Lecture31.pptx

Lecture31.pptx - 4/4/11 PHYS 360 Quantum Mechanics...

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Unformatted text preview: 4/4/11 PHYS 360 Quantum Mechanics Mon Apr 4, 2011 Lecture 31: How do you build a Stern ­Gerlach apparatus? 4.4.1 Spin ½ The electron has spin one ­half, and is the most important example. There are only two eigenstates: sm = sm = 11 22 =↑ 1 2 (− 1 ) = ↓ 2 HW #9 due Monday, Apr 11: Ch 4, #28, 29, 34, 49 & Ch 5, # 4 These are the only two basis funcUons in the enUre Hilbert space for spin ½. 1 2 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 eigenfuncUons 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 equaUon 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 = 3 3 2 ⎛ 1 0⎞ ⎜ ⎟ 4 ⎝ 0 1⎠ 4 Using the same basis funcUons, 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 ≡ ⎜ ⎟ , σ y ≡ ⎝ i 0⎠ , σ z ≡ ⎝ 0 − 1 ⎠ ⎜ ⎟ ⎜ ⎟ ⎝ 1 0⎠ Sx and Sz can be determined: Sx = ⎛ 0 1⎞ ⎛0 − i⎞ , Sy = ⎜ 2 ⎜ 1 0⎟ 2 ⎝ i 0⎟ ⎝ ⎠ ⎠ 5 6 1 4/4/11 Eigenvalue equaUon: 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 funcUons: ψ n = ∑ dmφ m = d1φ1 + d2φ2 + ... The operator: The eigenvalue equaUon: The (z) basis funcUons Sx = ⎛ 0 1 ⎞ 2⎜ 1 0 ⎟ ⎝ ⎠ The operator: The basis funcUons 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 eigenfuncUons: ⎛ ⎜ (x χ+ ) = ⎜ ⎜ ⎜ ⎝ 7 1⎞ ⎟ 2⎟ = 1⎟ ⎟ 2⎠ 1 2 {χ + + χ− , } ⎛ ⎜ (x χ− ) = ⎜ ⎜ ⎜ ⎝ 1⎞ ⎟ 2⎟ = −1 ⎟ ⎟ 2⎠ 1 2 {χ + − χ− } 8 Now we can write any arbitrary spin state in terms of the z eigenfuncUons or the x eigenfuncUons: ⎛ a + b⎞ x ⎛ a − b⎞ x χ = aχ + + bχ − = ⎜ ⎟ χ +⎜ ⎟χ , ⎝ 2⎠ + ⎝ 2⎠ − normalization: a + b = 1 2 2 4.4.2 Electron in a MagneUc Field Classically, we describe a current loop as having a magneUc dipole moment. Microscopically, this can be related to the angular momentum of the system. For the case of a “spinning” electron with angular momentum S, we have: Let a=1, b=0: ⎛1⎞ x ⎛1⎞ x χ = χ+ = ⎜ χ+ χ ⎝ 2⎟ + ⎜ 2⎟ − ⎠ ⎝ ⎠ µ =γS 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 StaUsUcal InterpretaUon of QM) 9 The raUo between spin and magneUc moment is called the gyromagneUc raUo. 10 A magneUc field exerts a torque on a magneUc moment, so it takes work to rotate the moment, hence the energy of the system depends on the angle: The Stern ­Gerlach experiment: We have noted that a magneUc field exterts a torque on a spin, which affects its energy. If the B field is non ­ uniform, however, it also exerts a force: F = ∇ µ •B H = − µ ⋅ B = − µ B cos θ This can also be wriken as: ( ) ) H = −γ B ⋅ S Consider: ˆ ˆ B( x , y , z ) = −α xi + B0 + α z k ( This causes a net force in the z direcUon: Fz = γα S z 11 12 2 4/4/11 Spin up 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 Spin down Sy -1/2 S sm = 1 2 If you want to prepare a beam of spin ­½ parUcles to be in a certain eigenstate, run them through a Stern ­ Gerlach apparatus and select one of the two beams. Sx (− 1 ) = ↓ 2 13 14 Z basis funcUons: ⎛1⎞ z χ+ = ⎜ ⎟ , ⎝ 0⎠ ⎛ 0⎞ z χ− = ⎜ ⎟ ⎝1⎠ X basis funcUons: ⎛ ⎜ (x χ+ ) = ⎜ ⎜ ⎜ ⎝ 1⎞ ⎟ 2⎟ , 1⎟ ⎟ 2⎠ ⎛ ⎜ (x χ− ) = ⎜ ⎜ ⎜ ⎝ 1⎞ ⎟ 2⎟ −1 ⎟ ⎟ 2⎠ An unpolarized beam of spin ½ parUcles is travelling in the +y direcUon. A Stern ­Gerlach apparatus is aligned to split this beam into spin up (+½) and spin down ( ­½) beams; we call this a Z ­Spliker. Since the incident beam was unpolarized, the two new beams have equal numbers. The spin down beam is blocked, and the spin up beam enters an X ­Spliker (a Stern ­Gerlach apparatus aligned in the x ­ direcUon), which produces two new beams. 1.  What are the possible spin components of these two beams? 2.  Do they have equal numbers of parUcles, or does one beam have all the parUcles? 15 16 TransformaUon: ( χ+ ) = x 1 2 { z z χ+ + χ− , } ( χ− ) = x 1 2 { z z χ+ − χ− } General wave funcUon: a + b⎞ x ⎛ a − b⎞ x z z χ = aχ + + bχ − = ⎜ χ+ χ, ⎝ 2⎟ + ⎜ 2⎟ − ⎠ ⎝ ⎠ normalization: a + b = 1 2 2 One of these new beams is blocked, and the other proceeds to another Z ­Spliker. X-Splitter 3. How many output beams are there? Sz Sx +z Z-Splitter +y +x Sz 4. What are the spin components of these beams? 5. Are there equal numbers in each beam? Sz X-Splitter Sz Sx Z-Splitter +y +x 17 18 Z-Splitter +z Z-Splitter 3 4/4/11 Let’s consider the case where we have two different HermiUan operators that “span” the same space. To be specific, let the first one be the Hamiltonian: Let the second operator be P: Hψ n = Enψ n Ψ (r, 0 ) = ∑ cnψ n =c1ψ 1 + c2ψ 2 + ... (Note: a complete set) ΨHΨ = Pφ m = pmφ m Φ(r, 0 ) = ∑ dmφ m =d1φ1 + d2φ2 + ... (Note: a complete set) ΦPΦ = ∑ cn*ψ n* H ∑ cnψ n ∑d φ ** mm P ∑d φ mm 2 = d12 p1 + d2 p2 + ... 2 = c12 E1 + c2 E2 + ... A measurement would yield E1 with probability |c1|2, etc… Measurement would yield p1 with probability |d1|2, etc… 19 20 Now, what if we know the state Ψ = ∑ cnψ n …and we do a measurement of P. What do you get? We know that only the eigenvalues p1, p2, p3… are possible, but with what probabiliUes? We must express the original wave funcUon in terms of the P eigenfuncUons. To simplify, start with a determinate state: Ψ = cnψ n Now, expand this in terms of the other complete set: ψ n = ∑ dmφ m = d1φ1 + d2φ2 + ... Pψ n = P(d1φ1 + d2φ2 + ...) = d1 p1φ1 + d2 p2φ2 + ... 2 2 P = d12 p1 + d2 p2 + d3 p3 + ... Measurement will yield p1 with probability |d1|2, p2 with probability |d2|2, etc… 21 22 Back to the spin ½ system…. A dialogue: 1.  A parUcle 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 parUcle’s spin angular momentum?” However, it is not in a determinate state of Sx. 3.  You say, “Well, half the Ume you would measure +ħ/2 and half the Ume  ­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. 23 24 4 ...
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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.

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