QHO Observables

Mean

Let all the variables in Quantum Harmonic Oscillator

⟨n∣x^∣n⟩=α2⟨n∣a^+a^†∣n⟩\braket{n|\hat{x}|n} = \tfrac{\alpha}{\sqrt{2}}\braket{n|\hat{a} + \hat{a}^\dagger|n}

By finding out that

N^(a^∣n⟩)=(n−1)a^∣n⟩\hat{N}(\hat{a}\ket{n}) = (n-1)\hat{a}\ket{n}

Since n−1n-1 is an eigenvalue here, we know that ∣n−1⟩\ket{n-1} must exist

a^∣n⟩=n∣n−1⟩a^†∣n⟩=n+1∣n+1⟩\hat{a}\ket{n}=\sqrt{n}\ket{n-1}\quad \hat{a}^\dagger \ket{n}=\sqrt{n+1}\ket{n+1}

so we can get

a^∣n⟩=cn∣n−1⟩\hat{a}\ket{n}=c_n\ket{n-1}

We find c1c_1 by doing

∥a^∣n⟩∥2=⟨n∣a^†a^∣n⟩=n\|\hat{a}\ket{n}\|^2 = \bra{n}\hat{a}^\dagger \hat{a}\ket{n} = n   ⟹  cn=n\implies c_n = \sqrt{n}

so we can get

⟨n∣x^∣n⟩=α2⟨n∣(n∣n−1⟩+n+1∣n+1⟩)=0\braket{n|\hat{x}|n} = \tfrac{\alpha}{\sqrt{2}}\bra{n}\left(\sqrt{n}\ket{n-1} + \sqrt{n+1}\ket{n+1}\right) = 0

it's equal to zero as ⟨n∣n±1⟩=0\braket{n|n\pm 1}=0

Mean squared

⟨n∣x^2∣n⟩=α22⟨n∣a^2+a^†2+a^a^†+a^†a^∣n⟩\braket{n|\hat{x}^2|n} = \tfrac{\alpha^2}{2}\braket{n|\hat{a}^2 + \hat{a}^{\dagger 2} + \hat{a}\hat{a}^\dagger + \hat{a}^\dagger \hat{a}|n}

We know that

a^2∣n⟩∝∣n−2⟩\hat{a}^2\ket{n}\propto\ket{n-2} a^†2∣n⟩∝∣n+2⟩{\hat{a}^{\dagger}}^2\ket{n}\propto\ket{n+2}

So

⟨n∣x^2∣n⟩=α22⟨n∣a^a^†+a^†a^∣n⟩\bra{n}\hat{x}^2\ket{n} = \tfrac{\alpha^2}{2}\braket{n|\hat{a}\hat{a}^\dagger + \hat{a}^\dagger \hat{a}|n}

We know

[a^,a^†]=1[\hat{a}, \hat{a}^\dagger]=1 ⇒a^a^†=1+a^†a^\Rightarrow\quad\hat{a}\hat{a}^\dagger=1+\hat{a}^\dagger\hat{a}

so

⟨n∣x^2∣n⟩=α22⟨n∣1+2a^†a^∣n⟩\bra{n}\hat{x}^2\ket{n}= \tfrac{\alpha^2}{2}\braket{n|1 + 2\hat{a}^\dagger \hat{a}|n}

via Number Operator

=α22(2n+1)= \tfrac{\alpha^2}{2}(2n+1)

Variance

Δx2(n)=α22(2n+1)\Delta x^2(n) = \tfrac{\alpha^2}{2}(2n+1)