## Class Notes, Spring 2002

 Week of Jan 14 Feb 4 Mar 4 Apr 1 Phy 360 page Jan 21 Feb 11 Mar 18 Apr 15 Jan 28 Feb 18 Mar 25 Apr 22 Feb 25

## Phy 360, Week 10

25 Mar Briefly review the oscillation modes of a string stretched between two fixed positions.

Apply the same ideas to the Schroedinger equation. As the first example, choose the one that is most like the string problem, the "particle in a box." The first significant difference is the presence of the "i" in the equation. This changes the time dependence of the result from a cosine to an exponential: e- i w t.

y ( x, t ) = e- i w t u ( x )

Then u must satisfy the equation

E u = ( - hbar2 / 2 m ) u'' + V u.

Here I use E for the combination hbar w = (h/2p) w.

The easiest case to examine is one in which the potential energy, V, is zero and the domain is 0 < x < L. The function will vanish at the endpoints 0 and L just as with the string, though it's not so obvious why that should be so in this case. To understand that will require looking at a slightly more complicated problem first and then coming back to this one.

The equation is now

E u = ( - hbar2 / 2 m ) u''

This should be a familiar equation, though presented in an unfamiliar context. It is the common differential equation for the simple harmonic oscillator but with different variables and constants. It says that the second derivative of a function is a negative number times the function itself. That leads to sines or cosines.

The solution is

u ( x ) = A sin ( C x + D ),     where     E = ( hbar2 / 2 m ) C2

Now use the boundary conditions that the function must vanish at two points.

u ( 0 ) = 0     and   u( L ) = 0

imply that D=0 and that CL=np. This in turn determines the energy levels:

En = ( (h/2p)2 / 2 m ) ( n2 p2 / L2 )

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