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Transcript
ã Measuring a single spin in an arbitrary direction ` A spin 1/2 quantized in an arbitrary direction given by spherical unit vector r is represented as Sr = Sx Sin@ΘD Cos@ΦD + Sy Sin@ΘD Sin@ΦD + Sz Cos@ΘD where Θ and Φ are the usual spherical angles. In terms of matrices expressed in units of Ñ/2, (Pauli matrices) this becomes Sr = Cos@ΘD Sin@ΘD ã-ä Φ Cos@ΘD Sin@ΘD HCos@ΦD - äSin@ΦDL = Sin@ΘD HCos@ΦD + ä Sin@ΦDL -Cos@ΘD Sin@ΘD ã+ä Φ -Cos@ΘD Diagonalization yields eigenvectors and eigenvalues, +> = - > = Cos@Θ 2D Sin@Θ 2D ã+ä Φ Sin@Θ 2D ã-ä Φ -Cos@Θ 2D for Λ = +1 for Λ = -1 In terms of the "spin up" and "spin down" basis with respect to a z-axis (as defined by a magnetic field), we write Θ Θ + > = CosB F Α > +SinB F ã+ä Φ 2 2 Β > Θ Θ - > = SinB F ã-ä Φ Α > -CosB F 2 2 Β > This says that, if we measure Hwith a Stern -Gerlach analyzerL the spin of a particle in state Θ + >, 2 we will get "spin up" Heigenvalue + 1 or along + zL with probability CosB F and "spin down" 2 Θ 2 Heigenvalue - 1 or along - zL with probability SinB F . Likewise, if we measure the spin of a particle in state - >, 2 Θ 2 Θ 2 we will get "spin up" with probability SinB F and "spin down" with probability CosB F . Note that if we measure an ensemble of many spins, 2 2 the following two possibilities exist : è if the ensemble of electron spins is randomly oriented, the result will be an average over all possible angles Θ which gives a 50-50 split between "up" and "down" since the average of the squares of a sine or cosine is 1/2. è if the ensemble is not randomly oriented, i.e. all elecrons are in the same "prepared" state, then the probability for "up" and "down" outcomes will depend on Θ as shown above, varying from 1 "up" and 0 "down" to 0 "up" and 1 "down". One can prepare such an ensemble by sending a beam of electrons through a "polarizing" Stern-Gerlach and allow only the "up" electrons to emerge from it. Angle Θ is the angle between the magnetic fields of the polarizer and the analyzer. ã Measuring the two-spin singlet state The two-spin singlet state, e.g. the electron-positron pair produced by the decay of a Π0 , has zero angular momentum and is expressed by the normalized state +e > - p > - -e > + p > , 2 where the subscripts "e" and "p" refer to the electron and positron respectively. We assume that the two particles are separated in space and move away from each other so that we can send each through its own Stern -Gerlach apparatus and measure its spin. Let the orientation ` ` Hz axis for each Stern -Gerlach apparatusL be given by unit vectors a and b for the electron and positron respectively. Clearly, the possible outcomes of the measurements are ± 1 for each particle, but what are the probabilities for each ? To answer this question, we note that, Χ > = Θa +e > = CosB 2 Θa F Αe > +SinB Θb + p > = CosB 2 F ãä Φa 2 Θb F Α p > +SinB 2 F ãä Φb Θa Βe >; -e > = SinB Β p >; - p > = SinB F ã-ä Φa 2 Θb 2 F ã-ä Φb Θa Αe > -CosB F Βe > 2 Θb Α p > -CosB Printed by Mathematica for Students F 2 Βp > 2 AboutSpins.nb Then, 1 Θa :B CosB Χ > = 2 2 Θa Θa F Αe > +SinB F ã-ä Φa -SinB 2 Θb F ãä Φa Βe >FBSinB 2 Θa Α p > -CosB 2 Θb Βe >FB CosB F Αe > +CosB Θb F ã-ä Φb 2 2 Θb F Α p > +SinB F Β p >F + B 2 F ãä Φb Β p >F> 2 After we distribute and collect like terms, we get 1 Χ > = 2 Θa :B CosB Θb F - SinB 2 Θa -B CosB Θa F SinB 2 2 Θb F CosB Θa F + SinB 2 Θb F CosB 2 Αe > Α p > +BCosB 2 Θb F SinB 2 Θa F ã-ä IΦa -Φb M F ã-ä Φb 2 F ã-ä IΦa -Φb M F Αe > Θa Β p > +BSinB Θb F SinB 2 Θb F SinB 2 Θa F - SinB 2 Θb F CosB 2 F ãä IΦa -Φb M + CosB Θa 2 F ãä IΦa -Φb M F ãä Φb 2 Θb F CosB 2 FF 2 Βe > Α p >> HaL The probability of measuring both spins "up" is obtained by multiplying the coefficient of Αe > Α p > with its complex conjugate. 1 PΑΑ = Θa B CosB 2 2 2 2 Θb F SinB F ã-ä IΦa -Φb M FB CosB 2 Θa + SinB F 2 Θb F CosB 2 Θa : CosB Θa F - SinB 2 1 = Θb F SinB 2 Θb 2 Θa 2 F CosB 2 Θa F CosB Θb F - SinB 2 Θa - 2 SinB F 2 Θb F SinB 2 F CosB 2 Θa Θb F SinB 2 F ãä IΦa -Φb M F 2 Θb F CosB 2 2 F Cos@Φa - Φb D> (b) The probability of measuring both spins "down" is the same. 1 P ΒΒ = Θa 2 2 Θa : CosB 2 Θa F - SinB 2 1 = Θb F SinB BCosB 2 2 Θb F SinB 2 Θb 2 Θa + SinB F F ãä IΦa -Φb M FBCosB F CosB Θb 2 Θb 2 2 F - SinB 2 Θa - 2 SinB F Θa F SinB 2 F CosB 2 Θa 2 Θa F CosB 2 Θb 2 Θb F SinB 2 F ã-ä IΦa -Φb M F F CosB Θb F CosB 2 2 F Cos@Φa - Φb D> HcL The probability of measuring electron "up" and positron "down" is 1 PΑΒ = Θa B CosB 2 2 1 Θa : CosB = Θb F CosB 2 2 Θb F SinB 2 2 Θb F CosB 2 Θa F + SinB 2 Θa + SinB F 2 F ã-ä IΦa -Φb M FB CosB Θb F CosB 2 2 Θb F SinB 2 Θa Θa + 2 SinB F 2 Θa F + SinB 2 Θa F CosB 2 2 Θb F SinB 2 Θb F SinB F ãä IΦa -Φb M F 2 Θb F CosB 2 2 F Cos@Φa - Φb D> (d) The probability of measuring electron "down" and positron "up" is the same, 1 P ΒΑ = Θa 2 1 2 2 Θa Θb : CosB = 2 Θb F SinB BSinB F CosB 2 F ãä IΦa -Φb M + CosB Θb F CosB 2 2 F 2 Θa Θa + SinB 2 2 Θb F SinB 2 Θa F 2 Θb F SinB FFBSinB 2 Θa + 2 SinB 2 Θa F CosB 2 F ã-ä IΦa -Φb M + CosB Θb FF 2 Θb F CosB 2 Θb F CosB 2 F SinB 2 Θa 2 F Cos@Φa - Φb D> The expressions are simplified with the use of directional cosines. We have unit vectors ` ` ` ` ` ` ` a = Sin@Θa D Cos@Φa D x + Sin@Θa D Sin@Φa D y + Cos@Θa D z = Cos@a1 D x + Cos@a2 D y + Cos@a3 D z ` ` ` ` ` ` ` b = Sin@Θb D Cos@Φb D x + Sin@Θb D Sin@Φb D y + Cos@Θb D z = Cos@b1 D x + Cos@b2 D y + Cos@b3 D z which allows us to identify the direction cosines, Cos@a1 D = Sin@Θa D Cos@Φa D; Cos@a2 D = Sin@Θa D Sin@Φa D; Cos@a3 D = Cos@Θa D Cos@b1 D = Sin@Θb D Cos@Φb D; Cos@b2 D = Sin@Θb D Sin@Φb D; Cos@b3 D = Cos@Θb D Then Printed by Mathematica for Students Βe > Βp > AboutSpins.nb Then 2 SinB Θa Θa F CosB 2 F SinB Θb 2 F CosB 2 Θb 2 1 F Cos@Φa - Φb D = 2 Sin@Θa D Sin@Θb D HCos@Φa D Cos@Φb D + Sin@Φa D Sin@Φb DL 1 = 2 HCos@a1 D Cos@b1 D + Cos@a2 D Cos@b2 DL Now we use the half angle trig formulas for the other terms to find that 2 Θa CosB Θb F SinB 2 = 2 2 Θa F = CosB 2 2 Θb F 1 - CosB 2 F = 2 1 + Cos@Θa D 1 - Cos@Θb D × 2 2 1 + Cos@a3 D 1 - Cos@b3 D 1 × = H1 + Cos@a3 D - Cos@b3 D - Cos@a3 D Cos@b3 DL 2 2 4 Likewise, 2 Θa SinB Θb F CosB 2 2 2 Θa CosB 1 4 Θb F SinB 2 H1 - Cos@a3 D + Cos@b3 D - Cos@a3 D Cos@b3 DL, so that 2 Θa F + SinB 2 1 PΑΑ = 2 Θb F CosB 2 Θa : CosB 2 = 2 F = 2 2 2 Θb F SinB F 2 1 F = 2 H1 - Cos@a3 D Cos@b3 DL. Therefore, Θa + SinB 2 2 2 Θb F CosB F Θa - 2 SinB 2 Θa F CosB 2 Θb F SinB 2 Θb F CosB 2 2 F Cos@Φa - Φb D> 1 1 1 : H1 - Cos@a3 D Cos@b3 DL - HCos@a1 D Cos@b1 D + Cos@a2 D Cos@b2 DL> 2 2 2 1 = 4 1 81 - @Cos@a1 D Cos@b1 D + Cos@a2 D Cos@b2 D + Cos@a3 D Cos@b3 DD< = ` ` J1 - a × bN = P ΒΒ 4 Similarly, we find 1 PΑΒ = P ΒΑ = ` ` J1 + a × bN. To summarize, the outcome probabilities are, 4 1 PΑΑ = P ΒΒ = 4 1 PΑΒ = P ΒΑ = ` ` J1 - a × bN spin product outcome + 1 ` ` J1 + a × bN spin product outcome - 1 4 Note that the sum of all probabilities is unity and that the probability of having "two spins the same" and "two spins opposite" is 1/2, i.e. there is spin - inversion symmetry. Also note that the probability for a given outcome depends only on the (cosine of the) angle between the two detectors and not on any other kind of angle. This makes great sense in retrospect. The spin of the singlet is zero and can only have the direction of the "zero vector" - no diection at all. We denote the expectation HaverageL value of products of outcomes from many measurements by PHa, bL. Then 1 1 ` ` ` ` ` ` ` ` PJa, bN = H+1L I PΑΑ + P ΒΒ M + H-1L IPΑΒ + P ΒΑ M = J1 - a × bN - J1 + a × bN = -a × b 2 2 ã Bell's Inequality If one postulates a "hidden variable" theory designed to enhance an assumed incompleteness of Quantum Mechanics, then the following inequality must always be obeyed. ` ` ` ` ` ` PJa, bN - PIa, cM £ 1 + PJb, cN According to Quantum Mechanics, the inequality will be violated with an appropriate selecton of orientations. For example, with Printed by Mathematica for Students 3 4 AboutSpins.nb According to Quantum Mechanics, the inequality will be violated with an appropriate selecton of orientations. For example, with ` ` ` ` ` ` ` a = Sin@Θa D Cos@Φa D x + Sin@Θa D Sin@Φa D y + Cos@Θa D z = Cos@a1 D x + Cos@a2 D y + Cos@a3 D z ` ` ` ` ` a = z; b = x ; c = 1 ` x+ 1 2 ` z, 2 1 1 ` ` ` ` ` ` ` ` ` ` ` ` PJa, bN = -a × b = 0; PIa, cM = -a × c = ; PJb, cN = -b × c = . Then 2 2 ` ` ` ` PJa, bN - PIa, cM = 1 1 ` ` = 0.707; 1 + PJb, cN = 1 = 0.293 2 2 and Bell's inequality is violated. So the obvious course of action is to do the experiment and see what Nature wants. All experimental indications so far show that Quantum Mechanics is correct and there are no hidden variables. Printed by Mathematica for Students