Unformatted text preview: 4/8/11 Z basis funcKons: PHYS 360 Quantum Mechanics Fri Apr 8, 2011 Lecture 33: What if you can't tell the diﬀerence between your two parKcles? X basis funcKons: ⎛1⎞ z χ+ = ⎜ ⎟ , ⎝ 0⎠ ⎛ 0⎞ z χ− = ⎜ ⎟ ⎝1⎠ ⎛ ⎜ (x χ+ ) = ⎜ ⎜ ⎜ ⎝ 1⎞ ⎟ 2⎟ , 1⎟ ⎟ 2⎠ ⎛ ⎜ (x χ− ) = ⎜ ⎜ ⎜ ⎝ 1⎞ ⎟ 2⎟ −1 ⎟ ⎟ 2⎠ TransformaKon: ( χ+ ) =
x 1 2 {χ z + z + χ− , } ( χ− ) =
x 1 2 {χ z + z − χ− } HW #9 due Monday, Apr 11: Ch 4, #29, 34, 49 & Ch 5, # 4 General wave funcKon: ⎛ a + b⎞ x ⎛ a − b⎞ x z z χ = aχ + + bχ − = ⎜ χ+ χ, ⎝ 2⎟ + ⎜ 2⎟ − ⎠ ⎝ ⎠ normalization: a + b = 1
2 2 1 2 More than one parKcle: AddiKon of Angular Momentum Consider two spin
½ parKcles. Here are 4 ways they might be ordered: ↑↑, ↑↓, ↓↑, ↓↓ ↑↑: ↑↓: ↓↑: ↓↓: m =1 m=0 m=0 m = −1 ⎫ ⎪ ⎪ ⎬ but this is wrong! ⎪ ⎪ ⎭ Construct an operator for the total angular momentum: 2 S ≡ S (1) + S ( ) Operate with it on the wave funcKon: S z χ1 χ 2 = S …because there seems to be “extra” state with m=0. (We know m has to change in units of Δm=1. Applying the lowering operator to the top state leads to: ( (1) z +S (2) z ) χ χ = (S χ ) χ
1 2 (1) z 1 2 + χ1 ( S χ 2 )
(2) z = m1χ1 χ 2 + m2 χ1 χ 2 = ( m1 + m2 )χ1 χ 2 This leads to: ↑↑: ↑↓: ↓↑: ↓↓: m =1 m=0 m=0 m = −1 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ ⎧ 11 = ↑↑ ⎪ 1 ⎪ ↑↓ + ↓↑ ⎨ 10 = 2 ⎪ ⎪ 1 − 1 = ↓↓ ⎩ ( ) ⎫ ⎪ ⎪ ⎬ s = 1 (triplet ) ⎪ ⎪ ⎭ ⎧ ⎫ 1 ↑↓ − ↓↑ ⎬ s = 0 ( singlet ) ⎨ 00 = ⎪ ⎪ 2 ⎩ ⎭ ( ) 3 (entangled states) 4 Applying operator algebra to the triplet state yields: ⎛ 3 2 3 2 2 ⎞ S2 1 0 = ⎜ + + 2 ⎟ 1 0 = 2 2 1 0 4 4⎠ ⎝4 More than one parKcle: AddiKon of Angular Momentum Consider two spin
½ parKcles. Construct an operator for the total angular momentum: 2 S ≡ S (1) + S ( ) Using raising & lowering operators etc., deduce the allowed eigenstates: This means that s=1, since S2=s(s+1)ħ2=2ħ2. The singlet state yields s=0: ⎛ 3 2 3 2 3 2 ⎞ S2 0 0 = ⎜ + −2 0 0 =0 4 4⎟ ⎝4 ⎠ ⎧ 11 = ↑↑ ⎪ 1 ⎪ ↑↓ + ↓↑ ⎨ 10 = 2 ⎪ ⎪ 1 − 1 = ↓↓ ⎩ ( ) ⎫ ⎪ ⎪ ⎬ s = 1 (triplet ) ⎪ ⎪ ⎭ 5 ⎧ ⎫ 1 ↑↓ − ↓↑ ⎬ s = 0 ( singlet ) ⎨ 00 = ⎪ ⎪ 2 ⎩ ⎭ ( ) 6 1 4/8/11 3 S 2 = s( s + 1) 2 = 1 ( 1 + 1) 2 = 4 2 22 Sz sm =
11 22 S= 3 0.866 2 =↑ Consider the singlet state: ⎧ ⎫ 1 ↑↓ − ↓↑ ⎬ s = 0 ( singlet ) ⎨ 00 = ⎪ ⎪ 2 ⎩ ⎭ +1/2 S Sy Sz ( ) 1/2 S sm =
1 2 Sx (− 1 ) = ↓ 2 triplet S sm = 11 =↑↑ NoKce this cannot be wriken as the product of two single
parKcle states (e.g. χ1χ2). One parKcle is simultaneously in this entangled state. +1 sm = 10 =
0 1 ↑↓ + ↓↑ 2 ( ) S S 2 = s( s + 1) 2 = 1(1 + 1) 2 = 2 2 S = 2 1.414 Sx 1 Sy S sm = 1(−1) =↓↓
7 8 Note that ConservaKon of Angular Momentum requires that if a parKcle with zero spin (and zero orbital angular momentum) decays into new parKcles, then the total spin of the new parKcles must also be zero. For example: This pion decay also releases energy, so the electron and positron are ejected in opposite direcKons with a lot of kineKc energy. This makes it easy to measure the properKes of just the electron, for example. π 0 → e− + e+ π 0 → e− + e+
1 ↑− ↓+ − ↓− ↑+ 2 Assume the neutral pion was at rest. It also has zero spin. The electron and the positron each are spin
½, but their total spin must be zero. Hence they must be in a spin singlet state: 1 ↑− ↓+ − ↓− ↑+ 2 ( ) ( )
9 10 π 0 → e− + e+
1 ↑− ↓+ − ↓− ↑+ 2 Chapter 5: IdenKcal ParKcles Two
par0cle systems The wave funcKon: ( ) 1. If you measure one parKcle to have +½, must the other be spin
½? 2. Were the parKcles already “born” with these spins? 3. Maybe they were born this way, but QM is missing this part of Nature. Could there be a “hidden” variable? 4. How does one parKcle ﬁnd out the other has been measured? 11 Ψ r1 , r2 , t ( ) The wave equaKon: i ∂Ψ = HΨ ∂t
2 2 2 2 ∇− ∇ + V r1 , r2 , t 2 m1 1 2 m2 2 The Hamiltonian: H= ( )
12 2 4/8/11 The probability “density”: If the potenKal is independent of Kme, apply separaKon of variables and solve for Kme
dependent eigenfuncKon as: Ψ r1 , r2 , t ( ) 2 d 3r1d 3r2 Ψ r1 , r2 , t = ψ r1 , r2 e− iEt / …where E is the total energy of the system (eigenstate). The corresponding Kme
independent equaKon is: ( ) ( ) This is the probability that parKcle 1 is found in volume element d3r1 and parKcle 2 is found in the volume element d3r2 at the Kme t. The normalizaKon condiKon is: −
2 ∫ Ψ (r , r , t ) 1 2 d 3r1d 3r2 = 1
13 2 2 2 ∇1ψ − ∇ 2ψ + V ψ = Eψ 2m 1 2m 2 2 14 Bosons and Fermions Put parKcle 1 in state a and parKcle 2 in state b: ψ + r1 , r2 = A ⎡ψ a ( r1 )ψ b ( r2 ) + ψ b ( r1 )ψ a ( r2 ) ⎤ ⎣ ⎦ ψ + r2 , r1 = A ⎡ψ a ( r2 )ψ b ( r1 ) + ψ b ( r2 )ψ a ( r1 ) ⎤ ⎣ ⎦
+ 1 2 + 2 1 ψ r1 , r2 = ψ a ( r1 )ψ b ( r2 )
Problem: this is not actually possible if you cannot tell the diﬀerence between the two parKcles. For indisKnguishable partcles we must write: ( ) () () ψ (r , r ) = ψ (r , r ) ψ ± r1 , r2 = A ⎡ψ a ( r1 )ψ b ( r2 ) ± ψ b ( r1 )ψ a ( r2 ) ⎤ ⎣ ⎦ ( ) ψ − r1 , r2 = A ⎡ψ a ( r1 )ψ b ( r2 ) − ψ b ( r1 )ψ a ( r2 ) ⎤ ⎣ ⎦ ψ − r2 , r1 = A ⎡ψ a ( r2 )ψ b ( r1 ) − ψ b ( r2 )ψ a ( r1 ) ⎤ ⎣ ⎦
− 1 2 − 2 1 () () ψ ( r , r ) = −ψ ( r , r ) 15 16 ψ ± r1 , r2 = A ⎡ψ a ( r1 )ψ b ( r2 ) ± ψ b ( r1 )ψ a ( r2 ) ⎤ ⎣ ⎦
This version doesn’t assume you can tell them apart. But which sign do we choose? Plus sign = bosons, minus sign = fermions ⎧all particles with integer spin are bosons, and ⎨ ⎩all particles with half integer spin are fermions ( ) Suddenly, we have the Pauli Exclusion Principle: ψ − r1 , r2 = A ⎡ψ a ( r1 )ψ a ( r2 ) − ψ a ( r1 )ψ a ( r2 ) ⎤ = 0 ⎣ ⎦ ( ) Two fermions cannot be in the same state! 17 3 ...
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 Spring '11
 DURBIN,STEPHEN
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