Constant of proportionality in probability of superposition of states

  • #1
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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?
 

Answers and Replies

  • #2
Gaussian97
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Quantum Mechanics is based in some postulates, one of those (usually called the 3rd postulate) says that given an observable ##A## (I will assume that has no degeneracy, that is your case if we assume ##a\neq b##) then the probability of measure a given output ##a_i## with corresponding eigenstate ##\left|\psi_i\right>## is $$\frac{\left|\left<\psi_i|\psi\right>\right|^2}{\left<\psi|\psi\right>}$$. Note that this "constant of proportionality" ##\left|\left<\psi|\psi\right>\right|^{-1}## does not depend on what outcome you measure.
 
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  • #3
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Quantum Mechanics is based in some postulates, one of those (usually called the 3rd postulate) says that given an observable ##A## (I will assume that has no degeneracy, that is your case if we assume ##a\neq b##) then the probability of measure a given output ##a_i## with corresponding eigenstate ##\left|\psi_i\right>## is $$\left|\frac{\left<\psi_i|\psi\right>}{\left<\psi|\psi\right>}\right|^2$$. Note that this "constant of proportionality" ##\left|\left<\psi|\psi\right>\right|^{-2}## does not depend on what outcome you measure.
Oh, I see. So the | ⟨ψ|ψ⟩|^2 term here is equivalent to (|α|^2 + |β|^2) in my expressions? And that's just because ψ is a superposition of eigenstates with coefficients alpha and beta. Is that correct?
 
  • #4
hilbert2
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The reason why the ##\left|\left<\psi\left|\right.\psi\right>\right|^2## is needed is that the given state ##\left|\right.\psi\left.\right>## is not normalized to 1.
 
  • #5
Gaussian97
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Oh, I see. So the | ⟨ψ|ψ⟩|^2 term here is equivalent to (|α|^2 + |β|^2) in my expressions? And that's just because ψ is a superposition of eigenstates with coefficients alpha and beta. Is that correct?

Exact, again one of the postulates of QM is that any observable ##A## is described by a hermitic operator, is not difficult to show that then the eigenstates of ##A## are orthogonal, then:
$$\left<\psi|\psi\right>=\left|\alpha\right|^2\left<A|A\right>+\alpha^*\beta\underbrace{\left<A|B\right>}_{0}+\alpha\beta^*\underbrace{\left<B|A\right>}_{0}+\left|\beta\right|^2\left<B|B\right>=\left|\alpha\right|^2+\left|\beta\right|^2$$
 
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  • #6
PeroK
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Problem 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?
If the constant of proportionality could be different for A and B, then the state you were given would be completely ambiguous and meaningless.

In an extreme case you could have ##k_B = 0## and then you'd have simply state A.
 
  • #7
PeroK
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Exact, again one of the postulates of QM is that any observable ##A## is described by a hermitic operator, is not difficult to show that then the eigenstates of ##A## are orthogonal, then:
$$\left<\psi|\psi\right>=\left|\alpha\right|^2\left<A|A\right>+\alpha^*\beta\underbrace{\left<A|B\right>}_{0}+\alpha\beta^*\underbrace{\left<B|A\right>}_{0}+\left|\beta\right|^2\left<B|B\right>=\left|\alpha\right|^2+\left|\beta\right|^2$$
Although, if you allow ##\psi## to be not normalized, then how do know ##A## and ##B## are normalized?
 
  • #8
Gaussian97
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Although, if you allow ##\psi## to be not normalized, then how do know ##A## and ##B## are normalized?
True, but this does not change the important point, that is that ##k=\left<\psi|\psi\right>^{-1}## doesn't depend on what outcome you get when measuring an observable. Also, I have used this inner product notation because it's the easiest and we don't have degeneracy, in general, you write the probabilities in terms of the projectors so we don't care about the norm of the basis.
 
  • #9
PeroK
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True, but this does not change the important point, that is that ##k=\left<\psi|\psi\right>^{-1}## doesn't depend on what outcome you get when measuring an observable. Also, I have used this inner product notation because it's the easiest and we don't have degeneracy, in general, you write the probabilities in terms of the projectors so we don't care about the norm of the basis.
If it doesn't say A and B are normalized then the question is ill-posed. Suppose, for example, that ##A = 2B##?

I would say the question just looks wrong with those coefficients. ##\psi## is not a valid state.

In any case, how can you assume that ##\langle A|A \rangle = 1## if states are not, by definition, normalized?
 
  • #10
Gaussian97
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Well, if ##\left|A\right>## and ##\left|B\right>## are not orthogonal, then the question doesn't make sense in any case, even if they are normalized, since in this case, the eigenvalues must be the same and the probability of getting anything is trivially 1. But ##\left|\psi\right>## would still be a valid state (if you consider non-normalized states valid).

In general, if ##\left|A\right>## is not normalized you simply compute the probability with the projector, i.e. a Hermitian ##P_A##operator such that ##P_A^2=P_A## and ##P_A\left|A\right>=\left|A\right>##. Then the probability of measuring the outcome ##a## is ##\frac{\left\|P_A\left|\psi\right>\right\|^2}{\left<\psi|\psi\right>}##. Here you don't need to supose ##\left|A\right>## normalized. Indeed, if ##a## is a non-degenerated eigenvalue then $$P_A=\frac{\left|A\right>\left<A\right|}{\left<A|A\right>}$$
 

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