In a non-relativistic quantum mechanical system in an infinite potential well. I try to measure the energy and the position of the system simultaneously. Since, the respective operators do commute according to Heisenberg's uncertainty relation I should be able to measure them both with infinite precision. Now, since I know that there is no potential energy in the well I can use $ E=\frac{p^2}{2m} $ since the potential energy is 0 and determine it's momentum provided I know it's mass. But I shouldn't be able know the momentum and position simultaneously with infinite precision! So where am I going wrong?

# [Physics] Measuring position and momentum at the same time

commutatorheisenberg-uncertainty-principleMeasurementsoperatorsquantum mechanics

###### Related Question

- [Physics] Particle in a box: simultaneously bounded momentum and position
- [Physics] What does it mean for 2 observables to be compatible
- [Physics] Is it true that we can’t measure position and momentum together
- Quantum Mechanics – Why It’s Impossible to Measure Position and Momentum Simultaneously with Arbitrary Precision
- [Physics] When an electron hits a fluorescent screen mounted on a spring, why can’t we get both position and momentum

## Best Answer

If you are considering a system of a single particle in a potential well with infinitely high walls and with finite width, the energy operator is $ H = \frac{p^2}{2m} + V(x) $ where $ V(x) $ is the potential energy operator, vanishing inside the well and infinite outside it. Being that $ \frac{p^2}{2m} $ does not commute with $ x $, how are you saying that the energy and position operators commute? Besides, if that were true than we could place a particle in any point inside the well and the particle would stay there forever.