Cohen-Tannoudji on mutually exclusive (?) events

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The discussion centers on the compatibility of observables in quantum mechanics, specifically referencing Cohen-Tannoudji's work. It examines the operators A and B, which commute as indicated by the relation [A,B]=0, and the implications for calculating probabilities of finding certain states. The probability for observing a specific value a_1 is derived from the amplitudes of the states |a_i b_j⟩, leading to the conclusion that the amplitudes are not mutually exclusive when considering quantum probabilities. The conversation highlights the distinction between classical and quantum probability calculations, emphasizing the role of interference terms in quantum mechanics.

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terra
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I was looking at what Cohen-Tannudji has to say on compatibility of observables.
Assumptions: ## A,B## are operators such that ##[A,B]=0 ## and we denote ## |a_i \,b_j\rangle## to be states for which ##A | a_i \, b_j \rangle= a_i | a_i \, b_j \rangle##, ##B | a_i \, b_j \rangle= b_j | a_i \, b_j \rangle##.
We start with the state ##|\psi \rangle= \sum_{i,j} c_{i,j} | a_i \, b_j \rangle ##.
The whole discussion starts with the following:
"The probability for finding ## a_1 ## is ## P(a_1)= \sum_{j} |c_{1,j}|^2 ##." (Page 232 in a 1977 edition.)
Have I forgotten something fundamental? I thought that the amplitudes ## \langle a_1 \, b_{j'} | \psi \rangle ## and ## \langle a_1 \, b_j | \psi \rangle ## are mutually exclusive for ##j' \neq j ##, so that according to quantum rules for probability
$$P(a_1)= \big| \sum_j \langle a_1 \, b_{j} | \psi \rangle \big|^2= \sum_j \sum_{j'} \langle a_1 \, b_j | \psi \rangle \langle \psi | a_1 \, b_{j'} \rangle. $$
I see no reason as to why ##j'= j ## should hold.
My apologies for the slightly dull question, but I'm a bit lost.
 
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I'm completely ignorant of the physics here. However when dealing with probabilities, mutually exclusive means the probability of both happening is 0.
 
mathman said:
I'm completely ignorant of the physics here. However when dealing with probabilities, mutually exclusive means the probability of both happening is 0.
Thanks for the reply, yeah.
Let ## a,b## be exclusive events. For classical physical things ## P(a\mathrm{ \, or \, }b)= P(a) + P(b)##. In quantum physics, however, we have ## A(a\mathrm{ \, or \, }b)= A(a) + A(b)## where ##A \in \mathbb{C}## is a 'probability amplitude' so that ## P(x)= A(X) A^*(X)##, star denoting a complex conjugate, for some event ## X##. This definition will bring so called interference terms when compared with the classical case (they disappear in the classical limit).
In my case, ## | a_i \, b_j \rangle ## are vectors.. some physical states, in fact, that have well-defined (=certain) values for some observables ##A,B## (so, a physical state always has some value or a distribution of values for both). We have different states for different ## i## and ## j##. A term ##\langle a_1 \, b_j | \psi \rangle## is actually just such an amplitude for the state ## | \psi \rangle## having the value ## a_1 ## for ## A## and the value ## b_j## for ## B##. I'm trying to determine the total probability to find that the value of ## A## for ## |\psi \rangle## is ## a_1##. As I see it, ##|\psi \rangle## can have that value while having ##b_1,b_2,b_3,... ## for ## B##.
 

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