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I've just started learning about the double slit experiment (just in the short appendix section in Schroeder's Thermal Physics), and I'm extremely confused by this one thing:

In it, out of basically nowhere he pulls out the De Broglie equation, that λ = h/p.

I've studied double slit diffraction before, and I've been trying to connect them in order to understand what this wavelength actually means.

In double slit diffraction, when the wavelength is larger, the diffraction "stripes" that form on the wall appear further apart. They also appear larger.

y = $\displaystyle{\frac{m\lambda L}{d}}$ (approx, considering the distance to the screen is really large and thus almost parallel rays (drawn out waves) can have a path difference and interfere)

If we were to make the wavelength extremely small, that would mean that anything a little off-center would interfere, so the smaller the wavelength, the closer together the "stripes" on the wall would be.

Now, when we connect the 2 equations, this means that the faster the electrons are moving (the smaller their wavelength) the more places they will interfere on the wall, and therefore there will be a lesser distance between adjacent places where the electrons hit (bright spots) and places where they don't (dark spots).

The way I'm interpreting this is that the smaller the "wavelength" of the electron, the more the probability it has to have been in different places at the same time, that is, the less we can know its position. That's why more stripes will appear on the detecting screen because there are more positions which the electron could've been in, and since its technically in all of them at the same time while it travels, it can interfere with itself more.

Is this interpretation correct? Does a faster momentum (a smaller wavelength) mean that the electron literally is at more places at the same time while it travels from the electron gun, through the slits, and to the wall? Thank you!

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  • $\begingroup$ The number of spots or stripes is not the way to gauge the uncertainty in position. Rather, the overall distribution of the spots or stripes tells you the position uncertainty. If you have two or three stripes covering a region 2 cm wide, or 50 stripes covering the same region, the uncertainty in position is the same. $\endgroup$ – S. McGrew Aug 19 '18 at 14:22
  • $\begingroup$ Ooook, so if they are more spread out that means more uncertainty or if they're closer together that means more uncertainty @S.McGrew ? $\endgroup$ – joshuaronis Aug 19 '18 at 14:31
  • $\begingroup$ I think you meant "Okaaaaay", not the simian sounding "ooook". $\endgroup$ – JEB Aug 19 '18 at 15:18
  • $\begingroup$ @JoshuaRonis, if all the the fringes (stripes or spots) together cover a larger total area, the position uncertainty is greater. The interference pattern you see on the screen is a map of the magnitude of the wavefunction: the probability density of finding a photon at that location. $\endgroup$ – S. McGrew Aug 19 '18 at 15:52
  • $\begingroup$ @JEB :) Yeaaaaaaa $\endgroup$ – joshuaronis Aug 19 '18 at 20:21
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The uncertainty that matters is transverse. Imagine an monochromatic plane wave (of infinite extent, wavenumber $\vec k = k \hat z = (2\pi/\lambda) \hat z$) impinging normally on the slit apparatus (one slit, width $w$ in the $x$-direction).

The uncertainty in the transverse momentum is:

$$ \Delta p_x = 0 $$

Now it goes through the slit. We have now localized an infinite plane wave within a region of extent $w$:

$$ \Delta x = w $$

It acquires an uncertainty in transverse momentum such that:

$$ \Delta p_x \Delta x \approx \hbar/2 $$

Or:

$$ \Delta p_x = \frac{\hbar}{2w} $$

Hence:

$$ \Delta k_x = \frac{1}{2w} $$

That means there is an angular spread in the wave emanating from the slit:

$$ \Delta\theta = \frac{\Delta k_x}{||k||} = \frac{\lambda}{\pi w}$$

That is, of course, diffraction. Diffraction can be viewed as a consequence of the uncertainty in position at the slit.

With 2 slits separated by $d$, the fringe-spacing (or rate of change of phase difference) can be computed with trig without appealing to the uncertainty principle.

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Does a faster momentum (a smaller wavelength) mean that the electron literally is at more places at the same time while it travels from the electron gun, through the slits, and to the wall?

The electron is an elementary particle in the standard model of particle physics and its study belongs to the quantum mechanical framework.

In quantum mechanics , particles and atoms and molecules follow the dynamical equations of quantum mechanics. The deBroglie wavelength is a useful handle for the quantum mechanical behavior of particles, as long as one understands that it is the probability distributions of the particle interactions that describe a wave behavior. The particles themselves when measured and in interactions have a specific (x,y,z,t) and appear as spots in single particle at a time experiments detecting them. They are not spread over space; one has to do the experiment many times with the same boundary conditions to record the probability distribution and observe the wave nature . The single electron at a time double slit is such an example.

A specific electron is not spread out over space, when detected it has a clear (x,y,z) footprint, characteristic of its particle nature. The wave nature is in the probability distribution.

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The de Broglie wavelength is the wavelength associated with a particle with a fully-specified spatial momentum (which is impossible in reality): namely, the position-space wave function of a moving particle in a momentum eigenstate $|\mathbf{p} = \mathbf{p}_\mathrm{given}\rangle$ is, at any given time,

$$\psi(P) = Ae^{i[\mathbf{p}_\mathrm{given} \cdot (P - O)]/\hbar}$$

where $O$ is the origin and $A$ describes the phase. The periodicity is in the direction of $\mathbf{p}_\mathrm{given}$ and has spatial length

$$\lambda = \frac{\hbar}{|\mathbf{p}_\mathrm{given}|}$$

which is the de Broglie wavelength.

For particles in more complicated states, the de Broglie wavelength is less evident, but it still pops up in many cases as a characteristic natural scale, particularly when we can treat a particle's momentum as relatively well-specified (having a relatively high information content).

The reason it appears in the double-slit experiment is it is useful to model the incoming particles, on the gun-facing side of the slit, as plane waves of this form.

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It appears that the question is seeking a connection between the wavelength and the size of a particle or its wavefunction. Actually, there is no such connection.

"Wavelength" corresponds to how quickly the phase of the particle changes along a trajectory. The square of the "Amplitude" at a point in space corresponds to the probability density for finding the particle at that point.

The size of the particle is independent of the wavefunction. For example, any measurement of the size of a proton or electron will give the same result regardless of the particle's wavelength or phase.

The size of the region over which it is likely that the particle can be detected corresponds to the region over which the amplitude of the particle's wavefunction is big enough to matter in the detection experiment. But: if the probability density over all that region is added up, (that is, if the integral of the squared amplitude over the region is calculated), the result is "1" or less. The total probability of finding the particle someplace can't be greater than "1".

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