QHO Wave Function

Let all the variables in Quantum Harmonic Oscillator

Ground State

Recall that ϕn(x)\phi_n(x) is the position-space wave function of the nn-th energy eigenstate.

We start with

a^∣0⟩=0\hat{a}\ket{0}=0

that implies

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

via Wave function definition

ϕn(x)=⟨x∣n⟩\phi_n(x)=\braket{x|n}

so

⟨x∣a^∣0⟩=⟨x∣0\bra{x}\hat{a}\ket{0}=\bra{x}0 ⟨x∣a^∣0⟩=0\bra{x}\hat{a}\ket{0}=0 ⟨x∣[12(x^α+iαℏp^)]∣0⟩=0\bra{x}\left[\frac{1}{\sqrt{2}}\left(\frac{\hat{x}}{\alpha} + \frac{i\alpha}{\hbar}\hat{p}\right)\right]\ket{0} = 0 12[1α⟨x∣x^∣0⟩+iαℏ⟨x∣p^∣0⟩]=0\frac{1}{\sqrt{2}}\left[\frac{1}{\alpha}\bra{x}\hat{x}\ket{0} + \frac{i\alpha}{\hbar}\bra{x}\hat{p}\ket{0}\right] = 0

Note that due to

⟨x∣x^=x⟨x∣\bra{x}\hat{x}=x\bra{x} ⟨x∣p^=−iℏddx⟨x∣\bra{x}\hat{p}=-i\hbar \frac{d}{dx}\bra{x}

Then

12[xα ϕ0(x)+iαℏ⋅(−iℏ)dϕ0dx]=0\frac{1}{\sqrt{2}}\left[\frac{x}{\alpha}\,\phi_0(x) + \frac{i\alpha}{\hbar}\cdot\left(-i\hbar\right)\frac{d\phi_0}{dx}\right] = 0

This ODE evaluates to

12(xα+αddx)ϕ0(x)=0\frac{1}{\sqrt{2}}\left(\frac{x}{\alpha} + \alpha\frac{d}{dx}\right)\phi_0(x) = 0

This is Gaussian as

(xα+αddx)ϕ0=0  ⟹  dϕ0dx=−xα2ϕ0\left(\frac{x}{\alpha} + \alpha \frac{d}{dx}\right)\phi_0 = 0 \implies \frac{d\phi_0}{dx} = -\frac{x}{\alpha^2}\phi_0

This is a separable ODE with an exponential of −x2/2α2-x^2/2\alpha^2 hence it fits within the Gaussian form A e−(x−x0)2/2σ2A\, e^{-(x-x_0)^2 / 2\sigma^2}.

First Excited State

The first excited state

∣1⟩=a^†∣0⟩\ket{1} = \hat{a}^\dagger \ket{0} ϕ1(x)=12(X^α−iαℏP^)ϕ0(x)\phi_1(x) = \tfrac{1}{\sqrt{2}}\left(\tfrac{\hat{X}}{\alpha} - \tfrac{i\alpha}{\hbar}\hat{P}\right)\phi_0(x) =12(xα−αddx)ϕ0= \tfrac{1}{\sqrt{2}}\left(\tfrac{x}{\alpha} - \alpha\tfrac{d}{dx}\right)\phi_0 =N1 x e−x2/2α2,N1= N_1\, x\, e^{-x^2/2\alpha^2}, \quad N_1 =2αN0= \tfrac{\sqrt{2}}{\alpha}N_0