Quantum Mechanics for Mathematicians

Such an operator satisfies the commutation relations

[N,a] = [a†a,a] =a†[a,a] + [a†,a]a=−a

and
[N,a†] =a†
If|c〉is a normalized eigenvector ofNwith eigenvaluec, one has

c=〈c|a†a|c〉=|a|c〉|^2 ≥ 0

so eigenvalues ofNmust be non-negative. Using the commutation relations of
N,a,a†gives

Na|c〉= ([N,a] +aN)|c〉=a(N−1)|c〉= (c−1)a|c〉

and
Na†|c〉= ([N,a†] +a†N)|c〉=a†(N+ 1)|c〉= (c+ 1)a†|c〉

This shows thata|c〉will have eigenvaluec−1 forN, and a normalized eigen-
function forNwill be

|c− 1 〉=

1

√

c

a|c〉

Similarly, since

|a†|c〉|^2 =〈c|aa†|c〉=〈c|(N+ 1)|c〉=c+ 1

we have

|c+ 1〉=

1

√

c+ 1

a†|c〉

If we start off with a state| 0 〉that is a non-zero eigenvector forNwith eigenvalue
0, we see that the eigenvalues ofNwill be the non-negative integers, and for
this reasonNis called the “number operator”.
We can find such a state by looking for solutions to

a| 0 〉= 0

| 0 〉will have energy eigenvalue^12 ~ω, and this will be the lowest energy eigenstate.
Acting bya†n-times on| 0 〉gives states with energy eigenvalue (n+^12 )~ω. The
equation for| 0 〉is

a| 0 〉=

(√

mω

2 ~

Q+i

√

1

2 mω~

P

)

ψ 0 (q) =

√

mω

2 ~

(

q+

~

mω

d

dq

)

ψ 0 (q) = 0

One can check that this equation has a single normalized solution

ψ 0 (q) = (

mω

π~

)

(^14)
e−
mω 2 ~q 2
which is the lowest-energy eigenfunction.