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≈ Lecture 13: The classical limit € € Phy851/fall 2009 ? Wavepacket Evolution • For a wavepacket in free space, we have already seen that x = x0 + p0 t M p = p0 – So that the center of the wavepacket obeys Newton’s Second Law (with no force): p d x = dt m d p =0 dt • Assuming that: – The wavepacket is very narrow – spreading is negligible on the relevant timescale • Would Classical Mechanics provide a quantitatively accurate description of the wavepacket evolution? – How narrow is narrow enough? • What happens when we add a potential, V(x)? – Will we find that the wavepacket obeys Newton’s Second Law of Motion? Equation of motion for expectation value • How do we find equations of motion for expectation values of observables? – Consider a system described by an arbitrary Hamiltonian, H – Let A be an observable for the system – Question: what is: d A dt ? • Answer: d d A = ψ (t ) A ψ (t ) dt dt d = ψ (t ) dt = ∂A A ψ (t ) + ψ (t ) ψ (t ) + ψ (t ) ∂t i i ∂A ψ (t ) HA ψ (t ) − ψ (t ) AH ψ (t ) + h h ∂t d i ∂A A = − [ A,H ] + dt h ∂t € d A ψ (t ) dt € Example 1: A free particle • Assuming that ∂A =0 ∂t • The basic equation of motion is: d i A = − [ A,H ] dt h P2 • For a free particle, we have: H = 2M € 1 X,H = X,P 2 ] [ ] [ [H,P ] = 0 2M 1 2 2 = XP − P X) ( 2M Very common 1 2 2 = XP − PXP€+ PXP − P X ) trick in QM to ( 2M become familiar 1 with = [ X,P ] P + P [ X,P ]) ( € 2M P = ih € M € P d i P X = − ih = dt h m m d P =0 dt Adding the Potential • Let P2 H= +V (X ) 2M 1 2 X,P + [ X,V (X)] [ X,H ] = [ ] 2M = 0 1 = X,P 2 ] [ 2M P = ih M € € [H,P ] = 1 P 2 ,P ] + [V (X),P ] [ 2M = 0 = −[ P,V (X)] € • How do we handle this commutator? € One possible approach [P,V ( X )] • We want an expression for: • We can instead evaluate: x [ P,V (X)] ψ Theorem: If 〈x|A|ψ〉=〈x|B|ψ〉 is true for any x and € |ψ〉, then it follows that A=B. • Will need to make use of x P ψ = −ih d xψ dx You should think of this as the defining equation for how to handle P in x-basis x [ P,V (X)] ψ = x PV (X) ψ − x V (X)P ψ = −ih x V ′( X ) ψ € – Where d V ′( x) = V ( x) dx • Thus we have: [P,V ( X )]= −ihV ′( X ) You will derive this in the HW Equations of motion for 〈X〉 and 〈P〉 • As long as: P2 H= +V (X ) 2M P d • The we will have: X = dt m – Not just true for a wavepacket • For the momentum we have: € d i P = − [ P,V (X)] dt h i = − − ihV ′( X ) h = − V ′( X ) = F(X ) F(X) is the Force operator F ( x) = − d V ( x) dx • The QM form of the Second Law is Thus: d2 M 2 X = F(X ) dt Difference between classical and QM forms of the 2nd Law of Motion • Classical: d2 M 2 x = F ( x) dt • Quantum: d2 M 2 X = F(X ) dt • Classical Mechanics would be an accurate description of the motion of the center of a wavepacket, defined as x (t ) = X , if: F(X) ≈ F ( X • So that € ) d2 M 2 X = F( X ) dt • This condition is satisfied in the limit as the width of the wavepacket goes to zero • ALWAYS TRUE for a constant (e.g. gravity) or linear force (harmonic oscillator potential), regardless of the shape of the wavefunction Narrow wavepacket • The expectation value of the Force is: F ( X ) = ∫ dx F ( x)ψ ( x) 2 • Let us assume that the force F(x) does not change much over the length scale σ F (x) ψ (x) 2 〈X 〉 = x0 x0 • In this case we can safely pull F(x) out of integral: F ( X ) ≈ F ( x0 ) ∫ dx | ψ ( x) |2 ≈ F ( x0 ) ≈ F( X ) • So CM is a valid description if the wavepacket is narrow enough x A More Precise Formulation F ( X ) = ∫ dx F ( x)ψ ( x) 2 • Expand F(x) around x = 〈X〉: F(X) = F ( X ) + F ′( X )( X − X ) + F ′′( X ) (X − X 2 ) 2 +K • Then take the expectation value: F(X) = F ( X )I + F ′( X ) X− X + F ′′( X ) (X − X 2 I =1 X − X = X − X =0 € (X − X € ) 2 = X 2 − 2X X + X = X2 − X := (ΔX ) € 2 2 2 This is known as the Quantum Mechanical Variance. We will study it more formally later. ) 2 +K • We have: F(X) = F ( X ) + F ′′( X ) (ΔX ) 2 2 +K • For CM to be a good approximation, it is therefore necessary that: € F( X ) ΔX ) ( >> F ′′( X ) 2 2 – The width of the wavepacket squared times the curvature of the force should be small € compared to the force itself • CAUTION: Even if the width of the wavepacket is small enough at one instant to satisfy this inequality, we also need to consider the rate of spreading Example 1: Baseball • Lets consider a baseball accelerating under the force of gravity: – m = 1 kg – let ΔX = 10-10 m • Will then be stable for 30 million years – The force is: F(X ) = − GMm X2 2GMm X3 6GMm F ′′( X ) = − X4 F ′( X ) = • Requirement for CM validity is: F( X ) >> F ′′( X ) (ΔX )2 GMm 6GMm 2 >> ΔX ( ) 2 4 X X € X 2 >> 6(ΔX ) 2 15 −14 • Since X ≈ Re = 6_107 m this gives 10 >> 10 € • So CM€should work pretty well for a baseball Example 2: Hydrogen atom • Consider the very similar problem of an electron orbiting a proton: – Use ground state parameters e2 F( x ) = − 4 πε0 x F ′′( X ) = 6F ( X X € 2 ) 4 • Thus applicability of classical mechanics requires: € X 2 >> 6(ΔX ) 2 • In the ground state we have: Identical result as the baseball, due to 1/r2 law for gravity and coulomb forces X →0 ΔX ~ a0 = 10 −10 m € • Which gives 0 >> 10 −20 – So CM will not be valid for an electron in the hydrogen ground state • Highly excited `wavepacket’ states can be described classically Rydberg States – Because 〈X〉 can get extremely large Rydberg States in Hydrogen • The coulomb potential is High quantum numbers: motion very classical. Kepplerian orbits. RYDBERG STATES Low quantum numbers: motion very quantum mechanical