Chapter 2. One-Dimensional Systems
2.1 Free Particles
The task of this class is to solve time-dependent Schrödinger equation to reveal physics of target
systems. For a time-independent Hamiltonian H ,
ψ(t) = exp(−iHt / ) ψ(0) = ∑ exp(−iEnt / )cn φn
(2.1)
n
where ψ(0) = ∑ cn φn , and En and φn are eigenvalues and eigenstates of H :
n
H φn = En φn .
(2.2)
Therefore we need solve the time-independent Schrödinger equation
⎡ P2
⎤
⎢
⎥ φ =E φ ,
+V(X
)
n
n
⎢ 2m
⎥ n
⎢⎣
⎥⎦
(2.3)
particularly, in real-space representation,
⎡ 2 d 2
⎤
⎢−
⎥ φ(x) = Eφ(x).
+V(x)
⎢ 2m dx 2
⎥
⎢⎣
⎥⎦
(2.4)
The simplest case is a free particle with mass m, V(x) = 0 , then we solve
−
2 d 2
d2
φ(x)
=
Eφ(x)
⇒
φ(x) + k 2φ(x) = 0,
2
2
2m dx
dx
where the wave number k =
(2.5)
2mE
. The solutions are Ae ikx and Be −ikx ; because they are

degenerate, the eigenfunction for wave number k is
φk (x) = Ae ikx + Be −ikx .
(2.6)
We then use the boundary and normalization conditions to determine A and B.
(1) Since there is no bound, the probability density should be a constant, i.e.,
2
2
2
const. = φ(x) = A + B + 2Re ⎡⎢AB *e 2ikx ⎤⎥ ,
⎣
⎦
thus, either A or B must be zero ⇒ φ(x) = Ae ikx (now we allow k be negative).
(2) Normalization condition:
25
(2.7)
δ(k ′ −k) = φk ′ φk = ∫
∞
−∞
then A =
1
2π
⇒ φk (x) =
e iϕ
2π
2
′
A e −ik xe ikx dx ,
(2.8)
e ikx . We can set the irrelevant phase ϕ = 0 , then we finally
reach the familiar eigenfunction for a 1D free particle:
1
φk (x) =
with eigenvalue (eigenenergy) Ek =
2π
e ikx
(2.9)
 2k 2
. In addition, this is also an eigenfunction of
2m
momentum, with eigenvalue pk = k .
2.2 Confined Particles
Consider a particle with mass m in an infinite potential well
with width L. We are solving exactly the same equation,
Eq. (2.5), obtaining the same general solution,
φk (x) = Ae
ikx
+ Be
−ikx
0
L/2
L
.
However, the boundary condition is different:
φk (0) = 0, and φk (L) = 0 ⇒ A + B = 0; Ae ikL + Be −ikL = 0
⇒ A = −B and sin(kL) = 0 ⇒ kL = nπ , with n = 1, 2, 3... (Question: can n be 0?)
Then kn = nπ / L and φn (x) = An sin(knx) . Normalization condition:
δn ′n = φn ′ φn = ∫
∞
−∞
A*n ′An sin(kn ′x)sin(knx)dx ⇒ An = 2 / L .
The eigenfunctions are
φn (x) =
⎛ nπ ⎞
2
sin ⎜⎜⎜ x ⎟⎟⎟ ,
L
⎝ L ⎟⎠
(2.10)
with eigenvalues
En =
(kn )2
2m
=
n 2π 2 2
.
2mL2
Question: are these states of H still eigenstates of momentum?
26
(2.11)
A few crucial remarks:
(1) Confinement leads to discrete energy levels. Only bound states are quantized, while the
unbound states have continuous eigenvalues.
(2) A bound state can’t be an eigenstate of momentum; therefore, it has no well-defined
momentum. Momentum conservation is not required for transitions involved bound states.
(3) In nanoscale systems (size < 10 nm), quantum confinement plays a major role.
(4) When L shrinks, the energy level differences increase; L → 0 , V(x) → −λδ(x) with λ > 0 ,
there is only one bound state:
mλ −|κ|x
mλ
 2κ 2
mλ 2
φ(x) =
e
with κ = 2 ; E = −
=− 2 .

2m

2
(2.12)
(5) A finite potential well has both bound and continuous states, depending on if E <V0 or
E >V0 .
(6) Finally, wavefunctions are either symmetric or anti-symmetric, while the potential well is
symmetric (w.r.t. the well center).
2.3 Particles in a Symmetric Potential
For a symmetric potential, V(x) =V(−x) , Question: are the eigenfunctions symmetric or
antisymmetric or neither?
2 d 2
−
φ(x) +V(x)φ(x) = Eφ(x) .
2m dx 2
(2.13)
2 d 2
φ(−x) +V(x)φ(−x) = Eφ(−x) .
2m dx 2
(2.14)
Replacing x with −x :
−
Define the symmetric and antisymmetric functions: ψS (x) ≡ φ(x) + φ(−x) and
ψA(x) ≡ φ(x)− φ(−x) , respectively.
Adding and subtracting Eqs. (2.13) and (2.14), we find both ψS (x) and ψA(x) are solutions.
However, if ψS (x) and ψA(x) are degenerate, then a linear combination of them,
ψ(x) = c1ψS (x) +c2ψA(x) , which is neither symmetric nor anti-symmetric, is also an
27
eigenfunction for a symmetric potential! This is the famous spontaneous symmetry breaking,
responsible for many fascinating phenomena, such as ferromagnetism and superconductivity.
2.4 Periodic Potential: the Bloch Wavefunction
Consider a particle in a periodic potential, e.g., in a
crystal as illustrated on the right,
V(x + d) =V(x) ⇒ V(x + nd) =V(x) , (2.15)
with n = 1, 2, 3 Question: does the wave function have
the same translational symmetry, i.e., φ(x + d) = φ(x) ?
The translation operator: T(d) x = x + d , T(d)φ(x) = φ(x −d) , and
T(d) = exp(−iPd / ) . ⇒ exp(−iPd / )φ(x) = φ(x −d) .
⇒ φ(x) = exp(iPd / )φ(x −d) .
V(x + d) =V(x) ⇒ H(x + d) = H(x), Hamiltonian is invariant under translation
⇒ T(d)H(x) = H(x)T(d), or [T(d),H ] = 0
Thus one could construct the common eigenstates φ(x) for Hamiltonian and translation
operator T(d) : [Note: though it could fail under certain boundary conditions…]
T(d)φ(x) = λφ(x), while T(d)φ(x) = φ(x −d)
⇒ λφ(x) = φ(x −d) ⇒ | λ |2 = 1, and λ is a function of d
⇒ λ = exp(−ikd), with k an arbitrary real number.
Therefore,
(2.16)
φ(x) = exp(ikd)φ(x −d).
Obviously φ(x + d) ≠ φ(x) , but the wave function must have certain periodic component.
Define φ(x) ≡ F(x)u(x), and u(x) is periodic, i.e., u(x) = u(x −d) .
Then φ(x −d) = F(x −d)u(x −d) ⇒ F(x)u(x) = exp(ikd)F(x −d)u(x −d)
⇒ F(x) = exp(ikd)F(x −d) , so that F(x) = exp(ikx) . We thus derived the Bloch theorem:
φk (x) = exp(ikx)uk (x) ,
uk (x + nd) = uk (x)
(2.17)
Quantum phase in the term exp(ikx) is relevant now, which defines the crystal momentum k
due to periodic crystal lattice. [Though an irrelevant arbitrary phase term eiδ can still be added.]
3D Bloch theorem ( R1 , R 2 and R 3 are crystal primitive lattice vectors):
φk (r) = exp(ik ⋅ r)u k (r), with u k (r) = u k (r + n1R1 + n2R 2 + n3R 3 ).
28
(2.18)
The Kronig-Penney Model:
⎪⎧0, (0 < x < a)
, with V(x + d) =V(x) .
V(x) = ⎪⎨
⎪⎪V0, (a < x < d)
⎩
Step 1: Solving wave function in one periodic interval, 0 < x < d .
1. 0 < x < a . φ(x) = A cos(αx) + B sin(αx) , with α = 2mE /  .
Here A and B are two complex numbers, but could be simplified to real by neglecting e iϕ .
2. a < x < d . We only study bound states, whose energy E < V0 .
φ (x) = C exp(− β x) + D exp( β x) , with β = 2m(V0 − E) /  .
Step 2: Extend to other intervals using Bloch’s theorem Eq. (2.17):
u(x + d) = u(x) , φ(x) = exp(ikx)u(x) .
Denote φ I (x) the wave function in the range of 0 < x < d , assuming φ I (x) is known.
We want to find φ− I (x) in the range of −d < x < 0 .
u(x + d) = exp[−ik(x + d)]φI (x + d) , and u(x) = exp(−ikx)φ−I (x)
⇒ exp[−ik(x + d)]φI (x + d) = exp(−ikx)φ−I (x) ⇒
φ−I (x) = exp(−ikd)φI (x + d)
(2.19)
For −nd < x < −(n −1)d, φ−N (x) = exp(−iknd)φI (x + nd);
(2.20)
Then
For (n −1)d < x < nd,
φN (x) = exp(iknd)φI (x −nd).
(2.21)
Boundary conditions:
φ (x) and
d
φ (x) are continuous at x = a and x = 0 , therefore we obtain four equations:
dx
A cos(αa) + B sin(αa) = C exp(−βa) + D exp(βa)
(2.22)
−Aα sin(αa) + Bα cos(αa) = −C β exp(−βa) + Dβ exp(βa)
(2.23)
A = exp(−ikd)[C exp(−βd) + D exp(βd)]
(2.24)
Bα = exp(−ikd)[−C β exp(−βd) + Dβ exp(βd)]
(2.25)
29
In order to find non-trivial solutions to Eqs. (2.22)-(2.25), the determinant must be zero:
cos(αa)
sin(αa)
− exp(−βa)
− exp(βa)
−α sin(αa) α cos(αa) β exp(−βa)
− βexp(βa)
det
=0
1
0
− exp(−βd −ikd) − exp(βd - ikd)
0
α
β exp(−βd −ikd) − βexp(βd - ikd)
⇒
⎛ β 2 − α2 ⎞⎟
⎟⎟ sinh(βb)sin(αa)
cos(kd) = cosh(βb)cos(αa) + ⎜⎜⎜
⎜⎝ 2αβ ⎟⎠
(2.26)
where b = d −a .
Because α and β both depend on energy E , the right-hand-side of Eq. (2.26) is a function
of E denoted by F(E) . Accordingly,
cos(kd) = F(E) ⇒ F(E) ≤ 1 .
(2.27)
This leads to the most important feature of periodic potentials:
continuous energy bands with forbidden energy gaps.
Periodicity in crystal potential leads to electronic band
structures, which determine electronic properties:
Small band gap: semiconductors; Large gap: insulators;
No gap: metals. But there are exceptions…
Silicon: a semiconductor
Aluminum: a metal
30
2.5 Simple Harmonic Oscillators
The simple harmonic oscillator (SHO) is one of the most
important models in quantum mechanics:
(1) Any realistic potential well near the bottom can be
approximated by a SHO.
(2) In particular, molecular vibrations, lattice vibrations in
crystals, nuclear structures, etc.
(3) Theoretically, the foundation of field quantization, leading to quantum field theory.
One example is the potential energy for a diatomic molecule,
V(r −r0 ) =V(r0 ) +
1 d 2V(r)
(r −r0 )2 +O(r −r0 )3 .
2
2 dr r=r
(2.28)
0
Then the basic Hamiltonian for a one-dimensional SHO is
H=
P2 1 2 P2 1
+ kX =
+ mω 2X 2 .
2m 2
2m 2
(2.29)
Classically, the Hamiltonian equation for SHO:
dx
∂H
p
dp
∂H
=
= ,
=−
= −mω 2x .
dt
∂p
m dt
∂x
(2.30)
dx 2
+ ω 2x = 0 .
2
dt
(2.31)
1
x(t) = A sin(ωt + ϕ); E = mω 2A2 .
2
(2.32)
Therefore,
Solution:
However, quantum mechanically the solutions are very different. We solve the timeindependent Schrödinger equation to obtain eigenfunctions and eigenenergies:
⎛ 2 d 2
⎞
1
2 2⎟
⎜⎜−
⎟⎟ ψ(x) = E ψ(x) .
+
mω
x
⎜⎜ 2m dx 2 2
⎟⎠
⎝
(2.33)
We can solve Eq. (2.33) using lowering (annihilation) and raising (creation) operators defined as
a≡
mω ⎛⎜
P ⎞⎟
mω ⎛⎜
P ⎞⎟
⎟⎟, then a † =
⎟⎟ ,
⎜⎜X + i
⎜⎜X −i
2 ⎝
mω ⎟⎠
2 ⎝
mω ⎟⎠
31
(2.34)
then
a †a =
H
i
H 1
+ [X,P ] =
− .
ω 2
ω 2
(2.35)
aa † =
H
i
H
1
− [X,P ] =
+ .
ω 2
ω 2
(2.36)
Similarly,
Eqs. (2.35) and (2.36) ⇒
⎡a,a † ⎤ = 1 ,
⎢⎣
⎥⎦
(2.37)
1
1
H = ω(a †a + ) = ω(N + ) ,
2
2
(2.38)
where the number operator N ≡ a †a .
The equation we want to solve is
1
⇒ ω(N + ) n = En n .
2
H n = En n
(2.39)
Obviously, H and N have the same eigenvectors n : N n = λn n with λn the eigenvalue.
Then a n and a † n are eigenvectors of N as well, with eigenvalues of λn −1 and λn + 1 ,
respectively. This is because
(
a †a a n
) = (aa
†
)(
) = aa a n
−a n = aλn n −a n = (λn −1) a n , i.e.,
(
) = (λ
(
−1 a n
N an
†
n
(
)
(
−1) a n ; N a † n
) = (λ
n
(
)
)
+ 1) a † n .
(2.40)
Summary: a † increases the eigenvalue, while a decreases the eigenvalue.
Question: are n eigenvectors of a † or a or both?
Consider the ground state 0 , N 0 = λ0 0 , and then no other eigenstates have eigenvalue
(
lower than λ0 . But a 0 has an eigenvalue of λ0 −1 , thus the state a 0
) simply doesn’t exist,
i.e., a 0 = 0 . This is a very important conclusion! Now we can easily derive En for a SHO.
a 0 = 0 ⇒ a †a 0 = 0 ⇒ λ0 = 0
(
)
(
) (
a †a a † 0 = (λ0 + 1) a † 0 = 1 a † 0
)
⇒ a † 0 = C 0 1 and λ1 = 1.
32
(2.41)
Similarly, a † 1 = C 1 2 , since state a † 1 has an eigenvalue of λ2 = 1 + λ1 = 2.
Thus
a † n −1 = C n−1 n ; and λn = n .
(2.42)
Eq. (2.42) leads to
⇒ H n = ω (n + 1 / 2) n ,
N n =n n
(2.43)
i.e., the eigenenergies of a SHO is
En = (n + 1 / 2)ω .
(2.44)
A summary of these extremely useful second-quantization operators:
The number operator: N = a †a , N n = n n , and H = (N + 1 / 2)ω .
Raising (creation) operator: a † n = C n n + 1 , ⇒ n a = n + 1 C n* , then
n aa † n = C n
2
2
n + 1 n + 1 = Cn .
(2.45)
On the other hand,
aa † = a †a + 1 ⇒ aa † n = (n + 1) n .
(2.46)
Combining Eqs. (2.45) and (2.46), we obtain
a† n = n + 1 n + 1 ;
C n = n + 1;
n a = n +1 n +1 .
(2.47)
Lowering (annihilation) operator: a n = Dn n −1 , similarly, we can obtain
Dn = n;
a n = n n −1 ;
n a † = n −1 n .
(2.48)
Based on Eqs. (2.47) and (2.48), one can easily verify that
a †a n = n n ; aa † n = (n + 1) n ; [a,a † ] = aa † −aa † = 1 .
In addition,
n =
1
n
a † n −1 =
0 =
1
1
a1 =
1
†
n(n −1)
1
1⋅2
(a )
2
(a )
33
2
1
n −2 ==
2 ==
1
n!
n!
n
(a )
(a )
n .
†
n
0 .
(2.49)
(2.50)
The (energy-level) number representation:
⎡ 0⎤
⎡0
⎢
⎢ ⎥
⎢
⎢ 0⎥
⎢0
⎢ ⎥
⎢
n = ⎢⎢ ⎥⎥ , a = ⎢0
⎢
⎢ ⎥
⎢0
⎢1 ⎥
⎢
⎢ ⎥
⎢
⎢⎣ ⎥⎦
⎢
⎢
⎣
1
0
0
2
0
0
0

0

⎡0
0
 ⎤⎥
⎢
⎥
⎢
0  ⎥
⎢ 1 0
⎥,
⎢
†
⎥
2
3  ⎥ a = ⎢⎢0
⎥
⎢
0  ⎥
0
⎢0
⎥
⎢
  ⎥
⎢

⎥⎦
⎣
0
0
0
0
0
0
0
3

0

⎡0
 ⎤⎥
⎢
⎥
⎢0
 ⎥
⎢
⎥,
⎢
 ⎥ N = ⎢0
⎥
⎢
⎢0
 ⎥⎥
⎢
⎥
⎢⎣
 ⎥
⎦
0
1
0
0

0
0
2
0

0
0
0
3






⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥
⎥⎦
Next, we want to work out the eigenfunctions for a SHO, ψn (x) = x n . We use the real
, and p0 ≡ mω , then Eq. (2.34) ⇒
mω
space presentation. Define x 0 ≡
a=
1 ⎛⎜ X
P⎞
1 ⎛⎜ x
 d ⎞⎟⎟
⎜⎜ + i ⎟⎟⎟ =
⎜⎜ +
⎟.
p0 ⎟⎟⎠
2 ⎜⎝ x 0
2 ⎜⎝ x 0 p0 dx ⎟⎟⎠
It is always a good idea to define a dimensionless variable y ≡
a=
(2.51)
d
d
x
= x0
⇒
⇒
dy
dx
x0
1 ⎛⎜
d ⎞⎟
1 ⎛⎜
d ⎞⎟
†
⎜⎜y + ⎟⎟; a =
⎜⎜y − ⎟⎟ .
dy ⎟⎠
dy ⎟⎠
2⎝
2⎝
(2.52)
The strategy is finding ψ0(x) first, then obtaining ψn (x) iteratively.
a 0 =0 ⇒ x a 0 =
dψ (y)
1 ⎛⎜
d ⎞⎟
⎜⎜y + ⎟⎟ ψ0(y) = 0 ⇒ 0 + yψ0(y) = 0 ⇒
dy
dy ⎟⎠
2⎝
dψ0(y)
ψ0(y)
= −ydy
(2.53)
Solving Eq. (2.53), wavefunction for the ground state is
ψ0(y) = A0 exp(−y 2 / 2) ,
(2.54)
1/4
⎛ mω ⎞⎟
⎟
ψ0(x) = A0 exp −x / 2x , where A0 = ⎜⎜
⎜⎝ π ⎟⎟⎠
(
2
2
0
)
1/4
⎛1⎞
= ⎜⎜ ⎟⎟⎟ x 0−1/2 .
⎜⎝ π ⎟⎠
(a )
=
†
Then, wavefunction for any state of a SHO is readily computed using n
(2.55)
n
n!
0 :
n
1 ⎡⎢ 1 ⎛⎜
d ⎞⎟⎤⎥
⎟⎟ A0 exp(−y 2 / 2) .
ψn (x) = x n =
y
−
⎜
⎢ ⎜⎝
dy ⎟⎠⎥⎥⎦
n ! ⎢⎣ 2
34
(2.56)
Eq. (2.56) involves Hermite Polynomials:
n
⎡ 1 ⎛
d ⎞⎤
H n (y) = exp(y / 2) ⎢⎢ ⎜⎜⎜y − ⎟⎟⎟⎥⎥ exp(−y 2 / 2) .
dy ⎟⎠⎥⎦
⎢⎣ 2 ⎝
2
(2.57)
Using Hermite polynomials, wavefunction for a SHO is
ψn (x) = AnH n (y)exp(−y 2 / 2) = AnH n (x / x 0 )exp(−x 2 / 2x 02 ) ,
(2.58)
where the normalization constant
1/4
1/4
⎛
1
mω ⎞⎟
⎟
An = ⎜⎜⎜ 2n
⎜⎝ π2 (n !)2  ⎟⎟⎠
⎛
⎞⎟
1
⎟ x −1/2 .
= ⎜⎜⎜ 2n
⎜⎝ π2 (n !)2 ⎟⎟⎠ 0
Question: How to evaluate 4 X 2 2 ? Answer: 4 X 2 2 = 3x 02 =
(2.59)
3
.
mω
One can easily verify that for the ground state 0 ,
(Δx)2 (Δp)2 =
1 2

4
1
⇒ Δx ⋅ Δp =  .
2
(2.60)
Question: for an arbitrary energy eigenstate n , what is Δx ⋅ Δp ? (Your homework)
Now let’s evaluate the optical transition matrix elements m exp(ikx) n between two
eigenstates of a charged SHO. Light is an electromagnetic wave; consider a monochromatic laser
with angular frequency ω , whose electric field is
E = E0 exp[i(k ⋅ r − ωt)]
(2.61)
The interaction between the charged SHO and the field is
H ′ = −qr ⋅ E
(2.62)
Based on perturbation theory (We will study it later this semester), the optical transition rate
between two eigenstates is proportional to m H ′ n
2
, which can be decomposed to a couple of
matrix elements in the form of m exp(ikx) n for 1D SHO.
Method 1: in the x-representation (Eigenstates of X as basis)
m exp(ikx) n = ∫
∞
−∞
ψ*m (x)e ikx ψn (x)dx .
Method 2: in the number-representation (Eigenstates of Hamiltonian as basis)
35
(2.63)
(ik)j
m Xj n .
j
!
j =0
∞
m exp(ikx) n = ∑
(2.64)
Method 3: Feynman’s approach
First, we write e ikx in terms of lowering and raising operators
e ikx = exp(iλa † + iλa) ,
where λ = kx 0 / 2 = k
(2.65)

. We use the Feynman-Glauber theorem to simplify Eq. (2.65).
2mω
The Feynman-Glauber theorem (You have proved it in HW#3) states that for two operators A
and B, if both of them commute with the commutator C ≡ [A,B] : [A,C ] = [B,C ] = 0 , then
e A+B = e Ae Be −C /2
(2.66)
Here A = iλa † , B = iλa , and C = (iλ)2[a †,a] = λ 2 , obviously [A,C ] = [B,C ] = 0 . Then
exp(iλa + iλa † ) = exp(iλa † )exp(iλa)exp(−λ 2 / 2)
2
†
(2.67)
†
Thus m exp(ikx) n = e −λ /2 m e iλa e iλa n , and we need to evaluate e iλa n and m e iλa .
Using Eq. (2.48), we can derive
n
(iλ)j j
(iλ)j
a n =∑
j!
j!
j =0
j =0
n!
n−j ,
(n − j)!
∞
e iλa n = ∑
me
iλa †
(iλ)l † l
= m ∑
(a ) = m −l
l!
l=0
∞
(iλ)l
∑
l!
l=0
m
m!
.
(m −l)!
(2.68)
(2.69)
Combining Eqs. (2.68) and (2.69):
m
m e iλa e iλa n = ∑
†
l=0
(iλ)l
l!
n
m!
(iλ)j
∑
(m −l)! j =0 j !
n!
m −l n − j
(n − j)!
(2.70)
Where m −l n − j = δm−l,n−j , thus for the non-zero terms,
m −l = n − j
⇒ l = m + j −n .
Without losing generality, we assume m ≥ n :
(iλ)m+j−n
m ! (iλ)j
j!
j =0 (m + j −n)! (n − j)!
n
m exp(ikx) n = e −λ /2 ∑
2
n
= e −λ /2(iλ)m−n m !n ! ∑
2
j =0
n!
(n − j)!
(iλ)2j
.
j !(m + j −n)!(n − j)!
36
(2.71)