Lecture 12: Harmonic Oscillator, Wednesday, Sept. 28
Simple Harmonic Oscillator, Creation and Annihilation Operators
Consider a simple one-dimensional harmonic oscillator with the following
hamiltonian
p̂2
1
Ĥ =
+ mω 2 x̂2 .
(29)
2m 2
Introduce the following creation and annihilation operators
a=
r
Ã
!
mω
ip̂
x̂ +
;
2h̄
mω
†
a =
r
Ã
mω
ip̂
x̂ −
2h̄
mω
!
(30)
Both operators are non-hermitian, but it is easy to see that a and a† are
hermitian conjugation of each other, and
[a, a† ] = 1
(31)
And the hamiltonian can be written as
1
Ĥ = h̄ω(N̂ + )
2
(32)
where N̂ = a† a is called the number operator. It is a positive definite operator
in the sense that the expectation value of N̂ between any state is positive
definite.
The number operator is hermitian. Assume the number operator has real
eigenvalue n (positive),
N̂ |ni = n|ni
(33)
It is easy to prove that [N̂ , a] = −a and [N̂ , a† ] = a† . Therefore a|ni is an
eigenstate of N̂ with eigenvalue n − 1, and a† |ni is an eigenstate of N̂ with
eigenvalue n + 1. Since N̂ is positive definite, n must be an integer, otherwise
keeping applying a on a state leads to negative eigenvalues. Let’s see why
when n is integer, we are safe.
Assume a|ni = c|n−1i. Then from the normalization condition, we know
that |c|2 = n, or
√
(34)
a|ni = n|n − 1i
When n is integer, after applying sufficient number of a, one gets a state |0i,
and then a simply annihilates the state.
8
It is easy to see that
a† |ni =
√
n + 1|n + 1i
(35)
Now, we find the ground state of the system |0i with energy E = 21 h̄ω,
corresponding to n = 0. Higher states have energy eigenvalue E = (n +
1/2)h̄ω. The normalized state can be constructed from a† ,
(a† )n
|ni = √ |0i
n!
(36)
Using the creation and annihilation operator formalism, one can obtain
the matrix element of x̂ and p̂ and any other operators, and calculate the
uncertainty relations. For instance,
2
x =
Ã
!
h̄
(a2 + a†2 + a† a + aa† )
2mω
(37)
Then in the ground state, only the last term contributes to the expectation
value,
h̄
h0|x2 |0i =
(38)
2mω
And likewise h0|p2 |0i = h̄mω/2. Therefore ∆x∆p = h̄/2, i.e., the ground
state of the oscillator obeys the minimum uncertainty relation. For any
excited state, one can show ∆x∆p = (n + 1)h̄/2
One can also recover the coordinate space wave function. Since a|0i = 0,
this is a differential equation in coordinate space
!
Ã
d
ψ0 (x) = 0
hx|a|0i ∼ x + b2
dx
(39)
q
where b = h̄/mω is the length scale of the oscillator. The normalized
solution is just
s
1
(40)
ψ0 (x) = √ exp(−x2 /(2b2 ))
πb
The wave functions of the excited states can be obtained through application
of a† ,
!n
Ã
µ
¶
1 2 2
1
1
2 d
√
exp − x /b
x−b
(41)
hx|ni =
dx
2
π 1/4 2n n! bn+1/2
9
where the differentiations lead to Hermite polynomial,
Hn (x) = (−1)n ex
2
dn −x2
e
dxn
(42)
with H0 = 1, H1 (x) = 2x, H2 = 4x2 − 2, H3 = 8x3 − 12x, .... Thus we write,
Ã
1
1 x2
ψn (x) = q √ √
Hn (x/b) exp − 2
2b
b π 2n n!
A useful
√ integration formula for Hermite polynomials is
n
2 n! π.
!
R∞
−∞
(43)
2
e−x Hn2 (x)dx =
Time development in Harmonic oscillator
Solving the Heisenberg equation of motion,
dp̂
= −mω 2 x̂;
dt
dx̂
p̂
=
.
dt
m
(44)
one finds,
p̂(0)
sin ωt ,
mω
p̂(t) = −mωx̂(0) sin ωt + p̂(0) cos ωt .
x̂(t) = x̂(0) cos ωt +
(45)
The result can also be derived from the Baker-Hausdorff lemma,
i2 λ 2
in λ n
e Ae
= A+iλ[G, A]+
[G, [G, A]]+...+
[G, [G, ...[G, A]...]+... ,
2!
n!
(46)
However, for a stationary state, the expectation values of x̂ and p̂ are zero.
Therefore to observe a classical type of oscillation, one must be in a superimposed state such as |ψi = c0 |0i + c1 |1h.
A type of superimposed quantum states which are closely related to the
classical oscillator states are the so-called coherent states, defined as eigenstates of the annihilation operators
iGλ
−iGλ
a|λi = λ|λi .
(47)
Since a is not hermitian, λ may not be real. [In fact, it is a complex eigenvalue!] Moreover, the eigenstates
10
corresponding to different λ are not orthogonal to each other. However,
these states do have certain interesting features as we shall see.
The above equation can be solved by expanding |λi in the basis of |ni
|λi =
∞
X
f (n)|ni
(48)
n=0
one arrives at a recursion relation for f (n).
f (n + 1) = √
λ
f (n)
n+1
(49)
√
The solution is f (n) = λn / n!f (0). The normalization condition is f (0) =
exp(−|λ|2 /2). Therefore,
λn
2
f (n) = √ e−|λ| /2
n!
which means the state can also be written as
|λi = e−|λ|
2 /2
(50)
†
eλa |0i .
(51)
It can be shown that
hxi =
moreover, ∆x =
q
s
2h̄
Reλ;
mω
hpi =
h̄/2mω, and ∆p =
eλa
† −λ∗ a
q
√
2mh̄ωIeλ .
mh̄ω/2, and ∆x∆p = h̄/2. Using
†
= eλa e−λa e−|λ|
we can write
|λi = eλa
(52)
† −λ∗ a
2 /2
.
|0i
(53)
(54)
q
Recall that the translation operator is T (x0 ) = e−ipx0 . Since p = i 2/mωh̄(a† −
q
a), we have T (x0 ) = exp( 2/mωh̄x0 (a† − a)). If we translate the oscillator
ground state by some finite distance, we can obtain the above state. Therefore coherent state satisfies the minimum uncertainty relation.
One can calculate the expectation value of x(t),
hλ|x(t)|λi =
s
2h̄
Re(λe−iωt )
mω
q
= |λ| 2h̄/mω cos(ωt − φ)
where φ is the phase angle of λ.
11
(55)