Consider a particle of mass m moving in a potential energy function U(x) =
½kx^{2}.

The force on the particle is given by F(x) = -dU(x)/dx = -kx.

The motion of the classical particle is governed by Newton’s second law md^{2}x/dt^{2}
= -kx.

The general solution to this equation is x = x_{M}cos(ωt + φ), with ω^{2}
= k/m.

We have dx/dt = -ωx_{M}sin(ωt + φ), T = ½m(dx/dt)^{2} =
½mω^{2}x_{M}^{2}sin^{2}(ωt + φ),

E = T + U = ½mω^{2}x_{M}^{2}sin^{2}(ωt + φ) +
½mω^{2}x_{M}^{2}cos^{2}(ωt + φ) = ½mω^{2}x_{M}^{2}.

The total energy is independent of time, we have a **conservative system**.

To make the transition to Quantum Mechanics we express E in terms of p
and x and then replace the classical quantities p and x by their
corresponding quantum mechanical operators P and X.

E = T + U = ½p^{2}/m + ½mω^{2}x^{2} --> H = ½P^{2}/m
+ ½mω^{2}X^{2}.

X

H = ½ħω[P

where H and H

X

[Xs,Ps] = (mω/ħ)

Let us define some new operators and investigate their properties. Define

a = (2)^{-1/2}(X_{s} + iP_{s}), and its adjoint, a^{†}
= (2)^{-1/2}(X_{s} - iP_{s}).

Then X_{s} = (2)^{-1/2}(a^{†} + a), P_{S}
= i(2)^{-1/2}(a^{†} - a).

Note that a and a^{†} do not commute.

[a,a^{†}] = ½[(X_{s} + iP_{s}), X_{s} - iP_{s})]
= ½[-i[X_{s},P_{s}] + i[P_{s},X_{s}] = ½(1 + 1)
= 1.

Now we study the product operator a^{†}a.

a^{†}a = ½(X_{s} - iP_{s})(X_{s} + iP_{s})
= ½(X_{s}^{2} + P_{s}^{2} + iX_{s}P_{s}
- iP_{s}X_{s}) = ½(X_{s}^{2} + P_{s}^{2}
- 1).

Similarly, aa^{†} = ½(X_{s}^{2} + P_{s}^{2}
+ 1).

We may therefore write H_{s} = a^{†}a + ½ = aa^{†} - ½ =
½(aa^{†} + a^{†}a).

Define the number operator N = a^{†}a.

Then H_{s} = N + ½.
H_{s} and N have common eigenstates, finding the
eigenstates of N is equivalent to finding the eigenstates of H_{s} and
H.

The operator N does not commute
with a and a^{†}.

[N,a] = [a^{†}a,a] = a^{†}[a,a] + [a^{†},a]a = -a.

[N,a^{†}] = [a^{†}a,a^{†}] = a^{†}[a,a^{†}]
+ [a^{†},a^{†}]a = a^{†}.

We will now study some of the **properties of the eigenstates **of N and therefore
the eigenstates of H_{s} and H.

(a) The eigenvalues of N are positive.

Proof: Let |Φ_{ν}> be an eigenstate of N
with eigenvalue ν.

N|Φ_{ν}> = ν|Φ_{ν}>.

Find the norm of the vector a|Φ_{ν}>.

|a|Φ_{ν}>|^{2} = <Φ_{ν}|a^{†}a|Φ_{ν}>
= <Φ_{ν}|N|Φ_{ν}> = ν<Φ_{ν}|Φ_{ν}> ≥
0,

since the norm is always positive,
and <Φ_{ν}|Φ_{ν}> ≥ 0 --> ν ≥ 0.

(b) If |Φ_{ν}> is an eigenstate of N
with eigenvalue ν ≥ 1 and <Φ_{ν}|Φ_{ν}> ≠ 0 then a|Φ_{ν}>
is an eigenstate of N with eigenvalue ν - 1.

If |Φ_{ν}> is an eigenstate of N with eigenvalue ν < 1 then ν = 0 and a|Φ_{ν}> = a|Φ_{0}> = 0.

Proof: |a|Φ_{ν}>|^{2} = ν > 0, unless ν = 0.

[N,a]|Φ_{ν}> = -a|Φ_{ν}>. Na|Φ_{ν}> = aNΦ_{ν}>
- a|Φ_{ν}> = (ν - 1)a|Φ_{ν}>.

a|Φ_{ν}>
is an eigenstate of N with eigenvalue ν - 1.

a|Φ_{ν}>
is non zero unless ν = 0.
But the eigenvalues have to be positive. Therefore if ν < 1 we need
a|Φ_{ν}> = 0 which requires ν = 0.

(c) If |Φ_{ν}> is an eigenstate of N
with eigenvalue ν ≥ 0 and <Φ_{ν}|Φ_{ν}> ≠ 0 then a^{†}|Φ_{ν}> is an eigenstate of N with eigenvalue
ν + 1.

Proof: | a^{†}|Φ_{ν}>|^{2} = <Φ_{ν}|aa^{†}|Φ_{ν}>
= <Φ_{ν}|a^{†}a + 1|Φ_{ν}> = (ν + 1)<Φ_{ν}|Φ_{ν}>
> 0.

[N,a^{†}]|Φ_{ν}> = a^{†}|Φ_{ν}>, Na^{†}|Φ_{ν}>
= a^{†}N|Φ_{ν}> + a^{†}|Φ_{ν}> =
(ν + 1) a^{†}|Φ_{ν}>.

(d) The eigenvalues of N are the non negative integers.

Proof: N has at least one non-zero eigenvector with eigenvalue ν.

(It is a Hermitian operator).

N|Φ_{ν}> = ν|Φ_{ν}>.

Then |Φ_{ν-1}> =
a|Φ_{ν}> is also an eigenvector if (ν - 1) > 0,

|Φ_{ν-n}>
= a^{n}|Φ_{ν}>is also an eigenvector if (ν - n) > 0.

But for some n’ = n + 1
the number ν - n’
will become smaller than zero.

Therefore |Φ_{ν-n}> must be zero.

But
a|Φ_{ν-n}> is
only zero if |Φ_{ν-n}> = |Φ_{0}> or (ν - n) = 0.

This means that n is a non-negative integer.

If |Φ_{n}> is a non-zero eigenvector of N with
integer eigenvalue n, then the |Φ_{n’}>
, n’ = 0, 1, 2, ... are also eigenvectors. (We can generate these eigenvectors by
applying a or a^{†} to |Φ_{n}>
. )

The eigenvectors of N are {|Φ_{n}>
} with eigenvalues {n}, where n is a non-negative integer.

The eigenvectors of H_{s} are {|Φ_{n}> } with eigenvalues {n
+ ½}.

The eigenvectors of H are {|Φ_{n}>
} with eigenvalues {(n + ½)ħω}.

The energy of the harmonic oscillator is quantized.

The ground state energy is
½ħω. The energy levels are evenly spaced.

(e) The ground state of N or H_{s} or H is not degenerate.

Proof: N|Φ_{0}> = 0, a|Φ_{0}> = 0, (2)^{-1/2}(X_{s} + iP_{s})|Φ_{0}>
= 0,

(2)^{-1/2}((mω/ħ)^{1/2}X + i(mωħ)^{-1/2}P)|Φ_{0}>
= 0.

(mωX + iP)|Φ_{0}> = 0. In the {|x>} representation
this becomes

mω<x|X|Φ_{0}> + i<x|P|Φ_{0}> = 0, mωx<x|Φ_{0}>
+ i((ħ/i)∂/∂x)<x|P|Φ_{0}> = 0,

(mωx/ħ)Φ_{0}(x) + ∂/∂x)Φ_{0}(x) = 0,

Φ_{0}(x)
= Cexp(-½mωx^{2}/ħ)

is the solution to this first
order differential equation. It is unique up to a multiplicative constant C.
Therefore
there exists only one normalized Φ_{0}(x) and
|Φ_{0}> with the eigenvalue 0.
The eigenstate
corresponding to the lowest eigenvalue of N or H_{s} or H**
** is not degenerate.

By
induction we can now proof that **all eigenstates of**
N or H_{s} or H** are not degenerate**.

The eigenvectors {|Φ_{n}> } of N or H_{s} or H**
**are a basis for the state space E_{x}
of a particle in a one dimensional problem. The |Φ_{n}> are orthogonal since all eigenvalues are not degenerate. We now want to choose a **normalized**
set {|Φ_{n}> }.

- Assume |Φ
_{0}> is normalized, <Φ_{0}|Φ_{0}> = 1.

The normalized vector |Φ_{1}> is |Φ_{1}> = c_{1}a^{†}|Φ_{0}>.

If <Φ_{1}|Φ_{1}> = 1, then |c_{1}|^{2}<Φ_{0}|aa^{†}|Φ_{0}> = |c_{1}|^{2}<Φ_{0}|a^{†}a + 1|Φ_{0}> = |c_{1}|^{2}= 1.

Therefore |Φ_{1}> = a^{†}|Φ_{0}>. - The normalized vector |Φ
_{n}> is |Φ_{n}> = c_{n}a^{†}|Φ_{n-1}>. - If <Φ
_{n}|Φ_{n}> = 1, then |c_{n}|^{2}<Φ_{n-1}|aa^{†}|Φ_{n-1}> = |c_{n}|^{2}<Φ_{n-1}|a^{†}a + 1|Φ_{n-1}> = n|c_{n}|^{2}= 1.

Therefore |Φ_{n}> = n^{-1/2}a^{†}|Φ_{n-1}>,

|Φ_{n}> = n^{-1/2}(n-1)^{-1/2}(a^{†})^{2}|Φ_{n-2}> = n^{-1/2}(n-1)^{-1/2}... 2^{-1/2}(a^{†})^{n}|Φ_{0}>The set {|Φ

_{n}> = (n!)^{-1/2}(a^{†})^{n}|Φ_{0}>} forms an orthonormal basis for E_{x}.

<Φ_{n}|Φ_{n'}> = δ_{n,n'}, ∑_{n}|Φ_{n}><Φ_{n}| = I.

Let {|Φ_{n}>} denote this orthonormal eigenbasis of the operator H = ½P^{2}/m + ½mω^{2}X^{2}.

H|Φ_{n}> = (n + ½)ħω|Φ_{n}>. - The results of operating with a or a
^{†}on |Φ_{n}> are given by

a|Φ_{n}> = √(n) |Φ_{n-1}>, a^{†}|Φ_{n}> = √(n+1) |Φ_{n+1}>.The matrices representing a and a

^{†}in the {|Φ_{n}>} basis therefore are

The order of the basis vectors is (|Φ_{0}>, |Φ_{1}>, |Φ_{2}>, ...),

<Φ_{n}|a|Φ_{n'}> = √(n') δ_{n,n'-1}, <Φ_{n}|a^{†}}Φ_{n''}> = √(n'+1) δ_{n,n'+1}. - In terms of a and a
^{†}the observables X an P are given by

X = (ħ/(2mω))^{1/2}(a^{†}+ a), P = i(mħω/2)^{1/2}(a^{†}- a).

The matrices representing X and P therefore are

X and P are Hermitian operators.