Measuring Energy and Linear Momentum Consecutively

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SUMMARY

The total energy and linear momentum of a particle in a constant potential field can be measured consecutively without uncertainty. The energy operator, represented as \(\hat{H} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V\), and the momentum operator, \(\hat{p} = -i\hbar\frac{\partial}{\partial x}\), commute, indicating they share the same eigenstates. Consequently, measuring one observable collapses the wavefunction into an eigenstate that is also valid for the unmeasured observable, allowing for precise measurement of both quantities.

PREREQUISITES
  • Quantum mechanics fundamentals
  • Understanding of operators in quantum mechanics
  • Knowledge of wavefunction collapse
  • Familiarity with eigenstates and eigenvalues
NEXT STEPS
  • Explore the implications of operator commutation in quantum mechanics
  • Study the concept of wavefunction collapse in measurement theory
  • Learn about the relationship between energy and momentum in quantum systems
  • Investigate the role of potential fields in quantum mechanics
USEFUL FOR

Students and professionals in physics, particularly those focused on quantum mechanics, as well as researchers interested in measurement theory and the behavior of particles in potential fields.

Domnu
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Problem
Can the total energy and linear momentum of a particle moving in one dimension in a constant potential field be measured consecutively with no uncertainty in the values obtained?

Solution
Yes, this is possible. The energy and linear momentum operators are, respectively

[tex]\hat{H} = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V[/tex]

[tex]\hat{p} = -i\hbar\frac{\partial}{\partial x}[/tex]

Now, note that these two operators commute. Because of this, they share the same eigenstates. Now, once the total energy or linear momentum is measured, the wavefunction of the state collapses into an eigenstate of the measured observable. Now, this eigenstate is also an eigenstate of the observable which hasn't been measured yet. Therefore, we can precisely measure this unmeasured observable as well. [tex]\blacksquare[/tex]

Are my arguments correct?
 
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You can also directly tell from the fact that the energy in a constant potential field is a function of momentum only. If you've measured momentum, you know the energy too. And yes, your argument is correct.
 

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