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wave_packets [2021/02/17 07:03] – adminwave_packets [2021/02/18 00:24] (current) – [In Class Activities] admin
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 \[\psi(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{ik(x-ct)}\,\mathrm{d}k = \psi_0(x-ct).\] \[\psi(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{ik(x-ct)}\,\mathrm{d}k = \psi_0(x-ct).\]
  
-In this case, the wave packet keeps the same shape over time and just moves to the right with velocity $c$.+In this case, the wave packet keeps the same shape over time and just moves to the right with velocity $c$, as illustrated below. 
 + 
 +{{ :nondispersive.png?direct&600 |}}
  
 Note that $\omega = kc$ implies $\hbar \omega = \hbar k c$, or $E = pc$, so in quantum mechanics only free massless particles, like photons, are nondispersive.  In general, we will have //**dispersion**//.  For example, for a nonrelativistic free massive particle we have Note that $\omega = kc$ implies $\hbar \omega = \hbar k c$, or $E = pc$, so in quantum mechanics only free massless particles, like photons, are nondispersive.  In general, we will have //**dispersion**//.  For example, for a nonrelativistic free massive particle we have
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   * The wave packet spreads in time: $\Delta x$ increases.   * The wave packet spreads in time: $\Delta x$ increases.
  
-This video illustrates the difference between phase and group velocity:+The difference between phase and group velocity is illustrated below. 
 +{{ :phaseandgroup.png?direct&600 |}} 
 +It is perhaps clearer to understand the difference from a video. 
 +{{ youtube>NLQs7bD6w98?medium }} 
 +Note that the video shows a situation in which we have a periodic envelope with faster oscillations within it.  However, the situation where the envelope is not periodic is similar.  The phase velocity can be slower, the same, or faster than the group velocity depending on the dispersion relation.
  
-and this video illustrates how a Gaussian wave packet spreads over time:+The video below illustrates how a Gaussian wave packet spreads over time.  This wave packet has an initial average position of zero and an initial average momentum of zero, so only the spreading effect is present.  In general, the wave packet would also move to the right or left in addition to spreading. 
 +{{ youtube>491hkK8VNCo?medium }}
  
 In general, the phase and group velocities are given by In general, the phase and group velocities are given by
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 Since, classically, $p=mv$, it is the group velocity, not the phase velocity, that corresponds to the velocity of a classical particle.  In other words, the motion of the overall envelope is classical (when the potential is constant).  Note, this is another reason why plane wave solutions are not physically realizable.  For a plane wave, the phase and group velocities are equal, and do not obey the classical equation $p=mv$. Since, classically, $p=mv$, it is the group velocity, not the phase velocity, that corresponds to the velocity of a classical particle.  In other words, the motion of the overall envelope is classical (when the potential is constant).  Note, this is another reason why plane wave solutions are not physically realizable.  For a plane wave, the phase and group velocities are equal, and do not obey the classical equation $p=mv$.
  
-Although we sometimes model a "classical" particle as a wave packet with $\Delta x, \Delta p \sim \hbar$, but it is not really classical because the wave packet spreads in time.+Although we sometimes model a "classical" particle as a wave packet with $\Delta x, \Delta p \sim \hbar$, it is not really classical because the wave packet spreads in time.
  
 Without going into details of the calculation, if we have an initial Gaussian wave packet Without going into details of the calculation, if we have an initial Gaussian wave packet
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 {{:question-mark.png?direct&50 |}} {{:question-mark.png?direct&50 |}}
 ====== In Class Activities ====== ====== In Class Activities ======
- 
   - Use   - Use
-  \begin{align*}v_p & = \frac{\omega}{k},& v_g & = \frac{\mathrm{d}\omega}{\mathrm{d}k}, \end{align*}+  \begin{align*} 
 +  v_p & = \frac{\omega}{k}, & v_g & = \frac{\mathrm{d}\omega}{\mathrm{d}k}, 
 +  \end{align*}
   together with   together with
-  \begin{align*}E & = \hbar \omega, & p & = \hbar k, \end{align*}+  \begin{align*} 
 +  E & = \hbar \omega, & p& = \hbar k, 
 +  \end{align*}
   to show that   to show that
-  \begin{align*}v_p & = \frac{E}{p}, & v_g & = \frac{\mathrm{d}E}{\mathrm{d}p}.\end{align*} +  \begin{align*} 
-  - Consider a particle with initial position uncertainty $\sigma_0 = 1\,\text{nm}$.  Using+  v_p & = \frac{E}{p}, & v_g & = \frac{\mathrm{d}E}{\mathrm{d}p} 
 +  \end{align*} 
 +  - Consider a particle with initial position uncertainty $\sigma_0 = 1\,\text{nm}$. 
 +  Using
   \[\sigma_t = \sigma_0\sqrt{1 + \frac{\hbar^2t^2}{4m^2\sigma_0^4}},\]   \[\sigma_t = \sigma_0\sqrt{1 + \frac{\hbar^2t^2}{4m^2\sigma_0^4}},\]
   determine how long it would take for the wave packet to have $\sigma_t = 1\,\text{m}$ if   determine how long it would take for the wave packet to have $\sigma_t = 1\,\text{m}$ if
-    - $m = 9.11\times 10^{-31}\,\text{kg}$ (the mass of an electron), +    - $m=9.11\times 10^{-31}\,\text{kg}$ (the mass of an electron), 
-    - $m = 70\,\text{kg}$ (the mass of a physics instructor).+    - $m=70\,\text{kg}$ (the mass of a physics instructor).