Quantum Mechanics: Schrödinger Wave Equation, Particle in a Box, Finite Potential Well, Tunneling, Harmonic Oscillator

1. Chapter 40 Quantum Mechanics

2. Goals for Chapter 40 • To introduce and interpret the Schrödinger wave equation for quantum-mechanical waves • To solve the Schrödinger equation for a one-dimensional “particle in a box” • To study the behavior of a quantum-mechanical particle in a finite potential well • To examine tunneling, in which quantum mechanics allows a particle to travel through a region that would be forbidden by Newtonian physics • To consider the quantum-mechanical harmonic oscillator, a model for molecular vibrations

3. The Wave Equation • Recall in chapter 15 on waves, we discussed the one-dimensional wave equation • We saw that the solution to this equation (as can be seen by direct substitution) is where A and B are two constants determined by initial conditions. This solution works as long as v = lf = w/k. • In the previous lecture, we introduced de Broglie’s idea that matter can also be thought of as wave-like in nature, with energy and momentum given by • However, for this case we cannot have such a simple relation as v = lf = w/k. Writing the total energy of a free particle (U = 0), we would have E = K = p2/2m, so • If matter behaves like a wave, we ought to be able to come up with a wave equation for matter, but it has to obey the above relation between w and k (for the case of a free particle).

4. The Wave Equation • Let us look for a wave equation (containing derivatives with respect to x and t) whose solution is that obeys equation 1: . • A second derivative with respect to x brings out a –k2, and a single derivative with respect to t brings out an w: • If we multiply the spatial derivative term by that will give the left side of equation 1. • We can almost get the right side of equation 1 by multiplying by , but the time derivative term is a little tricky because the first term has become a sine function while the second term became a negative cosine. We need a factor C such that

5. The Schrödinger Equation • Clearly A = –BC, and B = CA will do the trick. Thus, A = –C2A, which means C2 = –1, so . • The final wave equation that seems to work is . • This is the one-dimensional wave equation for a free particle, and is a special case of the Schrödinger equation, first developed by Erwin Schrödinger in 1926. • Since C = i, B = iA, so we can write the wave function (solution) as which is a complex function with real part , and imaginary part . This represents a particle moving to the right if k is positive, and to the left if k is negative. • Using Euler’s formula , the wave function can be written more compactly as

6. The Schrödinger equation in 1-D • We found that the one-dimensional Schrödinger equation for a free particle of mass m is • How do we interpret the complex solution ? This represents a distribution of “something” in space and time. Any real quantity, however, must have a real solution. Recall that we interpreted the interference intensity pattern as representing the square of the electric field, and individual photons land on a screen with a probability given by the intensity pattern (more land where the intensity is high, fewer land where it is low). Likewise, the quantity is the (real) probability in space and time where the particle will be found, where the * represents the complex-conjugate found by replacing i with -i. • The square of the abs. value of the wave function, |(x, t)|2, is the probability distribution function. It tells us the probability of finding the particle near position x at time t.

7. The Schrödinger equation in 1-D: Wave packets • If is a solution to the Schrödinger equation, any superposition of such waves is also a solution. This would be written: • A free-particle wave packet localized in space (see Figure 40.6 at right) is a superposition of states of definite momentum and energy. • The function itself is “wavy,” but the probability distribution function is not. • The more localized in space a wave packet is, the greater the range of momenta and energies it must include, in accordance with the Heisenberg uncertainty principle. • The Schrödinger equation discussed so far is only for a free particle (in a region where potential energy U(x) = 0). We will now add non-zero U(x).

8. The Schrödinger equation in 1-D: Stationary states • If a particle of mass m moves in the presence of a potential energy function U(x), the one-dimensional Schrödinger equation for the particle is • This equation can be thought of as an expression of conservation of energy, K + U = E. Inserting , the first term is K times . • Likewise, the term on the RHS is E times . • For a particle in a region of space with non-zero U(x), we have to add the term U(x)Y(x, t) on the left to include the potential energy. • Let’s write the wave function in separable form, where the lower-case y(x) is the time-independent wave function. • Further, we can write , so that .

9. The Schrödinger equation in 1-D: Stationary states • For such a stationary state the probability distribution function |(x, t)|2 = |(x)|2 does not depend on time, which you can see by • In this case, the time-independent one-dimensional Schrödinger equation for a stationary state of energy E simplifies to where the time derivative has been explicitly taken on the RHS. • We will spend most of the rest of the lecture on solving this equation to find the stationary states and their energies for various situations. • Note: The term stationary state does not refer to the motion of the particle it represents. The particles are not stationary, but rather their probability distribution function is stationary (does not depend on time), rather like a standing wave on a string. • Example 40.2: Stationary state: Consider the wave function for a free particle with U(x) = 0. Determine its energy by calculating

10. Particle in a box • As a first approximation to an electron bound to a hydrogen atom, consider a “particle in a box,” i.e. a particle of mass m confined to a region between x = 0 and x = L. The potential energy is zero inside the “box” and infinite outside (see figure at lower right). The equation inside the box will be the same as for a free particle (U(x) = 0), but if it has finite total energy, then it cannot exist outside the box. This changes the solutions due to the boundary conditions.

11. Particle in a box: Wave functions, energy levels • Because of the boundary conditions (x) = 0 at x = 0 and x = L, only certain wavelengths are permissible, exactly as we saw for standing waves in a closed pipe, or on a string with tied ends. The energy levels and associated stationary-state wave functions (x) for a particle in a box are shown. To see this, start with our earlier solution to the 1D time-independent Schrödinger equation: • The boundary conditions will require certain constraints on constants A1 and A2. We have (0) = A1 + A2 = 0 => A2 = –A1 and (L) = 2iA1sin kL = 0. Writing 2iA1 = C (a constant), the specific solution is (x) = Csin kL. But the sine function is zero for any k = np/L. Final solution: (x) = Csin npx/L . What are the energies? Note, n = 0 is not a solution (would imply y(x) = 0 everywhere.

12. Particle in a box: Probability and normalization • Note that this view of matter as waves leads directly to the idea of quantized energy levels! The figure at right shows the first three stationary-state wave functions (x) for a particle in a box (top) and the associated probability distribution functions |(x)|2 (bottom). There are locations where there is zero probability of finding the particle. What about the constant C in our solution (x) = Csin npx/L ? • Wave functions must be normalized so that the integral of the probability density function |(x)|2 over all x equals 1 (which means there is 100% probability of finding the particle somewhere). • We can evaluate this using the identity sin2q = ½ (1- cos 2q ) to get C = (2/L)½. The final, normalized solution is then

13. Particle in a finite potential well I • The infinite potential outside the box is not very realistic, so let us relax that assumption. A finite potential well is a region where potential energy U(x) is lower than outside the well, but U(x) is not infinite outside the well (see Figure 40.13 below). • In Newtonian physics, a particle whose energy E is less than the “height” of the well can never escape the well. In quantum mechanics such a trapped state is called a bound state. All states are bound for the infinite well. For a finite well, however, the particle energy can be greater than U0, in which case it is not bound. Also, the bound states for a finite well are subtly different. • We need a solution that solves everywhere. • For the potential well below, we have three regions x < 0, 0 < x < L, and x > L. The equations to solve are

14. Particle in a finite potential well I • We already saw that for the middle (well) region, the solution is with energy, , so we can write k in terms of E as , and the solution is • However, we cannot assume the same boundary conditions as for the infinite well. It turns out the wave function can exist outside the well! There we have the equation • For a bound particle, E < U0 so the quantity on the right is positive. That make the solution an exponential, not a wave! Writing , we have the general solution • Now, for x < 0, the second term grows to infinity, which is not a physical solution, so we must have D = 0. Likewise, for x > L the first term grows to infinity, so in that region we must have C = 0. • Joining all of these together will allow us to find the values of A, B, C and D.

15. Particle in a finite potential well I • For a single solution to the entire problem, we must have the solution at the boundaries join smoothly, i.e. at x = 0, and at x = L, both the functions and their first derivatives must have the same values. We will not go into the details of this matching procedure, which requires solving transcendental equations by numerical approximation, but this matching can only be done for specific values of the energy E. • The figures below show some solutions for the wave-function (at left) and the probability distribution function (at right) for three energy levels. Note that there is a finite probability that the particle will be outside the potential well (the exponential tails of the probability distribution function). http://en.wikipedia.org/wiki/Finite_potential_well

16. Potential barriers and tunneling • Figure 40.19 (below left) shows a potential barrier. In Newtonian physics, a particle whose energy E is less than the barrier height U0 cannot pass from the left-hand side of the barrier to the right-hand side. • Figure 40.20 (below right) shows the wave function (x) for such a particle. The wave function is nonzero to the right of the barrier, so it is possible for the particle to “tunnel” from the left-hand side to the right-hand side. • The tunneling probability T is given (in the limit of small T) as: • As U0 or L decreases, the probability of tunneling goes up.

17. Potential barriers and tunneling • Example 40.7: A 2.0-eV electron encounters a barrier 5.0 eV high. What is the probability that it will tunnel through the barrier if the barrier width is (a) 1.00 nm and (b) 0.50 nm? • E/U0 = 2/5, so G = 16(2/5)(3/5) = 3.8. Also • When L = 1.00 nm, T = Ge-2kL = 3.8e-17.8, = 7.110-8. • When L = 0.50 nm, T = Ge-2kL = 3.8e-8.9, = 5.210-4.

18. Applications of tunneling • A scanning tunneling microscope measures the atomic topography of a surface. It does this by measuring the current of electrons tunneling between the surface and a probe with a sharp tip. As the tip nears an atom, the barrier gets thinner, so the current indicates how close the atom is to the tip (the height of the surface) • An alpha particle inside an unstable nucleus can only escape via tunneling. The reverse happens in fusion, such as what goes on in the Sun.

19. Harmonic Oscillator • The finite square-well potential is also not very realistic, so let us think about another shape that appears often in nature—a potential that varies quadratically (U(x) = ½k′x2). Notice that this is the relation for potential energy of a spring, and in fact is called the harmonic oscillator potential for that reason. Note that we use k′ here to show that it is not the wave-number, but rather is the usual spring constant (i.e. the proportionality between distance and force, F = -k′x). • Now the Schrödinger equation becomes: • The solutions to this should be finite within the well, and asymptotically fall to zero outside the well. The energies turn out to be quantized according to:

20. A comparison of Newtonian and quantum oscillators • Figure 40.26 (below, top) shows the first four stationary-state wave functions (x)for the harmonic oscillator. A is the amplitude of oscillation in Newtonian physics. • Figure 40.27 (below, bottom) shows the corresponding probability distribution functions |(x)|2. The blue curves are the Newtonian probability distributions.