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Wave Packet Motion

Friday, February 11, 2022

Traveling Waves

The equation of a traveling wave with wave number kk and angular frequency ω\omega (ω=2πf\omega=2\pi f) is

y(x,t)=Acos⁡(kx−ωt)y\left(x, t\right)=A\cos{\left(kx-\omega t\right)}

And a combined traveling wave is

y(x,t)=A1cos⁡(k1x−ω1t)+A2cos⁡(k2x−ω2t)y\left(x, t\right)=A_1\cos{\left(k_1x-\omega_1 t\right)}+A_2\cos{\left(k_2x-\omega_2 t\right)}

Wave speed is found by v=λfv=\lambda f, or v=ω/kv=\omega/k. This is also known as the phase speed. As the wave moves, the peaks of the individual component waves line up in a way that makes the peak of the combined wave move faster than the peak of either component wave.

The equations for the waves (with equal amplitude and similar wave number and angular frequency) can be combined using trigonometric identities:

y(x,t)=2Acos⁡(Δk2x−Δω2t)cos⁡(k1+k22x−ω1+ω22t)y\left(x, t\right)=2A\cos{\left(\frac{\Delta k}{2}x-\frac{\Delta \omega}{2}t\right)}\cos{\left(\frac{k_1+k_2}{2}x-\frac{\omega_1+\omega_2}{2}t\right)}

The first term in the equation above dictates the overall shape of the wave and the second term determines the fluctuations in that shape. Using v=ω/kv=\omega/k, the group speed of the combined wave is calculated as:

vgroup=dωdkv_{group}=\frac{d\omega}{dk}

Group speed of de Broglie waves

For a particle with a group of de Broglie waves, the energy is E=hf=ℏωE=hf=\hbar\omega so dE=ℏ dωdE=\hbar~d\omega, and the momentum is p=h/λ=ℏkp=h/\lambda=\hbar k so dp=ℏ dkdp=\hbar~dk. Therefore, the group speed of the de Broglie wave is

vgroup=dωdk=dE/ℏdp/ℏ=dEdpv_{group}=\frac{d\omega}{dk}=\frac{dE/\hbar}{dp/\hbar}=\frac{dE}{dp}

And for a classical particle with only kinetic energy, E=K=p2/2mE=K=p^2/2m so

dEdp=ddp(p22m)=pm=v\frac{dE}{dp}=\frac{d}{dp}\left(\frac{p^2}{2m}\right)=\frac{p}{m}=v

Therefore, the group speed and particle speed are identical.

For classical particles, vgroup=vparticlev_{group}=v_{particle}.

The spreading of a moving wave packet

Imagine a particle passing through a single-slit apparatus. It has initial position uncertainty of Δx0\Delta x_0 and initial momentum uncertainty of Δpx0\Delta p_{x0}, moving with velocity vxv_x. Therefore, its initial velocity uncertainty will be Δvx0=Δpx0/m\Delta v_{x0}=\Delta p_{x0}/m. Over time, its xx position will be x=vxtx=v_xt and its velocity will be vx=vx0±Δvx0v_x=v_{x0}\plusmn \Delta v_{x0}. Using both of these uncertainties for the xx position gives the following relationship:

Δx=(Δx0)2+(Δvx0t)2=(Δx0)2+(Δpx0t/m)2\Delta x=\sqrt{\left(\Delta x_0\right)^2+\left(\Delta v_{x0}t\right)^2}=\sqrt{\left(\Delta x_0\right)^2+\left(\Delta p_{x0}t/m\right)^2}

But since Δpx0=ℏ/Δx0\Delta p_{x0}=\hbar /\Delta x_0 due to the uncertainty principle,

Δx=(Δx0)2+(ℏt/mΔx0)2\Delta x=\sqrt{\left(\Delta x_0\right)^2+\left(\hbar t/m\Delta x_0\right)^2}

This leads to the property that the more successful we are at confining a wave packet (making Δx0\Delta x_0 small), the faster it expands with time. In terms of the single-slit experiment, the smaller the slit, the faster the wave expands.