Transcript Lecture8
Ben Gurion University of the Negev
www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter
Physics 3 for Electrical Engineering
Lecturers: Daniel Rohrlich, Ron Folman
Teaching Assistants: Daniel Ariad, Barukh Dolgin
Week 8. Quantum mechanics – raising and lowering operators,
1D harmonic oscillator • harmonic oscillator eigenvalues and
eigenfunctions • matrix representation • motion of a minimumuncertainty wave packet • 3D harmonic oscillator • classical limit
Sources: Merzbacher (2nd edition) Chap. 5 Sects. 1-4;
Merzbacher (3rd edition) Chap. 5 Sects. 1, 3 and Chap. 10 Sect. 6;
Tipler and Llewellyn, Chap. 6 Sect. 5.
We already stated that Schrödinger’s
wave equation, with its continuous
solutions, and Heisenberg’s matrix
algebra, with its quantum jumps, are
equivalent.
Let’s see and compare how these two
different methods apply to the quantum
harmonic oscillator.
The 1D harmonic oscillator
x
0
The 1D harmonic oscillator
t1
t2
t3
t4
t5
t6
x
0
x
0
x
0
x
0
x
0
x
0
The 1D harmonic oscillator
x
t1
0
x
t2
0
F(t2)
x
t3
0
F(t3)
x
t4
0
F(t4)
x
t5
0
x
t6
F(t6)
0
The 1D harmonic oscillator
x
0
d
F ( x) kx V ( x)
dx
1 2
V ( x) kx
2
This system is a model for many systems, e.g. molecules made
of two atoms.
2 d 2 k 2
Eψ( x)
x ψ( x )
2
2
2m dx
The 1D harmonic oscillator
x
0
d
F ( x) kx V ( x)
dx
1 2
V ( x) kx
2
Any system with a potential minimum at some x = x0 may
behave like a harmonic oscillator at low energies:
2 d 2 k 2
( x) 0)2 V′′(x
0) + ….
x ψ( x )
V(x) = V(x0)E+ψ(x–x
2
2
2m dx
x0
The 1D harmonic oscillator
x
0
d
F ( x) kx V ( x)
dx
1 2
V ( x) kx
2
Also, a mode of the electromagnetic field of frequency ν
behaves like a 1D harmonic oscillator of frequency ν. Its
energy levels nhν correspond
2 tod n2 photons
frequency ν.
k of
Eψ( x)
x ψ( x )
2
2
2m dx
2
The 1D harmonic oscillator
x
0
d
F ( x) kx V ( x)
dx
1 2
V ( x) kx
2
Schrödinger’s equation:
2 d 2 k 2
Eψ( x)
x ψ( x )
2
2
2m dx
Solving Schrödinger’s equation Schrödinger’s way:
m
Define a new variable
x , where k / m.
In terms of ξ , Schrödinger’s equation is
d2
2E
2
.
2 ψ 0 , where
d
Solutions: Try ψ e
2 / 2
H ( ) , where H(ξ) is a polynomial.
ψ is a solution when H(ξ) is one of the Hermite polynomials:
H0(ξ) = 1, H1(ξ) = 2ξ, H2(ξ) = 4ξ2–2, H3(ξ) = 8ξ3–12ξ, ….
Solving Schrödinger’s equation Heisenberg’s way:
Define new variables
aˆ
m
pˆ
xˆ i
, aˆ
2
m
m
pˆ
xˆ i
,
2
m
where k / m.
Let’s prove that
1
ˆ
H aˆ aˆ ,
2
aˆ , aˆ 1 , Hˆ , aˆ aˆ
ˆ , Bˆ , Cˆ ) that
Try also to prove (for any three operators A
Aˆ Bˆ , Cˆ Aˆ Bˆ , Cˆ Aˆ, Cˆ Bˆ
.
.
Raising and lowering operators
To prove:
ˆ ˆ 1
ˆ
H a a ,
2
aˆ , aˆ 1 , Hˆ , aˆ aˆ
m
pˆ
pˆ 1
ˆ ˆ 1
a a
xˆ i
xˆ i
2
m
m 2
2
pˆ 2 m 2 2
ˆ
ˆ
ˆ
x i x , p
2m
2
2
2
pˆ 2 m 2 2
xˆ Hˆ .
2m
2
.
Raising and lowering operators
To prove:
ˆ ˆ 1
ˆ
H a a ,
2
aˆ , a
ˆ
aˆ , aˆ 1 , Hˆ , aˆ aˆ
m
2
pˆ
xˆ i
m
i
2 pˆ , xˆ
2
i
2 i 1 .
2
pˆ
, xˆ i
m
.
Raising and lowering operators
To prove:
ˆ ˆ 1
ˆ
H a a ,
2
aˆ , aˆ 1 , Hˆ , aˆ aˆ
.
1
ˆ
ˆ
ˆ
ˆ
H , a a a , aˆ
2
aˆ
aˆ (aˆ aˆ ) aˆ aˆ aˆ
(aˆ aˆ 1) aˆ aˆ aˆ aˆ
.
Suppose ψ is an eigenstate of Ĥ with eigenvalue E. Then
Hˆ aˆ ψ aˆ Hˆ aˆ ψ E aˆ ψ
,
using Hˆ , aˆ aˆ . Therefore, aˆ ψ is an eigenstate of Ĥ
with eigenvalue E . We call â a raising operator.
Homework: Show that â is a lowering operator, i.e.
Hˆ aˆ ψ aˆHˆ aˆ ψ E aˆ ψ
.
Suppose ψ is an eigenstate of Ĥ with eigenvalue E. Then
Hˆ aˆ ψ aˆ Hˆ aˆ ψ E aˆ ψ
,
using Hˆ , aˆ aˆ . Therefore, aˆ ψ is an eigenstate of Ĥ
with eigenvalue E . We call â a raising operator.
The harmonic oscillator must have a ground state – call it ψ0(x)
or 0 – with minimum energy. For 0 we have aˆ 0 0
which means
pˆ
d
0 xˆ i
ψ 0 ( x) x
ψ 0 ( x) ,
m
m dx
and therefore
m x 2 / 2
ψ0 ( x) A0 e
.
1
Since Hˆ aˆ aˆ and aˆ 0 0, the ground-state energy
2
is E0
, and the energy of the state n , defined by
2
1
1 / 2 ˆ n
n (n!)
a
0 , is En n .
2
1
1
ˆ
ˆ
ˆ
From a a n H n En n n n
2
2
we learn that aˆ aˆ n n n . Thus we call aˆ aˆ the number
operator. Since n aˆ aˆ n n we can write aˆ n n n 1 .
Likewise, since n aˆ aˆ n n aˆ aˆ 1 n n 1, we can
write aˆ n n 1 n 1 .
1
Since Hˆ aˆ aˆ and aˆ 0 0, the ground-state energy
2
is E0
, and the energy of the state n , defined by
2
1
1 / 2 ˆ n
n (n!)
a
0 , is En n .
2
Normalization: 1
| ψ 0 ( x) |2 dx
| A0 |
| A0 |
2
2
e
m
m x 2 /
dx
.
So the ground state normalization is | A 0 |
all n, n n 1. (Prove it!)
4
m
. Then for
Harmonic oscillator eigenvalues and eigenfunctions
What are the lowest eigenstates of the harmonic oscillator?
m m x 2 / 2
ψ 0 ( x)
e
1/ 4
m
ˆ
ψ1 ( x) a ψ 0 ( x) 1/4
1/2
2
3/ 4
m x 2 / 2
xe
1 2
2m 2 m x 2 / 2
1/2 m
ψ 2 ( x)
aˆ ψ 0 ( x) 2
x 1e
2
1/ 4
1 3
1 m
ψ 3 ( x)
aˆ ψ 0 ( x) 1/4
6
3
3/ 4
2m 3
m x 2 / 2
x 3 x e
Note that the eigenfunctions ψn(x) are even or odd in x. Why?
Harmonic oscillator eigenvalues and eigenfunctions
This Figure is taken from
here.
Harmonic oscillator eigenvalues and eigenfunctions
This Figure is taken from
here.
If the harmonic oscillator
represents a mode of the
electromagnetic field, an
energy level En (n ½ )
represents n photons each
having energy , plus
additional “zero-point
energy” of 2 per mode.
Matrix representation
In the basis of harmonic-oscillator eigenvectors, we can
represent the operators Hˆ , aˆ, aˆ , etc., and even xˆ and pˆ , as
matrices. Since m Hˆ n (n ½ ) if m = n and vanishes
otherwise, we can represent Ĥ as an infinite matrix:
1
0
Hˆ 2 0
0
...
0
3
0
0
...
0
0
5
0
...
0
0
0
7
...
...
...
...
...
...
Matrix representation
In the basis of harmonic-oscillator eigenvectors, we can
represent the operators Hˆ , aˆ, aˆ , etc., and even xˆ and pˆ , as
matrices. Since m aˆ n n if m = n–1 and vanishes
otherwise, we can represent â as an infinite matrix:
0
0
â 0
0
...
1
0
0
0
...
0
2
0
0
...
0
0
3
0
...
...
...
...
...
...
Matrix representation
In the basis of harmonic-oscillator eigenvectors, we can
represent the operators Hˆ , aˆ, aˆ , etc., and even xˆ and pˆ , as
matrices. Since m aˆ n n 1 if m = n+1 and vanishes
otherwise, we can represent â as an infinite matrix:
0
1
â 0
0
...
0
0
2
0
...
0 0
0 0
0 0
3 0
... ...
...
...
...
...
...
The normalized harmonic-oscillator eigenvectors themselves
are the basis vectors:
1
0
0 0 ,
0
...
0
1
1 0 ,
0
...
0
0
2 1 ,
0
...
0
0
3 0 , ... .
1
...
The transpose of a matrix M̂ is written M̂ T and defined by
Mˆ T ij i Mˆ T j j Mˆ i Mˆ ji :
a b ...
a c ...
T
ˆ
ˆ
M c d ... , M b d ... .
... ... ...
... ... ...
The adjoint of a matrix M̂ is written M̂ and defined by
*
*
ˆ
ˆ
ˆ
ˆ
M ij i M j j M i M ji :
a b ...
a * c * ...
ˆ
ˆ
M c d ... , M b * d * ... .
... ... ...
... ... ...
The adjoint of a matrix M̂ is written M̂ and defined by
*
*
ˆ
ˆ
ˆ
ˆ
M ij i M j j M i M ji :
a b ...
a * c * ...
ˆ
ˆ
M c d ... , M b * d * ... .
... ... ...
... ... ...
Any observable  is self-adjoint, i.e. Aˆ Aˆ .
“Â is Hermitian” and “Â is self-adjoint” mean the same thing.
The adjoint of a matrix M̂ is written M̂ and defined by
*
*
ˆ
ˆ
ˆ
ˆ
M ij i M j j M i M ji :
a b ...
a * c * ...
ˆ
ˆ
M c d ... , M b * d * ... .
... ... ...
... ... ...
Any observable  is self-adjoint, i.e. Aˆ Aˆ .
“Â is Hermitian” and “Â is self-adjoint” mean the same thing.
The raising and lowering operators â and â are not selfadjoint, but are adjoints of each other:
aˆ
ˆ
a
,
a
ˆ
aˆ .
Hˆ , xˆ and pˆ are manifestly self-adjoint:
1 0 0
0 3 0
ˆ
H 2
0 0 5
... ... ...
...
...
,
...
...
Hˆ , xˆ and pˆ are manifestly self-adjoint:
xˆ
ˆa aˆ
2m
0
1
2m 0
...
0
m 1
m
aˆ aˆ i
pˆ i
2 0
2
...
1
0
2
...
0
2
0
...
...
...
,
...
...
0
1
0 2
0
2
...
...
...
...
.
...
...
Vectors, too, have adjoints. For any ψ we have
ψ
ψ
,
and, for example,
1
0
1 0 0 ... .
0
...
Try to prove, for any two operators Aˆ , Bˆ , the rule
( Aˆ Bˆ ) Bˆ Aˆ
.
Motion of a minimum-uncertainty wave packet
The ground state wave function ψ0(x) is a minimum-uncertainty
wave function: we can calculate
m
, x p
.
2
2
, p
2m
x
In general, the time evolution of any initial wave function
Ψ(x,0) can be obtained from the expansion of Ψ(x,0) in the
basis of energy eigenstates. If the initial wave function is
(0)
cn n
,
n 0
for given cn, then the wave function at time t is
(t )
n 0
cn e iEnt / n e it / 2
n 0
cn e int n
.
Motion of a minimum-uncertainty wave packet
(t )
cn e iEnt / n e it / 2
n 0
cn e int n
.
n 0
The period of a classical harmonic oscillator having angular
frequency ω is T = 2π /ω . If we add T to t in Ψ(x,t), the wave
function does not change, up to an overall phase factor –1:
(t T ) (t )
.
Schrödinger even found a solution Ψ(x,0) that moves between
x = R and x = –R while the probability distribution |Ψ(x,0)|2
keeps its (minimum uncertainty) shape:
1/ 4
m
2
| ( x, t ) |
e
m ( x R cos t ) 2 / 2
.
Motion of a minimum-uncertainty wave packet
Schrödinger thought his solution might be typical, but it is not.
Usually the probability distribution spreads over time. Prove
that a free 1D wave packet spreads, that if it is initially
( x,0) (2 )
1 / 4
(x)
1 / 2 i p x / x 2 / 4(x) 2
e
e
,
then at time t the wave packet is
( x, t )
e
i p 2 t / 2 m i p x / ( x p t / m) 2 /[ 4( x ) 2 2it / m]
e
( 2 )1 / 4 x it 2 m x 1 / 2
Show that Δp = 2x is constant in time, but
x(t )
x 2 p 2 t 2 / m 2
.
.
3D harmonic oscillator
The Schrödinger equation for a general 3D harmonic oscillator,
2 2 kx 2 k y 2 kz 2
Eψ( x, y, z )
x
y
z ψ ,
2
2
2
2m
allows for a harmonic potential with different strengths along
the three axes.
Show that the eigenfunctions are products of 1D harmonic
oscillator eigenfunctions. What are the lowest energies and
their degeneracies as a function of kx, ky and kz?
Classical limit
A classical particle in a square well has equal probability to be
at any point inside. How about a quantum particle in its ground
state? a quantum particle in a highly excited state?
The probability P(x) for classical harmonically oscillating
particle to be at any point x is inversely proportional to its speed
at that point: P(x) ~ (2E/m – ω2x2)–1/2, where E is total energy.
n=0
.
n = 10