## Thursday, May 9, 2013

### Solutions to Quantum 'Box' Problems

Inside the box the potential V = 0 and outside, V =  ¥

The particle cannot be at locations, x > L or less than 0, else it would have infinite energy.

Therefore, outside the interval (0,L) one has the quantum wave form: y*y = 0

The Schrodinger equation to solve is: dy2/dx2  + K2 y = 0

where K =  Ö [2mE] / ħ

And this second order differential equation has the solution:

y = C sin Kx + D cos Kx

(C, D constants)

We set boundary conditions thus:
At x = 0+, y (0+) = D (since sin(0) = 0)

At x = 0, y (0) = 0

The wave function must be continuous (since we want y*y to represent something physically observable).

So: y (0+) = y (0-) therefore D = 0

y = C sin Kx

Now, at x = L+, y (L+) = 0

At x = L-, y (L-) = C sin K(L)

For continuity: y (L+) = y
(L-)

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

Therefore: sin K(L-) = 0 Þ  sin (Ö[2mE] / ħ) L = 0,

and
((Ö[2mE] / ħ) L = n π
Therefore:   L = nπ /K and n = + 1, + 2, ..etc.

2mE = n2 π2 ħ2     Then   E = { n2 π2 ħ2 } / 2mL2

From which we obtain the energy eigenvalues for the levels n = 1 and n=2.

For n = 1:      En=1 =    π2 ħ2/ 2mL2,

For energy level n = 2:   En=2 =   4 π2 ħ2/ 2mL2

Now m = 9.1 x 10-31 kg (mass of electron) and L = 0.2nm = 2 x 10-10 m

Also:  ħ  = h/ 2 π  = 6.62 x 10-34 J-s/  2 π   = 1.054 x 10-34 J-s

Then the actual energy at the first level (n= 1) is:

π2 ħ2/ 2mL2  =   (3.1416)2 (1.054 x 10-34 J-s)2 / 2(9.1 x 10-31 kg) (2 x 10-10 m)
=  1.43 x 10  16 J

Then the actual energy at the 2nd level (n= 2) is:

4 En=1 =    4( 1.43 x 10  16 J) =   5.72 x 10  16 J

(2)  What is the probability the particle will be found between x = 0 and x = L/4.

A particle confined in a box of length L has a wave function that can be expressed in simplest form as:

y = Ö2/ ÖL   [sin (2npx/L)]

Note that if: y = C sin Kx  then normalization requires we do the following:

ò Lo   y2  dx =

ò Lo   ‖C‖2  sin 2Kx  dx =  ‖C‖2   L/ 2 = 1

So that:  ‖C‖2     = 2/L   and C = Ö2/ ÖL

Consider the simplest energy level : n = 1

Then: P ab  =     òL/4 0   (Ö2/ ÖL )2   sin2 (2px/L) dx

Let q = 2px/L  and use the trig identity:  sin2 q= ½ (1 – cos 2q)

P ab  =     2/ L òL/4 0    ½ [1 - cos (4px/L)] dx

P ab  =     2/ L [½  òL/4 0    dx  - ò L/4 0   cos (4px/L)] dx

And: P ab  =     x/ L  -  1/2p  sin (4px/L)] L/4 0

P ab  =      ¼ - 0 = 0.25