Read More
Date: 29-8-2016
1212
Date: 28-7-2016
833
Date: 18-8-2016
1034
|
Shallow Square Well II
A particle of mass m is confined to move in one dimension by a potential V(x) (see Figure 1.1):
(1)
Figure 1.1
a) Derive the equation for the bound state.
b) From the results of part (a), derive an expression for the minimum value of V0 which will have a bound state.
c) Give the expression for the eigenfunction of a state with positive energy E > 0.
d) Show that the results of (c) define a phase shift for the potential, and derive an expression for the phase shift.
SOLUTION
a) For the bound state we can write the eigenvalue as E = -h2α2/2m, where α is the decay constant of the eigenfunction outside the square well. Inside the square well we define a wave vector k by
(1)
The infinite potential at the origin requires that all eigenfunctions vanish at x = 0. So the lowest eigenfunction must have the form
(2)
At the point x = a, we match the eigenfunctions and their derivatives:
(3)
(4)
We eliminate the constants A and B by dividing these two equations:
(5)
Earlier we established the relationship between k and α. So the only unknown variable is α, which is determined by this equation.
b) To find the minimum bound state, we take the limit α → 0 as in the eigenvalue equation. From (1) we see that k goes to a nonzero constant, and the eigenvalue equation only makes sense as α → 0 if tan ka → ∞, which happens at ka → π/2. Using (1) gives π2/4 = k2a2 = 2mV0a2/h2. Thus, we derive the minimum value of V0 for a bound state:
(6)
c) For a positive energy state set where is the wave vector outside the square well. Inside the square well we again define a wave vector k according to
(7)
(8)
Again we have the requirement that the eigenfunction vanish at x = 0. For x > a we have an eigenfunction with two unknown parameters B and δ. Alternatively, we may write it as
(9)
in terms of two unknowns C and D. The two forms are equivalent since C = B cos δ, D = B sin δ. We prefer to write it with the phase shift δ. Again we match the two wave functions and their derivatives at x = a:
(10)
(11)
Dividing (10) by (11), we obtain
(12)
Since k is a known function of the only unknown in this equation is δ, which is determined by this equation.
d) From (12) we derive an expression for the phase shift:
(13)
|
|
علامات بسيطة في جسدك قد تنذر بمرض "قاتل"
|
|
|
|
|
أول صور ثلاثية الأبعاد للغدة الزعترية البشرية
|
|
|
|
|
مستشفى العتبة العباسية الميداني في سوريا يقدّم خدماته لنحو 1500 نازح لبناني يوميًا
|
|
|