Discussion
If quantum mechanical operators of two observables of a system do not commute, then :
Answer: D
If the operators do not commute, it is impossible to know the precise values of the observables simultaneously.
Two observables can be measured simultaneously with arbitrary precision only if they share a common set of eigenstates, which is the case exactly when their operators commute. For non-commuting operators the generalised uncertainty principle sets a lower limit:
\[ \begin{aligned} \Delta A\, \Delta B &\ge \frac{1}{2}\left|\langle C \rangle\right| \\ C &= [\hat{A}, \hat{B}] \end{aligned} \]
Position and momentum, with \([x, p] = i\hbar\), are the standard example. Commutation says nothing about the sign of the energy, the parity of the wave function, or whether the observables depend on time, so the other options are unrelated.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development