- #1
Phys12
- 351
- 42
- Homework Statement
- Assume that the measurement of the energy on a state |A> always yields the value 'a' and the measurement of the energy on a state |B> always yields the value 'b'. Now consider a quantum system in the superposition state (1+2i)|A> + (1-i)|B>. What are the probabilities Pa and Pb to measure energy and 'a' and 'b' respectively?
- Relevant Equations
- For a state Ψ = α|A> + β|B>;
Pa ∝ |α|^2
and Pb ∝ |β|^2
Using the fact that
Pa ∝ |α|^2 and Pb ∝ |β|^2, we get:
Pa = k|α|^2 and Pb = k|β|^2
Since the probability of measuring the two states must add up to 1, we have Pa + Pb = 1 => k = 1/(|α|^2 + |β|^2). Substituting this in Pa and Pb, we get:
Pa = |α|^2/(|α|^2 + |β|^2)
and Pb = |β|^2/(|α|^2 + |β|^2)
And using these equations, I could get the correct answer. However, I assumed that the constant of proportionality is the same for calculating Pa and Pb and in fact, it is. But I am not sure why that is the case...why do they have to be the same?
Pa ∝ |α|^2 and Pb ∝ |β|^2, we get:
Pa = k|α|^2 and Pb = k|β|^2
Since the probability of measuring the two states must add up to 1, we have Pa + Pb = 1 => k = 1/(|α|^2 + |β|^2). Substituting this in Pa and Pb, we get:
Pa = |α|^2/(|α|^2 + |β|^2)
and Pb = |β|^2/(|α|^2 + |β|^2)
And using these equations, I could get the correct answer. However, I assumed that the constant of proportionality is the same for calculating Pa and Pb and in fact, it is. But I am not sure why that is the case...why do they have to be the same?