Section 2-5
Section 2-5 The Free Particle in One Dimension
We shall see that the correlation of wavefunction symmetries in such a manner as this is a powerful technique in understanding and predicting chemical behavior.
The splitting of energy levels resulting from barrier penetration is an extremely pervasive phenomenon in quantum chemistry. It occurs regardless of whether the barrier separates identical or nonidentical potential regions, i.e., regardless of whether the final system is symmetric or unsymmetric. When two atoms (N and N, or C and O) interact to form a molecule, the original atomic wavefunctions combine to form molecular wavefunctions in much the same way as was just described. One of these molecular wavefunctions may have an energy markedly lower than those in the corresponding atoms. Electrons having such a wavefunction will stabilize the molecule relative to the separated atoms.
Another case in which energy level splitting occurs is in the vibrational spectrum of ammonia. Ammonia is most stable in a pyramidal configuration, but is capable of inverting through a higher-energy planar configuration into an equivalent “mirror image” pyramid. Thus, vibrations tending to flatten out the ammonia molecule occur in a potential similar to the double well, except that in ammonia the potential is not discontinuous. The lowest vibrational energy levels are not sufficiently high to allow classical inversion of ammonia. However, these vibrational levels are split by interaction through barrier penetration just as quantum mechanics predicts. The energy required to excite ammonia from the lowest of these sublevels to its associated sublevel can be accurately measured through microwave spectroscopy. Knowledge of the level splittings in turn allows a precise determination of the height of the barrier to inversion in ammonia (see Fig. 2-14).
It is easy to anticipate the appearance of the solutions for the square well with central barrier for energies greater than the partition height. They will be sinusoidal waves, symmetric or antisymmetric in the well, and vanishing at the walls. Their wavelengths will be somewhat longer in the region of the partition than elsewhere because some of the kinetic energy of the particle is transformed to potential energy there. A sketch of the final results is given in Fig. 2-15.
EXAMPLE 2-5 Fig. 2-15 shows energy levels for states when the barrier has finite height. When the barrier is made infinitely high, the levels at E1 and E2 merge into one level. Where does the energy of that one level lie—below E1, between E1 and E2, or above E2—and why?
SOLUTION It lies above E2. When the barrier is finite, there is always some penetration, so λ is always at least a little larger than is the case for the infinite barrier. If λ is larger, E is lower.
2-5 The Free Particle in One Dimension
Suppose a particle of mass m moves in one dimension in a potential that is everywhere zero. The Schr¨odinger equation becomes −h2 d2ψ =Eψ
(2-39)
8π 2m dx2
Chapter 2 Quantum Mechanics of Some Simple Systems
Figure 2-14
Sketch of potential for inversion vibrational mode in ammonia. The lowest levels are split by tunneling. The low energy transition E1 is visible in the microwave region whereas the second transition E2 is visible in the infrared. E1 = 0.16 × 10−22 J; E2 = 7.15 × 10−22 J.
which has as solutions
√
ψ = A exp(±2πi 2mEx/ h)
(2-40)
or alternatively, trigonometric solutions
√
√
ψ = A sin(2π 2mEx/ h), ψ = A cos(2π 2mEx/ h)
(2-41)
As is most easily seen from the exponential forms (2-40), if E is negative, ψ will blow up at either +∞ or −∞, and so we reject negative energies. Since there are no boundary conditions, it follows that E can take on any positive value; the energies of the free particle are not quantized. This result would be expected from our earlier results on constrained particles. There we saw that quantization resulted from spatial constraints, and here we have none.
The constants A and A of Eqs. (2-40) and (2-41) cannot be evaluated in the usual way, since the solutions do not vanish at x = ±∞. Sometimes it is convenient to evaluate them to correspond to some experimental situation. For instance, suppose that one was working with a monoenergetic beam of electrons having an intensity of one electron every 10−6 m. Then we could normalize ψ of Eq. (2-40) so that 10−6 m |ψ|2dx=1
0
There is a surprising difference in the particle distributions predicted from expres sions (2-40) and (2-41). The absolute square ψ*ψ of the exponentials is a constant (A*A), whereas the squares of the trigonometric functions are fluctuating functions
Section 2-5 The Free Particle in One DimensionFigure 2-15 Wavefunctions for the infinite square well with finite partition.
of x. It seems sensible for a particle moving without restriction to have a constant probability distribution, and it seems absurd for it to have a varying probability distribution. What causes this peculiar behavior? It results from there being two independent solutions for each value of E (except E = 0). This degeneracy with respect to energy means that from a degenerate pair, ψ and ψ, one can produce any number of new eigenfunctions, ψ = aψ + bψ (Problem 2-11). In such a situation, the symmetry proof of Section 2-2 does not hold. However, there will always be an independent pair of degenerate wavefunctions that will satisfy certain symmetry requirements. Thus, in the problem at hand, we have one pair of solutions, the exponentials, which do have the proper symmetry since their absolute squares are constant. From this pair we can produce any number of linear combinations [one set being given by Eq. (2-41)], but these need not display the symmetry properties anymore.
The exponential solutions have another special attribute: A particle whose state is described by one of the exponentials has a definite linear momentum, whereas, when described by a trigonometric function, it does not. In Section 1-9, it was shown that the connection between classical and wave mechanics could be made if one related the classical momentum, px, with a quantum mechanical operator (h/2πi)d/dx. Now, for a particle to have a definite (sharp) value p for its momentum really means that, if we measure the momentum at some instant, there is no possibility of getting any value other than p. This means that the particle in the state described by ψ always has momentum p, no matter where it is in x; i.e., its momentum is a constant of motion, Chapter 2 Quantum Mechanics of Some Simple Systems just as its energy is. This corresponds to saying that there is an eigenvalue equation for momentum, just as for energy. Thus h dψ =pψ
(2-42)
2π i dx
The statement made earlier, that the exponential solutions correspond to the particle having sharp momentum, means that the exponentials (2-40) must be solutions to Eq. (2-42). This is easily verified:
√
√
h
d
±2πi 2mEx √
±2πi 2mEx
A exp = ± 2mE A exp 2π i dx
h
h
Thus, the positive and negative exponential solutions correspond to momentum values
√
√
of + 2mE and − 2mE, respectively, and are interpreted as referring to particle motion toward +∞ and −∞ respectively. Since energy is related to the square of the momentum, these two solutions have identical energies. (The solution for E = 0 corresponds to no momentum at all, and the directional degeneracy is removed.) A mixture of these states contains contributions from two different momenta but only one energy, so linear combinations of the exponentials fail to maintain a sharp value for momentum but do maintain a sharp value for energy.
EXAMPLE 2-6 An electron is accelerated along the x axis towards x = ∞ from rest through a potential drop of 1.000 kV.
a) What is its final momentum?
b) What is its final de Broglie wavelength?
c) What is its final wavefunction?
√
SOLUTION a) px = 2mE = [2(9.105 × 10−31 kg)(1.602 × 10−16 J)]1/2 = 1.708 × 10−23 kg m s−1 b) λ = h =
6.626×10−34 J s = 3.879 × 10−11 m
p
1.708×10−23 kg m s−1
√
c) ψ = A exp(+2πi 2mEx/ h)(choose + because moving towards x = +∞) = A exp(2πip/ h)x = A exp(2πix/λ) = A exp[2πi(2.578 × 1010 m−1)x].
2-6 The Particle in a Ring of Constant Potential Suppose that a particle of mass m is free to move around a ring of radius r and zero potential, but that it requires infinite energy to get off the ring. This system has only one variable coordinate—the angle φ. In classical mechanics, the useful quantities and relationships for describing such circular motion are those given in Table 2-1.
Comparing formulas for linear momentum and angular momentum reveals that the variables mass and linear velocity are analogous to moment of inertia and angular velocity in circular motion, where the coordinate φ replaces x. The Schr¨odinger equation for circular motion, then, is −h2 d2ψ (φ) =Eψ(φ)
(2-43)
8π 2I
dφ2
Section 2-6 The Particle in a Ring of Constant PotentialTABLE 2-1
Quantity
Formula
Units
Moment of inertia
I = mr2 g cm2 or kg m2
Angular velocity ω = φ/t = v/r
s−1
Angular momentum (linear mvr = I ω g cm2/s or erg s or J s momentum times orbit radius) which has, as solutions A exp(±ikφ)
(2-44)
or alternatively
A sin(kφ)
(2-45)
and
A cos(kφ)
(2-46)
where [substituting Eq. (2-45) or (2-46) into (2-43) and operating]
√
k = 2π 2I E/ h
(2-47)
Let us solve the problem first with the trigonometric functions. Starting at some arbitrary point on the ring and moving around the circumference with a sinusoidal function, we shall eventually reencounter the initial point. In order that our wavefunction be single valued, it is necessary that ψ repeat itself every time φ changes by 2π radians.
Thus, for φ given by Eq. (2-45), sin(kφ) = sin[k(φ + 2π)]
(2-48)
Similarly, for ψ given by Eq. (2-49)
cos(kφ) = cos(kφ + 2kπ)
(2-49)
Either of these relations is satisfied only if k is an integer. The case in which k = 0 is not allowed for the sine function since it then vanishes everywhere and is unsuitable.
However, k = 0 is allowed for the cosine form. The normalized solutions are, then,
√
ψ = (1/ π) sin(kφ), k = 1, 2, 3, . . .
√
ψ = (1/ π) cos(kφ), k = 1, 2, 3, . . .
√
ψ = (1/ 2π) (from the k = 0 case for the cosine)
(2-50)
Now let us examine the exponential form of ψ (Eq. 2-44). The requirement that ψ repeat itself for φ → φ + 2π gives A exp(±ikφ) = A exp[±ik(φ + 2π)] = A exp(±ikφ) exp(±2πik)
Chapter 2 Quantum Mechanics of Some Simple Systems or exp(±2πik) = 1
Taking the positive case and utilizing Eq. (2-3), we obtain cos(2π k) + i sin(2πk) = 1 or cos(2π k) = 1 and sin(2π k) = 0
Again, k must be an integer. (The same result arises by requiring that dψ/dφ repeat for φ → φ + 2π.) Thus, an alternative set of normalized solutions is
√
ψ = 1/ 2π exp(ikφ) k = 0, ±1, ±2, ±3, . . .
(2-51)
The energies for the particle in the ring are easily obtained from Eq. (2-47): E = k2h2/8π2I, k = 0, ±1, ±2, ±3, . . .
(2-52)
The energies increase with the square of k, just as in the case of the infinite square well potential. Here we have a single state with E = 0, and doubly degenerate states above, whereas, in the square well, we had no solution at E = 0, and all solutions were nondegenerate. The solution at E = 0 means that there is no finite zero point energy to be associated with free rotation, and this is in accord with uncertainty principle arguments since there is no constraint in the coordinate φ.
The similarity between the particle in a ring and the free particle problems is strik ing. Aside from the fact that in the ring the energies are quantized and the solutions are normalizable, there are few differences. The exponential solutions (2-51) are eigenfunctions for the angular momentum operator (h/2π i)d/dφ. The two angular momenta for a pair of degenerate solutions correspond to particle motion clockwise or counterclockwise in the ring. (The nondegenerate solution for E = 0 has no angular momentum, hence no ability to achieve degeneracy through directional behavior.) The particle density predicted by the exponentials is uniform in the ring, while that for the trigonometric solutions is not. Since the trigonometric functions tend to localize the particle into part of the ring, thereby causing φ = ∞, it is consistent that they are impure momentum states ( ang. mom. = 0). (Infinite uncertainty in the coordinate φ means that all values of φ in the range 0–2π are equally likely.)
EXAMPLE 2-7 Demonstrate that any two degenerate exponential eigenfunctions for a particle in a ring are orthogonal.
SOLUTION Such a pair of degenerate wavefunctions can be written as ψ+ = 1
√
exp(ikφ),
2π
ψ
2π
− = 1
√
exp(−ikφ). These are orthogonal if
2π
0
ψ∗
+ψ− dφ = 0 where we must use the complex conjugate of either one of the wavefunctions since ψ is complex. But ψ∗ + = ψ−, so
2π
2π
2π
ψ−ψ− dφ = 1 exp(−2iφ) dφ = 1 [cos(2φ) − i sin(2φ)]dφ
0
2π 0
2π 0
= 1 sin(2φ)|2π − i(− cos(2φ)|2π
2π
0
0
= 1 [sin(4π) − sin(0) + i cos(4π) − i cos(0)]
2π = 1 [0 − 0 + i − i] = 0.
2π
We shall see that the correlation of wavefunction symmetries in such a manner as this is a powerful technique in understanding and predicting chemical behavior.
The splitting of energy levels resulting from barrier penetration is an extremely pervasive phenomenon in quantum chemistry. It occurs regardless of whether the barrier separates identical or nonidentical potential regions, i.e., regardless of whether the final system is symmetric or unsymmetric. When two atoms (N and N, or C and O) interact to form a molecule, the original atomic wavefunctions combine to form molecular wavefunctions in much the same way as was just described. One of these molecular wavefunctions may have an energy markedly lower than those in the corresponding atoms. Electrons having such a wavefunction will stabilize the molecule relative to the separated atoms.
Another case in which energy level splitting occurs is in the vibrational spectrum of ammonia. Ammonia is most stable in a pyramidal configuration, but is capable of inverting through a higher-energy planar configuration into an equivalent “mirror image” pyramid. Thus, vibrations tending to flatten out the ammonia molecule occur in a potential similar to the double well, except that in ammonia the potential is not discontinuous. The lowest vibrational energy levels are not sufficiently high to allow classical inversion of ammonia. However, these vibrational levels are split by interaction through barrier penetration just as quantum mechanics predicts. The energy required to excite ammonia from the lowest of these sublevels to its associated sublevel can be accurately measured through microwave spectroscopy. Knowledge of the level splittings in turn allows a precise determination of the height of the barrier to inversion in ammonia (see Fig. 2-14).
It is easy to anticipate the appearance of the solutions for the square well with central barrier for energies greater than the partition height. They will be sinusoidal waves, symmetric or antisymmetric in the well, and vanishing at the walls. Their wavelengths will be somewhat longer in the region of the partition than elsewhere because some of the kinetic energy of the particle is transformed to potential energy there. A sketch of the final results is given in Fig. 2-15.
EXAMPLE 2-5 Fig. 2-15 shows energy levels for states when the barrier has finite height. When the barrier is made infinitely high, the levels at E1 and E2 merge into one level. Where does the energy of that one level lie—below E1, between E1 and E2, or above E2—and why?
SOLUTION It lies above E2. When the barrier is finite, there is always some penetration, so λ is always at least a little larger than is the case for the infinite barrier. If λ is larger, E is lower.
2-5 The Free Particle in One Dimension
Suppose a particle of mass m moves in one dimension in a potential that is everywhere zero. The Schr¨odinger equation becomes −h2 d2ψ =Eψ
(2-39)
8π 2m dx2
Chapter 2 Quantum Mechanics of Some Simple Systems
Figure 2-14
Sketch of potential for inversion vibrational mode in ammonia. The lowest levels are split by tunneling. The low energy transition E1 is visible in the microwave region whereas the second transition E2 is visible in the infrared. E1 = 0.16 × 10−22 J; E2 = 7.15 × 10−22 J.
which has as solutions
√
ψ = A exp(±2πi 2mEx/ h)
(2-40)
or alternatively, trigonometric solutions
√
√
ψ = A sin(2π 2mEx/ h), ψ = A cos(2π 2mEx/ h)
(2-41)
As is most easily seen from the exponential forms (2-40), if E is negative, ψ will blow up at either +∞ or −∞, and so we reject negative energies. Since there are no boundary conditions, it follows that E can take on any positive value; the energies of the free particle are not quantized. This result would be expected from our earlier results on constrained particles. There we saw that quantization resulted from spatial constraints, and here we have none.
The constants A and A of Eqs. (2-40) and (2-41) cannot be evaluated in the usual way, since the solutions do not vanish at x = ±∞. Sometimes it is convenient to evaluate them to correspond to some experimental situation. For instance, suppose that one was working with a monoenergetic beam of electrons having an intensity of one electron every 10−6 m. Then we could normalize ψ of Eq. (2-40) so that 10−6 m |ψ|2dx=1
0
There is a surprising difference in the particle distributions predicted from expres sions (2-40) and (2-41). The absolute square ψ*ψ of the exponentials is a constant (A*A), whereas the squares of the trigonometric functions are fluctuating functions
Section 2-5 The Free Particle in One DimensionFigure 2-15 Wavefunctions for the infinite square well with finite partition.
of x. It seems sensible for a particle moving without restriction to have a constant probability distribution, and it seems absurd for it to have a varying probability distribution. What causes this peculiar behavior? It results from there being two independent solutions for each value of E (except E = 0). This degeneracy with respect to energy means that from a degenerate pair, ψ and ψ, one can produce any number of new eigenfunctions, ψ = aψ + bψ (Problem 2-11). In such a situation, the symmetry proof of Section 2-2 does not hold. However, there will always be an independent pair of degenerate wavefunctions that will satisfy certain symmetry requirements. Thus, in the problem at hand, we have one pair of solutions, the exponentials, which do have the proper symmetry since their absolute squares are constant. From this pair we can produce any number of linear combinations [one set being given by Eq. (2-41)], but these need not display the symmetry properties anymore.
The exponential solutions have another special attribute: A particle whose state is described by one of the exponentials has a definite linear momentum, whereas, when described by a trigonometric function, it does not. In Section 1-9, it was shown that the connection between classical and wave mechanics could be made if one related the classical momentum, px, with a quantum mechanical operator (h/2πi)d/dx. Now, for a particle to have a definite (sharp) value p for its momentum really means that, if we measure the momentum at some instant, there is no possibility of getting any value other than p. This means that the particle in the state described by ψ always has momentum p, no matter where it is in x; i.e., its momentum is a constant of motion, Chapter 2 Quantum Mechanics of Some Simple Systems just as its energy is. This corresponds to saying that there is an eigenvalue equation for momentum, just as for energy. Thus h dψ =pψ
(2-42)
2π i dx
The statement made earlier, that the exponential solutions correspond to the particle having sharp momentum, means that the exponentials (2-40) must be solutions to Eq. (2-42). This is easily verified:
√
√
h
d
±2πi 2mEx √
±2πi 2mEx
A exp = ± 2mE A exp 2π i dx
h
h
Thus, the positive and negative exponential solutions correspond to momentum values
√
√
of + 2mE and − 2mE, respectively, and are interpreted as referring to particle motion toward +∞ and −∞ respectively. Since energy is related to the square of the momentum, these two solutions have identical energies. (The solution for E = 0 corresponds to no momentum at all, and the directional degeneracy is removed.) A mixture of these states contains contributions from two different momenta but only one energy, so linear combinations of the exponentials fail to maintain a sharp value for momentum but do maintain a sharp value for energy.
EXAMPLE 2-6 An electron is accelerated along the x axis towards x = ∞ from rest through a potential drop of 1.000 kV.
a) What is its final momentum?
b) What is its final de Broglie wavelength?
c) What is its final wavefunction?
√
SOLUTION a) px = 2mE = [2(9.105 × 10−31 kg)(1.602 × 10−16 J)]1/2 = 1.708 × 10−23 kg m s−1 b) λ = h =
6.626×10−34 J s = 3.879 × 10−11 m
p
1.708×10−23 kg m s−1
√
c) ψ = A exp(+2πi 2mEx/ h)(choose + because moving towards x = +∞) = A exp(2πip/ h)x = A exp(2πix/λ) = A exp[2πi(2.578 × 1010 m−1)x].
2-6 The Particle in a Ring of Constant Potential Suppose that a particle of mass m is free to move around a ring of radius r and zero potential, but that it requires infinite energy to get off the ring. This system has only one variable coordinate—the angle φ. In classical mechanics, the useful quantities and relationships for describing such circular motion are those given in Table 2-1.
Comparing formulas for linear momentum and angular momentum reveals that the variables mass and linear velocity are analogous to moment of inertia and angular velocity in circular motion, where the coordinate φ replaces x. The Schr¨odinger equation for circular motion, then, is −h2 d2ψ (φ) =Eψ(φ)
(2-43)
8π 2I
dφ2
Section 2-6 The Particle in a Ring of Constant PotentialTABLE 2-1
Quantity
Formula
Units
Moment of inertia
I = mr2 g cm2 or kg m2
Angular velocity ω = φ/t = v/r
s−1
Angular momentum (linear mvr = I ω g cm2/s or erg s or J s momentum times orbit radius) which has, as solutions A exp(±ikφ)
(2-44)
or alternatively
A sin(kφ)
(2-45)
and
A cos(kφ)
(2-46)
where [substituting Eq. (2-45) or (2-46) into (2-43) and operating]
√
k = 2π 2I E/ h
(2-47)
Let us solve the problem first with the trigonometric functions. Starting at some arbitrary point on the ring and moving around the circumference with a sinusoidal function, we shall eventually reencounter the initial point. In order that our wavefunction be single valued, it is necessary that ψ repeat itself every time φ changes by 2π radians.
Thus, for φ given by Eq. (2-45), sin(kφ) = sin[k(φ + 2π)]
(2-48)
Similarly, for ψ given by Eq. (2-49)
cos(kφ) = cos(kφ + 2kπ)
(2-49)
Either of these relations is satisfied only if k is an integer. The case in which k = 0 is not allowed for the sine function since it then vanishes everywhere and is unsuitable.
However, k = 0 is allowed for the cosine form. The normalized solutions are, then,
√
ψ = (1/ π) sin(kφ), k = 1, 2, 3, . . .
√
ψ = (1/ π) cos(kφ), k = 1, 2, 3, . . .
√
ψ = (1/ 2π) (from the k = 0 case for the cosine)
(2-50)
Now let us examine the exponential form of ψ (Eq. 2-44). The requirement that ψ repeat itself for φ → φ + 2π gives A exp(±ikφ) = A exp[±ik(φ + 2π)] = A exp(±ikφ) exp(±2πik)
Chapter 2 Quantum Mechanics of Some Simple Systems or exp(±2πik) = 1
Taking the positive case and utilizing Eq. (2-3), we obtain cos(2π k) + i sin(2πk) = 1 or cos(2π k) = 1 and sin(2π k) = 0
Again, k must be an integer. (The same result arises by requiring that dψ/dφ repeat for φ → φ + 2π.) Thus, an alternative set of normalized solutions is
√
ψ = 1/ 2π exp(ikφ) k = 0, ±1, ±2, ±3, . . .
(2-51)
The energies for the particle in the ring are easily obtained from Eq. (2-47): E = k2h2/8π2I, k = 0, ±1, ±2, ±3, . . .
(2-52)
The energies increase with the square of k, just as in the case of the infinite square well potential. Here we have a single state with E = 0, and doubly degenerate states above, whereas, in the square well, we had no solution at E = 0, and all solutions were nondegenerate. The solution at E = 0 means that there is no finite zero point energy to be associated with free rotation, and this is in accord with uncertainty principle arguments since there is no constraint in the coordinate φ.
The similarity between the particle in a ring and the free particle problems is strik ing. Aside from the fact that in the ring the energies are quantized and the solutions are normalizable, there are few differences. The exponential solutions (2-51) are eigenfunctions for the angular momentum operator (h/2π i)d/dφ. The two angular momenta for a pair of degenerate solutions correspond to particle motion clockwise or counterclockwise in the ring. (The nondegenerate solution for E = 0 has no angular momentum, hence no ability to achieve degeneracy through directional behavior.) The particle density predicted by the exponentials is uniform in the ring, while that for the trigonometric solutions is not. Since the trigonometric functions tend to localize the particle into part of the ring, thereby causing φ = ∞, it is consistent that they are impure momentum states ( ang. mom. = 0). (Infinite uncertainty in the coordinate φ means that all values of φ in the range 0–2π are equally likely.)
EXAMPLE 2-7 Demonstrate that any two degenerate exponential eigenfunctions for a particle in a ring are orthogonal.
SOLUTION Such a pair of degenerate wavefunctions can be written as ψ+ = 1
√
exp(ikφ),
2π
ψ
2π
− = 1
√
exp(−ikφ). These are orthogonal if
2π
0
ψ∗
+ψ− dφ = 0 where we must use the complex conjugate of either one of the wavefunctions since ψ is complex. But ψ∗ + = ψ−, so
2π
2π
2π
ψ−ψ− dφ = 1 exp(−2iφ) dφ = 1 [cos(2φ) − i sin(2φ)]dφ
0
2π 0
2π 0
= 1 sin(2φ)|2π − i(− cos(2φ)|2π
2π
0
0
= 1 [sin(4π) − sin(0) + i cos(4π) − i cos(0)]
2π = 1 [0 − 0 + i − i] = 0.
2π