Chemistry 460
Spring 2013
Dr. Jean M. Standard
February 21, 2013
Harmonic Oscillator: Raising and Lowering Operators
I.
Preliminaries
Recall the basic definitions of the harmonic oscillator. The potential energy is defined by
V (x) =
1
2
kx 2 ,
(1)
where k is the force constant. The Hamiltonian operator therefore has the following form,
€
2 d 2
1
Hˆ = −
+
k x2 .
2
2m dx
2
(2)
In order to make the equations easier to work with, we defined dimensionless variables Pˆ and Qˆ ,
€
⎛ mk ⎞1/ 4
Qˆ = ⎜ 2 ⎟ xˆ ,
⎝  ⎠
€
€
€
(3)
d
Pˆ = − i
.
dQ
(4)
These scaled operators have the property that [ Pˆ , Qˆ ] = −i . In these new variables, the Hamiltonian operator
has the form
€
hν 0 ˆ 2 ˆ 2
Hˆ =
P +Q .
2
(
€
)
(5)
The Schrödinger equation for the harmonic oscillator is
€
Hˆ ψ = Eψ ,
or
hν 0 ˆ 2 ˆ 2
P + Q ψ = Eψ .
2
(
)
(6)
Dividing by hν 0 , and defining the dimensionless energy ε as ε = E / hν 0 , the equation becomes
€
1 ˆ2 ˆ2
P + Q ψ = εψ .
2€
(
€
)
(7)
The scaled Schrödinger equation is then
€
hˆ ψ = ε ψ ,
€
(8)
2
where the scaled Hamiltonian operator hˆ is given by
1 ˆ2
hˆ =
P + Qˆ 2 .
2
(
€
II.
)
(9)
Definitions of Raising and Lowering
Operators
€
We now define two new operators, called raising and lowering operators, for reasons that will become apparent
later on. Define
1 ˆ
P − i Qˆ ,
2
(
a =
)
(10)
1
2
a+ =
(Pˆ + i Qˆ ) .
+
The operator a is called the lowering operator, and the operator a is called the raising operator (the "hats" have
+
€
been left off these operators for convenience).
Using the definitions of a and a , it is easy to show that the
+
commutator is a, a = 1.
€
€
[
III.
]
Scaled
Hamiltonian Operator
€
+
The scaled Hamiltonian operator hˆ for the harmonic oscillator may be expressed in terms of a and a in a
number of different ways.
+
+
Form 1. First, consider
€ the expression aa + a a . Using the definitions, we get
1 ˆ
1 ˆ
aa + + a + a =
P − i Qˆ Pˆ + i Qˆ +
P + i Qˆ Pˆ − i Qˆ
2
2
€
= Pˆ 2 + Qˆ 2 .
(
)(
)
(
)(
€
)
(11)
Thus, we can write the scaled Hamiltonian operator as
€
1
hˆ =
a a+ + a+a .
2
(
)
(12)
+
Form 2. Next, consider simply the
€ expression a a . Again, substituting the definitions of the operators, the
expression can be written
1 ˆ
€
a+a =
P + i Qˆ Pˆ − i Qˆ
2
1 ˆ2 ˆ2
i ˆ ˆ
=
P +Q
−
P, Q
2
2
1
= hˆ −
.
2
€
(
)(
)
(
)
[
]
(13)
3
So, the scaled Hamiltonian operator can also be written in the form:
1
hˆ = a + a +
.
2
(14)
Form 3. Starting from form 2, we can use
€ the commutator to express the scaled Hamiltonian operator in one
+
more way. This form can also be derived in a similar fashion to form 2 by starting with the expression a a .
From form 2 of the scaled Hamiltonian operator,
1
hˆ = a + a +
.
2
€
(15)
Using the definition of the commutator,
€
[a, a+ ]
= 1,
or a a + − a + a = 1.
(16)
€
+
Solving for a a yields
€
a + a = a a + − 1.
(17)
€
Substituting this into the expression for hˆ gives
€
1
hˆ = a a + −
.
2
€
IV. More on Commutators
(18)
€
+
The commutators of a and a with hˆ are useful expressions which are easily evaluated.
+
A. Commutator of a with hˆ
€
€
Using form 2 of the Hamiltonian operator,
€
€
a + , hˆ
[
]
⎡
1 ⎤
= ⎢ a + , a + a + ⎥
⎣
2 ⎦
[
= a+ , a+a
]
⎡
1 ⎤
+ ⎢ a + , ⎥
⎣
2 ⎦
The second commutator is zero since the commutator of any operator with a constant is zero.
€
(19)
4
Using the identity
[ Aˆ ,
Bˆ Cˆ
]
= Bˆ Aˆ , Cˆ
[
]
+
[ Aˆ , Bˆ ] Cˆ ,
(20)
+
[a+ , a+ ] a .
(21)
[a+ , Hˆ ] = [a+ , a+ a]
= a + [ a + , a]
€
[
+
The commutator of any operator with itself is zero, so the second term above is zero. Since a, a
+
− 1 . Therefore,
we know that a , a = €
[
]
= 1,
]
[a+ , hˆ]
€
B. Commutator of a with hˆ
€
= − a+ .
(22)
€
In a similar manner, the commutator of a and hˆ can be evaluated to yield
€
a, hˆ = a .
€
[
]
(23)
+
V. Proof that a is a raising operator
€
+
Many of the relations derived in previous sections are useful in demonstrating that a is a raising operator. We
start€with the scaled Schrödinger equation,
hˆψ = ε ψ .
€
(24)
+
Apply a to both sides,
€
a + hˆ ψ = ε a + ψ .
(25)
€
Use the commutator,
€
[ a+ , hˆ]
= − a+
a + hˆ − hˆ a + = − a + .
(26)
a + hˆ = hˆ a + − a + .
(27)
+
Solving for a hˆ ,
€
€
Inserting this relation into Eq. (25) yields,
€
€
( hˆ a
+
)
− a+ ψ = ε a+ ψ .
(28)
5
Moving the second term on the left to the right side,
hˆ a + ψ = a + ψ + ε a + ψ
( )
hˆ a + ψ
( )
= (ε + 1) a + ψ .
(29)
+
Eq. (29) is the Schrödinger Equation with the eigenfunction a ψ and eigenvalue ε + 1. It is easy to show by
+
analogy that applying a to the€Schrödinger Equation n times yields the equation
⎛
hˆ ⎜ a +
⎝
( )
€
n
⎞
ψ ⎟€ =
⎠
(ε
n€ ⎞
⎛
+ n) ⎜ a + ψ ⎟ .
⎝
⎠
( )
(30)
This procedure yields a whole series of eigenvalues and eigenfunctions of the Schrödinger Equation:
€
Eigenfunction
(a )
+


€
Eigenvalue
ψ
ε + n


3
(a ) ψ
(a ) ψ
+
€
n
+
ε+3
2
€
a+ ψ
ψ
€
ε+2
ε +1
ε
€
€
The set of eigenfunctions
and eigenvalues is sometimes€called the "ladder of solutions" of the Schrödinger
€
Equation for the harmonic
oscillator. The raising and lowering
operators are therefore sometimes referred to as
€
€
"ladder operators".
+
If we write the Schrödinger Equation for the nth state, ψ n , we have hˆ ψ n = ε ψ n . Applying a to this
equation as previously described yields
(
hˆ €
a +ψ n
)
(
)
= (ε n €
+ 1) a +ψ n .
€
(31)
+
Thus, we see that the function a ψ n is an eigenfunction of the Schrödinger Equation with eigenvalue ε n + 1 .
The eigenvalue has been raised€by 1; therefore, it corresponds to the eigenvalue for the n+1st state,
ε n+1 = ε n + 1. Therefore, the eigenfunction a + ψ n must be proportional to the eigenfunction for the n+1st
state,
€
€
€
a + ψ n = c+ ψ n+1 ,
€
where the coefficient c+ is yet to be determined.
€
€
(32)
6
VI. Proof that a is a lowering operator
Again, we start with the scaled Schrödinger equation,
hˆ ψ = ε ψ .
(33)
a hˆ ψ = ε a ψ .
(34)
Apply a to both sides,
€
Using the commutator,
€
[ a, hˆ]
= a
a hˆ − hˆ a = a .
(35)
Solving for a hˆ ,
€
a hˆ = hˆ a + a .
(36)
€
Substituting this relation into Eq. (34),
€
( hˆ a + a) ψ
= ε aψ .
(37)
Moving the second term on the left to the right side,
€
hˆ a ψ = − a ψ + ε a ψ
hˆ ( a ψ ) = (ε − 1) ( a ψ ) .
(38)
Eq. (38) is the Schrödinger Equation with the eigenfunction a ψ and eigenvalue ε − 1. It can be shown that
applying a to the Schrödinger Equation
n times yields the equation
€
⎛
⎞
hˆ ⎜ a + ψ ⎟ €=
⎝
⎠
( )
(ε
⎛ n €⎞⎟
− n) ⎜ ( a) ψ
.
⎝
⎠
A ladder of solutions of the Schrödinger Equation is generated by this procedure:
€
Eigenfunction
ψ
aψ
a2 ψ
€
€
€
a3ψ


€
€
€
an ψ
Eigenvalue
ε
ε −1
ε−2
ε−3


ε−n
€
€
€
€
(39)
7
Writing the Schrödinger Equation for the nth state, ψ n , we have hˆ ψ n = ε n ψ n . Applying a to this equation
as before yields
hˆ ( a ψ n ) €= (ε n − 1) ( a ψ n ) .
€
(40)
The function a ψ n is therefore an eigenfunction of the Schrödinger Equation with eigenvalue ε n − 1 . The
eigenvalue has been lowered by 1; therefore,
it corresponds to the eigenvalue for the n–1st state,
€
ε n−1 = ε n − 1. Therefore, the eigenfunction a ψ n must be proportional to the eigenfunction for the n–1st
state,
€
€
€
a ψ n = c− ψ n−1 ,
€
(41)
where the coefficient c− must be determined.
€
+
VII. Non-Hermitian
character of a and a
€
+
The operators a and a are not Hermitian. Recall that an operator Aˆ is Hermitian if
€
f Aˆ g = Aˆ f g .
€
€
(42)
By starting with the left side of this relation and considering the operator to be the lowering operator a, we can
show that it does not equal the right side
€ and therefore a is not Hermitian.
The left side of the Hermitian equation for the operator a is given by
f a g .
(43)
Using the definition of the operator a, we have
f a g
€
=
=
Pˆ − i Qˆ g
1
2
f
1
2
f Pˆ g −
i
2
f
Qˆ g .
(44)
ˆ
The operators Pˆ and Q are Hermitian (this can be proved using a method similar to the one that was used in
Problem Set 1). Thus, €
from the definition of a Hermitian operator
€
€
and
f Pˆ g
=
Pˆ f g ,
f Qˆ g
=
Qˆ f g .
(45)
Substituting these expressions into Eq. (44) yields
€
€
f a g
=
1
2
f Pˆ g −
=
1
2
Pˆ f g −
i
2
i
2
f Qˆ g
Qˆ f g .
(46)
8
Since i* = –i, the second term in Eq. (46) can be rewritten as
=
1
2
f a g
=
f a g
Pˆ f g +
1
2
i Qˆ f g .
(47)
Recombining the two terms on the right,
€
Since by definition a + =
€
1
2
(Pˆ + i Qˆ ) f
1
2
(48)
g .
(Pˆ + i Qˆ ) , the equation above can be written
f a g
a+ f g .
=
(49)
€
Since f a g
≠
a f g , the operator a is NOT Hermitian.
€
VIII.
Determination of the coefficients c+ and c−
€
In order to make effective use of the raising and lowering operators, it is necessary to determine the constants c+
and c− in the expressions
€
€
a + ψ n = c+ ψ n+1
a ψ n = c− ψ n−1 .
€
A.
€
(50)
Determination of c−
To determine c− , we start with €
consideration of the integral
a ψn a ψn
€
.
(51)
€
Using the equation above, this integral can be rewritten as
€
a ψn a ψn
c− ψ n−1 c− ψ n−1
=
= c−* c−
a ψn a ψn
c−* c−
=
ψ n−1 ψ n−1
,
(52)
assuming that the eigenfunctions are normalized.
€
We can also write Eq. (51) in another way, using the non-Hermitian character of the raising and lowering
operators; that is, that
f a g
=
a+ f g .
(53)
Eq. (51) can therefore be rewritten as
€
a ψn a ψn
€
=
a+a ψn ψn .
(54)
9
Next, the relation a + a = hˆ −
1
2
can be substituted,
a ψn a ψn
=
€
=
a ψn a ψn
(hˆ − 12 ) ψn ψn
(ε n − 12 ) ψn ψn
= εn −
1
2
.
*
From Eq. (52), the left side of the above equation is also equal to c− c− ; therefore,
€
c−* c− = ε n − 12 .
€
1
Finally, since the scaled eigenvalue ε n equals n + 2 , we have
(55)
(56)
€
c−* c− = n .
(57)
€
*
Choosing c− to be real allows c − = c − . This then leads to
€
c− = n ,
(58)
€
€
€
and therefore the equation for the lowering operator becomes
€
a ψn =
n ψ n−1 .
(59)
B. Determination of c+
€
The coefficient c+ is determined in a similar fashion. This time, we start with consideration of the integral
a+ ψn a+ ψn
€
.
(60)
€
This integral can be rewritten using the definition of the raising operator,
€
a+ ψn a+ ψn
=
c+ ψ n+1 c+ ψ n+1
= c++ c+
+
+
a ψn a ψn
=
c+* c+
ψ n+1 ψ n+1
.
(61)
Using the non-Hermitian character of the raising and lowering operators, the integral given in Eq. (60) can
be written as
€
a+ ψn a+ ψn
€
=
a a+ ψn ψn .
(62)
10
Next, the relation a a + = hˆ +
1
2
can be substituted,
a + ψ n a +ψ n
€
=
=
a+ ψn a+ ψn
(hˆ + 12 ) ψn ψn
(ε n + 12 ) ψn ψn
= εn +
1
2
.
(63)
*
From Eq. (61), the left side of the above equation is also equal to c+ c+ ; therefore,
€
c+* c+ = ε n + 12
=€n + 12 +
1
2
c+* c+ = n + 1.
(64)
*
Choosing c+ to be real allows c + = c + . This then leads to
€
c+ = n + 1 ,
(65)
€
€
and therefore the equation for the raising operator becomes
€
€
a+ ψn =
n + 1 ψ n+1 .
(66)