QHO States killed

Let all the variables in Quantum Harmonic Oscillator

Let

∣n⟩≜∣ϕn⟩∀n∈Z+\ket{n}\triangleq\ket{\phi_n}\quad\forall \quad n\in \mathbb{Z}^+

Note that ϕ(x)\phi(x) is a wave function and, due to the definition, that

ϕ0(x)=⟨x∣0⟩\phi_0(x)=\braket{x|0}

We must find ϕ0(x)\phi_0(x) as we don't know it. We do this by projecting. Project into position basis by x^→x,p^→−iℏddx\hat{x}\rightarrow x, \hat{p}\rightarrow -i\hbar \frac{d}{dx}

12(xα+iαℏ⋅ℏiddx)ϕ0(x)=0\frac{1}{\sqrt{2}}\left(\frac{x}{\alpha} + \frac{i\alpha}{\hbar}\cdot\frac{\hbar}{i}\frac{d}{dx}\right)\phi_0(x) = 0 ⇒12(xα+αddx)ϕ0=0\Rightarrow\quad\frac{1}{\sqrt{2}}\left(\frac{x}{\alpha}+\alpha\frac{d}{dx}\right)\phi_0=0 ⇔dϕ0dx=−xα2ϕ0(x)\Leftrightarrow\quad \frac{d\phi_0}{dx} = -\frac{x}{\alpha^2}\phi_0(x)

This is seperable first-order ODE. Integrating gives

ϕ0(x)=N0 e−x2/2α2,N0=1(πα2)1/4\phi_0(x) = N_0\, e^{-x^2/2\alpha^2}, \qquad N_0 = \frac{1}{(\pi\alpha^2)^{1/4}}

This means there there is a unique solution and therefore no Degeneracy.

Note

  1. Note
a^†∣0⟩=0+1 ∣1⟩=∣1⟩\hat{a}^\dagger\ket{0} = \sqrt{0+1}\,\ket{1} = \ket{1}
  1. Note
a^†∣1⟩=1+1 ∣2⟩=2 ∣2⟩\hat{a}^\dagger\ket{1} = \sqrt{1+1}\,\ket{2} = \sqrt{2}\,\ket{2}
  1. Generally
a^†∣n⟩=n+1 ∣n+1⟩,a^∣n⟩=n ∣n−1⟩\hat{a}^\dagger\ket{n} = \sqrt{n+1}\,\ket{n+1}, \qquad \hat{a}\ket{n} = \sqrt{n}\,\ket{n-1}
  1. Note
∣n⟩=1n! (a^†)n∣0⟩\ket{n} = \frac{1}{\sqrt{n!}}\,(\hat{a}^\dagger)^n\ket{0}
  1. Note
a^†a^∣n⟩=n∣n⟩,⟨m∣n⟩=δmn\hat{a}^\dagger\hat{a}\ket{n} = n\ket{n}, \qquad \braket{m|n} = \delta_{mn}