64y6 − 480y4 + 720y2 − 120) exp(−y2/2) SOLUTION a) is unacceptable because exp(y2/2) blows up at large |y|.
b) is unacceptable because the polynomial contains terms of both even and odd powers.
c) is acceptable.
3-5 Quantum-Mechanical Average Value ofthe Potential Energy We showed in Section 3-2 that the classical harmonic oscillator stores, on the average, half of its energy as kinetic energy, and half as potential. We now make the analogous comparison in the quantum-mechanical system for the ground (n = 0) state.
The wavefunction is ψ0(x) = (β/π)1/4 exp(−βx2/2)
(3-75)
and the probability distribution of the particle along the x coordinate is given by ψ2(x).
0
The total energy is constant and equal to E0 = 1 hν = (h/4π) k/m
(3-76)
2
Chapter 3 The One-Dimensional Harmonic Oscillator and the potential energy as a function of x is V (x) = 1 kx2
(3-77)
2
The probability for finding the oscillating particle in the line element dx around some point x1 is ψ2(x
0
1) d x, since ψ0 of Eq. (3-75) is normalized. Hence, the average value for the potential energy is simply the sum of all the potential energies due to all the elements dx, each weighted by the probability for finding the particle there:
∞
+∞
¯V = [prob. to be in dx][V at dx] dx =
ψ2 0 (x)V (x) d x
(3-78)
−∞
−∞
This is
+∞
¯
V = (β/π)1/2 · 1 k exp −βx2 x2 dx = β/π · 1 k · 1 π/β3
(3-79)
2
−∞
2
2
where we have referred to Appendix 1 to evaluate the integral. Using the definition of β2 (Eq. 3-17) we have
√
¯V = k/4β = (k/4) · h/(2π mk) = (h/8π) k/m
(3-80)
which is just one half of the total energy [Eq. (3-76)]. This means that the average value of the kinetic energy must also equal half of the total energy, since ¯
V + ¯ T = E.
We thus arrive at the important result that the ratio of average potential and kinetic energies is the same in the classical harmonic oscillator and the ground state of the quantum-mechanical system. This result is also true for the higher states. For other kinds of potential, the storage need not be half and half, but whatever it is, it will be the same for the classical and quantum-mechanical treatments of the system. We discuss this point in more detail later when we examine the virial theorem (Chapter 11 and Appendix 8).
3-6 Vibrations of Diatomic Molecules
Two atoms bonded together vibrate back and forth along the internuclear axis. The standard first approximation is to treat the system as two nuclear masses, m1 and m2 oscillating harmonically with respect to the center of mass. The force constant k is determined by the “tightness” of the bond, with stronger bonds usually having larger k.
The two-mass problem can be transformed to motion of one reduced mass, µ, vibrat ing harmonically with respect to the center of mass. µ is equal to m1m2/(m1 + m2).
The force constant for the vibration of the reduced mass remains identical to the force constant for the two masses, and the distance of the reduced mass from the center of mass remains identical to the distance between m1 and m2. Thus we have a very convenient simplification: We can use the harmonic oscillator solutions for a single oscillating mass µ as solutions for the two-mass problem. All of the wavefunctions and energy formulas are just what we have already seen except that m is replaced by µ. The practical consequence of this is that we can use the spectroscopically measured energy spacings between molecular vibrational levels to obtain the value of k for a molecule.
3-4.J Summary of Solution of Harmonic-OscillatorSchr ¨odinger Equation
The detailed solution is so long that the reader may have lost the broad outline.
The basic steps were:
1. Determine the asymptotic behavior of the Schr¨odinger equation and of ψ. This produces a gaussian factor exp(−y2/2) times a function of y, f (y).
2. Obtain a differential equation for the rest of the wavefunction, f (y).
3. Represent f (y) as a power series in y, and find a recursion relation for the coefficients in the series. The symmetries of the wavefunctions are linked to the symmetries of the series.
4. Force the power series to be finite (i.e., polynomials) so as not to spoil the asymptotic behavior of the wavefunctions. This leads to a relation between α and β that produces uniformly spaced, quantized energy levels.
5. Recognize the polynomials as being Hermite polynomials, and utilize some of the known properties of these functions to establish orthogonality and normalization constants for the wavefunctions.
EXAMPLE 3-3 Which of the following expressions are, by inspection, unacceptable eigenfunctions for the Schr¨odinger equation for the one-dimensional harmonic oscillator?
a) (64y6 − 480y4 + 720y2 − 120) exp(y2/2) b) (64y6 − 480y5 + 720y3 − 120) exp(−y2/2) c) (64y6 − 480y4 + 720y2 − 120) exp(−y2/2) SOLUTION a) is unacceptable because exp(y2/2) blows up at large |y|.
b) is unacceptable because the polynomial contains terms of both even and odd powers.
c) is acceptable.
3-5 Quantum-Mechanical Average Value ofthe Potential Energy We showed in Section 3-2 that the classical harmonic oscillator stores, on the average, half of its energy as kinetic energy, and half as potential. We now make the analogous comparison in the quantum-mechanical system for the ground (n = 0) state.
The wavefunction is ψ0(x) = (β/π)1/4 exp(−βx2/2)
(3-75)
and the probability distribution of the particle along the x coordinate is given by ψ2(x).
0
The total energy is constant and equal to E0 = 1 hν = (h/4π) k/m
(3-76)
2
Chapter 3 The One-Dimensional Harmonic Oscillator and the potential energy as a function of x is V (x) = 1 kx2
(3-77)
2
The probability for finding the oscillating particle in the line element dx around some point x1 is ψ2(x
0
1) d x, since ψ0 of Eq. (3-75) is normalized. Hence, the average value for the potential energy is simply the sum of all the potential energies due to all the elements dx, each weighted by the probability for finding the particle there:
∞
+∞
¯V = [prob. to be in dx][V at dx] dx =
ψ2 0 (x)V (x) d x
(3-78)
−∞
−∞
This is
+∞
¯
V = (β/π)1/2 · 1 k exp −βx2 x2 dx = β/π · 1 k · 1 π/β3
(3-79)
2
−∞
2
2
where we have referred to Appendix 1 to evaluate the integral. Using the definition of β2 (Eq. 3-17) we have
√
¯V = k/4β = (k/4) · h/(2π mk) = (h/8π) k/m
(3-80)
which is just one half of the total energy [Eq. (3-76)]. This means that the average value of the kinetic energy must also equal half of the total energy, since ¯
V + ¯ T = E.
We thus arrive at the important result that the ratio of average potential and kinetic energies is the same in the classical harmonic oscillator and the ground state of the quantum-mechanical system. This result is also true for the higher states. For other kinds of potential, the storage need not be half and half, but whatever it is, it will be the same for the classical and quantum-mechanical treatments of the system. We discuss this point in more detail later when we examine the virial theorem (Chapter 11 and Appendix 8).
3-6 Vibrations of Diatomic Molecules
Two atoms bonded together vibrate back and forth along the internuclear axis. The standard first approximation is to treat the system as two nuclear masses, m1 and m2 oscillating harmonically with respect to the center of mass. The force constant k is determined by the “tightness” of the bond, with stronger bonds usually having larger k.
The two-mass problem can be transformed to motion of one reduced mass, µ, vibrat ing harmonically with respect to the center of mass. µ is equal to m1m2/(m1 + m2).
The force constant for the vibration of the reduced mass remains identical to the force constant for the two masses, and the distance of the reduced mass from the center of mass remains identical to the distance between m1 and m2. Thus we have a very convenient simplification: We can use the harmonic oscillator solutions for a single oscillating mass µ as solutions for the two-mass problem. All of the wavefunctions and energy formulas are just what we have already seen except that m is replaced by µ. The practical consequence of this is that we can use the spectroscopically measured energy spacings between molecular vibrational levels to obtain the value of k for a molecule.