QuantumPhysics.dvi

ways,

|Φ〉 =

1

√

2

|a+;a−〉−

1

√

2

|a−;a+〉

|Φ〉 =

1

√

2

|b+;b−〉−

1

√

2

|b−;b+〉

|Φ〉 =

1

√

2

|c+;c−〉−

1

√

2

|c−;c+〉 (17.44)

The probability amplitude for measuring|Φ〉in a state|a+;b+〉is given by

〈a+;b+|Φ〉=

1

√

2

〈a+;b+|a+;a−〉−

1

√

2

〈a+;b+|a−;a+〉 (17.45)

The last term vanishes because it involves the inner product〈a+|a−〉= 0 for the first particle.
The first term also simplifies by using〈a+|a+〉= 1 for the first particle, and we are left with,

〈a+;b+|Φ〉=

1

√

2

〈b+|a−〉 (17.46)

where the inner product is for the particle 2. Hence, the probability we seek is given by

P(a+,b+) =|〈a+;b+|Φ〉|^2 =

1

2

|〈b+|a−〉|^2 (17.47)

To compute the last inner product, we express the states|bβ〉in terms of the states|aα〉. This
is the same formula as for expressing a general state|n±〉in terms of|z±〉, but now the angleθ
betweennandzbecomes the angleθabbetween the vectorsaandb,

|b+〉 = cos

θab

2

|a+〉+eiφsin

θab

2

|a−〉

|b−〉 = cos

θab

2

|a−〉−e−iφsin

θab

2

|a+〉 (17.48)

Here, we have omitted an overall phase which is immaterial inevaluating the probabilities, and we
find the probability amplitude to be,

〈a+|b−〉=−e−iφsin

θab

2

(17.49)

and the probability,

P(a+,b+) =

1

2

sin^2

θab

2

(17.50)

The phaseφis immaterial in calculating this probability. Note that, in the limit whereb→a,
the probabilityP(a+,b+)→0. This is consistent with the fact that the spin must be perfectly
anti-correlated when measured along the same axis byAandB. Also, we haveP(a+,b+)→ 1 /2 as