This is an example of a problem that my teacher posted. I can\'t figure out what
ID: 1894924 • Letter: T
Question
This is an example of a problem that my teacher posted. I can't figure out what he is doing or why."If a polarized photon is in the state 3|t> + 4|a> , where |t> is a state of transmission and |a> is a state of absorption (with respect to a polarizer), what is the probability of observing the photon in the state:
4|t> + 3|a>?"
Answer:
"Normalize 3/5|t> + 4/5|a> ; 4/5|t> + 3/5|a>
Probability = |(4/5|t> + 3/5|a>)(3/5|t+4/5|a>)^2
=(12/25 + 12/25)^2 = (24/25)^2 = 0.92"
Please help!
Explanation / Answer
The polarized photon is initially in the state 3|t> +4|a>. According to the normalization condition, the norm of this state should be 1. The norm of any state |x> is .
This should be 1 because we have performed an experiment and found the photon to be in this state thereby collapsing the wavefunction to this state of polarization.
However, when we want to know the probability of the photon to be in some other state, we must take an overlap of that state with the state that we have already measured i.e. we should take an inner product of the two states and then take the square of their modulus.
However, while taking the inner product, we must take care of the orthogonality of states.
For example, the state |t> is orthogonal to |a> hence their inner product will be zero while and will be one.
When we normalize the given state 3|t> + 4|a>, we see that this state does not give 1 on its own on taking the probability. Hence the coefficients must be modified to 3/5 and 4/5
(easy to see as 32+42=25=52 so when we divide by 5, the sum of their squares is 1)
Thus the probability that the photon will be found in the other state 4|t> + 3|a> is (remember first normalize this state by dividing by 5 again so that this state norm is 1) take the overlap with the initial state
((4/5<t| + 3/5<a|)(3/5|t> + 4/5|a>))2 = (12/25 + 12/25)^2 = (24/25)^2 = 0.92
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