The Particle in a Box

Transcription

1 Page 324 Lecture 17: Relation of Particle in a Box Eigenstates to Position and Momentum Eigenstates General Considerations on Bound States and Quantization Continuity Equation for Probability Date Given: 2008/11/07 Date Revised: 2008/11/07

2 The Particle in a Box Section 5.2 Simple One-Dimensional Problems: The Particle in a Box Page 325 Relation to { x } and { p } Basis States We make the obvious point that our energy eigenstates are not position eigenstates: position operator eigenstates are independent of the Hamiltonian, and our states are just not the same as the position operator eigenstates in the { x }-basis representation, so they are not the same in any representation. We can also easily see that, even though the eigenstates of the Hamiltonian have a definite energy, they do not have a definite momentum. The momentum eigenstates are not dependent on the form of the Hamiltonian; it is always true that ψ p,x (x) = x p = 1 2 π e i p x (5.41) Note that we consider momentum eigenstates for the entire real line because the inner product space we began with was for functions on the entire real line: while the wavefunction vanishes outside the box, it is still a well-defined function there. The position basis matrix representation ψ En,x (x) = x ψ En of our eigenstate ψ En is different from the above position-basis matrix representations of the p basis elements in two ways:

3 The Particle in a Box (cont.) Section 5.2 Simple One-Dimensional Problems: The Particle in a Box Page 326 Since the ψ En,x (x) are sines and cosines, we would need to take linear combinations of states at p and p to obtain them. h i More importantly, ψ En,x (x) only equal to ψ p,x (x) over the interval L 2, L, 2 even though both ψ En,x (x) and ψ p,x (x) are defined on the entire real line. Hence, our energy eigenstates are simply not equal to momentum eigenstates. These facts are not surprising and arise simply from the fact that the Hamiltonian makes dynamics happen. In order for { x } basis states to be eigenstates of the Hamiltonian, a particle, once placed at a point, must never move from that point or spread out in position. Not even the free particle Hamiltonian allows that! And, while the free particle Hamiltonian s eigenstates are also momentum eigenstates, that clearly holds because there is no potential to affect the momentum of the particle. Once one includes any kind of potential, there is a force that can change the particle momentum and thus eigenstates of the Hamiltonian simply cannot be eigenstates of momentum.

4 General Considerations on Bound States and Quantization Section 5.3 Simple One-Dimensional Problems: General Considerations on Bound States and Quantization Page 327 Whence Quantization? We made the point above that the quantization of energies for the particle in a box arises because of boundary conditions imposed by the potential energy function, not by the postulates. This argument holds generally for the bound states of any potential. We present a less detailed version of the argument given in Shankar Section 5.2 on this point. Bound states are states whose energy E is less than the asymptotic value of the potential at ±. Classically, this is the case where the particle simply does not have enough energy to escape to ±. Quantum mechanically, the wavefunction must fall off at ±. To make the argument, we need to count up the number of free parameters and see how they are determined by the boundary conditions. For an arbitrary potential, one can think of breaking it up into small intervals of size ɛ. As ɛ 0, the potential can be treated as piecewise constant. We thus have our exponential solutions in any interval, with the argument being imaginary or real depending on whether E is greater than or less than the value of V (x) in the interval. There are four coefficient degrees of freedom for each of these intervals (the real and imaginary parts of the A and B coefficients).

5 General Considerations on Bound States and Quantization (cont.) Section 5.3 Simple One-Dimensional Problems: General Considerations on Bound States and Quantization Page 328 We have matching of both ψ x (x) and d ψx (x) at all the boundaries (the derivative dx must now match also because the steps in the potential are finite, as opposed to the particle-in-a-box case). That imposes four conditions (two equations each with real and imaginary parts) at each edge of the interval. Now, let us cascade the conditions from left to right. Suppose the four coefficient degrees of freedom in the interval have been set. That gives the four conditions that the wavefunction in the first finite interval must meet. The four coefficent degrees of freedom in the first finite interval thus are set. This procedure cascades through the last finite interval, which sets the four conditions at the last boundary. This provides enough information to set the four coefficient degrees of freedom in the + infinite interval. So, once the four coefficient degrees of freedom in the interval and the energy E are set, the rest of the wavefunction is determined. Now, let s consider how these first four coefficient degrees of freedom are set depending on whether we have a free or bound state.

6 General Considerations on Bound States and Quantization (cont.) Section 5.3 Simple One-Dimensional Problems: General Considerations on Bound States and Quantization Page 329 For a free state, we are allowed to keep both the e ±i k x solutions since there is no worry about either blowing up at ±. That means we have both the A and B coefficients, and so we really have four degrees of freedom. For free states, we will always have two energy-degenerate states with the same energy, a right-going state and a left-going state. One can see this by noting that the energy eigenvalue of the Schrödinger Equation does not care about the sign of the argument of the imaginary exponential because two derivatives are taken: d 2 dx 2 e±i k x = k 2 e ±i k x So, for a free particle, there will always be two independent, energy-degenerate solutions in the ± regions. Since they are independent, their overall phases are arbitrary and independent. We make an arbitrary choice for this phase (e.g., take A and B to be real in the region) and that sets the phase of the two independent solutions. The amplitude of A and B in the region will then be determined by normalization, as the two states must individually be normalized (to delta functions, since they do not decay away at ± ). Once we have set A and B in the region, the matching conditions cascade through to fully determine the wavefunction on the rest of the real line out to +. Thus, in the end, we are able to meet all the conditions on the wavefunction using only the coefficient degrees of freedom. No quantization is required.

7 General Considerations on Bound States and Quantization (cont.) Section 5.3 Simple One-Dimensional Problems: General Considerations on Bound States and Quantization Page 330 For a bound state, the wavefunction must become real exponential outside of some x for a given value of E because E is less than V somewhere: recall that the solution in any region where the potential is piecewise constant is So, when V > E, κ is real. r 2 m e ±κ x with κ = (V E) 2 Once we are forced into the real exponential solution, we acquire an additional condition we did not have in the free-particle case: the wavefunction may not blow up at ± in order for it to be normalizable. This eliminates two of the four coefficient degrees of freedom in each infinite interval. Essentially, we get four additional conditions that we didn t have in the free particle case, which sets A = 0 in the region and B = 0 in the + region. In the interval, this presents no problem: rather than having two independent solutions, we have a single solution, but we still pick an arbitrary phase and normalization for the wavefunction in this region as before. Again, the matching conditions cascade through. However, now in the + region, instead of having four coefficient degrees of freedom to use to meet the four matching conditions, we only have the A coefficient degree of freedom because B = 0 is necessary to prevent the wavefunction from blowing up at +. The only way to meet the four matching conditions is to restrict the freedom in E. Hence, we obtain energy quantization.

8 General Considerations on Bound States and Quantization (cont.) Section 5.3 Simple One-Dimensional Problems: General Considerations on Bound States and Quantization Page 331 The final case to consider is when the energy is such that the particle is bound on one side and free on the other. In such cases, there will be no energy quantization, but there will also not be independent left- and right-going solutions. The easiest way to see this is to begin defining the wavefunction on the bound side, which we will take to be the side without loss of generality. The argument goes through as above, resulting in four matching conditions at the left edge of the + interval. Here, we can satisfy the four matching conditions without quantizing E simply be allowing the four matching conditions to set A and B in this region. A and B get tied together, tying the nominally independent e ±i k x solution together. So, being bound on one side removes energy degeneracy but does not result in quantization. Classically, the way to see this is that, if you start out with a left-going state heading into this potential, it will encounter the potential barrier at and be turned around into a right-going state. Since the eigenstates of the Hamiltonian are time-independent, they must contain both these behaviors. Thus, the solution is a superposition of left-going and right-going in the unbound region. (As an aside, this is also a way to think about the solutions in the bound regions for which E > V : since the particle is bound and thus bounces off the potential barriers on the two sides, it can be in neither a pure left-going or right-going state. Thus, our solution in bound regions is always the sum of the left- and right-going waves.)

9 General Considerations on Bound States and Quantization (cont.) Section 5.3 Simple One-Dimensional Problems: General Considerations on Bound States and Quantization Page 332 How did this all of this function in our particle in a box case? One can think of it in terms of keeping V 0 finite, considering only the bound state solutions, and then letting V 0 go to infinity. With finite V 0, the above generic bound state explanation would apply, resulting in energy quantization for bound states. Then, letting V 0 would not change this, but would simply eliminate any free states. We did not do the problem in this fashion because, by letting V 0 a bit earlier, we could conclude that the wavefunction vanished in the ± regions. Rather than setting four conditions at the left side of the box using the wavefunction and its derivative, we instead set two conditions at each side of the both on the wavefunction alone. This was still four total conditions on the region II solution, fully specifying it. It would have been a bit more painful to do it in the generic way because we would have to carry along the non-vanishing region I and III solutions a bit longer, resulting in more algebra.

10 The Continuity Equation for Probability Section 5.4 Simple One-Dimensional Problems: The Continuity Equation for Probability Page 333 Analogy to Electromagnetism Postulate 3 of QM tells us to interpret x ψ(t) as the probability P(x, t) that the position of the particle is in the interval x to x + dx at time t; P(x, t) is a probability density for finding the particle. This is similar to the idea of a charge density in electromagnetism, ρ(x). In the case of electromagnetism, we have the following important results: The charge in an infinitesimal interval dx or volume d 3 x is dq(x, t) = ρ(x, t) dx or dq( x, t) = ρ( x, t) d 3 x The total charge in an interval [a, b] or volume V is Z b Q(t) = a The total charge over all space is conserved. Z dx ρ(x, t) or Q(t) = d 3 x ρ( x, t) V

11 The Continuity Equation for Probability (cont.) Section 5.4 Simple One-Dimensional Problems: The Continuity Equation for Probability Page 334 The electric current density is defined as j(x, t) = ρ(x, t) v(x, t) or j( x, t) = ρ( x, t) v( x, t) where v or v is the velocity of the charges currently at position x or x at time t. s The charge density satisfies the continuity equation ρ(x, t) t + j(x, t) x = 0 or ρ( x, t) t + j(x, t) = 0 Note the use of partial derivatives now. If one thinks of the charge density as a density smoothed over a large number of point charges, those charges are moving their position is a function of t. By, we mean do not move along t with the charge that moves at velocity v (or v), just sit at a point x and watch the charges flow by and measure the rate of change of the density at the point x and by, we mean look at the gradient in x at a fixed point x, again do x not move in x with the charges.

12 The Continuity Equation for Probability (cont.) Section 5.4 Simple One-Dimensional Problems: The Continuity Equation for Probability Page 335 The integral version of the continuity equation is or Z b dx ρ(x, t) + [j(b, t) j(a, t)] = 0 t a Z Z d 3 x ρ( x, t) + d 2 x ˆn( x) j( x, t) = 0 t V S V In the one-dimensional case, we initially had an integral over the interval [a, b] of the perfect differential j(x,t), which we simply integrated to get the x difference of the values of j(x, t) at the boundaries. In the three-dimensional case, we used Gauss s law to convert the volume integral of the divergence of j( x, t) to a surface integral; ˆn is the outward surface normal at x. Note that, in both cases, the boundary is fixed in time. The continuity equation says that charge must flow from one point to another in a smooth fashion no sudden appearance or disappearance of charge is possible. Given that overall charge is conserved, we then have that the electrical current density must either vanish at infinity or the total current flux must vanish.

13 The Continuity Equation for Probability (cont.) Section 5.4 Simple One-Dimensional Problems: The Continuity Equation for Probability Page 336 We can prove total conservation and the continuity equation for quantum-mechanical probability density, which is P(x, t) = x ψ(t) 2 = ψ x (x, t) 2 or P( x, t) = x ψ(t) 2 = ψ x ( x, t) 2 (5.42) We shall see that the approprate definition for the associated probability current is or j(x, t) = i 2 m i j( x, t) = 2 m ψx (x, t) x «ψx (x, t) ψx (x, t) x ψ x (x, t) ψ x ( x, t) ψ x ( x, t) ψ x ( x, t) ψ x ( x, t) (5.43)

14 The Continuity Equation for Probability (cont.) Section 5.4 Simple One-Dimensional Problems: The Continuity Equation for Probability Page 337 First, let us prove that total probability is conserved. This is just a matter of using the fact that the Schrödinger Equation implies unitary evolution of the state when H is time-independent: Z V Z d 3 x P( x, t) = d 3 x ψ(t) x x ψ(t) = ψ(t) ψ(t) V = ψ(0) U (t)u(t) ψ(0) = ψ(0) ψ(0) Hence, if we take a time derivative of the integrated probability, we get zero: the total probability is conserved:. Z d d 3 x P( x, t) = 0 (5.44) dt V

15 The Continuity Equation for Probability (cont.) Section 5.4 Simple One-Dimensional Problems: The Continuity Equation for Probability Page 338 Now, let us prove the other important result from E&M, the continuity equation. We prove this by explicitly taking the time derivative of the probability density using the Schrödinger Equation. We will prove this for the three-dimensional case; the restriction of the proof to one dimension will be clear. Note also that we immediately restrict to the position-space representation ψ x ( x, t) = x ψ(t) because we are really only interested in the probability density and current in this representation. It would be interesting to discuss the probability current in other representations (momentum, energy, etc.), but is not relevant here.

16 The Continuity Equation for Probability (cont.) Section 5.4 Simple One-Dimensional Problems: The Continuity Equation for Probability Page 339 We have P( x, t) t = ψx ψx ( x, t) ( x, t) + ψ x ( x, t) ψ x ( x, t) t t = ψx ( x, t)» i» i H( x, t)ψx ( x, t) + ψ x ( x, t) = i ψ x ( x, t)» 2 2 ψ x ( x, t) + V ( x, t)ψ x ( x, t) 2 m + i» ψx ( x, t) 2 2 m H ( x, t)ψx ( x, t) 2 ψx ( x, t) + V ( x, t)ψx ( x, t) = i h ψx ( x, t) 2 ψ x ( x, t) ψ x ( x, t) 2 ψx ( x, t) 2 m where we used the Schrödinger Equation and its conjugate to get from the first line to the second, wrote out the Hamiltonian in the third line (using the fact that the potential V ( x, t) must be real for the Hamiltonian to be Hermitian and generalizing d 2 /dx 2 to 2 for three dimensions), and then canceled the V ( x, t) term to get the last line. i

17 The Continuity Equation for Probability (cont.) Section 5.4 Simple One-Dimensional Problems: The Continuity Equation for Probability Page 340 Finally, pull one to the front of the expression and manipulate a bit: P( x, t) t» = i 2 m» i 2 m = j( x, t) h ψx ( x, t) ψ x ( x, t) ψ x ( x, t) ψ i x ( x, t) h i ψ x ( x, t) ψx ( x, t) ψx ( x, t) ψ x ( x, t) and we have our desired result for the continuity equation: P( x, t) t + j( x, t) = 0 (5.45)