0
$\begingroup$

In Quantum Mechanics for scientists and engineers by David Miller, he explains the diffraction by two slits (page $13$) and considers a beam of electrons whose wave function is $\psi(x)= \exp(ikr)$ where $r$ is the radius from the source point (the slit). So as there are two slits, superposition of those two waves occurs. The resultant wave function is $$\psi(x)=\exp(i\alpha)\cos\left(\frac{\pi sx}{\lambda z_o}\right) \tag 1$$ (we get that after doing some algebra and using taylor expansion $\sqrt{(1+a)}=1+\frac{a}{2}$ for small $a$'s)

where $s$ is distance between the slits. $z_o$ is distance between the source and the screen. $x$ is the distance from the point on the screen corresponding to the middle point of the slit to the point on the screen which we want to calculate the wave function at $\alpha = k(z_o+x^2/2z_o+s^2/8z_o)$

He then says that $\alpha$ is just a phase factor (which I understand as saying it's a constant) as so we can conclude :

$$|\psi(x)|^2 \propto \cos^2(\pi sx/\lambda z_o) \tag 2$$

My question is: Why do we consider this $\alpha$ as being a constant whereas it contains the independent variable $x$?

If my understanding of the word "phase factor" is incorrect, then what does it mean? And how do we move from $(1)$ to $(2)$ (such movement seems to treat $e^{i\alpha}$ as if it were a constant)

$\endgroup$
  • 3
    $\begingroup$ A phase is any complex number of unit norm, i.e. anything of the form $\exp(\mathrm{i}\phi)$. $\endgroup$ – ACuriousMind Sep 12 '15 at 16:03
  • $\begingroup$ Addition for the pedantic: where $\phi \in \mathbb{R}$. ;) $\endgroup$ – Sebastian Riese Sep 12 '15 at 16:41
0
$\begingroup$

A step has been missed out.

$\psi(x) = A e^{i\alpha} \cos (\pi s x/\lambda z_0)$
$|\psi(x)|^2 = \psi^* \psi = A e^{i\alpha}\cos (\pi s x/\lambda z_0) \times A e^{-i\alpha} \cos (\pi s x/\lambda z_0) = A^2 \cos^2 (\pi s x/\lambda z_0)$

because $e^{i\alpha} e^{-i\alpha}=1$ for all values of $\alpha(x)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.