This simple example is more realistic than the infinite potential well of Chap. 5, Vol. 1:
It allows for bound and free motion. We have a potential of the form (cf. Fig.15.5):
V =
−V0 for −L <x<L; V0>0
0 otherwise (15.25)
and thus the three SEq’s:
region 1:x<−L Eϕ1= −2 2mϕ1 region 2: −L <x<L Eϕ2= −2
2mϕ2−V0ϕ2. region 3:x>L Eϕ3= −2
2mϕ3
(15.26)
Depending on the magnitude ofE, we have to make a case distinction: ForE <0, there are only bound states, while forE>0 there are only scattering states.9In any
8In the classical case, we wouldalwayshave transmission 1 forE>V0.
9This is an example of a spectrum that has both a discrete and a continuous part.
Fig. 15.5 Finite potential well
-V0
Region 3 Region 2
Region 1
x L
-L
Potential, energy
case, we can already give the solution in region 2:
region 2:ϕ2(x)=Bei kx+Ce−i kx; k2 =2m
2 (V0+E) >0. (15.27)
15.3.1 Potential Well, E < 0
We first consider energies with−V0 <E <0, that is, bound motion. With κ2= 2m
2 |E| (15.28)
the solutions are
region 1:ϕ1(x)= Aeκx+Ae−κx
region 3:ϕ3(x)=Deκx+De−κx. (15.29) Since physically reasonable solutions must be bounded, we must choose A=0 andD=0. The other constants are defined by the matching conditions at the two discontinuities.
15.3.1.1 Matching at the Discontinuities
At the discontinuityx= −Lwe have the two equations
Ae−κL =Be−i k L+Cei k L and κAe−κL=i k Be−i k L−i kCei k L (15.30)
15.3 Finite Potential Well 13
and at the discontinuityx = +Lthe two equations
De−κL =Bei k L+Ce−i k L and −κDe−κL =i k Bei k L−i kCe−i k L. (15.31) This is a homogeneous system of four equations with four unknowns. In order that the system have non-trivial solutions, the determinant of the coefficient matrix for A, B,C,Dmust be equal to zero. Instead of calculating the determinant, we can also multiply the first equation in (15.30) byκand subtract the two equations.
We obtain:
0=κBe−i k L + κCei k L− i k Be−i k L +i kCei k L =(κ−i k)e−i k LB + (κ+i k)ei k LC (15.32) and analogously from (15.31):
0=κBei k L+κCe−i k L+i k Bei k L−i kCe−i k L =(κ+i k)ei k LB+(κ−i k)e−i k LC. (15.33) With (15.32) and (15.33), we have a homogeneous system for the two unknowns BandC. It can be solved if the coefficient determinant forB,Cvanishes, i.e. for
(κ−i k)2e−2i k L−(κ+i k)2e2i k L=! 0. (15.34) This equation gives the allowed energy values. We insert
κ±i k=
κ2+k2e±iarctank/κ (15.35)
into (15.34) and obtain sin
2k L+2 arctank κ
=! 0. (15.36)
The solution is evidently
2k L+2 arctank κ
=! Nπ; N =1,2,3, . . . (15.37)
where N numbers consecutively the valid solutions in such a way that the smallest energy eigenvalue has the index 1. A closer inspection of this equation is given below; here we will first determine the constants as far as possible. From (15.32) and ((15.33) would give the same information), it follows with (15.35) and because of (15.37) that:
C= −Bκ−i k
κ+i ke−2i k L= −Be−2iarctank/κe−2i k L
= −Be−i Nπ=B(−1)N+1 (15.38)
and thus from (15.30) and (15.31) A=
2BeκLcosk L
−2i BeκLsink L and D=
2BeκLcosk L
2i BeκLsink L forN odd
even. (15.39)
15.3.1.2 Energy Eigenvalues
Equation (15.37) is not solvable in closed form. In order to obtain solutions for specific values of E andV0, we must proceed numerically. Nevertheless, we can make general statements with the help of estimates. For this purpose, we rearrange (15.37) as follows:
2 arctank
κ=Nπ−2k L; N =1,2,3, . . . (15.40)
Sincekandκare positive, we can estimate 0<2 arctank
κ<π. (15.41)
Thus we obtain the inequality
0<Nπ−2k L<π or (N−1)π<2k L<Nπ. (15.42) Since there are no negative terms, we can square. Substitutingk2=2m(V0−|E|)/2 and subsequent rearranging yields
V0− 2 2m
N 2Lπ
2
<|E|<V0− 2 2m
N−1 2L π
2
. (15.43)
This equation allows us to draw several conclusions:
1. For N=1 we have
V0− 2 2m
π 2L
2
<|E|<V0. (15.44) It follows that there is always a solution. As we shall see below, it is symmetrical.
This ‘lowest’ solution is also called theground state.
2. For N=2, we have V0− 2
2m 2π
2L 2
<|E|<V0− 2 2m
π 2L
2
. (15.45)
It follows that there is a second state (thefirst excited state), ifV0 > 2m2 π
2L
2
. As we shall see below, this state is antisymmetric.
15.3 Finite Potential Well 15
3. Similarly, one sees that theN-th state exists ifV0> 2m2 N−1
2L π2
.
4. There is anN0, from which onwards the right side of the inequality is no longer satisfied. It follows that in each potential well of the kind we are considering, there is only a finite number of energy levels (see exercises).
15.3.1.3 Eigenfunctions
We now know that there is a finite number of solutions of the (15.36). We number them from 1 to N0 (sincek andκdepend on E, they also depend on N, which is indicated by a corresponding index). For each of these solutions, there is an eigenfunction; we distinguish these according to the parity of the energy quantum numberN. With (15.38) and (15.39), it follows that
ϕ1,N(x)=2BeκNLcoskNL·eκNx ϕ2,N(x)=2B·coskNx
ϕ3,N(x)=2BeκNLcoskNL·e−κNx
forN odd (15.46)
and ϕ1,N(x)= −2i BeκNLsinkNL·eκNx ϕ2,N(x)=2i B·sinkNx
ϕ3,N(x)=2i BeκNLsinkNL·e−κNx
forNeven. (15.47)
The solutions in region 2, i.e. within the potential well, are evidently standing waves.
The parity of the eigenfunctions alternates on climbing up the ‘energy ladder’, with the ground state symmetric. By the way, this parity property is a consequence of the symmetry of the problem.
15.3.2 Potential Well, E > 0
We now consider the case E>0, i.e. free motion. With k2= 2m
2 E (15.48)
the solutions read
region 1:ϕ1(x)=Aei kx+Ae−i kx
region 3:ϕ3(x)=Dei kx+F e−i kx (15.49) and from above we have
region 2:ϕ2(x)=Bei kx+Ce−i kx. (15.50)
If we require that the incoming wave of amplitudeF is incident on the potential well from the right, we haveA=0, because in region 1, a wave coming from the left cannot be present. To determine the other constants, we use the matching conditions at the two discontinuities. They are
Aei kL =Be−i k L+Cei k L
−kAei kL =k Be−i k L−kCei k L (15.51)
and
Bei k L+Ce−i k L=Dei kL+F e−i kL
k Bei k L−kCe−i k L=kDei kL−kF e−i kL. (15.52) These are four equations for four unknowns (the amplitudeFcan be chosen at will).
Their solution reads
A= −4kk
N e−2i kLF B =2k
k − k
N ei k Le−i kLF C = −2k
k +k
N e−i k Le−i kLF D=2isin(2k L)
k2 − k2
N e−2i kLF
(15.53)
with
N =e2i k L k−k2
−e−2i k L k+k2
=2i
k2+k2
sin(2k L)−4kkcos(2k L) . (15.54)
The transmission and reflection coefficients are given by T =|A|2
|F|2 and R= |D|2
|F|2 (15.55)
and this leads to T = 16k2k2
|N|2 and R=2(k2−k2) (1−cos(4k L))
|N|2 =1−T (15.56)
with
|N|2=2k4+12k2k2+2k4−2(k2−k2)2cos(4k L). (15.57)
15.3 Finite Potential Well 17
Fig. 15.6 Transmission coefficient (15.60) for scattering by the potential well withμ=15 as a function ofz=VE0
0 0.5 1 1.5 2
0 0.5 1
Because of
k2 =2m
2 E and k2= 2m
2(E+V0) (15.58)
and with
z= E V0; μ=
2m
2 V0L2, (15.59)
it follows for the transmission coefficient:
T = z(z+1)
z(z+1)+1−cos 4k L8 = z(z+1)
z(z+1)+1−cos 4μ8√z+1 . (15.60) We see thatT =1 for cos 4k L =1, which means that 4k L =2mπ. Withλ= 2kπ, the intuitively-clear conditionmλ2 =2Lfollows, i.e. a kind of resonance condition, as is seen very nicely in Fig.15.6.