Particle in a Box Problem 2

Lecture #10 1

Lecture 10

Objectives: 1. Be able to solve the particle in a box problem 2. Compare the classical and quantum harmonic oscillators 3. Compare the classical and quantum rotors

Example: Particle in a box

Consider a particle trapped in a one-dimensional box, of length L. The Schrödinger equation for this system (considering only the spatial function) is h2 d2ψ − + Uψ = Eψ, 8π2m dx2 where U = 0 when x is inside the box and U = ∞ for x outside the box. This is a second order ordinary differential equation. The general solution to this equation is ψ = α sin γx + β cos γx with γ = (8π2m(E − U)/h2)1/2 We need two boundary conditions to fix α and β. What are they? ψ(0) = 0, and ψ(L) = 0. These BCs give ψ(0) = β = 0 and ψ(L)= α sin γL + β cos γL = 0 hence, either α = 0(trivial solution) or sin γL = 0. Taking the latter we find nπ γL = nπ → γ = L Hence, nπx ψ = α sin n L ∞ ∗ The values of αn can be computed by recalling that 0 ψ ψdx = 1 so that ∞ L L ∗ ∗ ∗ nπxR ψ ψdx = ψ ψdx = α α sin2 = 1 n n L Z0 Z0 Z0 Recalling that sin2udu = u/2 − sin(2u)/4+ C we have that nπ ∗ L R u sin(2u) ∗ L nπ sin(2nπ) αnαn − = αnαn − = 1 nπ 2 4 0 nπ 2 4 1/2 2 Therefore, αn = i L . We can now solve for the energy levels by from the equation L L L ∗ ∗ ∗ ψ Hψdx = ψ Eψdx = E ψ ψdx = E Z0 Z0 Z0 L 2 2 L ∗ h nπ 2 nπx ψ Hψdx = − − sin2 dx = E 8π2m L L L n Z0 ! Z0 2 2 2 nh 1 n h Therefore, En = L 8m = 8mL2 . Lecture #10 2

The Harmonic Oscillator

The Classical Harmonic Oscillator A vibrating body subject to a restoring force, which increases in proportion to the displacement from equilibrium, will undergo harmonic motion at constant frequency and is called a harmonic oscillator. Figure 1(a) shows one example of a harmonic oscillator, where a body of mass m is connected to a ﬁxed support by a spring with a force constant, k. We will assume that gravitational forces are absent.

Figure 1: (a) Harmonic Oscillator Consisting of a Mass Connected by a Spring to a Fixed Support; (b) Potential Energy, V ,and Kinetic Energy, EK For the Harmonic Oscillator.

When the system is at equilibrium, the mass will be at rest, and at this point the displacement, x, from equilibrium has the value zero. As the mass is pulled to the right, there will be a restoring force, f, which is proportional to the displacement. For a spring obeying Hooke’s law, d2x f = −kx = m (1) dt2 The minus sign in equation 1 is related to the fact that the force will be negative, since the mass will tend to be pulled toward the −x direction when the force is positive. From Newton’s second d2x law, the force will be equal to the mass multiplied by the acceleration, dt2 . The equation of motion is a second order ordinary diﬀerential equation, obtained by rearranging equation 1 as d2x k + x = 0, (2) dt2 m and has a general solution given by

x(t)= A sin ωt + B cos ωt, (3) where

ω =(k/m)1/2. (4)

The units of ω are radians s−1, and since there are 2π radians/cycle, the frequency ν = ω/2π cycles s−1. [Note: The student should check this solution by substituting equation 3 back into equation 2]. We again require two boundary conditions to specify the constants A and B. We choose the mass to be at x = 0 moving with a velocity v0 at time= 0. The ﬁrst condition gives x(t =0)= A · 0+ B · 1= B = 0 (5) Lecture #10 3 so that

x(t)= A sin ωt. (6)

Using this result, the second boundary condition can be written as dx v0 = v(t =0)= = Aω cos ωt|t=0 = Aω, (7) dt t=0

from which we see that A = v0/ω. As the spring stretches, or contracts, when the mass is undergoing harmonic motion, the potential energy of the system will rise and fall, as the kinetic energy of the mass falls and rises. The change in potential energy, dV , is

dV = −fdx = kxdx, (8) so that upon integrating, 1 V = kx2 + constant. (9) 2 The constant of integration may be set equal to zero. This potential energy function is shown as the parabolic line in Figure 1(b). The kinetic energy of this harmonic oscillator is given by

1 1 dx 2 m E = mv2 = m = (Aω)2 cos2 ωt (10) K 2 2 dt 2 This function is also plotted in Figure 1(b). The total constant energy, E, of the system is given by 1 m 1 m E = V + E = kx2 + (Aω)2 cos2 ωt = kA2 sin2 ωt + (Aω)2 cos2 ωt, (11) K 2 2 2 2 where we have substituted equation 6 for x. Substituting equation 4 for ω2, we can write 1 1 E = A2k sin2 ωt + cos2 ωt = A2k (12) 2 2 The total energy is thus a constant and is shown as a horizontal line in Figure 1(b). At the limits of oscillation, where the mass is reversing its direction of motion, its velocity will be momentarily equal to zero, and its momentary kinetic energy will therefore also be zero, meaning that the potential energy will be maximized and equal to the total energy of the system at the two turning points. As the oscillator begins to undergo acceleration away from the turning points, the kinetic energy will increase, and the potential energy will decrease along the curve, V , as shown in Figure 1(b). If the spring constant, k, should not be constant, but should vary slightly from a constant value as x changes, we would be dealing with an anharmonic oscillator. In most cases, an anharmonic oscillator may be closely approximated by the harmonic oscillator equations for small displacements from equilibrium, x. Soon we will be comparing the quantum harmonic oscillator with the classical harmonic oscillator, and the probability of finding the mass at various values of x will be of interest. We now calculate this probability for the classical harmonic oscillator. The probability of finding the mass, m, at any given value of x is inversely proportional to the velocity, v, of the mass. This is reasonable, since the faster the mass is moving, the less likely it is to observe the mass. Hence, we expect that the probability of observing the mass will be have Lecture #10 4 a minimum at x = 0, where the velocity is at a maximum, and conversely will exhibit maxima when x = ±A. From equation 7 we see that dx v(x)= = Aω cos ωt, (13) dt so that dx P(x)dx ∝ , (14) Aω cos ωt where P(x)dx is the probability of finding the mass between x and x + dx. Note that since x is a continuous variable that we define P(x)dx as the probability and P(x) is called the probability density, which is the probability per unit length, in this case. We note that at the turning points π 3π of the oscillation, when 1/4 or 3/4 of a cycle has occurred, that t = 2ω or t = 2ω and at these points 1 cos ωt goes to zero, with v going to infinity. We know that the probability of finding the mass at the end points must be a maximum, but must not be infinite. The reason that P(x) remains finite is that dx in equation 14 always has finite width, therefore, P(x) is not defined exactly at a π point. For example, at t = 2ω , P(x) is evaluated over the range −Aground state we will find that P(x) has a maximum at x = 0.

v(x)

P(x)

0 -A +A

Figure 2: Probability Density, P(x), for Classical Harmonic Oscillator at Various Displacements, x. P(x) is plotted as the dashed line and the velocity, v(x) is plotted as the solid curve. The two 1 vertical lines give the limits of the oscillator motion. Note that P(x) ∝ v(x) .

The Quantum Harmonic Oscillator The quantum harmonic oscillator is a very important problem in quantum mechanics. For example, it serves as a ﬁrst-order approximation for the bond vibrational problem in diatomic and (with Lecture #10 5 coupling) polyatomic molecules. We will examine the quantum harmonic oscillator in some detail, comparing it with what we know about the classical harmonic oscillator from the previous section. The potential energy function for the quantum harmonic function is the same as for the classical harmonic oscillator, namely, V = 1/2kx2. Thus, in the quantum Hamiltonian is

Hop = EKop + Vop (15) and we may write the Schr¨odinger equation as −¯h2 ∂2ψ 1 + kx2ψ = Eψ (16) 2m ∂x2 2 The general solution to this problem (which we will not derive) can be written as

1/4 1/2 α 1 −y2/2 ψn(x)= n Hn(y)e , (17) π 2 n! mω 1/2 where n = 0, 1, 2,... is the quantum number, α = ¯h , y = α x, and Hn(y) is a Hermite polynomial of order n. The ﬁrst few Hermite polynomials are

H0(y) = 1 (18)

H1(y) = 2y (19) 2 H2(y) = 4y − 2 (20) 3 H3(y) = 8y − 12y (21) 4 2 H4(y) = 16y − 48y + 12 (22) Hermite polynomials of any order can be calculated from the recursion relation

Hn+1(y) = 2yHn(y) − 2nHn−1(y). (23) The allowed energies (eigen energies) for the quantum harmonic oscillator are 1 E = n + ¯hω (24) n 2 and since ω = 2πν, 1 E = n + hν. (25) n 2 Using equation 4 for ω, we may write 1 k 1/2 E = n + ¯h . (26) n 2 m The ground state wavefunction for the quantum harmonic oscillator can be obtained by substituting 1/2 H0(y) from equation 18, using y = α x, into equation 17,

1/4 1/4 2 α − 2 mω − mωx ψ (x)= e αx /2 = e 2¯h . (27) 0 π ¯hπ Likewise, the ﬁrst and second excited state wavefunctions are

3 1/4 4α −αx2/2 ψ1(x) = xe (28) π ! 1/4 α − 2 ψ (x) = 2αx2 − 1 e αx /2. (29) 2 4π Lecture #10 6

n=2; E=5/2hn

|Y | Y

n=1; E=3/2hn

0 2

|Y |

n=0; E=1/2hn

|Y |

2 Figure 3: Allowed Energy Levels, ψn, and |ψn| for the Quantum Harmonic Oscillator with Quan- tum Numbers n = 0, 1, 2. The harmonic potential, V (x) = 1/2kx2, is shown, along with the wave functions and probability densities for the ﬁrst three energy levels.

Figure 3 shows the first few allowed energy levels for the quantum harmonic oscillator. Also 2 shown are the wavefunctions, ψn and the probability densities, |ψn| for the levels n = 0, 1, 2. The equally spaced set of allowed vibrational energy levels observed for a quantum harmonic oscillator is not expected classically, where all energies would be possible. The quantization of the energy levels of the harmonic oscillator is similar in spirit to the quantization of the energy levels for the particle in a box, except that for the harmonic oscillator, the potential energy varies in a parabolic manner with the displacement from equilibrium, and the walls of the “box” therefore are not vertical. We might say, in comparison to the “hard” vertical walls for a particle in a box, that the walls are “soft” for the harmonic oscillator. In addition, the spacing between the allowed energy levels for the harmonic oscillator is a constant value, hν, whereas for the particle in a box, the spacing between levels rises as the quantum number, n, increases. There is another interesting feature seen in Figure 3. For the lowest allowed energy, when n = 0, 1 we see that the quantum harmonic oscillator possesses a zero-point energy of 2 hν This too is reminiscent of the particle in a box, which displays a finite zero-point energy for the first allowed quantum number, n = 1. This lowest allowed zero-point energy is unexpected on classical grounds, since all vibrational energies, down to zero, are possible in the classical oscillator case. Recall that we developed an expression for the probability of observing a classical harmonic oscillator between x and x + dx and found that this probability is inversely proportional to the velocity of the oscillator (see equation 14). The corresponding probability for a quantum harmonic ∗ 2 oscillator in state n is proportional to ψnψn = |ψn| . We now compare the probability densities of classical and quantum harmonic oscillators. Recall that the ground state for the classical oscillator has zero energy (and zero motion), whereas the quantum oscillator in the ground state has an 1 energy of 2 hν. Therefore, it does not make sense to compare classical and quantum oscillators in their ground states. We can however directly compare probability densities by comparing quantum and classical oscillators having the same energy. Equating the quantum and classical oscillator Lecture #10 7 energies, we have 1 1 E = n + hν = A2k, (30) n 2 2 where A is the classical amplitude, or limit of motion. Solving for A, we have

2hν 1/2 A = (2n + 1) (31) k That is, a classical oscillator with energy En will oscillate between x = ±A, with A given by equation 31. The quantum and classical probability densities for n = 0, 2, 5, and 20 are plotted in Figure 4. 2 We see from equation 27 that |ψ0| is a Gaussian function function with a maximum at x = 0. This is plotted in the upper left panel of Figure 4. Contrast this behavior with the classical harmonic oscillator, which has a minimum in the probability at x = 0 and maxima at the turning points. Also note that the limits of oscillation are strictly obeyed for the classical oscillator, shown by the vertical lines. In contrast, the probability density for the quantum oscillator “leaks out” beyond the x = ±A classical limits. The quantum harmonic oscillator penetrates beyond the classical turning point! This phenomenon is akin to the quantum mechanical penetration of a finite barrier seen previously. Thus, the probability densities for the quantum and classical oscillators for n = 0 have almost opposite shapes and very different behavior. Next, we compare the classical and quantum oscillators for n = 2 (top right panel in Figure 4). Note that the probability density for the quantum oscillator now has three peaks. In general, the quantum probability density will have n+1 peaks. In addition to having n+1 maxima, the probability density also has n minima. Remarkably, these minima correspond to zero probability! This means that for a particular quantum state n there will be exactly n forbidden locations where the wave function goes to zero (nodes). This is very different from the classical case, where the mass can be at any location within the limits −A ≤ x ≤ A. Note also that the middle peak centered at x = 0 has a smaller amplitude that the outer two peaks. Thus, for n = 2 we are beginning to see behavior that is closer in spirit to the classical probability density, that is, the probability of observing the oscillator should be greater near the turning points than in the middle. The classical probability density is essentially the same for all energies, but is just “stretched” to larger amplitudes for higher energies. For n = 5, shown in the lower left panel of Figure 4, we see the continued trend that the peaks near x = 0 are smaller than the peaks near the edges. Note that the probability densities continue to extend past the classical limits of motion, but die off exponentially. Finally, for n = 20 note that the gaps between the peaks in the probability density become very small. At large energies the distance between the peaks will be smaller than the Heisenberg uncertainty principle allows for observation. In other words, we will no longer be able to distinguish the individual peaks. The probability will be smeared out. You should be able to see that for n = 20 an appropriate the average of the quantum probability density closely approximates the classical behavior probability density. The region of non-zero probability outside the classical limits drops very quickly for high energies, so that this region will also be unobservable as a result of the uncertainty principle. Thus, the quantum harmonic oscillator smoothly crosses over to become a classical oscillator. This crossing over from quantum to classical behavior was called the “correspondence principle” by Bohr. Lecture #10 8

n=2

n=0

| |

0 0

ProbabilityDensity

n=20 n=5

2 | |

5 | |

0 0

Position

Figure 4: Probability Densities for Quantum and Classical Harmonic Oscillators. The probability 2 densities for quantum harmonic oscillators, |ψn(x)| , are plotted as solid lines for the quantum numbers n = 0, 2, 5, 20. The probability densities of the classical harmonic oscillators having the same energies as the quantum oscillators are plotted as dashed lines. The classical limits of motion are shown by the vertical lines. Lecture #10 9

The Rigid Rotor

The Classical Rigid Rotor—1 Dimensional We will consider a mass, m, coupled to a fixed point by a weightless rigid rod of length, r, which moves in a fixed plane and is not influenced by gravity. Figure 5 shows this mechanical situation and the physical parameters that are of importance. At all values of the angle, φ, the potential energy for the rotor is zero, i.e., V (φ) = 0. The moment of inertia is I = mr2. The kinetic energy 2 −1 of the rotor is Erot = Iω /2, where ω is the angular speed expressed in radians s . Classically, all values of Erot are allowed. The angular momentum of the rotor is pφ = Iω.

Figure 5: The Rigid Rotor—Planar. The coordinate appropriate to describe the motion of this system is the angle, φ

The Quantized Rigid Rotor—Planar We may write the Schr¨odinger equation for the planar rigid rotor in exactly the fashion we used previously for problems involving one coordinate, except that the coordinate of importance is the angle, φ, and instead of the mass, m, we employ the moment of inertia, I. −¯h2 ∂2ψ rot = Eψ . (32) 2I ∂φ2 rot One solution to this problem is

imℓφ ψ = Nrote (33) where Nrot is the normalization constant and mℓ is a quantum number, as will be shown later when we consider the periodicity of the rotor. Substituting equation 33 into equation 32, we ﬁnd that −¯h2 −m2eimℓφ = E eimℓφ (34) 2I ℓ mℓ or, m2¯h2 E = ℓ , m = 0, ±1, ±2,... (35) mℓ 2I ℓ Lecture #10 10

Note that if we solve equation 35 for mℓ, we obtain

(2IE )1/2 m = ± mℓ (36) ℓ ¯h where mℓ has both positive and negative values. The two signs for the quantum number refer to the two directions of rotation of the rigid rotor. Note that in this quantum mechanical problem, in contrast to the previous problems we have considered, the zero point energy for the rotor is zero. The normalized wave functions for the planar rigid rotor may be obtained as shown before. Thus,

2π 2 ∗ Nrot ψrotψrotdφ = 1 (37) Z0 Using the wave function given in equation 33, and integrating, we ﬁnd that

1/2 2 1 Nrot · 2π = 1; Nrot = (38) 2π and the normalized wave function is therefore 1/2 1 imℓφ ψrot = e (39) 2π The quantization feature of this problem may be found by considering the periodic behavior of the wave function. We recognize that

eimℓφ = eimℓ(φ+2π) (40)

Converting equation 40 to the transcendental form through Euler’s formula, we obtain

cos mℓφ + i sin mℓφ = cos mℓ(φ + 2π)+ i sin mℓ(φ + 2π) (41)

This equality will hold only for integer values of mℓ, as may be seen in the following equations derived from equation 41.

mℓ = 0 ⇒ cos0=cos0=1 (42)

mℓ = +1 ⇒ cos φ + i sin φ = cos(φ + 2π)+ i sin(φ + 2π) (43)

mℓ = −1 ⇒ cos φ − i sin φ = cos(φ + 2π) − i sin(φ + 2π) (44)

Therefore, from these periodic equations, we see that mℓ = 0, ±1, ±2,... will satisfy the periodic character of the wave function. [The student should examine equations 42–44 to show that non-integral values of mℓ will not satisfy these equalities]. It is instructive to examine the normalized plane rotor wave functions as a function of the quantum number, mℓ. These are shown for the ﬁrst three allowed quantum states of the rotor in Figure 6. These wave functions are plotted together with the quantized energy levels for the rotor. Let us now consider the value of ψ∗ψ for the planar rigid rotor, by considering the normalized 1/2 1 imℓφ ∗ 1 −imℓφ+imℓφ 1 wave function ψ = 2π e . We note that ψ (φ)ψ(φ) = 2π e = 2π , which is a constant for all values ofmℓ and all values of φ! This means that the probability of ﬁnding dφ the rotor in the range φ to φ + dφ for any angle, φ, is constant and is 2π , no matter what mℓ may be. Lecture #10 11

Figure 6: Wave Functions and Allowed Energy Levels for the Rigid Rotor—Planar Lecture #10 12

The Quantized Rigid Rotor—Three-Dimensional Our reason for studying the idealized rigid rotor is to ultimately apply our understanding to molecules that rotate in a quantized fashion. At this stage of our development, the student should be thinking about a diatomic molecule that is rotating like a dumbbell in space. A schematic of a carbon monoxide molecule is shown in Figure 7. This diatomic molecule possesses a single moment 2 2 of inertia, I = m1r1 + m2r2, where m1 and m2 are the masses of the atoms, and r1 and r2 are the distances of these masses from the center of mass of the molecule. In the three dimensional case, the diatomic molecule can tumble in space, or more speciﬁcally its plane of rotation can occur in any plane in space. The physical condition we have previously considered in which rotation occurs in a ﬁxed plane (planar rotor) does not apply. We will not derive the equations related to the three-dimensional rigid rotor but will give the result in equation 45 ¯h2 E3D = J(J + 1) (45) rot 2I where the quantum number J = 0, 1, 2,...

rC rO

Figure 7: Schematic of a Carbon Monoxide Molecule. The distance of each atom from the center of mass is shown, where rC is the distance from the carbon atom (light sphere) to the center of mass and rO is the distance from the oxygen atom (dark sphere) to the center of mass.

A plot of the allowed rotational energy levels for a diatomic molecule is shown in Figure 8 as a function of the quantum number, J. It may be seen that as J increases, the spacing between allowed rotational quantum states increases. If we let J be the rotational quantum number of a particular state, with (J − 1) the rotational quantum number of the next lower allowed state, then 2 2J¯h one may calculate that the spacing between adjacent levels increases by the amount 2I , which will be called 2JB where B is the rotational constant, given by ¯h2 B = . (46) 2I This will become important when we discuss spectroscopic transitions between neighboring rotational states. [The student should prove ∆E = EJ − EJ−1 = 2JB]. Lecture #10 13

Figure 8: Allowed Energy Levels for the Rotations of a Diatomic Molecule.