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

[bon]

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!

 

[bon]

anyone?

 

[bon]

hellooo?

 

[Bendavid2]

lal²+lbl²=1

where lal² is aa*

 

[bon]

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)
..

 

[Bendavid2]

That is if you have two particles.

 

[bon]

aha okay thanks

 

[Bendavid2]

No 0 e^PIi squared is 1 so a=0

 

[bon]

No 0 e^PIi squared is 1 so a=0

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

 

[bon]

3/4?

 

[Bendavid2]

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.

 

[bon]

noice thinking..you = quantum god.

 

[bon]

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