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xboy
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In Modern Quantum Mechanics page 44-46, Sakurai compares two experiments:
1. Here three observables A,B,C are measured. The probability of obtaining the results a,b and c respectively are then :(|<c|b>|^2)(|<b|a>|^2). Summing up over all values of b we'll find [tex]\Sigma[/tex][tex]_{}b/tex](<c|b>|)^2(|<b|a>|)^2
2. But if we don't measure B,the probability of getting c if we have already got a becomes|<c|a>|^2
So the two probabilities are different.According to Sakurai the two probabilities are equal is [A,B]=0 or [B,C]=0.
I'm trying to prove this but haven't been able to find a good proof yet.All help will be appreciated.
1. Here three observables A,B,C are measured. The probability of obtaining the results a,b and c respectively are then :(|<c|b>|^2)(|<b|a>|^2). Summing up over all values of b we'll find [tex]\Sigma[/tex][tex]_{}b/tex](<c|b>|)^2(|<b|a>|)^2
2. But if we don't measure B,the probability of getting c if we have already got a becomes|<c|a>|^2
So the two probabilities are different.According to Sakurai the two probabilities are equal is [A,B]=0 or [B,C]=0.
I'm trying to prove this but haven't been able to find a good proof yet.All help will be appreciated.