Download Measuring a single spin in an arbitrary direction A spin 1/2

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
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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
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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
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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.
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