Questioning Quantum Rules: Is A(A or B) = A(A) + A(B)?

In summary, the conversation is discussing the quantum rule for calculating the probability of an event happening, where A(A or B) = A(A) + A(B) and then the squared sum of the amplitudes is used to find p(A or B). The conversation also touches on the concept of interference and the probability of finding an object in a specific location. The participants in the conversation are seeking clarification and assistance with their calculations.
  • #1
bon
559
0

Homework Statement



See q attached

Homework Equations





The Attempt at a Solution



So basically... my questions is this:

I thought that the rule in quantum for p(A or B) is that A(A or B) = A(A) + A(B) then you square [A(A) + A(B)] to find p(A or B) (where A(X) is amplitude of X)

But then surely the sum of the amplitudes all squared must = 1 rather than sum of amplitudes squared added..
i.e. in this example surely it would be that |(a+b)^2| = 1 rather than |a^2| + |b^2| =1 ?
but i think the latter is the right method...

please help!
 

Attachments

  • quantq.pdf
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  • #2
anyone?
 
  • #3
hellooo?
 
  • #4
lal2+lbl2=1

where lal2 is aa*
 
  • #5
so why is the 'quantum rule' that you add amplidues for different ways of an event happening..i.e.
A(A or B) = A(A) + A(B) then you square [A(A) + A(B)] to find p(A or B)

so p(A or B) = p(A) + p(B) + I (interference term)
..
 
  • #6
That is if you have two particles.
 
  • #7
aha okay thanks
 
  • #8
No 0 e^PIi squared is 1 so a=0
 
  • #9
Bendavid2 said:
No 0 e^PIi squared is 1 so a=0

sorry - posted the wrong q - meant this one...part b
 

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  • question.pdf
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  • #10
3/4?
 
  • #11
Yup I think so. Chance of finding it in well 0 is 1/2 due symmetrie both side are even so 1/4 + 1/2 is 3/4 so yup.
 
  • #12
noice thinking..you = quantum god.
 
  • #13
Bendavid2 said:
Yup I think so. Chance of finding it in well 0 is 1/2 due symmetrie both side are even so 1/4 + 1/2 is 3/4 so yup.

sorry to be a pain. thanks for you help.. also stuck on q11 c
 

Attachments

  • another q!.pdf
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FAQ: Questioning Quantum Rules: Is A(A or B) = A(A) + A(B)?

What does A(A or B) mean in the context of quantum rules?

A(A or B) refers to the probability that either event A or event B will occur in a quantum system. This is known as a disjunction, where the system can be in one of the two possible states.

Is A(A) + A(B) the same as A(A or B)?

No, A(A) + A(B) is not the same as A(A or B). A(A) + A(B) refers to the probability that both event A and event B will occur simultaneously in a quantum system. This is known as a conjunction, where the system must be in both states at the same time.

Can A(A or B) be greater than 1?

No, A(A or B) cannot be greater than 1. In quantum mechanics, probabilities are represented by numbers between 0 and 1, where 0 means impossible and 1 means certain. A(A or B) represents the probability of at least one of the events A or B occurring, so it cannot exceed 1.

How does A(A or B) relate to the superposition principle?

A(A or B) relates to the superposition principle in that it represents the probability of a quantum system being in a superposition of states. A superposition is a state in which a quantum system is simultaneously in multiple states, with each state having a certain probability of being observed.

Can A(A or B) be calculated using classical probability rules?

No, A(A or B) cannot be calculated using classical probability rules. In quantum mechanics, probabilities are calculated using complex numbers and the principles of linear algebra, which are different from classical probability rules. This is due to the probabilistic nature of quantum systems and the concept of superposition.

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