# B Does precise momentum guarantee precise energy?

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1. Jul 12, 2017

### referframe

In non-relativistic QM, wave functions are represented as a superposition of complex exponentials in which the energy/frequency is a function only of momentum/wave number (dispersion relation). This seems to imply that under the HUP, precise momentum guarantees precise energy. Are there any experimental situations in which that would not be true? Thanks in advance.

2. Jul 12, 2017

### hilbert2

This is true only if potential energy is independent of position (i.e. in the case of a free particle).

3. Jul 12, 2017

### NFuller

It sounds like you are referring to the eigenfunctions of the Hamiltonian, which do have a precise energy. This is validated by looking at the expectation value of the Hamiltonian
$$<\psi_{n}|H|\psi_{n}>=<\psi_{n}|E_{n}|\psi_{n}>=<\psi_{n}|\psi_{n}>E_{n}=E_{n}$$.
In general however, the wavefunction is a weighted sum of these eienfunctions which means there is a spread of momentum and energy values.

4. Jul 12, 2017

### dextercioby

„Precise momentum” doesn't exist, neither theoretically, nor technically (from an experimental point of view).

5. Jul 12, 2017

### muscaria

To my understanding, dispersion relations give the energies of the collective modes of a system. Each of these modes have a definite energy. If the system is in one of these modes (momentum or quasi-momentum), it stays there until it interacts with another system since no energy is exchanged between modes. Each mode has a definite energy. So precise quasi-momentum gives precise energy, as you say. But I don't see what this has to do with the uncertainty principle?

6. Jul 12, 2017

### referframe

What about a particle in a confined space, like an infinite square well? It will have discreet energy eigenstates and do not those have associated sharp momentum states?

7. Jul 12, 2017

### NFuller

It could, but probably not. A more realistic case would be that the particle is a Gaussian wave packet. This would be composed of many eigenstates and would also have a Gaussian distribution of momentum values. This is because the Fourier transform of a Gaussian distribution is a Gaussian distribution.

8. Jul 13, 2017

### dextercioby

The infinite square well is a non-realistic model.

9. Jul 13, 2017

### referframe

Is it because the momentum eigenstates are not normalizable? If so, what about energy eigenstates? They are also not normalizable.

10. Jul 14, 2017

### atyy

Precise momentum and energy can be simultaneously obtained if momentum and energy commute. This is true for a free particle, but not in other cases. This is the same as what hilbert2 said in post #2.

11. Aug 8, 2017

### jfizzix

One can prove that a confined particle (whose probability density is exactly zero outside some region of space) cannot have a well-defined momentum $(\hat{p}_{x},\hat{p}_{y},\hat{p}_{z})$, regardless of whether the particle is in an energy eigenstate.

This is because the Fourier transform of any confined position wavefunction in position space, is a momentum wavefunction with nonzero values extending over all momentum space.

From this it follows that the momentum probability density also extends over all momentum space, and that the momentum uncertainties $\sigma_{p_{x}}$, $\sigma_{p_{y}}$, and $\sigma_{p_{z}}$ are nonzero for all confined wavefunctions.

Alternatively, if the particle is only bound, but not totally confined, where the position uncertainties $(\sigma_{x},\sigma_{y},\sigma_{z})$ are some nonzero values, the position-momentum uncertainty principle requires that the momentum uncertainties be nonzero as well, regardless of whether the particle is in an energy eigenstate.

Furthermore, if a particle is bound, some energy eigenstate will have a nonzero minimum momentum uncertainty. The only way to get a wavefunction with a smaller momentum uncertainty would be a superposition of multiple energy eigenstates, meaning a larger energy uncertainty.

Precise momentum does not imply precise energy, and vise versa

Last edited: Aug 8, 2017