Probability of finding a state

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quietrain
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according to born's rule, |<a/b>|2=probability

so in the case of (1/2 + i/2) |a> = |b>,

the probability of finding |a> in system prepared in |b> = |<a/b>|2

so i should modulus the (1/2 + i/2) first and then square it right?

so i get 1/2 + 1/2 = 1, squaring gives 1 again, but my lecturer gives the answer as 1/2...

did i do anything wrong here? thanks a lot
 
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also, another probability of e[tex]\pi[/tex]i/8 / 2 gives a probability of 1/4. why?

if i square it and mod i get e[tex]\pi[/tex]i/4 /4 which doesn't give me 1/4.

is my concept wrong ?
 
The Born's rule in the form you have quoted it assumes that |a> and |b> are normalized - their norms should be 1. In general, for unnormalized states, it should read as

[tex]\frac{|\langle a,b\rangle|^2}{\langle a,a\rangle\,\langle b,b\rangle}[/tex]
 
I think you are not converting amplitudes to probabilities correctly. An amplitude of (1/2 + i/2) becomes a probability as follows: take the modulus, which is sqrt((1/2)^2 + (1/2)^2) = sqrt(1/2). Square to get 1/2. Alternatively, you can accomplish the same thing by multiplying the amplitude by its complex conjugate: (1/2 + i/2)(1/2 - i/2) = 1/2. (Ex: show these methods are equivalent)

If you have an amplitude of (1/2)exp(i*pi/8), take the modulus to get 1/2 and square that to get 1/4. Alternatively, do the other method: (1/2)exp(i*pi/8) * (1/2)exp(-i*pi/8) = (1/2)^2 = 1/4.
 
ic thanks everyone!