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Lecture 10 The one and three-dimensional particle in a box are prototypes of bound systems. As we move on in our study of quantum chemistry, we'll be considering bound systems that are more and more complex and which represent more and more complex chemical systems. Another system that is important in chemistry is the harmonic oscillator. This problem is important because the harmonic oscillator is the most basic model with which we treat the vibrations of molecules. The properties of these vibrations are critical in interpreting infrared and Raman spectra, for the understanding of chemical dynamics and understanding the heat capacities of gas phase and liquid phase systems. One feature of quantum mechanics that we've learned is that in order to properly set up a problem we need to know how the analogous classical problem is set up. So we'll begin our study of the harmonic oscillator by considering the classical case. The classical harmonic oscillator is usually represented by a mass m suspended from a wall by a massless spring. In this problem we ignore the effect of gravity. [Draw]
Initially, the spring is at some length at which the spring exerts no force on the mass. We call this position 0. As the mass is moved from this position, the spring is distorted and a force is exerted on the mass by the spring. The force is given by F = -kx The negative sign means that the force is exerted in the direction opposite from the initial displacement, i.e., when the spring is stretched, the force is in a direction to compress the spring to its initial length, while if the spring is compressed, the force is in a direction to stretch the
spring back to position 0. Such a force is called a restoring force. The constant k is called the force constant. If a spring is weak the force constant will be small, while if the spring is stiff the force constant will be large. According to Newton's second law, F ma m d xdt^2 2 -kx.
This gives us a differential equation for the position of the mass m as a function of time,
m d xdt^2 2 kx or d xdt^22 m kx
Note that this latter form is almost identical to the differential equation we solved for the particle in a box, and will therefore have the same type of solution,
x(t)= A sin t + B cos t,
where = ( m k )1/ 2 is the fundamental vibrational frequency in units of radians/s. is related
to the frequency , by = 2 .
To determine the value of the constants A and B in this problem, we need to specify a set of initial conditions. For this problem the conditions usually specified are the initial
displacement x and the initial velocity, dxdt or x'. A common case is that where we stretch the
spring to some initial displacement x 0 and release it. For this case the initial displacement x(0) = x 0 , while the initial velocity, x'(0) = 0. If we apply the first boundary condition by setting t to 0 and x to x 0 we get x(0) x (^) 0 A sin 0 B cos 0 B x. 0
So far we've discussed the harmonic oscillator as a single mass hanging from a wall. [Draw] The simplest model of a vibrating molecule is two masses, m 1 and m 2 , connected by a spring. [Draw] It turns out that this is completely equivalent to the problem we have just solved
if we replace the mass m by the reduced mass = m + mm m 1 1 22. We can do this because the
diatomic molecule is another example of two particles linked by a central force. Let's do three quick examples of calculating a reduced mass. The reduced mass in AMU of H^35 Cl is given by
= 1+ 35 1* 35=.9722^ AMU. The reduced mass of H^79 Br in AMU is^ = 1+79 1*79=.9875 AMU. The
reduced mass of 79 Br^35 Cl is = 35+79 35*79= 24.25AMU. Note that if we have a light atom bonded
to a heavy atom, that changing the mass of the heavy atom doesn't change the reduced mass very much. It is important to note that if k is in SI units, N m-1, the reduced mass must be in kg rather than in AMU in order to obtain intelligible values for the frequency. We convert between kg and AMU using the conversion factor 1 AMU = 1.66053 x 10-27^ kg. For diatomic molecules our differential equation for the motion of a harmonic oscillator now becomes
d xdt^2 2 -kx
and the vibrational frequency 21 ( k ) 1 2/.
These are the basics of the classical harmonic oscillator. Let’s turn now to the quantum mechanical case. To determine the wavefunction and energies of the quantum mechanical
harmonic oscillator, we need to solve the Schrödinger equation, H = E . To do this we need
to write down our Hamiltonian H ˆ. Remember that for a single dimension, the Hamiltonian is given by
H = - 2m^ ^2^^ dxd^2 2 +V(x)
For the harmonic oscillator, we replace the mass m with the reduced mass , and the potential
V(x) with 1/2 kx^2. This yields
H 2 ^2 dxd^22 ^12 kx^2
as our Hamiltonian. Even adding so simple a potential as 1/2 kx^2 makes this Schrödinger equation much harder to solve, in fact hard enough that we won't attempt it. I'll just present and analyze the solutions. First, the eigenvalues of the Hamiltonian are
E v = (^ v +^^21 )h ,^ ^ v = 0,1,2,3,....
where the frequency is given by = 21 ( k )1/ 2 , just as in the classical case. The integers v
are called vibrational quantum numbers. Note that the energies are quantized, and that like the case of the particle in a box, we have a potential energy function that corresponds to a bound system. The lowest energy, for v=0, is called the zero point energy and is equal to 1/2h. Note that as was the case for the particle in a box, this lowest energy is greater than zero. The next energy is for v = 1 and equals 3/2 h. The next energy is for v = 2 and equals 5/2 h. Notice
that for the harmonic oscillator, the spacing between energy levels is constant at h . The wavefunction for the harmonic oscillator has the formula
Note that in the lowest state the most probable position for the oscillator is at the minimum of the potential energy curve. For a molecule, this position corresponds to a bond length. Because this is the bond length that the molecule will be spending most of its time in when the molecule is in its vibrational ground state, this bond length is called the equilibrium bond length , and is symbolized by re. This quantum result that the molecule in v = 0 spends most of its time at re is in direct contradiction to our classical result, which says that the molecule will spend most of its time at the bond lengths corresponding to the classical turning points, the points where the kinetic energy is zero. However, notice that as the energy increases, the probability density moves out toward the edges of the potential curve. Once again, we see that as the quantum number increases, we approach the classical limit. Note also that when we plot the probability density vs. r, for all states there is some probability that the molecule will be found at positions outside of the classical turning points. The probability is greatest for v= 0, where the molecule spends fully 16% of its time outside of the turning points. This would be impossible classically, since the kinetic energy would have to be negative. This phenomenon, in which a particle is in a state that is energetically forbidden, is called quantum mechanical tunneling. It is very important in the understanding of many chemical phenomena, including electron transfer reactions, and the rates of chemical reactions. Let's end this discussion by calculating the vibrational frequencies of two molecules. Note that for this we need only know the reduced mass and the force constant. Let's do this for
HI and HCl. Our formula is = 21 ( k )1/ 2. For HCl we've already shown that = 0.
AMU, but for the purposes of our calculations we need in kg. The conversion is 1.66 x 10-
kg/AMU, so = 1.613 x 10-27^ kg. The force constant k for HCl = 515.7 N m-1, so
= 21 ( 1.613x 10515.7 -27)1/ 2 = 8.993 x 1013 s-1. As noted earlier, in spectroscopy we more
commonly use units of wavenumbers, cm-1, instead of frequency units. We calculate
wavenumbers from frequency by using ^ c , where c is the speed of light in cm/s, so for this
case, 8.993 103 10 x 10^ x^ cm s^13 / s^^ ^1 2998 cm ^1. For HI, = 1+127 1* 127x1. 66 10 x ^27 = 1. 647 10 x ^27 kg. The
force constant for HI is 293 N m-1. Plugging these into our formula for yields = 6.71 x 10^13 s-1^ or = 2237 cm-1. So we see that even though the reduced masses are very similar, the difference in force constants results in a substantially different vibrational frequency. We will find as well that when two molecules have similar force constants but different reduced masses, the vibrational frequencies will differ as well. Thus observed vibrational frequencies depend on both reduced mass and force constant.
linear velocity by v = r. If we substitute this in our definition for kinetic energy, we find that
T =^12 m v^2 =^12 mr^2 ^2.
Since the angular velocity is analogous to the linear velocity, it is logical that the term mr^2 serves a function similar to the mass for circular motion, i.e., just as mass represents the tendency to resist linear acceleration, the quantity mr^2 represents the tendency to resist angular acceleration. We call this quantity the moment of inertia , and for this simple case define it by I
= mr^2. Thus we can now express the kinetic energy with T =^12 m v^2 = 2 1 I ^2. Since there is a
circular analog for the velocity and a circular analog for the mass, it seems logical that there should be a circular analog for the linear momentum, p. This circular analog, the angular momentum , is constructed by taking the product of the angular velocity and the moment of inertia, L = I, just as the linear momentum is the product of mass and velocity. Thus we can express the kinetic energy for the rotation of a mass around a fixed axis in terms of the angular
velocity or in terms of the angular momentum, T =^12 I 2 = 2IL^2. The quantities I, and L are
fundamental quantities for the dynamics of angular motion. A rotating molecule can be represented by two masses bound by a rigid massless rod of length R rotating around the center of mass of the system. While this system may appear more complicated than a single mass rotating around a fixed axis, it turns out not to be. To treat our rotating molecule, we need only replace the mass of the single particle by the reduced mass
= m + mm m 1 1 22 of the diatomic molecule. The only change in our results on circular motion is that
the definition of angular momentum becomes I = r^2. Of course anywhere else where we refer to the mass it will be replaced by the reduced mass as well. To treat the rigid rotator quantum mechanically, we need to write down our Hamiltonian. WHAT ARE THE TWO TERMS IN EVERY HAMILTONIAN OPERATOR? The potential energy function is simple. A molecule rotating in a vacuum encounters no resistance, so our potential is 0. Thus the
Hamiltonian is simply H = - 2^ ^2 ^2_._ However,
for a problem with rotational symmetry it is most useful to express the Hamiltonian in terms of spherical coordinates. For those of you who are in need of refreshing about spherical coordinates, suppose we have a point at distance r 1 = x 1 , y 1 , z 1 from
the origin. This point can also be represented by two angular variables, and , and the radius r
from the origin. The angle is the angle from the z axis to the line between 0 and R, and the
angle is the angle between the projection of the vector on the x-y plane and the x-axis. Two
y
x
z
R
Spherical Coordinates
If we factor out the R^2 , then our denominator contains the factor R^2 , which we replace with the moment of inertia, I, to get finally
H (^) 2I^ ^2 ( (^) sin^1 2^ (^) ^2 2 (^) sin^1^ ( sin (^) )). The solution to the Schrödinger equation for this problem is a function of both and , and has the symbol Y( , ). The Schrödinger equation thus becomes
- 2I^ ^2 ( (^) sin^1 2^ (^) ^2 2 + (^) sin^1^ ( sin (^) ))Y( , )= EY( , )
The solutions to this equation are called spherical harmonics. HOW MANY QUANTUM NUMBERS DO WE EXPECT TO BE A PART OF EACH EIGENFUNCTION? Each solution includes two quantum numbers called l and m, and the solutions are labeled (^) Ylm (, ). The values of the quantum
numbers l and m are not independent, but are related. l can take on any value from 0 to . m can take on only values from - l to l****.
Some of the first few solutions are (^) Y 00 = ( 41 )1/ 2 , a spherical solution with no nodes,
Y 10 = ( 43 )1/ 2 cos^ , which has a single planar
node, and Y 20 = ( 165 ) 1/ 2(3 cos^2 -1) , which
has two planar nodes. DO THESE LOOK FAMILIAR TO YOU? So you see that the probability densities for the eigenfunctions of the rigid rotor are the same as the
angular portion of the hydrogen atom wavefunction. The eigenvalues of the rigid rotor depend only on the value of the quantum number l
and are given by (^) E = l l(l +1)2I^ ^2_._ The first few eigenvalues are E (^) 0 0 , (^) E = (^12) 2I^ ^2 , and
E = (^312) 2I^ (^2). Notice that the spacing between the energy levels increases with increasing l.
Note also that since the energy depends only on l , and since for each quantum number l , there are 2 l + 1 wavefunctions with different values of m but the same l , we can conclude that each energy level of the quantum mechanical rigid rotor is 2 l + 1 fold degenerate. For example, E 1 will be threefold degenerate, with wavefunctions Y 10 , Y 11 and Y 1 ^1 having the same energy. E 2
will be fivefold degenerate with wavefunctions Y 2 ^2 , Y 2 ^1 , Y 20 , Y 21 , Y 22 having the same energy
and so on. This will be important in interpreting the intensities of the infrared spectra of gas phase molecules.
r = x i + y j + z k p = px i + py j + pz k. These definitions lead to a convenient definition of the angular momentum in terms of the components of r and p, L = r x p = (ypz - zpy) i + (zpx - xpz) j + (xpy - ypx) k These lead in turn to the concept of components of angular momentum, Lx, Ly and Lz. These are the components of the classical angular momentum in the x, y and z directions and are given by Lx = ypz - zpy Ly = zpx - xpz Lz = xpy - ypx. Note that even though angular momentum is a vector, the magnitudes of the x, y and z components are scalar quantities. In quantum mechanics we are interested in the operators that correspond to these observables, and since angular momentum is particularly important in systems of spherical or cylindrical symmetry, we will express the operators for these three components in spherical coordinates,
L ^ x i ( sin cot cos )
L ^ y i (- cos cot sin )
L z -i
It is interesting to note that the operators for the three components of the angular momentum do not commute with each other. Remember that this means that the uncertainty principle applies to these quantities - we can measure one of them at a time with infinite precision, but not the other two. By convention, we usually say that Lz is the one that we can know with infinite precision, but not Lx or Ly. In the language of eigenfunctions and eigenvalues we say that if a wavefunction is an eigenfunction of Lz then it is not an eigenfunction of Lx or Ly. This has an important implication. Remember that L = Lx i + Ly j + Lz k. But we've just said that we cannot measure the three quantities Lx, Ly and Lz with infinite precision at the same time. This means that the angular momentum itself can never be known with infinite precision in quantum mechanics. In other words, angular momentum is not an eigenvalue of any wavefunction. We can, however, learn more about the angular momentum of a system than just Lz. It turns out that even though L cannot be determined exactly, its square, L^2 = L L, can be determined exactly. We can figure out what the operator for the angular momentum squared is by looking at our Hamiltonian for the rigid rotor. Remember that this Hamiltonian is the same as the operator for the kinetic energy, since the potential is zero for this problem. The kinetic
energy can be expressed in terms of the angular momentum, T = 2IL^2. If we compare this with
the Hamiltonian,
H^ = T = - 2I^ ^2^^ ( sin^1 2^ ^22 + sin^1^ ( sin )).
^ ^
we see that the operator for the angular momentum squared must be
THE MAXIMUM ANGULAR MOMENTUM IN THE Z DIRECTION? So we see that L > Lz, max, and you can never have the entire angular momentum along the z-axis. In other words, the angular momentum always has to point a little bit away from the z-axis. In addition to finding out the angular momentum and the angular momentum along the z- axis, we can find out something about the angular momentum along the x- and y-axes. Remember that we know L^2 and Lz, and that L^2 = Lx^2 + Ly^2 + Lz^2. This means that Lx^2 + Ly^2 = L^2 - Lz^2. What this equation tells us is that we can learn the magnitude of the vector i Lx + j Ly, but we can't know how the angular momentum is distributed between Lx and Ly. As I said earlier, the rigid rotor is the simplest model for the rotations of a molecule. When we discuss the rotational energies of molecules we use the letter J rather than l as the quantum number for the total angular momentum. Thus the energy of a rotating molecule is given by
J
2 E = 2I^ J(J +1)
Similarly if we wish to find other quantities like the square of the angular momentum or the angular momentum in the z direction, we simply substitute J for l in the equations we've already worked out. At this point we've applied quantum mechanics to three simple physical systems, the particle in a box, the harmonic oscillator and the rigid rotor. Each of these allows us to take a measurement of a frequency or wavelength and extract information about the molecule. In the case of the particle in a box, we learn the length in which an electron can move freely in a linear conjugated system. In the case of the harmonic oscillator we can extract k, the force constant for
the bond, which tells us how strong the bond is. The key thing that we learn from a rotational spectrum is the bond length of the molecule. We can learn this because the energy of a given
level depends only on , J and I. Once we know J, we can figure out I, and since I = r^2 , we can
figure out the bond length R. For example, suppose we find that the rotational energy of HI in its J = 5 state is 8.952 x 10-21^ J. What is its bond length?
J
2 E = J(J +1) 2I^
Solving for I gives
I = J(J +1) 2E^ ^2^ (^) J= 5x6x(1.055x2x8.95x 10^10 21 ^34 JJs ) 2 = 1.865x 10 ^47 kgm^2
I = r^2 , so r = ( I )1/ 2. Now we need to calculate . The reduced mass of HI, which we
calculated before, is given by,
HI = m + m m mHH II= (1+127) (1)(127)x1.66x 10 ^27 = 1.647x 10 ^27 kg
Therefore r, the bond length of HI, is
r = ( I ) 1/ 2^ = ( 1.865x1.647x^1010 ^4727 )1/ 2 = 1.604x 10 ^10 m
