- #1
terra
- 27
- 2
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.
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.