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Unformatted text preview: 4/1/11 PHYS 360 Quantum Mechanics Fri Apr 1, 2011 Lecture 30: If a spin points in the z direcEon, what do you get in the x direcEon? We assume that spin angular momentum follows all of the rules that we found for the algebraic (operator) theory of angular momentum: ⎡ Lx , Ly ⎤ = iLz ; [ Ly , Lz ] = iLx ; [Lz , Lx ] = iLy ⎣ ⎦
⎡ L2 , Lx ⎤ = 0, ⎡ L2 , Ly ⎤ = 0 , ⎡ L2 , Lz ⎤ = 0 ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ L2 f l m = 2 l l + 1 f l m ; Lz f l m ; = mf l m ( ) HW #8 due Monday: Ch 4, #24, 26, 27, 28 Last year’s ﬁnal exam is posted. 1 l = 0, 1 / 2, 1, 3 / 2, ...; m = − l , − l + 1, ... , l − 1, l Note: the eigenfuncEons are actual funcEons of the coordinates (xyz) or (rθϕ). 2 ⎡ S x , S y ⎤ = iS z , ⎡ S y , S z ⎤ = iS x , ⎡ S z , S x ⎤ = iS y ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ 4.4.1 Spin ½ The electron has spin one
half, and is the most important example. There are only two eigenstates: sm =
11 22 S 2 s m = s s + 1 2 s m ; S z s m = m s m ( ) =↑ S ± sm = s s + 1 − m m ± 1 s m ± 1
13 s = 0, ,1, , ...; m = − s, − s + 1, ..., s − 1, s 22 ( ) ( )( ) sm = 1 2 (− 1 ) = ↓ 2 These are the only two basis funcEons in the enEre Hilbert space for spin ½. Note: the eigenfuncEons are NOT funcEons of the coordinates (xyz) or (rθϕ)! 3 4 3 S 2 = s( s + 1) 2 = 1 ( 1 + 1) 2 = 4 2 22 Sz sm =
11 22 S= 3 0.866 2 =↑ +1/2 S We write the eigenfuncEons as a 2
element column matrix, called a spinor: ⎛ a⎞ χ = ⎜ ⎟ = a χ + + bχ − ⎝ b⎠
⎛1⎞ χ+ = ⎜ ⎟ ⎝ 0⎠ ⎛ 0⎞ χ− = ⎜ ⎟ ⎝1⎠ Sy 1/2 The eigenvalue equaEon for S2 is: S2χ+ = 32 3 χ + and S 2 χ − = 2 χ − 4 4 S sm =
1 2 Sx (− 1 ) = ↓ 2 Hence we can solve for S2: S2 =
5 3 2 ⎛ 1 0⎞ ⎜ ⎟ 4 ⎝ 0 1⎠
6 1 4/1/11 Using the same basis funcEons, consider Sz: Solve for Sz: Sz χ+ = χ , S χ = − χ− 2+ z− 2 These can be represented with ⎛ ⎞ S = ⎜ ⎟σ ⎝ 2⎠ Sz = ⎛ 1 0⎞ ⎜ ⎟ 2 ⎝ 0 − 1⎠ …using the Pauli spin matrices: By introducing raising & lowering operators… ⎛ 0 1⎞ ⎛ 0 0⎞ S+ χ − = ⎜ , S χ = ⎜ ⎝ 0 0⎟ − + ⎠ ⎝1 0 ⎟ ⎠ ⎛ 0 1⎞ ⎛0 − i⎞ ⎛ 1 0⎞ σx ≡ ⎜ , σy ≡ ⎜ , σz ≡ ⎜ ⎝ 1 0⎟ ⎠ ⎝ i 0⎟ ⎠ ⎝ 0 − 1⎟ ⎠ Sx and Sz can be determined: Sx = ⎛ 0 1⎞ ⎛0 − i⎞ , Sy = ⎜ ⎜ ⎟ 2 ⎝ 1 0⎟ 2 ⎝ i 0⎠ ⎠ 7 8 Eigenvalue equaEon: Sz χ± = ± χ± 2 ⎛ 1 0⎞ ⎜ ⎟ 2 ⎝ 0 − 1⎠
⎛ 0⎞ χ− = ⎜ ⎟ ⎝1⎠ Now, what if we have the Sz eigenstate, but we measure Sx. We need to express the eigenstates of the Sx operator in terms of our basis funcEons: The operator: The eigenvalue equaEon: The (z) basis funcEons
Sx = ⎛ 0 1 ⎞ 2⎜ 1 0 ⎟ ⎝ ⎠ The operator: The basis funcEons The eigenvalues: ⎛ 1 0⎞ ⎛ 1 ⎞ = + χ+ 2 ⎜ 0 − 1⎟ ⎜ 0 ⎟ 2 ⎝ ⎠⎝ ⎠ Sz = ⎛1⎞ χ+ = ⎜ ⎟ ⎝ 0⎠ x x Sx χ ± = ± χ ± 2
⎛1⎞ χ+ = ⎜ ⎟ ⎝ 0⎠ ⎛ 0⎞ χ− = ⎜ ⎟ ⎝1⎠ Sz χ+ = Sz χ− = ⎛ 1 0⎞ ⎛ 0 ⎞ =  χ− 2 ⎜ 0 − 1⎟ ⎜ 1 ⎟ 2 ⎝ ⎠⎝ ⎠ The eigenfuncEons: ⎛ ⎜ (x χ+ ) = ⎜ ⎜ ⎜ ⎝
9 1⎞ ⎟ 2⎟ = 1⎟ ⎟ 2⎠ 1 2 {χ + + χ− , } ⎛ ⎜ (x χ− ) = ⎜ ⎜ ⎜ ⎝ 1⎞ ⎟ 2⎟ = −1 ⎟ ⎟ 2⎠ 1 2 {χ + − χ− }
10 Now we can write any arbitrary spin state in terms of the z eigenfuncEons or the x eigenfuncEons: ⎛ a + b⎞ x ⎛ a − b⎞ x χ = aχ + + bχ − = ⎜ χ+ χ, ⎝ 2⎟ + ⎜ 2⎟ − ⎠ ⎝ ⎠ normalization: a + b = 1
2 2 A dialogue: 1. A parEcle is prepared in a state χ+. This is a determinate state, meaning that a measurement of Sz will deﬁnitely yield the value +ħ/2. 2. You are asked: “What is the x
component of the parEcle’s spin angular momentum?” However, it is not in a determinate state of Sx. 3. You say, “Well, half the Eme you would measure +ħ/2 and half the Eme
ħ/2.” 4. “I thought you knew what state you had, χ+. Here, I’ll do a measurement. Look, it turns out that Sx was actually +ħ/2. So now we know both Sz and Sx. 5. But then you say, “Actually, now I don’t know what Sz is. Your measurement put it in a determinate state of Sx.” When you know Sz, for example, it does not have a well
deﬁned Sx. 11 12 Let a=1, b=0: ⎛1⎞ x ⎛1⎞ x χ = χ+ = ⎜ ⎟ χ +⎜ ⎟χ ⎝ 2⎠ + ⎝ 2⎠ − What are the allowed results of an Sz measurement? Only +ħ/2. What are the allowed results of an Sx measurement? Either +ħ/2 or
ħ/2. (Generalized StaEsEcal InterpretaEon of QM) 2 ...
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This note was uploaded on 04/23/2011 for the course PHYS 360 taught by Professor Durbin,stephen during the Spring '11 term at Purdue UniversityWest Lafayette.
 Spring '11
 DURBIN,STEPHEN
 Momentum

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