Selected Solutions - Intro. to Quantum Mechanics (3)

3) The wave function for a particle confined to moving in a 1D box is given by:

y(x)  =   A [sin (npx/L)]  

Use the normalization condition on y(x)   to show the constant A is given by: 

A =  Ö2/ ÖL  

(Refer to the normalization condition applied to the system in 5(b), i.e. solving for A)


4) From the information and the Heisenberg Uncertainty Principle:

DP x (D x)  > h/ 2p  = 1.05 x 10^-34 J-s  = 1.05 x 10^-27 erg-s   (cgs units)

The uncertainty in the x-component is 0.5 angstrom where 1 A = 1.0 x 10^-8 cm

Then the uncertainty in the x-component is: Dx =  0.5A = 5.0 x 10^-9 cm

The uncertainty in the x-component of the  momentum of the electron is:

DP x»    (h/ 2p)  / D x  =    1.05 x 10^-27 erg-s/  5.0 x 10^-9  cm

Or, 2 x 10^-19 g cm-s

5 (a)) In classical theory the probability that the particle would be in the region between x and x + L/3 is 0.33.  Since no other information is given then it is equally probable that it’s in any place in the region – so one third of the time will be spent over any given line segment, i.e. which is 0ne-third the total length.

b) For a quantum mechanical interpretation, define the wave function (See e.g. http://brane-space.blogspot.com/2013/05/solutions-to-quantum-box-problems.html:

y = A sin 2kx

sin 2kL = 0  and k (wave number ) =   n p/L

y = A sin (n px/L)

 P(x) =  y ²  = A² sin ² (n px/L)

For normalization:

1 =  =     ò ^L ₀   P(x)   dx  =   A²  ò ^L ₀    sin ² (n px/L) dx

A²  (L/ n p) [ u/2 – sin 2u/ 4] ₀ ^{n p}           = 1

 A²  (L/ n p) (n p/2) = 1

 Or:  A²  =  2/L   so that:   A =  Ö2/ ÖL  

Then the probability it will lie between 0 and L/3 is:

ò ^L/3 ₀    sin² (npx/L) dx

=  2/ n p [n p/ 6  -  ¼ (sin 2 p/ 3 )]  = 


2/ p [p/ 6  -  ¼ (sin 2 p/ 3 )]      = 0.19

(Since n = 1 is the lowest energy state)

c) Since n = 2 is the second lowest state the probability the particle is between 0 and L/3 is:

  (2/L) L/ 2 p  [p/ 3  -  ¼ (sin 4 p/ 3 )]  =  0.40