Physics Friday 50

Quantum Mechanics and Momentum
Part 7: Plane Waves

Last week, we introduced the time-dependent Schrödinger equation , and we noted that for an energy eigenstate ψn(x) with energy En (), we have time-dependent solution , with .
Now, consider the case of a free particle of mass m. Here, the potential is everywhere zero, and the energy is given entirely by the kinetic energy. Recalling the (non-relativistic) formula for kinetic energy in terms of momentum, we have . Working in three dimensions, and replacing the classical momentum with the quantum momentum operator , we have , and thus the free particle Hamiltonian:

Note that a momentum eigenstate with momentum will also be an energy eigenstate with energy . This can be confirmed by applying to the momentum eigenfunction , where . Now, examining the corresponding solution to the time-dependent Schrödinger equation, we have
This, we see, is a wave propagating in the direction of , with (angular) frequency ω and (angular) wavenumber . The wavefronts of this wave, the surfaces of constant phase, can be seen to be planes perpendicular to . Thus, this kind of solution is known as a plane wave.

We have .
Now, from this we find that the group velocity of our plane wave is

which is just the classical velocity of the particle of mass m and momentum p (from p=mv).


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2 Responses to “Physics Friday 50”

  1. Physics Friday 57 « Twisted One 151’s Weblog Says:

    […] However, this does not mean that the current itself must be zero. Consider the three-dimensional plane wave . Then So So . Note that as is the particle’s velocity, the probability current of the […]

  2. Physics Friday 88 « Twisted One 151’s Weblog Says:

    […] The orbital states available can be specified by the wavenumber vector k of the wavefunction (see here). As before, our grand partition sum factors: . As we are considering fermions, each orbital state […]

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