Going from 1+1=2 to Quantum Mechanics Chaimongkol, 2026 A free particle in 1 dimension is described by
H ^ = p ^ 2 2 m \hat{H} = \frac{\hat{p}^2}{2m} H ^ = 2 m p ^ 2 What are eigenvectors of H ^ \hat{H} H ^
p ^ ∣ k ⟩ = ℏ k ∣ k ⟩ \hat{p}\ket{k}=\hbar k\ket{k} p ^ ∣ k ⟩ = ℏ k ∣ k ⟩ then
H ^ ∣ k ⟩ = ( ℏ k ) 2 2 m ∣ k ⟩ \hat{H}\ket{k} = \frac{(\hbar k)^2}{2m}\ket{k} H ^ ∣ k ⟩ = 2 m ( ℏ k ) 2 ∣ k ⟩ Time evolution operator in spectral form yields
U ^ ( t ) = e − i H ^ t / ℏ = ∫ − ∞ ∞ e − i ℏ k 2 2 m t ∣ k ⟩ ⟨ k ∣ d k \hat{U}(t) = e^{-i\hat{H}t/\hbar} = \int_{-\infty}^{\infty} e^{-i\frac{\hbar k^2}{2m}t}\,\ket{k}\bra{k}\,dk U ^ ( t ) = e − i H ^ t /ℏ = ∫ − ∞ ∞ e − i 2 m ℏ k 2 t ∣ k ⟩ ⟨ k ∣ d k Plane wave packet
ψ ( x , 0 ) = e i k 0 x ⟶ ψ ( x , t ) = e i ( k 0 x − ω 0 t ) , ω 0 = ℏ k 0 2 2 m \psi(x,0) = e^{ik_0 x} \longrightarrow \psi(x,t) = e^{i(k_0 x - \omega_0 t)}, \quad \omega_0 = \frac{\hbar k_0^2}{2m} ψ ( x , 0 ) = e i k 0 x ⟶ ψ ( x , t ) = e i ( k 0 x − ω 0 t ) , ω 0 = 2 m ℏ k 0 2 Gaussian wave packet
Recall Gaussian Wave Packet .
This is the superposition of plane waves.
We chose the Gaussian as the wave function ϕ ( x ) \phi(x) ϕ ( x ) Heisenberg Uncertainty Relation .
ψ ( x , 0 ) = N e − ( x − x 0 ) 2 4 σ 2 e i k 0 x \psi(x,0) = N\,e^{-\frac{(x-x_0)^2}{4\sigma^2}}\,e^{ik_0 x} ψ ( x , 0 ) = N e − 4 σ 2 ( x − x 0 ) 2 e i k 0 x Perform a FT
ψ ~ ( k , 0 ) = N ~ e − σ 2 ( k − k 0 ) 2 \tilde\psi(k,0) = \tilde N\,e^{-\sigma^2(k-k_0)^2} ψ ~ ( k , 0 ) = N ~ e − σ 2 ( k − k 0 ) 2 Time-evolve each component by e i ℏ k 2 t / ( 2 m ) e^{i\hbar k^2 t/(2m)} e i ℏ k 2 t / ( 2 m )
ψ ~ ( k , t ) = N ~ e − i ℏ k 2 2 m t e − σ 2 ( k − k 0 ) 2 \tilde\psi(k,t) = \tilde N\,e^{-i\frac{\hbar k^2}{2m}t}\,e^{-\sigma^2(k-k_0)^2} ψ ~ ( k , t ) = N ~ e − i 2 m ℏ k 2 t e − σ 2 ( k − k 0 ) 2 Perform an inverse FT
ψ ( x , t ) = N ~ e − [ x − x c ( t ) ] 2 4 σ c ( t ) 2 e i ( ⋯ ) \psi(x,t) = \tilde N\,e^{-\frac{[x - x_c(t)]^2}{4\sigma_c(t)^2}}\,e^{i(\cdots)} ψ ( x , t ) = N ~ e − 4 σ c ( t ) 2 [ x − x c ( t ) ] 2 e i ( ⋯ ) Where i ( . . . ) i(...) i ( ... )
i ( ⋯ ) = i k 0 x − i ω 0 t + i τ [ x − x c ( t ) ] 2 4 σ 2 ( 1 + τ 2 ) − 1 2 i arctan ( τ ) i(\cdots) = i k_0 x - i\omega_0 t + i\frac{\tau\,[x-x_c(t)]^2}{4\sigma^2(1+\tau^2)} - \tfrac{1}{2}i\arctan(\tau) i ( ⋯ ) = i k 0 x − i ω 0 t + i 4 σ 2 ( 1 + τ 2 ) τ [ x − x c ( t ) ] 2 − 2 1 i arctan ( τ ) and
x c ( t ) = x 0 + ℏ k 0 m t = x 0 + v 0 t , v 0 = ℏ k 0 m x_c(t) = x_0 + \frac{\hbar k_0}{m}t = x_0 + v_0 t, \qquad v_0 = \frac{\hbar k_0}{m} x c ( t ) = x 0 + m ℏ k 0 t = x 0 + v 0 t , v 0 = m ℏ k 0 and
σ c 2 ( t ) = σ 2 + ℏ 2 4 m 2 σ 2 t 2 \sigma_c^2(t) = \sigma^2 + \frac{\hbar^2}{4m^2\sigma^2}t^2 σ c 2 ( t ) = σ 2 + 4 m 2 σ 2 ℏ 2 t 2 Moving Gaussian
If we square the wave function then we can see its probability density
∣ ψ ( x , t ) ∣ 2 = ∣ N ~ ∣ 2 e − ( x − x c ( t ) ) 2 2 σ c ( t ) 2 |\psi(x,t)|^2 = |\tilde N|^2\,e^{-\frac{(x - x_c(t))^2}{2\sigma_c(t)^2}} ∣ ψ ( x , t ) ∣ 2 = ∣ N ~ ∣ 2 e − 2 σ c ( t ) 2 ( x − x c ( t ) ) 2 We can see that the Gaussian probability cloud drifts and broadens. The center x c ( t ) x_c(t) x c ( t ) σ c ( t ) \sigma_c(t) σ c ( t )
This is similar to diffusive brownian motion of a classical particle. A particle buffeted by random collision usually has a spread like:
σ classical ( t ) ∼ D t \sigma_\text{classical}(t) \sim \sqrt{Dt} σ classical ( t ) ∼ D t but a quantum particle spreads at
σ c ( t ) ≈ ℏ t 2 m σ \sigma_c(t) \approx \frac{\hbar t}{2m\sigma} σ c ( t ) ≈ 2 mσ ℏ t The quantum particle spreads linear in t t t t \sqrt{t} t
Wave packet spreads faster if the initial position is more localized/confined -- that means σ \sigma σ
Particle in Box