Problem Set 2 - Siavash Davani's Website

Exercise 1 (Wavefunction dynamics)

We saw in the lectures that the dynamics of a wavefunction $\Psi(x)$ of a particle with mass $m$ is described using the Schrodinger equation

\pdv{\Psi}{t} = \frac{i\hbar}{2m}\ppdv{\Psi}{x} - \frac{i}{\hbar} V(x) \Psi.

(1)

However, the Schrodinger equation is not the only differential equation that describes the dynamics of a quantum mechanical system. For example, massless particles like a photon can be described using the Klein-Gordon equation as

\frac{1}{c^2}\ppdv{\Psi}{t} = \ppdv{\Psi}{x},

(2)

where $c$ is the speed of light. We will work out some properties of such particles in this problem.

Show that any wavefunction of the form
$\Psi(x,t) = \psi(x-ct)$
(3)
is a solution to the Klein-Gordon equation.
Solution
The form suggests that the value of the wavefunction is only a function of the difference $x-ct$ as opposed to individual parameters $x$ and $t$ . Mathematically speaking, the wavefunction is a function of a variable we can define as $\gamma(x,t)=x-ct$ as
$\psi(\gamma)$
(4)
This means the partial derivatives are
$\begin{align} \pdv{\psi(\gamma)}{x} &= \pdv{\psi}{\gamma}\pdv{\gamma}{x} = \pdv{\psi}{\gamma} \\ \pdv{\psi(\gamma)}{t} &= \pdv{\psi}{\gamma}\pdv{\gamma}{t} = -c \, \pdv{\psi}{\gamma} \end{align}$
(5)
and second derivatives will lead to
$\begin{align} \ppdv{\psi(\gamma)}{x} &= \ppdv{\psi}{\gamma} \\ \ppdv{\psi(\gamma)}{t} &= c^2 \, \ppdv{\psi}{\gamma} \end{align}$
(6)
which proves that $\psi(\gamma)$ is a solution to the Klein-Gordon equation.
Using the wavefunction form in part $(a)$ and assuming that the expectation value of the position at time $t=0$ was
$x_0 = \int_{-\infty}^{+\infty} x \, \abs{\Psi(x,0)}^2 \, dx,$
(7)
calculate the expectation value of the position at any arbitrary time $t$
$\langle x(t) \rangle = \int_{-\infty}^{+\infty} x\, \abs{\Psi(x, t)}^2 \, dx.$
(8)
Hint: note that $\Psi(x,t)$ is by definition a normalized wavefunction for any $t$ .
Solution
$\begin{align} \langle x(t) \rangle &= \int_{-\infty}^{+\infty} x\, \abs{\Psi(x, t)}^2 \, dx \\ &= \int_{-\infty}^{+\infty} x\, \abs{\psi(x-ct)}^2 \, dx \\ \intertext{Now we apply a change of variables $x'=x-ct$} &= \int_{-\infty}^{+\infty} (x'+ct)\, \abs{\psi(x')}^2 \, dx' \\ &= \int_{-\infty}^{+\infty} x'\, \abs{\psi(x')}^2 \, dx' + ct \int_{-\infty}^{+\infty} \abs{\psi(x')}^2 \, dx' \\ &= \int_{-\infty}^{+\infty} x'\, \abs{\Psi(x', 0)}^2 \, dx' + ct \int_{-\infty}^{+\infty} \abs{\Psi(x',0)}^2 \, dx' \\ &= x_0 + ct \end{align}$
(9)
If we define the velocity as the change of the expectation value of position over time
$\langle v \rangle = \dv{ \langle x(t) \rangle}{t},$
(10)
calculate the velocity of this particle using part $(b)$ .
Solution
In part $(b)$ , we calculated
$\langle x(t) \rangle = x_0 + ct$
(11)
which indicates that the expectation value of position is starting at $x_0$ and moving in the direction of the $x$ axis as time passes. The velocity is
$\langle v \rangle = c,$
(12)
suggesting the particle described using this wavefunction is moving with the speed of light.

Exercise 2 (de Broglie relation)

The de Broglie relation suggests that a wavelength $\lambda=\frac{h}{p}$ can be assigned to any moving particle. Let us see if we can make sense of this relation and argue why the relation is not always readily recognized in our everyday experience.

Calculate the de Broglie wavelength of a car with the mass of $1500\,kg$ moving with the speed of $80\,m/s$ .
Solution
There was a mistake in the problem sheet where the speed was given as $80\,m/s$ where it was intended to be $80\,km/h$ ! But let us do the calculations using $80\,m/s$ !
The momentum of the car is
$p=mv = 1.2 \times 10^5 \, kg\cdot m/s$
(13)
Therefore, using $h=6.626\times 10^{-34} \,J\cdot s$ , the de Broglie wavelength is
$\lambda=\frac{h}{p} = 5.52 \times 10^{-39} \, m$
(14)
If we would calculate using the velocity $v=80\,km/h = 22.2\,m/s$ , we would have
$\lambda=\frac{h}{p} = 19.87 \times 10^{-39} \, m$
(15)
Calculate the de Broglie wavelength of an electron moving at the speed of $\frac{1}{100}\,c$ , where $c$ is the speed of light. (You can, but not necessarily have to take relativistic effects into account for the calculation.)
Solution
The relativistic momentum is
$p=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}.$
(16)
However, we neglect the relativistic effects and calculate similar to previous part.
Taking $m_e=9.1\times 10^{-31} \,kg$ and $c=3\times10^{8} \,m/s$ , we have
$p=mv = 2.73 \times 10^{-24} \, kg\cdot m/s$
(17)
Therefore,
$\lambda =\frac{h}{p} = 2.42 \times 10^{-10} \, m = 0.242 \,nm$
(18)
Compare the wavelengths of the objects you calculated with the wavelength of a photon in the typical visible range. Do you see any significant difference? What do you think can be the reason that the wave nature of a photon is exposed to us easier than the other objects? (In fact, you can search in the literature and find the double-slit experiment done with electrons which reveals the wave nature of electrons. We believe no one has tried the experiment with cars though!)
Solution
We know that the wavelength of photons in the visible spectrum are in the range of $400nm-700nm$ . If we compare these wavelengths with the ones calculated in parts $(a)$ and $(b)$ , we can see that the de Broglie wavelengths corresponding to a usual car or a moving electron are much smaller than visible photons. It is much harder to observe the wave-like behavior of objects having too small wavelengths.
In order to measure the wave nature of an object, we have to put it through an experiment which exposes this wave nature. One example of such experiments is the double-slit experiment. Using rough approximations and assuming our analysis holds for all objects of interest, the spacing between the fringes of the interference pattern in the double-slit experiment is given via
$w=\frac{\lambda z}{d},$
(19)
where λ is the wavelength of the wave, $z$ is the propagation distance from the slits to the screen, and $d$ is the distance between the slits. We can see from this relation that for very small λ, we have to either greatly increase the propagation distance (large apparatus) or bring the slits extremely close to each other. In any case, it will be more difficult to measure the wave nature of objects with extremely small wavelengths.
However, if this analysis is somewhat correct, the wavelength corresponding to the electron is not extremely far-fetched. Thus it is possible to observe the wave-like behavior from them in a double-slit experiment. Check out a video recording of such an experiment.