QuantumPhysics.dvi

Next, we show that the operatorH is bounded from below by^12 ̄hω. Indeed, take any

normalized state|ψ〉, and compute the expectation value ofH,

〈ψ|H|ψ〉=

1

2

̄hω+ ̄hω〈ψ|a†a|ψ〉=

1

2

̄hω+ ̄hω||aψ||^2 ≥

1

2

̄hω (5.75)

Hence all eigenvalues ofHmust be greater than or equal to^12 hω ̄.

5.6.2 Constructing the spectrum

Now here is the crucial step that will allow us to solve this system without ever solving a

differential equation. Consider a normalized state|n〉with energyEn. Sincealowers the

energy by ̄hω, it follows thatak|n〉should have energyEn−k ̄hω. But for sufficiently large

k, this energy would dip below the bound^12 hω ̄ , which is not possible. Thus, it must be that

for some values ofk >0, the stateak|n〉is actually zero, and the spectrum must contain a

state which is annihilated bya. We shall denote this state by| 0 〉,

a| 0 〉= 0 ⇒ E 0 =

1

2

̄hω (5.76)

The state| 0 〉is the ground state; it is the unique state of lowest energyE 0 , which actually

saturates the lower bound on the expectation value of the Hamiltonian. All other states are

then obtained by multiple application of the raising operatora†, and we find,

|n〉∼

(

a†

)n

| 0 〉 ⇒ En= ̄hω

(

n+

1

2

)

(5.77)

which is indeed the entire harmonic oscillator spectrum forn≥0.

Since the states|En〉for differentnbelong to different eigenvalues of the self-adjoint op-

eratorH, they are automatically orthogonal. We shall also assume that theyare normalized,

〈m|n〉=δm,n. The matrix elements of the operatorsaanda†are as follows,

a†|n〉 =

√

n+ 1|n+ 1〉

a|n〉 =

√

n|n− 1 〉 (5.78)

Or equivalently, we have〈m|a|n〉=

√

nδm,n− 1.

5.6.3 Harmonic oscillator wave functions

The wave functionψn(x) =〈x;X|n〉for a state|n〉may also be computed simply. We start

with the ground state, which satisfies

〈x;X|a| 0 〉=

1

√

2 M ̄hω

〈x;X|(MωX+iP)| 0 〉= 0 (5.79)