Figure 2.7a shows the fi rst four allowed energies for the particle in the infi nite potential well, and Figure 2.7b,c shows the corresponding wave functions and prob- ability functions. We may note that as the energy increases, the probability of fi nding the particle at any given value of x becomes more uniform.
The general solution to this equation can be written in the form
1(x) A1e jk1x B1e jk1x (x 0) (2.41) where the constant k 1 is
k1
_____ 2mE _ 2 (2.42)The fi rst term in Equation (2.41) is a traveling wave in the x direction that repre- sents the incident wave, and the second term is a traveling wave in the x direction
Figure 2.7 | Particle in an infi nite potential well: (a) four lowest discrete energy levels, (b) corresponding wave functions, and (c) corresponding probability functions.
(From Pierret [10].) (a) 15
n 1 n 2 n 4
n 3 10
5
冊 冉
E units of 22 2ma2(b)
n
x 0 x a
n2
(c) x 0 x a
Incident particles
Region I Region II V(x)
V0
x 0
Figure 2.8 | The step potential function.
that represents a refl ected wave. As in the case of a free particle, the incident and refl ected particles are represented by traveling waves.
For the incident wave, A 1 ⴢ A 1 * is the probability density function of the incident parti- cles. If we multiply this probability density function by the incident velocity, then v i ⴢ A 1 ⴢ A 1 * is the fl ux of incident particles in units of #/cm 2 -s. Likewise, the quantity v r ⴢ B 1 ⴢ B 1 * is the fl ux of the refl ected particles, where v r is the velocity of the refl ected wave. (The parameters v i and v r in these terms are actually the magnitudes of the velocity only.)
In region II, the potential is V V0. If we assume that E V0, then the differen- tial equation describing the wave function in region II can be written as
22(x)
__ x2 2m _
2 (V0 E)2(x) 0 (2.43) The general solution may then be written in the form
2(x) A2e k2x B2e k2x (x 0) (2.44) where
k2
__________ __ 2m(V02 E) (2.45)One boundary condition is that the wave function 2(x) must remain fi nite, which means that the coeffi cient B2 0. The wave function is now given by
2(x) A2e k2x (x 0) (2.46) The wave function at x 0 must be continuous so that
1(0) 2(0) (2.47)
Then from Equations (2.41), (2.46), and (2.47), we obtain
A1 B1 A2 (2.48) Since the potential function is everywhere fi nite, the fi rst derivative of the wave function must also be continuous so that
_ x1 x0
_ x2 x0 (2.49) Using Equations (2.41), (2.46), and (2.49), we obtain
jk1A1 jk1B1k2A2 (2.50) We can solve Equations (2.48) and (2.50) to determine the coeffi cients B 1 and A 2 in terms of the incident wave coeffi cient A 1 . The results are
B1 ( k 22 2jk1k2 k 12 )
____
k 22 k 12 ⴢ A1 (2.51a) and
A2 2k1(k1 jk2)
___ k 22 k 12 ⴢ A1 (2.51b)
2 . 3 Applications of Schrodinger’s Wave Equation 41
The refl ected probability density function is given by B1ⴢ B 1* ( k 22 k 12 2jk1k2)( k 22 k 12 2jk1k2)
______
( k 22 k 12 ) 2 ⴢ A1ⴢ A 1* (2.52) We can defi ne a refl ection coeffi cient, R , as the ratio of the refl ected fl ux to the incident fl ux, which is written as
R __ v r ⴢ B1ⴢ B 1*
v1ⴢ A1ⴢ A 1* (2.53) where i and r are the incident and refl ected velocities, respectively, of the particles.
In region I, V 0 so that E T , where T is the kinetic energy of the particle. The kinetic energy is given by
T 1 _
2 mv2 (2.54)
so that the constant k 1 , from Equation (2.42) may be written as
k1
__________ 2m _ 2 1 _ 2 mv2 _____m2 _ v22 _ mv (2.55)The incident velocity can then be written as
v i _ m ⴢ k1 (2.56) Since the refl ected particle also exists in region I, the refl ected velocity (magnitude) is given by
v r _ m ⴢ k1 (2.57) The incident and refl ected velocities (magnitudes) are equal. The refl ection coef- fi cient is then
R v r ⴢ B1ⴢ B 1*
__ v i ⴢ A1ⴢ A 1* B__ 1ⴢ B 1*
A1ⴢ A 1* (2.58) Substituting the expression from Equation (2.52) into Equation (2.58), we obtain
R B1ⴢ B 1*
__ A1ⴢ A 1* k 22 k 12 4 k 12 k 22
___ k 22 k 12 1.0 (2.59) The result of R 1 implies that all of the particles incident on the potential barrier for E V 0 are eventually refl ected. Particles are not absorbed or transmitted through the potential barrier. This result is entirely consistent with classical physics and one might ask why we should consider this problem in terms of quantum me- chanics. The interesting result is in terms of what happens in region II.
The wave solution in region II was given by Equation (2.46) as 2(x) A2e k2x. The coeffi cient A 2 from Equation (2.48) is A2 A1 B1, which we derived from the boundary conditions. For the case of E V0, the coeffi cient A 2 is not zero.
If A 2 is not zero, then the probability density function 2(x) ⴢ 2* (x) of the particle being found in region II is not equal to zero. This result implies that there is a fi nite
probability that the incident particle will penetrate the potential barrier and exist in region II. The probability of a particle penetrating the potential barrier is another difference between classical and quantum mechanics: The quantum mechanical penetration is classically not allowed. Although there is a fi nite probability that the particle may penetrate the barrier, since the refl ection coeffi cient in region I is unity, the particle in region II must eventually turn around and move back into region I.
2 . 3 Applications of Schrodinger’s Wave Equation 43
EXAMPLE 2.4 Objective: Calculate the penetration depth of a particle impinging on a potential barrier.
Consider an incident electron that is traveling at a velocity of 1 105 m/s in region I.
■ Solution
With V ( x ) 0, the total energy is also equal to the kinetic energy so that
E T 1 _
2 mv2 4.56 1021 J 2.85 102 eV
Now, assume that the potential barrier at x 0 is twice as large as the total energy of the inci- dent particle, or that V 0 2 E . The wave function solution in region II is 2 ( x ) A 2e −k2 , where the constant k 2 is given by k 2 _____________
2m(V0 E)2 .
In this example, we want to determine the distance x d at which the wave function magnitude has decayed to e −1 of its value at x 0. Then, for this case, we have k 2 d 1 or
1 d
___________2m(2E E)
___ 2 d
_____ 2mE _ 2The distance is then given by
d
_____ _ 2mE2 ______ _________________________1.054 10342(9.11 1031) (4.56 1021) 11.6 1010 m or
d 11.6 Å ■ Comment
This penetration distance corresponds to approximately two lattice constants of silicon. The numbers used in this example are rather arbitrary. We used a distance at which the wave func- tion decayed to e1 of its initial value. We could have arbitrarily used e2, for example, but the results give an indication of the magnitude of penetration depth.
■ EXERCISE PROBLEM
Ex 2.4 The probability of fi nding a particle at a distance d in region II compared with that at x 0 is given by exp (2k2d). Consider an electron traveling in region I at a velocity of 105 m/s incident on a potential barrier whose height is three times the kinetic energy of the electron. Find the probability of fi nding the electron at a dis- tance d compared with x 0 where d is ( a ) 10 Å and ( b ) 100 Å into the potential barrier. %] 9 10 2.43 8.680%; (b) P a) P [Ans. (
The case when the total energy of a particle, which is incident on the potential barrier, is greater than the barrier height, or E V0, is left as an exercise at the end of the chapter.