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Suppose, say, I have the following wave functionenter image description here

It represents the wave function of a free particle. I would want to calculate the probability of finding the particle with energy ħk and energy 2ħk. However, how do I do this?

Qmechanic
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1 Answers1

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For a free particle, energy eigenfunctions are also momentum eigenfunctions, since $\hat{H} = \frac{\hat{p}^2}{2\,m}$.

The wavefunction is a linear superposition of four orthogonal momentum eigenfunctions: two counterpropagating plane waves of wavenumber $\pm k$ and a further two of wavenumber $\pm 2\,k$, all equally weighted. Thus there is equal probability that the energy is $(\hbar\,k)^2/(2\,m)$ and $(2\,\hbar\,k)^2/(2\,m)$.

Selene Routley
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  • Is it a general rule that, if a wave function is a superposition of orthonormal basis, then the probability of finding a particle with energy E is the sum of the probabilities in finding the wave function in the basis with eigenvalue E? (can you also explain the logic behind how this conclusion is reached if you don't mind) –  Dec 21 '17 at 10:12
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    @delickcrow123 well, we can't really say "orthonormal" here, because the eigenfunctions in question are not normalizable. You have the general idea, though: what you state is part of the meaning of "orthogonal". In this case, you need a somewhat delicate limiting procedure to account for the fact that the eigenfunctions aren't normalizable. But we don't really need to go through all the machinations here because we can see that the four momentum eigenfunctions are equally weighted. – Selene Routley Dec 21 '17 at 10:19
  • May I ask, once I've measured the system to be, say, in the state with energy ħk, what state is the system in? Is it in the state with wavenumber +k or with wavenumber -k? –  Dec 21 '17 at 12:07