4.2 Bound and extended states
4.2.1 Rectangular box
We consider the same potential as in the above scattering problem:
V(x) =
V0 x <−a 0 −a≤x≤a V0 a < x
(4.66)
only shifted by a constant such thatV (x) = 0 in the center. In distinction to the situation of an incoming potential entering from one side, we are now only interested in the analysis of the solutions of the stationary Schr¨odinger equation.
Then we can use the fact that for a potential
V(x) =V (−x) (4.67)
follows that the wave function behaves as
ψ(x) =±ψ(−x). (4.68)
The latter occurred already in the solution of the infinite potential well as well as for the harmonic oscillator. It can be generally seen from the following argument: Introduce the parity operator
P ψb (x) =ψ(−x). (4.69)
It is an Hermitian operator ˆ
dxψ(x)∗P ψb (x) = ˆ
dxψ(x)∗ψ(−x)
= ˆ
dxψ(−x)∗ψ(x)
= ˆ
dx
P ψb (x)∗
ψ(x). (4.70)
54 CHAPTER 4. ADDITIONAL ONE-DIMENSIONAL PROBLEMS It also commutes with a Hamiltonian withV(x) =V (−x)
PbHψb (x) = Pb
ThusHb and Pb share the same eigenfunctions. It holds
Pb2ψ(x) =P ψb (−x) =ψ(x) (4.72) The eigenvalues of Pb2 are 1. Thus, the eigenvalues of Pb are ±1. Thus, the eigenfunctions of the Hamiltonian are either even or odd.
IfE > V0, it holds though there are two boundaries, the assumed reflection symmetry implies that it is sufficient to analyze only one. Doing so eliminates two of the three remaining constants. The final constant is being determined by the condition that the wave function is normalized inside some large volumeLa. We could proceed and do this analysis, but not much of interest happens. The key issue is that for E > V0a continuum of states (strictly a continuum only forL→ ∞) occurs.
It is more interesting to analyzeE < V0. In this case we have either ψ(x) =
for even wave functions or ψ(x) = for odd wave functions. Here we introduced
k =
4.2. BOUND AND EXTENDED STATES 55 We first analyze even functions. It holds
e−κa = C Acoska κe−κ0a = kC
Asinka (4.78)
Thus
κ =ktanka (4.79)
In case of odd wave functions follows
−e−κa = C Asinka κe−κa = kC
Acoska (4.80)
which yields
κ =−kcotka (4.81)
For convenience we introduce dimensionless quantities ξ = ka
η = κa (4.82)
Sincek andκboth depend on energy we find ξ2+η2 = a2
2mE
~2
+2m(V0−E)
~2
= V0
~2 2ma2
≡γ2 (4.83)
Given the potential strength and width, we know γ. The above equation is determines a circle of radius γ. Since we know that ξ > 0 and κ >0 we are only interested in the upper right quarter of the radius. In addition to Eq.4.83 we also need to solve
η=
ξtanξ ifψ(x) = +ψ(−x)
−ξcotξ ifψ(x) =−ψ(−x) (4.84) Forγ < π2 only one solution exists as cotξ <0 only for ξ > π2. Thus for
V0<π 2
2 ~2
2ma2 (4.85)
only one bound state exists that is even. For π2 < γ < πwe have one additional bound state that is odd. For γ < mπ2, corresponding to
V0<
mπ 2
~2
2ma2 (4.86)
56 CHAPTER 4. ADDITIONAL ONE-DIMENSIONAL PROBLEMS we have a total of mbound states. Obviously for γ→ ∞all states are bound states. In this limit we know the eigenvalues
En∞=nπ 2
2 ~2
2ma2 (4.87)
Thus, our condition for the existence of bound states is essentially that bound states disappear if
En∞> V0. (4.88)
One can also eliminate the variableη as follows:
ξ2+η2=ξ2 1 + tan2ξ
= ξ2
cos2ξ (4.89)
or
ξ2+η2=ξ2 1 + cot2ξ
= ξ2
sin2ξ (4.90)
Thus, it must hold ξ=
γcosξ ifψ(x) = +ψ(−x)
γsinξ ifψ(x) =−ψ(−x) (4.91) and the roots of this equation determine the eigenvalues (one must however keep in mind to only accept solutions with cotξ < 0 in case of odd functions and tanξ >0 in case of odd functions).
In the limit γ1, where we have only one bound state we have to solve.
ξ=γcosξ (4.92)
for smallγ. Thus, we expect a solution for very smallξ. We expand the cosine ξ=γ
1−1
2ξ2
(4.93) which is solved for
ξ=
p1 + 2γ2−1
γ (4.94)
Expandingξ2for smallγ yields
ξ2'γ2−γ4 (4.95)
which we can insert to obtain the eigenvalue of the single bound state:
E0=~2k2
2m = ~2ξ2
2ma2 =V0−2ma2
~2
V02. (4.96)
Chapter 5
Unitary transformations and symmetries
An operator ˆU in unitary, if
Ub−1=Ub† (5.1)
which implies
Dψ Ub
ψ0E
=D
Ub−1ψ|ψ0E
(5.2) Consider two states|ψiand|ϕi. If we act on both with a unitary transformation, i.e.
|ψ0i = Uˆ|ψi
|ϕ0i = Uˆ|ϕi, (5.3)
it follows that the scalar product of the two states does not change hϕ0|ψ0i = D
U ϕ|b U ψb E
= D
ϕ|Ub†U ψb E
= D
ϕ|Ub−1U ψb E
=hϕ|ψi. (5.4)
An interesting unitary operator is
Uˆ(t) =e−itc~H. (5.5)
It is easy to see that this operator is indeed unitary, as ˆU†(t) =eitc~H such that Uˆ†(t) ˆU(t) = 1. Let us apply this wave function to a state|ψi:
|ψ(t)i=e−itc~H |ψi (5.6) 57
58 CHAPTER 5. UNITARY TRANSFORMATIONS AND SYMMETRIES Let us take the time derivative of this state
∂
∂t|ψ(t)i= ∂
∂te−itc~H
|ψi (5.7)
It holds immeadiately
∂
∂te−itc~H =−i
~
Heb −itc~H (5.8)
and it follows
i~
∂
∂t|ψ(t)i=Heb −itc~H |ψi=Hb|ψ(t)i. (5.9) Thus,|ψ(t)iis indeed the correct time dependent Schr¨odinger equation. Thus, the operator ˆU(t) is the time-translation or time-evolution operator.
Lets now consider a system in a quantum stateψ(r, t). Suppose we translate this state by a vector δr we obtain a new state, which we call ψ0(r, t). In particular it holds
ψ0(r+δr, t) =ψ(r, t) (5.10) For example, if ψ(r, t) is maximal at r=r0, then ψ0(r, t) is maximal at r = r0+δr. This operation is often referred to as an active transformation (a formation of the wave function) as opposed to a passive transformation (a trans-formation of the coordinate system used to describe the same state).
We are searching for the operator that leads to a spatial translation, i.e.
ψ0(r, t) =Ur(δr)ψ(r, t) =ψ(r−δr, t) (5.11) In order to determine the operator explicitly, we expand
ψ(r−δr, t) = ψ(r, t)−δxα ∂
∂xα
ψ(r, t) +1
2δxαδxβ
∂2
∂xα∂xβψ(r, t) +· · ·
= e−δr·∇ψ(r, t) (5.12)
Using the fact that the momentum operator is
p=b ~
i∇ (5.13)
we find that the operator for spatial translations is determined by the momen-tum operator
Ur=e−iδr·b~p (5.14)
Sincepb is a Hermitian operator it follows immediately that the operatorUr is unitary:
Ur−1=eiδr·~pb =
e−iδr·pb
†
~
†
=Ur†. (5.15)
59 The stateψ(r, t) obeys the Schr¨ı¿œdinger equation
i~∂tψ(r, t) =Hψ(r, t). (5.16) We ask under what condition is the time evolution of the states ψ0(r, t) and ψ(r, t) the same.
i~∂tψ0(r, t) = i~∂tUrψ(r, t) =i~Ur∂tψ(r, t)
= UrHUr−1ψ0(r, t) (5.17)
Thus, the condition for identical time evolution is that
H =UrHUr−1 (5.18)
which implies that the translation operatorUrcommutes with the Hamiltonian
[H, Ur] = 0. (5.19)
This is fulfilled if the momentum operator commutes with H, i.e. if the mo-mentum is a conserved quantity.
[H,p] = 0.ˆ (5.20)
The logic can easily be inverted, i.e. a vanishing commutator implies that the spatially translated state ψ0(r, t) is a solution of the same Schr¨ı¿œdinger equation as the un-translated state. If for a given system space is homogeneous, we find that the momentum is a conserved quantity. Momentum conservation and homogeneity of space are once again tied to one another.
60 CHAPTER 5. UNITARY TRANSFORMATIONS AND SYMMETRIES
Chapter 6
Angular momentum and spin
6.1 Particle on a circular orbit
Before we start to develop the formal apparatus of the angular momentum theory we analyze a simple problem, the motion of a particle on a circular orbit.
We start from the classical Lagrange function L(ϕ,ϕ) =˙ mR2
2 ϕ˙2−V(ϕ), (6.1)
with radiusR of the orbit. The canonical momentum conjugated toϕis:
pϕ= ∂L(ϕ,ϕ)˙
∂ϕ˙ =mR2ϕ˙ (6.2)
such that the classical Hamiltonian function is H(pϕ, ϕ) = p2ϕ
2mR2 +V(ϕ) (6.3)
Using the quantization rules:
V (ϕ) → V (ϕ) pϕ → ~
i
∂
∂ϕ (6.4)
gives the Hamilton operator
H =− ~2 2mR2
∂2
∂ϕ2 +V (ϕ). (6.5)
Take a system without potential gives the Schr¨odinger equation
− ~2 2mR2
∂2
∂ϕ2ψ(ϕ) =Eψ(ϕ). (6.6)
61
62 CHAPTER 6. ANGULAR MOMENTUM AND SPIN Since we consider a motion on a ring, we must obey the boundary condition
ψ(ϕ+ 2π) =ψ(ϕ). (6.7)
The solution is
ψ(ϕ) = 1
√
2πeimϕ. (6.8)
The boundary condition implies that
1 =eim2π, (6.9)
i.e. mis an integer. The eigenvalues of the momentumpϕconjugated to ϕare
~m. The energy eigenvalues are:
Em= ~2
2mR2m2. (6.10)
The same result can be obtained if one writes the Laplace operator in spher-ical coordinates
x = x1=rcosϕsinθ y = x2=rsinϕsinθ
z = x3=rcosθ (6.11)
The Laplacian,∇2=P
α
∂2
∂x2α, in spherical coordinates is
∇2= ∂2
∂r2+2 r
∂
∂r+ 1 r2
1 sinθ
∂
∂θsinθ ∂
∂θ + 1 sin2θ
∂2
∂ϕ2
(6.12) The motion on a ring can be understood by fixingr=R andθ= π2. Then
∇2= 1 R2
∂2
∂ϕ2 (6.13)
and we obtain the above Hamiltonian.