PHYS 424 Notes

Use diagram with regions marked

Now we work with a square potential that does not rise to infinity. There are three regions:

Region A	[x<-a]:	V = 0
Region B	[-a<x<a]:	V = -V0
Region C	[x>a]:	V = 0

Note that the total width of this potential has been chosen to be 2a, not a as we had for the infinite square well. We will look first for bound states, E<0. It is useful to define the three constants

k = Ö(- 2 m E) / hbar   and k = Ö(+ 2 m E) / hbar
l = Ö[2 m (E + V₀)] / hbar

when E<0, k is real and we ignore k; later, when we choose E>0, k will be real and we will ignore k.

For E<0, the Schroedinger equation and y can be written as

Region A:	d2y/dx2 = k2y	y = Ae -k x + Bek x
Region B:	d2y/dx2 = - l2y	y = C sin(lx) + D cos(lx)
Region C:	d2y/dx2 = k2y	y = E'ek x + Fe -k x

In region A, e ^{-k x} blows up at - ¥, so normalizability requires A=0, while in region C, e^{+k x} blows up at + ¥, so E'=0. There are four constants left, and we have four boundary conditions to satisfy: y and dy/dx must be continuous at both x=a and x=-a. Tabulating,

Region A:	y = Bek x	dy/dx = Bkek x
Region B:	y = C sin(lx) + D cos(lx)	dy/dx = C l cos(lx) - D l sin(lx)
Region C:	y = Fe -k x	dy/dx = -Fke -k x

Now V(-x) = V(x), so the solutions y must be either even or odd. If y is even, B=F and C=0; if y is odd, B = - F and D = 0. In either case, satisfying the boundary conditions at x=a will automatically satisfy the boundary conditions at x=-a. Let us seek the even solution (the problem set contains the determination of the odd one). We have (at x=a)

D cos(la) - F e ^-ka = 0
-D l sin (la) + F k e ^-ka = 0

We can duck some complications by noting that, to have a nontrivial solution to these equations,

   |     cos(la)    - e -ka  |
   |                        |  =  0 .
   | - l sin(la)     k e -ka |

Hence k = l tan(la). Recovering the definitions of k and l,

This equation can be used to determine E; there are one to many discrete values of E. As always with bound states, the boundary conditions imply the existence of discrete values for the energy.

The solution is done well in the text. Do we need to look at it in class?

For E>0, the Schroedinger equation and y can be written as

Region A:	d2y/dx2 = - k2y	y = Aeikx + Be -ikx
Region B:	d2y/dx2 = - l2y	y = C sin(lx) + D cos(lx)
Region C:	d2y/dx2 = - k2y	y = E'e -ikx + Feikx

The terms multiplied by A and F are waves travelling to the right; the terms multiplied by B and E' are waves travelling to the left. A reasonable physical situation is a wave coming toward the potential from the left, which may be partly reflected at x = -a, and which may partly penetrate to give a wave travelling to the right beyond x=a. There would be no wave coming from the right for x>a. In this situation, A is set by the experimenter and E' is zero. There are five constants, four of which must be determined, and we have four boundary conditions to satisfy: y and dy/dx must be continuous at both x=a and x=-a. The equations are inhomogeneous because of the existence of A, so the four unknown constants can be determined in terms of A. Tabulating,

Region A:	y = Aeikx + Be -ikx	dy/dx = ikAeikx - ikBe -ikx
Region B:	y = C sin(lx) + D cos(lx)	dy/dx = C l cos(lx) - D l sin(lx)
Region C:	y = Feikx	dy/dx = ikFeikx

The four boundary conditions are

x = -a:	Ae -ika + Be ika = - C sin(la) + D cos(la)	ikAe -ika - ikBe ika = C l cos(la) + D l sin(la)
x = +a:	C sin(la) + D cos(la) = Feika	C l cos(la) - D l sin(la) = ikFe ika

The coefficients C and D are usually unobservable and so should be eliminated first. Then F and B can be determined in terms of A. In particular, the text gives

The transmission coefficient is defined to be the number of particles arriving per unit time beyond the well divided by the number of particles per unit time incident on the well from the left. The number per unit time is the density times the velocity; since the velocity to the left and right of the well is the same, the transmission coefficient is just the ratio of the probability densities, T = |F|² / |A|². If the velocity left and right of the well are not the same, the transmission coefficient will be more complicated. For this problem, the transmission coefficient is the rather unilluminating

There is a plot of the transmission coefficient as a function of energy in the text, and you should examine it. The only obvious physics in the expression is that, as shown in the text, the transmission coefficient is 1 for those energies where the infinite square well would have a bound state. This is a kind of resonance situation.