Hi. I will give you a question I have looked at and then tell you where I am confused.(adsbygoogle = window.adsbygoogle || []).push({});

The wavefunction for a particle of mass m is ψ(x) = sin(kx)exp(-iωt) where k is a constant.

(i) Is this particle in a state of defined momentum ? If so , determine its momentum.

(ii) Is this particle in a state of defined energy ? If so , determine its energy.

For (i) i thought if k is a constant then p=[itex]\hbar[/itex]k must also be a constant ie a definite momentum. The solution says to apply the momentum operator to ψ which shows that ψ is not an eigenfunction of momentum. Does p=[itex]\hbar[/itex]k not always apply ? If ψ is not an eigenfunction of momentum what happens when momentum is measured ?

For (ii) i just assumed E=[itex]\hbar[/itex]ω which in this case turns out to be right although according to the solution i should have used the energy operator i[itex]\hbar[/itex]dψ/dt which gives an eigenvalue of E=[itex]\hbar[/itex]ω. Could i instead apply the Hamiltonian operator to give an energy eigenvalue of [itex]\hbar[/itex]^2k^2/2m ? Are these 2 methods equivalent ?

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Momentum and energy eigenvalues

Loading...

Similar Threads - Momentum energy eigenvalues | Date |
---|---|

I Why is position-space favored in QM? | Jan 19, 2018 |

A Bound states and the energy-momentum relation... | Sep 23, 2017 |

B Does precise momentum guarantee precise energy? | Jul 12, 2017 |

I How can the total orbital angular momentum be zero? | Jun 15, 2017 |

I Momentum and energy in QM and QFT | Apr 15, 2017 |

**Physics Forums - The Fusion of Science and Community**