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:      E_n=1 =    π² ħ²/ 2mL²,

For energy level n = 2:   E_n=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 E_n=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 ^L_o   ‖y‖²  dx = 

 ò ^L_o   ‖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  =     ò^L/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 ò^L/4 ₀    ½ [1 - cos (4px/L)] dx

P _ab  =     2/ L [½  ò^L/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