Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Both sides previous revisionPrevious revision
Next revision		Previous revision
wave_packets [2021/02/15 20:21] – [In Class Activities] adminwave_packets [2021/02/18 00:24] (current) – [In Class Activities] admin
Line 1: Line 1:
 ====== 1.ix.1 Wave Packets and Fourier Transforms ====== ====== 1.ix.1 Wave Packets and Fourier Transforms ======
  
-A wavefunction of the form $\psi(x,t) \propto e^{i(kx - \omega t)}$ has a well-defined momentum, but is spread out over all space.  It is also not normalizable because+A wavefunction of the form $\psi(x,t) \propto e^{i(kx - \omega t)}$ has a well-defined momentum, but is spread out over all space.  See the graph below for an illustration of what this wavefunction looks like for $k = \pi$, $t=0$. 
 +{{ :momentumstate.png?direct&700 |}} 
 + 
 +The wavefunction $\psi(x,t) \propto e^{i(kx - \omega t)}$ is also not normalizable because
 \[\int_{-\infty}^{+\infty} |\psi(x)|^2\,\mathrm{d}x = \int_{-\infty}^{+\infty} \,\mathrm{d}x = \infty,\] \[\int_{-\infty}^{+\infty} |\psi(x)|^2\,\mathrm{d}x = \int_{-\infty}^{+\infty} \,\mathrm{d}x = \infty,\]
 so it does not give us a well-defined probability distribution. Therefore, it cannot be a physically realizable wavefunction. so it does not give us a well-defined probability distribution. Therefore, it cannot be a physically realizable wavefunction.
  
-Similarly, if we want to describe a particle with a well-defined position $x = x_0$, we would use $\psi(x,t) = \delta(x-x_0)$, where $\delta(x - x_0)$ is a **//Dirac delta function//**.  These will be defined more rigorously in the next module, but for now you can think of it as a "function" that is zero everywhere except at $x=x_0$, where it suddenly jumps to infinity.+Similarly, if we want to describe a particle with a well-defined position $x = x_0$, we would use $\psi(x,t) = \delta(x-x_0)$, where $\delta(x - x_0)$ is a **//Dirac delta function//**.  These will be defined more rigorously in the next module, but for now you can think of it as a "function" that is zero everywhere except at $x=x_0$, where it suddenly jumps to infinity.  The Dirac delta function $\delta(x-2)$ is illustrated below. 
 +{{ ::positionstate.png?direct&700 |}}
  
 By the uncertainty principle, since $\Delta x = 0$, we must have $\Delta p =\hbar/2\Delta x = \infty$, so these wavefunctions are spread out over all possible momenta.  Dirac delta functions are also not normalizable, so again they cannot be physically realizable wavefunctions. By the uncertainty principle, since $\Delta x = 0$, we must have $\Delta p =\hbar/2\Delta x = \infty$, so these wavefunctions are spread out over all possible momenta.  Dirac delta functions are also not normalizable, so again they cannot be physically realizable wavefunctions.
Line 15: Line 19:
 where $\phi(k)$ is called the **//amplitude//** of wave number $k$. where $\phi(k)$ is called the **//amplitude//** of wave number $k$.
  
-The factor $1/\sqrt{2\pi}$ is arbitrary because we can absorb a constant into $\phi(k)$, but it is chosen to make the Fourier transform formulas look more symmetric.  For simplicity, we will focus on the situation at $t=0$ and write $\psi_0(x) = \psi(x,o)$.  Then, we have+The factor $1/\sqrt{2\pi}$ is arbitrary because we can absorb a constant into $\phi(k)$, but it is chosen to make the Fourier transform formulas look more symmetric.  For simplicity, we will focus on the situation at $t=0$ and write $\psi_0(x) = \psi(x,0)$.  Then, we have
 \[\psi_0(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{ikx}\,\mathrm{d}k.\] \[\psi_0(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{ikx}\,\mathrm{d}k.\]
 Fourier transform theory tells us that the amplitude $\phi(k)$ is given by Fourier transform theory tells us that the amplitude $\phi(k)$ is given by
Line 34: Line 38:
  
 As we will see later in the course As we will see later in the course
-\[\int_{-\infty}^{+\infty}|\psi(k)|^2\,\mathrm{d}k = 1,\]+\[\int_{-\infty}^{+\infty}|\phi(k)|^2\,\mathrm{d}k = 1,\]
 so it is natural to interpret $|\phi(k)|^2$ as a probability density for wave number.  We can then say that the wave number is localized close to $k=0$. so it is natural to interpret $|\phi(k)|^2$ as a probability density for wave number.  We can then say that the wave number is localized close to $k=0$.
  
Line 48: Line 52:
  
 Don't worry if you do not follow the details of this, as we will do these calculations in more detail in the next module. Don't worry if you do not follow the details of this, as we will do these calculations in more detail in the next module.
 +
 +The typical situation for a wavefunction in which both $x$ and $p$ are well-localized is illustrated below.
 +{{ :gaussians.png?direct&1000 |}}
  
 ====== 1.ix.2 Gaussian States ====== ====== 1.ix.2 Gaussian States ======
Line 59: Line 66:
 We can now try to compute the momentum amplitude $\tilde{\phi}(p)$ by taking the Fourier transform. We can now try to compute the momentum amplitude $\tilde{\phi}(p)$ by taking the Fourier transform.
 \begin{align*} \begin{align*}
-\tilde{\phi}(p) & = \frac{1}{\sqrt{2\pi\hbar} \left ( \frac{1}{2\pi\sigma^2}\right )^{\frac{1}{4}}} \int_{-\infty}^{+\infty} e^{-\left ( \frac{x^2}{4\sigma^2} + \frac{ipx}{\hbar} \right )}\,\mathrm{d}x+\tilde{\phi}(p) & = \frac{1}{\sqrt{2\pi\hbar}} \left ( \frac{1}{2\pi\sigma^2}\right )^{\frac{1}{4}} \int_{-\infty}^{+\infty} e^{-\left ( \frac{x^2}{4\sigma^2} + \frac{ipx}{\hbar} \right )}\,\mathrm{d}x
 & = \left ( \frac{1}{2^3\pi^3\hbar^2 \sigma^2}\right )^{\frac{1}{4}} \int_{-\infty}^{+\infty} e^{-\left ( \frac{x^2}{4\sigma^2} + \frac{ipx}{\hbar} \right )}\,\mathrm{d}x. & = \left ( \frac{1}{2^3\pi^3\hbar^2 \sigma^2}\right )^{\frac{1}{4}} \int_{-\infty}^{+\infty} e^{-\left ( \frac{x^2}{4\sigma^2} + \frac{ipx}{\hbar} \right )}\,\mathrm{d}x.
 \end{align*} \end{align*}
Line 68: Line 75:
 \[\tilde{\phi}(p) = \left ( \frac{1}{2^3\pi^3\hbar^2 \sigma^2}\right )^{\frac{1}{4}} e^{-\frac{p^2 \sigma^2}{hbar^2}} \int_{-\infty}^{+\infty} e^{-\left ( \frac{x}{2\sigma} + \frac{ip\sigma}{\hbar}\right )^2 } \,\mathrm{d}x.\] \[\tilde{\phi}(p) = \left ( \frac{1}{2^3\pi^3\hbar^2 \sigma^2}\right )^{\frac{1}{4}} e^{-\frac{p^2 \sigma^2}{hbar^2}} \int_{-\infty}^{+\infty} e^{-\left ( \frac{x}{2\sigma} + \frac{ip\sigma}{\hbar}\right )^2 } \,\mathrm{d}x.\]
  
-Making the substitution $y = \frac{x}{2\sigma} + \frac{ip\sigma}{hbar}$ gives $\mathrm{d}y = \frac{\mathrm{d}x}{2\sigma}$, and+Making the substitution $y = \frac{x}{2\sigma} + \frac{ip\sigma}{\hbar}$ gives $\mathrm{d}y = \frac{\mathrm{d}x}{2\sigma}$, and
 \[\tilde{\phi}(p) = \left ( \frac{2\sigma^2}{\pi^3\hbar^2}\right )^{\frac{1}{4}} e^{-\frac{p^2 \sigma^2}{hbar^2}} \int_{-\infty}^{+\infty} e^{-y^2}\,\mathrm{d}y.\] \[\tilde{\phi}(p) = \left ( \frac{2\sigma^2}{\pi^3\hbar^2}\right )^{\frac{1}{4}} e^{-\frac{p^2 \sigma^2}{hbar^2}} \int_{-\infty}^{+\infty} e^{-y^2}\,\mathrm{d}y.\]
  
Line 91: Line 98:
 \[\psi(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{ik(x-ct)}\,\mathrm{d}k = \psi_0(x-ct).\] \[\psi(x,t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{+\infty} \phi(k) e^{ik(x-ct)}\,\mathrm{d}k = \psi_0(x-ct).\]
  
-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
Line 100: Line 109:
   * 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
Line 122: Line 136:
 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
Line 131: Line 145:
 {{: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).