## Wednesday, January 19, 2011

### Solution of a Simple Quantum Mechanical System (1)

When one takes a standard quantum mechanics course, one must learn to "walk" before one can run. This means that before a student can approach the Schrodinger solution for the hydrogen atom - whereby we use spherical harmonic functions to obtain the energy eigenvalues, etc. - one has to play around with much simpler quantum systems first.

Probably the simplest system of all is the 1-dimensional "infinite" square well shown in the accompanying top panel.

One notes the box limits are from 0 to a and this is putatively the x -direction (though granted it looks like y!)

Inside the box the potential V = 0 and outside, V = oo, hence the term, "infinite square well".

From these constraints, one sees that any particle cannot be at locations, x > a or less than 0, else it would have infinite energy. As we know, energies in QM are quantized, so no infinities are possible.

Therefore, outside the interval (0,a) one has the quantum wave form:

U*U = 0

The Schrodinger wave equation is written (for one dimension):

H^ U = E U

where H^ denotes the Hamiltonian operator:

H^ = = [-iħ d/dx]^2/2m

Then the Schrodinger equation to solve becomes:

-ħ^2/ 2m ( dU^2/dx^2 ) = EU

Thence:

dU^2/dx^2 + 2m/ ħ^2 (EU) = 0

Or, in more concise form:

dU^2/dx^2 + K^2 U = 0

where K = [2mE]^½/ ħ

And this second order differential equation has the solution:

U = C sin Kx + D cos Kx

(C, D constants)

We set boundary conditions, thus:

At x = 0+, U(0+) = D (since sin(0) = 0)

At x = 0, U(0) = 0

As a cautionary note, one must postulate that the wave function is continuous (since we want U*U to represent something physically observable).

Then:

U(0+) = U(0-) therefore D = 0

and

U = C sin Kx

Now, at x = a+, U(a+) = 0

At x = a-, U(a-) = C sin K(a)

For continuity: U(a+) = U(a-)

Therefore: C sin K(a-) = 0 but C can't = 0, or no particle!

Therefore: sin K(a-) = 0

-> sin ([2mE]^½/ ħ) a = 0, and

([2mE]^½/ ħ) a = n π

Therefore: a = nπ/K and n = +/- 1, +/- 2, ..etc.

Thus:

2mE = n^2 π^2 ħ^2

E = {n^2 π^2 ħ^2}/ 2ma^2

which yields the energy eigenvalues.

In general, E = {n^2 π^2 ħ^2}/ 2ma^2 (E(n) = 1, 2, 3, .....n)

Thus, E(n) is restricted to:

π^2 ħ^2/ 2ma^2, 4 π^2 ħ^2/ 2ma^2, 9 π^2 ħ^2/ 2ma^2 etc.

Obviously, this system is very elementary, but we challenge audacious readers to consider the 3-dimensional box with particle in it, for next time. (Lower graphic)