PHYS 424 Notes

V(x)=0, 0<=x<=a, and V(x)=infinity, otherwise. Draw

[hbar²/(2m)] d²y(x)/dx² + Ey(x) = 0 for 0<=x<=a ,
and y = 0 elsewhere. Justify

d²y(x)/dx² + (2mE/hbar²)y(x) = 0

Set k = Ö(2mE)/hbar, then

d²y(x)/dx² + k²y = 0

Try y = A sin ax + B cos ax. Substituting gives

A(-a² + k²)sin ax + B(-a² + k²) cos ax = 0

which must be true for all x. Thus a= ± k. However, cos is even and sin is odd, so -k gives no new solutions. It is adequate to take a = +k . Hence the solution is

y = Asin kx + B cos kx

A and B must be determined from boundary conditions, so we must know what the appropriate boundary condition would be. At every point, d²y/dx² must exist. Hence, at every point, dy/dx must exist, which implies that y is continuous. If |V| < infinity, d²y/dx² must be finite and hence dy/dx must be continuous. So y is continuous everywhere, and dy/dx is continuous everywhere that |V| < infinity.

Since y must be continuous and vanishes left of x=0 and right of x=a,

y(0)=0 and y(a)=0.

The first of these gives B = 0 . Then the second gives A sin ka =0. Here we get a surprise; if we attempt to determine A from the boundary condition we get A=0 and have no solution at all. The only way to satisfy the boundary condition is to demand that

There is a separate state for each value of n. The corresponding states are

y_n = A sin npx/a

and A is not yet determined. However, we must have a 100% chance of finding the particle somewhere in the well, hence
1 = ò y^*y dx
or 1 = |A|² ò sin² (npx/a) dx.
The integral is just a/2, so
|A| = Ö(2/a) or A = e^ifÖ(2/a)
where f is an arbitrary phase. As a result

y_n = e^if Ö(2/a) sin npx/a

Since the factor e^if is a constant, it will drop out of any expectation value, even those involving p=-i hbar d/dx. We may therefore choose it for our convenience, and f=0 is the most convenient choice. Putting everything together,

Plot a few of these.

Properties of the Wave Function

As a shorthand, instead of writing y_n(x), let us write |n>. [technically, we should write <x|n> to show that we are thinking about y_n(x) and not its Fourier transform <p|n> = y_n(p); we will use the shorter form and not worry about technicalities until we have to.] Also use <n| = y_n^* .

We can see two things by looking at the graphs of the lowest |n> :

1. |n> is even or odd around the center of the well when n is odd or even, respectively.
2. |n> has n-1 internal (|x|<>a) nodes

3. ò y_m^*(x) y_m(x) dx = d_mn
   = 0 if m<>n and 1 if m=n .
   In our shorthand notation, <m|n> = d_mn; note how the integration is implied rather than written out explicitly.

4. for any arbitrary y(x), there exist coefficients such that
   y(x) = S_n c_ny_n(x), c_n = ò y_n^* y dx .
   In shorthand, |y> = S_n c_n |n> , c_n=<n|y> In this notation |y> = S_n |n><n|y>

We have already proven #3 for n=m; it is just the normalization condition. We can prove the relation for n¹m by simply substituting the form of the wavefunctions:

<m|n> = ò Ö(2/a)sin(npx/a) Ö(2/a)sin(mpx/a) dx

and using sin ax sin bx = [cos(a-b)x - cos(a+b)x]/2,

<m|n> = (2/a)ò[(1/2)cos(n-m)px/a -(1/2)cos(n+m)px/a] dx

Since neither n-m nor n+m is zero, each part of the integral is an integration of a cosine over an integer number of cycles and vanishes. This proves the theorem.

We can't prove #4, but we can determine the form of the coefficients. We have

|y> = S_n c_n |n> ,

Multiply both sides by <m| and integrate:

<m|y> = S_n c_n <m|n> = S_n c_n d_mn = c_m

as previously quoted.