Fock states and a more mature development of the algebraic solution

A more canonical approach to the previous section is as follows. We write the Hamiltonian as

$$H = \frac{1}{2m}P^2 + \frac{m\omega^2}{2}Q^2$$ (4.6)

This can be written dimensionlessly with the introduction of substitute operators,

$$q=\sqrt{\frac{m\omega}{\hbar}}Q$$ (4.7)

$$p=\sqrt{\frac{1}{m\hbar \omega}}P$$ (4.8)

$$H = \frac{\hbar \omega}{2}(p^2 + q^2)$$ (4.9)

In parallel with the previous section, we consider the operators,

$$a=\frac{q+ip}{\sqrt{2}}$$ (4.10)

$$a^{\dagger}=\frac{q - ip}{\sqrt{2}}$$ (4.11)

$$H = \frac{\hbar \omega}{2}(aa^{\dagger}+a^{\dagger}a)$$ (4.12)

From basic quantum mechanics we know that $[Q,P]=i\hbar$, thus $[q,p]=i$, and so $[a,a^{\dagger}]=1$, whereby

$$H = \frac{1}{2}\hbar \omega(aa^{\dagger} + a^{\dagger}a)=\hbar\omega(aa^{\dagger} - \frac{1}{2}) = \hbar \omega (a^{\dagger}a + \frac{1}{2})$$ (4.13)

The Hamiltonian thus shares its eigenvalue spectrum with $N=a^{\dagger}a$, and

$$[N,a] = a^{\dagger}aa - aa^{\dagger}a=[a^{\dagger},a]a=-a$$ (4.14)

$$[N,a^{\dagger}]= a^{\dagger}aa^{\dagger} - a^{\dagger}a^{\dagger}a=a^{\dagger}[a,a^{\dagger}]=a^{\dagger}$$ (4.15)

Thus we find,

$$N\vert n\rangle=n\vert n\rangle \ \Longrightarrow Na\vert n\rangle =a(N-1)\vert n\rangle = (n-1)a\vert n\rangle$$ (4.16)

$$Na^b\vert n\rangle=(n-b)a^b\vert n\rangle$$ (4.17)

Thus $a^b\vert n\rangle$ is an eigenvector of N with eigenvalues $n-b$. We require these eigenvalues be positive for the following reason. An operator is required to have a real expectation value since the expectation value is what we would measure, i.e. we require that for an operator R,

$$\langle\psi\vert R\vert\psi\rangle=\langle\psi\vert R\vert\psi\rangle^*$$

This implies R is Hermitian, and given $R\vert\psi\rangle = r\vert\psi\rangle$ then we have,

$$\langle\psi\vert R\vert\psi\rangle=r\langle\psi\vert\psi\rangle=\langle\psi\vert R\vert\psi\rangle^*=r^*\langle\psi\vert\psi\rangle$$

Since the eigenvalues are real, then the squared norm of these eigenvectors follow the form:

$$(\langle n\vert a^{\dagger})(a\vert n\rangle)=\langle n \vert N \vert n \rangle = n\langle n \vert n \rangle \geq 0$$ (4.18)

and so we conclude that there must be a $n-b=0$ cutoff

As well,

$$Na^{\dagger}\vert n\rangle = a^{\dagger}(N+1)\vert n\rangle = (n+1)a^{\dagger}\vert n\rangle$$ (4.19)

$$(\langle n \vert a)(a^{\dagger}\vert n\rangle)=(n+1)\langle n \vert n \rangle$$ (4.20)

We compare equation 22 with the fact that $N\vert n+1\rangle = (n+1)\vert n+1\rangle$ and conclude that

$$a^{\dagger}\vert n\rangle = \beth \v... ...le)=(n+1) \ \therefore \ a^{\dagger}\vert n\rangle = \sqrt{n+1}\vert n+1\rangle$$ (4.21)

A similar argument yields $a\vert n\rangle = \sqrt{n}\vert n-1\rangle$. It follows that

$$\vert n\rangle=\frac{a^{\dagger}^n\vert\rangle}{\sqrt{n!}}$$ (4.22)

These are the so-called Fock states.