Switching observers in a quantum measurement

Keru
Messages
20
Reaction score
1
Homework Statement

observables.png

(c is a constant)

The attempt at a solution

-In the first measure we got a1, so the state of the system would be psi1.
-In the second measure, there's no information about what eigenvalue we got. Would the state of the system still be psi1? Psi1 is written in terms of B eigenvectors, and as we don't know which one we measured, the state should be described as a superposition of both vectors, which is precisely psi1. Is that correct?
-In the third measure, i wrote the "betas" in terms of the "psis", so i have something like:
(A and B being constants, not the observables)
psi1 = A beta1 + B beta2 = C psi1 + D psi2 + E psi 1 + F psi2 = G psi1 + H psi 2

Is it H the coefficient that tells me the probabilities of getting the a2 eigenvalue, or did I do something wrong?
 

Attachments

  • observables.png
    observables.png
    17.3 KB · Views: 817
Physics news on Phys.org
Keru said:
-In the first measure we got a1, so the state of the system would be psi1.
Correct.
Keru said:
-In the second measure, there's no information about what eigenvalue we got. Would the state of the system still be psi1? Psi1 is written in terms of B eigenvectors, and as we don't know which one we measured, the state should be described as a superposition of both vectors, which is precisely psi1. Is that correct?
No. I bolded the part which is wrong. It contradicts the postulate that after a measurement, the system is in an eigenstate of the corresponding observable.

If you perform a measurement and don't have information about the outcome, you can't use a single state vector to describe the situation after the measurement. You need to do a case-by-case analysis in order to calculate the probabilities.
 
Ok I think i got it. So, for one of the two possible cases it would continue like this?

medidaas.png
 

Attachments

  • medidaas.png
    medidaas.png
    8.3 KB · Views: 368
Yes. I haven't checked the calculations but conceptually, it is correct now.
 
  • Like
Likes Keru
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top