Question if expectation value is considered a measurement?

Click For Summary
SUMMARY

The discussion centers on the nature of expectation values in quantum mechanics and their relationship to measurements. It is established that expectation values are not considered measurements themselves, as they do not correspond to eigenvalues of an operator. For instance, the expectation value of the X component of the spin operator can yield a result of 0, which is not an eigenvalue of the operator, despite being the weighted average of the eigenvalues -1/2 and 1/2. This distinction is crucial for understanding quantum measurement theory.

PREREQUISITES
  • Understanding of quantum mechanics postulates
  • Familiarity with operators and eigenvalues in quantum systems
  • Knowledge of spin operators and their eigenstates
  • Concept of expectation values and their calculation
NEXT STEPS
  • Study the mathematical formulation of quantum mechanics postulates
  • Learn about the role of operators in quantum measurements
  • Explore the concept of eigenvalues and eigenstates in greater detail
  • Investigate the implications of expectation values in quantum mechanics
USEFUL FOR

Students and researchers in quantum mechanics, physicists interested in measurement theory, and anyone seeking to deepen their understanding of quantum operators and expectation values.

JordanGo
Messages
72
Reaction score
0
Hello,

I was just curious about expectation values. One of the postulates of quantum mechanics state:
The only possible results of a measurement is an eigenvalue of the operator.

Now, is the expectation value considered a measurement, thus considered an eigenvalue?

Thanks!
 
Physics news on Phys.org
An expectation value is a weighted average of all of the eigenvalues. So it may or may not actually be a valid eigenvalue itself.

As an example, the X component of the spin operator has two eigenvalues: -1/2 and 1/2. However, if you take the X expectation value of a Z state (which is a linear combination of both of those eigenstates), the expectation value will be 0, since both -1/2 and 1/2 states appear in equal combination. 0 is not actually a valid eigenvalue of the X operator, but it's the average of the two actual eigenvalues.
 

Similar threads

Replies
1
Views
2K
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
38
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
3
Views
2K