Quantum Harmonic Oscillator
with Ladder Operators
Kenneth Taliaferro
January 29, 2010
In Quantum Mecahnics, the state of a particle ∫is given by a complex
valued function Ψ of postion x and time t such that |Ψ(x, t)|2 dx = 1.
I’m looking for a solution Ψ to the Schrödinger Equation
ih̄
∂Ψ
p2 Ψ
=
+VΨ
∂t
2m
where h̄, k, and m are constants and
p = −ih̄
∂
∂x
1
V = kx2
2
(momentum operator)
(potential energy operator)
After separation of variables (Ψ(x, t) = ψ(x)ϕ(t)), this boils down to the
eigenvalue problem of finding twice differentiable functions ψ in L2 (R) such
that
Hψ = Eψ
where
1 2 mω 2 2
p +
x
2m
2
√
k
ω=
m
H=
(right side of Schrödinger equation)
1
x is self adjoint because, for f, g ∈ D(x) ⊂ L2 (R),
∫
⟨xf, g⟩ = xf (x)g(x)dx
∫
= f (x)xg(x)dx
= ⟨f, xg⟩.
By integration by parts, p is self-adjoint. Moreover, by the product rule,
[x, p] := xp − px = ih̄.
Define the ladder operators
√
)
(
mω
i
a+ =
x−
p
(raising operator)
mω
2h̄
√
(
)
mω
i
a− =
x+
p
(lowering operator)
mω
2h̄
Then a+ and a− are adjoints of each other. Using the fact that [x, p] = ih̄,
it is straightforward to show that
[a− , a+ ] = 1
h̄ω
H=
(2a+ a− + 1)
2
[H, a+ ] = h̄ωa+
(1)
(2)
Suppose that ψ is an eigenfunction of H with eigenvalue E. Then
Ha+ ψ = (Ha+ − a+ H + a+ H)ψ
= ([H, a+ ] + a+ H)ψ
= (h̄ωa+ + a+ E)ψ
= (h̄ω + E)a+ ψ
i.e. a+ ψ (if nonzero) is an eigenfunction with eigenvalue E + h̄ω. Similarly,
a− ψ (if nonzero) is an eigenfunction with eigenvalue E − h̄ω. This is why a+
and a− are called the raising and lowering operators.
Lemma 1: The only possible eigenvalues of H are h̄ω(n+ 12 ), n = 0, 1, 2, 3, ...
2
Proof: Use the raising and lowering operators. See the last page of this
document or [3], page 594.
Let E0 be the smallest eigenvalue of H. Let ψ0 be a corresponding normalized eigenfunction. Then
a− ψ0 = 0
(3)
because otherwise a− ψ0 would be an eigenfunction with eigenvalue E − h̄ω <
E0 . Thus, by (2),
0 = h̄ωa+ a− ψ0
)
(
h̄ω
ψ0
= H−
2
(
)
h̄ω
= E0 −
ψ0
2
so E0 = h̄ω
. An eigenfunction with eigenvalue En = (n + 12 )h̄ω is given by
2
n
n
a+ ψ0 (if a+ ψ0 ̸= 0). Let ψn be a corresponding normalized eigenfunction
with eigenvalue En . Then, for some constant Cn+1 ,
ψn+1 = Cn+1 a+ ψn .
But,
1 = ⟨ψn+1 , ψn+1 ⟩
= |Cn+1 |2 ⟨a+ ψn , a+ ψn ⟩
= |Cn+1 |2 ⟨ψn , a− a+ ψn ⟩
)
(
1
H
2
= |Cn+1 | ⟨ψn ,
+
ψn ⟩
h̄ω 2
)
(
(n + 12 )h̄ω 1
2
= |Cn+1 | ⟨ψn ,
+
ψn ⟩
2
h̄ω
so
Cn+1 = (n + 1)−1/2 .
It follows that
a+ ψn =
√
Similarly,
a− ψn =
n + 1ψn+1 .
√
3
nψn−1 .
(4)
(5)
Therefore
1
ψn = √ an+ ψ0 .
(6)
n!
This is sufficient for computing expectation values of powers of x and p
of the eigenfunctions because
√
h̄
x=
(a+ + a− )
√2mω
h̄mω
p=i
(a+ − a− )
2
For example, since eigenfunctions of a self-adjoint operator corresponding
to different eigenvalues are orthogonal and [a− , a+ ] = 1, the expectation value
of x2 on the state ψn is given by
(
)
h̄
2
⟨ψn , x ψn ⟩ = ⟨ψn ,
(a+ + a− )(a+ + a− )ψn ⟩
2mω
h̄
=
⟨ψn , (a2+ + a2− + a+ a− + a− a+ )ψn ⟩
2mω
h̄
=
⟨ψn , (a2+ + a2− + a+ a− + (a+ a− − a+ a− ) + a− a+ )ψn ⟩
2mω
h̄
=
⟨ψn , (a2+ + a2− + 2a+ a− + 1)ψn ⟩
2mω
h̄
⟨ψn , (2n + 1)ψn ⟩
=
2mω
h̄
=
(2n + 1)
2mω
√ ψ0 can be found explicitly by solving the separable ODE (3). Let α =
mω/h̄. In terms of the variable y = αx, (3) says
)
(
d
y+
ψ0 (y) = 0
dy
For any constant N0 , a solution is
ψ0 (y) = N0 e−y
i.e.
ψ0 (x) = N0 e−α
4
2 /2
.
2 x2 /2
.
By (6),
(
ψn (x) =
α
√ n
π2 n!
)1/2
e−α
2 x2 /2
Hn (αx)
where Hn is the Hermite polynomial of order n. By separation of variables,
Ψn (x, t) = e(−iEn /h̄)t ψn (x).
Since the position distribution |Ψn |2 doesn’t depend on time, these are called
stationary states.
References
[1] Agnolet, G. Physics 412 Lecture Notes.
[2] Bransden, B.H. and C.J. Joachain. Quantum Mechanics.
[3] Kreyszig, E. Introductory Functional Analysis with Applications.
5
Proof of Lemma 1: As derived earlier,
[a− , a+ ] = 1
h̄ω
H=
(2a+ a− + 1)
2
[H, a+ ] = h̄ωa+
(7)
(8)
Suppose ψ is an eigenfunction of H with eigenvalue E. Let
Ẽ =
E
1
− .
h̄ω 2
(9)
Then, by (8),
a+ a− ψ = Ẽψ.
Applying a− to each side,
a− a+ (a− ψ) = Ẽ(a− ψ).
By (7), a− a+ = a+ a− + 1, so
a+ a− (a− ψ) = (Ẽ − 1)a− ψ.
Repeating the last two steps j − 1 times yields
a+ a− (aj− ψ) = (Ẽ − j)aj− ψ.
(10)
Now aj− ψ = 0 for sufficiently large j; because, otherwise, by (10),
j+1
j
j
⟨aj+1
− ψ, a− ψ⟩ = ⟨a+ a− (a− ψ), a− ψ⟩
= (Ẽ − j)⟨aj− ψ, aj− ψ⟩
i.e.
Ẽ − j =
2
∥aj+1
− ψ∥
≥0
∥aj− ψ∥2
(11)
for all j. Thus ∃n ∈ N such that an− ψ ̸= 0 but an+1
− ψ = 0. By (11), Ẽ −n = 0.
By (9),
(
)
1
E = h̄ω n +
for some n ∈ {0, 1, 2, 3, ...}.
2
6