Physics Friday 50

Quantum Mechanics and Momentum 
Part 7: Plane Waves

Last week, we introduced the time-dependent Schrödinger equation $\imath\hbar\frac{\partial}{\partial{t}}\Psi=\hat{H}\Psi$ , and we noted that for an energy eigenstate ψ_n(x) with energy E_n ( $\hat{H}\psi_n=E_n\psi_n$ ), we have time-dependent solution $\Psi(x,t)=\psi(x)e^{-i\omega_n{t}}$ , with $\omega_n\equiv\frac{E_n}{\hbar}$ .
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 $E=T=\frac{p^2}{2m}$ . Working in three dimensions, and replacing the classical momentum with the quantum momentum operator $\hat{p}=\frac{\hbar}{\imath}\vec{\nabla}$ , we have $\hat{p}^2=\hat{p}_x^2+\hat{p}_y^2+\hat{p}_z^2=\left(\frac{\hbar}{\imath}\right)^2\vec{\nabla}\cdot\vec{\nabla}=-\hbar^2\nabla^2$ , and thus the free particle Hamiltonian:
$\hat{H}=\frac{\hat{p}^2}{2m}=-\frac{\hbar^2}{2m}\nabla^2$ .

Note that a momentum eigenstate with momentum $\mathbf{\vec{p}}$ will also be an energy eigenstate with energy $\frac{\left|\mathbf{\vec{p}}\right|^2}{2m}$ . This can be confirmed by applying $\hat{H}=-\frac{\hbar^2}{2m}\nabla^2$ to the momentum eigenfunction $\psi(\mathbf{\vec{x}})=Ae^{\imath\mathbf{\vec{k}}\cdot\mathbf{\vec{x}}}$ , where $\mathbf{\vec{p}}=\hbar\mathbf{\vec{k}}$ . Now, examining the corresponding solution to the time-dependent Schrödinger equation, we have
$\begin{eqnarray}\Psi(x,t)&=&\psi(x)e^{-i\omega{t}}\\&=&Ae^{\imath\mathbf{\vec{k}}\cdot\mathbf{\vec{x}}}e^{-i\omega{t}}\\&=&Ae^{\imath(\mathbf{\vec{k}}\cdot\mathbf{\vec{x}}-\omega{t})}\end{eqnarray}$ .
This, we see, is a wave propagating in the direction of $\mathbf{\vec{k}}$ , with (angular) frequency ω and (angular) wavenumber $\left|\mathbf{\vec{k}}\right|$ . The wavefronts of this wave, the surfaces of constant phase, can be seen to be planes perpendicular to $\mathbf{\vec{k}}$ . Thus, this kind of solution is known as a plane wave.

We have $\omega=\frac{E}{\hbar}=\frac{p^2}{2m\hbar}=\frac{\hbar{k^2}}{2m}$ .
Now, from this we find that the group velocity of our plane wave is
$\begin{eqnarray}v_g&=&\frac{\partial\omega}{\partial{k}}\\&=&\frac{\partial}{\partial{k}}\frac{\hbar{k^2}}{2m}\\&=&\frac{\hbar{k}}{m}\\&=&\frac{p}{m}\\\end{eqnarray}$
which is just the classical velocity of the particle of mass m and momentum p (from p=mv).

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Tags: Free Particle, Friday Physics, Group Velocity, Momentum, physics, Plane Wave, Quantum Mechanics, Schrödinger Equation

This entry was posted on December 12, 2008 at 3:21 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to “Physics Friday 50”

Physics Friday 57 « Twisted One 151’s Weblog Says:
January 30, 2009 at 12:12 am | Reply
[...] 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 [...]
Physics Friday 88 « Twisted One 151’s Weblog Says:
September 29, 2009 at 2:56 pm | Reply
[...] 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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