Download Heisenberg's Uncertainty Principle - Quantum Mechanics | PHYS 410 and more Study notes Physics in PDF only on Docsity! Lecture 5 Outline - Uncertaintly • de Broglie’s formula [1.6] • Heisenberg’s uncertainty principle [1.6] • 1-D Schrödinger eqn stationary states [2.1] Ehrenfest’s Theorem • Quantum mechanical expectations obey classical laws: • eg. velocity: 〈v〉 = d〈x〉/dt • eg. momentum: 〈p〉 = m d〈x〉/dt • and finally there is also the equivalent of Newtonian force: 〈 − ∂V ∂x 〉 = d〈p〉 dt ≡ m d2〈x〉 dt2 V(x,t) a b 2(x,t)Ψ Wavefunction Standard Deviations (σj) • Position variance: σ2x = 〈x 2〉 − 〈x〉2 • The general form for variance... σ2Q = ∫ Ψ∗ ( Q̂− 〈Q〉 )2 Ψ dx ≡ 〈Q2〉 − 〈Q〉2 • Thus momentum variance is σ2p = ∫ Ψ∗ ( p̂− 〈p〉 )2 Ψ dx = 〈p2〉 − 〈p〉2 • ie. need 4 integrals to find σx and σp ... • (in the process also gives us 〈T 〉 = 〈p2〉/2m) Wavefunction o,, 0)
e Example problem: Wavepacket and Gaussian
Integrations...
Time-dependent Schrödinger eqn • following de Broglie p = ~k and E = ~ω • and we also have the relations E = p2 2m = ~ 2k2 2m = ~ω so ω = ~k2 2m • assuming that a particle has well defined wavelength Ψ(x, t) = Aei(kx−ωt) = Aei(px−Et)/~ • from which we can take two derivatives... ∂Ψ ∂t = − iE ~ Ψ and ∂2Ψ ∂x2 = − p2 ~2 Ψ • and then we can use the E = p 2 2m relation i~ ∂ ∂t Ψ(x, t) = − ~ 2 2m ∂2 ∂x2 Ψ(x, t) Stationary States 3/3 dϕ dt = − iE ~ ϕ • Since we have the form df/dg ∝ f , then ϕ(t) = e− iE ~ t . Ψ(x, t) = ψ(x) ϕ(t) = ψ(x) e−iEt/~ . • Thus we have a stationary wavefunction: |Ψ(x, t)|2 = Ψ∗Ψ = ψ∗(x) eiEt/~ ψ(x) e−iEt/~ = |ψ(x)|2 . 〈Q(x, p)〉 = ∫ ψ∗ eiEt/~ Q̂ ( x,−i~ ∂ ∂x ) ψ e−iEt/~ dx • Thus for all expectation values d〈Q〉/dt = 0. Time-independent Schrödinger Eqn • is just the one ordinary differential equation... − ~ 2 2m d2ψ dx2 + V (x) ψ = Eψ . • but from classical mechanics, the total energy (kinetic + potential) comes from the Hamiltonian: H(x, p) = p2 2m + V (x) . • The quantum mechanical Hamiltonian operator is: Ĥ = − ~ 2 2m ∂2 ∂x2 + V (x) , • time-independent Schrödinger Equation Ĥψ = Eψ. Hamiltonian Operator Ĥ = − ~ 2 2m ∂2 ∂x2 + V (x) , • To calculate the total energy we evaluate: 〈H〉 = ∫ Ψ∗(x, t) Ĥ Ψ(x, t) dx = ∫ ψ∗(x) Ĥ ψ(x) dx = ∫ ψ∗(x) E ψ(x) dx = E ∫ |ψ(x)|2 dx = E
