Gaussian Wave Packet

Given a Wave Packet, we can play around with ϕ(x)\phi(x) to give us the probability density function of the probability P(a≤x≤b)=∫ab∣ψ(x)∣2 dxP(a \le x \le b) = \int_a^b |\psi(x)|^2\,dx that the particle will be exactly between a≤x≤ba\leq x\leq b.

Let's assume the probability distribution is a Gaussian

ϕ(x)=1(2πσ2)1/4 e−x24σ2\phi(x) = \frac{1}{(2\pi\sigma^2)^{1/4}}\,e^{-\frac{x^2}{4\sigma^2}}

Note that, because of a Gaussian,

E[x^]=0Δx2=E[x^2]=σ2\mathbb{E}[\hat{x}]=0\quad\Delta x^2=\mathbb{E}[\hat{x}^2]=\sigma^2

Fourier Transform of it is

ψ~(k)=N∫−∞∞e−x24σ2+ik0x−ikx dx=N~ e−σ2(k−k0)2\tilde{\psi}(k) = N \int_{-\infty}^{\infty} e^{-\frac{x^2}{4\sigma^2} + ik_0 x - ikx}\,dx = \tilde{N}\,e^{-\sigma^2(k-k_0)^2}

which is also Gaussian.

Note that

E[p^]=ℏk0Δp2=E[p^2]−E[p^]2=ℏ24σ2\mathbb{E}[\hat{p}]=\hbar k_0\quad \Delta p^2=\mathbb{E}[\hat{p}^2]-\mathbb{E}[\hat{p}]^2=\frac{\hbar^2}{4\sigma^2}

∣ψ(x)∣2|\psi(x)|^2 has width σ\sigma and ∣ψ^(k)∣2|\hat{\psi}(k)|^2 has width 1/σ1/\sigma

This causes (Note the Δ\Delta notation means standard deviation)

Δx Δp=ℏ2\Delta x\,\Delta p = \frac{\hbar}{2}

This means that Gaussians are minimum uncertainty wave packets because they saturate the Heisenberg uncertainty relation. This means that they are the closest quantum analog to a classical particle -> they have the most defined E[x]\mathbb{E}[x] and E[p]\mathbb{E}[p]