How Do You Calculate the Probabilities of Measured Values for a Quantum State?

Click For Summary
SUMMARY

The discussion focuses on calculating the probabilities of measured values for a quantum state represented by a Hermitian operator A with eigenvectors |A1>, |A2>, and |A3>. The state |psi> is given as a linear combination of these eigenvectors, specifically |psi> = ((3)^(-1/2))|A1> + 2((3)^(-1/2))|A2> + ((5/3)^(1/2))|A3>. To find the probabilities of measuring the eigenvalues, the coefficients of the eigenvectors in |psi> must be squared, but normalization of |psi> is necessary to ensure the probabilities sum to one.

PREREQUISITES
  • Understanding of Hermitian operators in quantum mechanics
  • Knowledge of eigenvalues and eigenvectors
  • Familiarity with quantum state normalization
  • Basic grasp of probability theory in quantum mechanics
NEXT STEPS
  • Learn about the normalization process for quantum states
  • Study the properties of Hermitian operators in quantum mechanics
  • Explore the concept of probability amplitudes in quantum mechanics
  • Investigate the implications of measurement in quantum mechanics
USEFUL FOR

Students and professionals in quantum mechanics, particularly those studying quantum states and measurements, will benefit from this discussion. It is also useful for anyone interested in the mathematical foundations of quantum theory.

bluebandit26
Messages
8
Reaction score
0

Homework Statement



Suppose that a Hermitian operator A, representing measurable a, has eigenvectors |A1>, |A2>, and |A3> such that A|Ak> = ak|Ak>. The system is at state:

|psi> = ((3)^(-1/2))|A1> + 2((3)^(-1/2))|A2> + ((5/3)^(1/2))|A3>.

Provide the possible measured values of a and corresponding probabilities.


Homework Equations



(A)(psi) = sum[anCnPsin]

The Attempt at a Solution


It would seem that A1 = (3)^(-1/2), A2 = 2((3)^(-1/2), and A3 = (5/3)^(1/2), but the state is not normalized so the probabilities don't add up to one... so I am confused about how to handle this.
 
Physics news on Phys.org
So normalise |psi>.
BTW, the probabilities of receiving the eigenvalue a_k is the square of the coefficient of |A_k> in |psi>.
 
In general, if [tex]|\psi>=\sum c|n>[/tex] the the probability to find |psi> in the state |n> (and measuring its eigenvalue) is |c|^2=cc* where c* is the complex conjugate of c.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
16
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K