Computing Expectation Values: What Makes Sense?

Click For Summary
SUMMARY

The discussion centers on the computation of expectation values for observables within a limited interval versus the entire space. It is established that calculating the expectation value of momentum within a restricted range, such as [0, a/3] for an infinite potential well defined on [0, a], yields nonsensical results. The consensus is that integration must occur over the entire space to ensure accurate representation of the wavefunction, which is defined across all real numbers despite the system's confinement.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly wavefunctions.
  • Familiarity with the concept of expectation values in quantum mechanics.
  • Knowledge of infinite potential wells and their implications on wavefunctions.
  • Proficiency in performing integrals over defined intervals in mathematical physics.
NEXT STEPS
  • Study the mathematical foundations of expectation values in quantum mechanics.
  • Explore the implications of wavefunction behavior in infinite potential wells.
  • Learn about the significance of boundary conditions in quantum systems.
  • Investigate the role of integration limits in quantum mechanical calculations.
USEFUL FOR

Students and professionals in quantum mechanics, physicists analyzing wavefunctions, and anyone interested in the mathematical treatment of quantum observables.

gulsen
Messages
215
Reaction score
0
How much sense does it make to compute expectation value of an observable in a limited interval? i.e.

[tex]\int_a^b \psi^* \hat Q \psi dx.[/tex]
rather than
[tex]\int_{-\infty}^{\infty} \psi \hat Q \psi dx[/tex]

Apparently, it shouldn't make any sense for it gives weird results when you compute e.v. of momentum for a part of infinite potentital well (say well is [0,a] and you do the e.v. integral from [0,a/3]]). Why do we have to integrate over all the space then?
 
Physics news on Phys.org
In the position representation, [itex]\psi (x)[/itex] is well defined on all real axis, even though the system might be constrained to "move" in a box.

Daniel.
 
you must integrate over entire space!
For infinite potential, there in no leak for
wavefunction beyond the potential boundary.
 

Similar threads

Replies
1
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K