Probabilities for system of two bosons

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The discussion centers on calculating the probabilities of two indistinguishable bosons in quantum states |0> and |1>. The three possible states of the system are identified as |00>, |11>, and |S> = (1/sqrt2)(|01> + |10>). The user queries whether these states are equally probable, suggesting that if each boson has an equal chance of being in either state, the probability of both being in |0> should be 1/3. However, the general state of the system is expressed as a|00> + b|11> + c|S>, leading to the conclusion that if a = b = c, then their values must be 1/sqrt3, factoring in the normalization condition |a|^2 + |b|^2 + |c|^2 = 1.

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ajl1989
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If there are two indistinguishable bosons that can either be in the |0> or |1> state, what is the probability that both will be in the |0> state? (ie the system will be in the |0>|0> state)

I know there are only three possibilities for the total state of the system: |0>|0>, |1>|1>, and (1/sqrt2)(|0>|1>+|1>|0>), but are these states equally probable? (I'm assuming that a single boson has an equal probability of being in either |0> or |1>) Would the probability of being in |0>|0> just be 1/3? or is it more complicated than that?
 
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Call the three states

|00\rangle,\;|11\rangle,\;\hbox{and}\;|S\rangle \equiv{\textstyle{1\over\sqrt2}}\bigl(|01\rangle+|10\rangle\bigr).

Then the most general state of the two particles is

a|00\rangle+b|11\rangle+c|S\rangle

with |a|^2+|b|^2+|c|^2=1.

What can you deduce about a, b, and c from the requirement that each particle be equally likely to be in |0\rangle or |1\rangle?
 
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Well, I guess if a=b=c, then they'd each have to be 1/sqrt3... unless we also have to take into account the 1/sqrt2 factor in front of |S>?
 

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