Uncertainty Principle and Single-Slit Diffraction

[peter0302]

I'm trying to calculate the uncertainty in momentum/direction resulting purely from the certainty in position caused by a small slit, as a function of the slit width. Can anyone tell me if I'm doing this right?

Starting with the HUP:
dX * dP >= h/4pi

dX = the slit width w

P = h/lambda * sin(theta)
where theta is the angle that the photon might be traveling after the slit relative to the center of the slit

so

dP = 2 * h/lambda * sin(theta)
A factor of 2 because the uncertainty can extend in either direction from the center of the slit

so

w * 2 * h/lambda * sin(theta) >= h / 4pi
sin(theta) >= lambda / (8pi * w)

Is that right?

My only concern with the calculation is that "w" could be so small that sin(theta) could be greater than 1, which is of course impossible, so that's why I think I made an error somewhere.

Thanks for any ideas!

 

[lbrits]

I guess the resolution to the problem is the assumption that
$$|p| = \frac{\hbar}{\lambda}$$
after the diffraction after the diffraction. In fact, for the particle to pass through the narrow slit, it's momentum is not a single value, but given by a probability distribution. The width of that distribution is what you'd call $$\Delta p_x$$. If the slit is made very narrow, $$\Delta p$$ can grow arbitrarily large, and so $$\langle \sqrt{|p|^2}\rangle$$ can grow arbitrarily large.

 

[peter0302]

Ok .. so how do I calculate the probability distribution after going through the slit?

 

[lbrits]

Either solve the Schrödinger equation or use Huygen's principle. Incidentally, in the far field region, the spatial wave-function is well approximated by the spatial Fourier transform of the aperture. The uncertainty principle applies to Fourier transforms also.

 

[peter0302]

Show me how to solve the Schrodinger equation for a photon.

Also I thought Huygen's principle doesn't work when the slit is smaller than a wavelength. When a slit is larger than a wavelength, you get single-slit diffraction which is not what I want.

In fact, for the particle to pass through the narrow slit, it's momentum is not a single value, but given by a probability distribution.

The magnitude of the momentum is always given by h/lambda. It's only the vector components, i.e., direction, that change due to Heisenberg Uncertainty. Otherwise light would change color after going through a slit and all lasers would be white.

 

[peter0302]

Ah, can I just use the Fraunhofer diffraction formula:

$$\psi$$($$\theta$$)=sinc($$\pi$$a$$\theta$$/$$\lambda$$)

?

 

[Lojzek]

Show me how to solve the Schrodinger equation for a photon.

Also I thought Huygen's principle doesn't work when the slit is smaller than a wavelength. When a slit is larger than a wavelength, you get single-slit diffraction which is not what I want.


The magnitude of the momentum is always given by h/lambda. It's only the vector components, i.e., direction, that change due to Heisenberg Uncertainty. Otherwise light would change color after going through a slit and all lasers would be white.

Momentum is always given by h/lambda, but lambda is sometimes not well defined. Far away from the slit waves will be nearly straight, so the momentum distribution will be delta(p-p0). But this is not true close to the slit. The momentum operator is defined by -i*2*pi*h* grad. As soon as you have more than one point source of waves, the sum of gradients can have different magnitude on diferent points.

The uncertainity principle alone can prove that the magnitude of p has a distibution with non-zero width when the uncertainity in position aproaches zero (since uncertainity in momentum aproaches infinity and becomes greater than the average momentum).
Actually your calculation was quite usefull for a slit with w>>lambda. I think you could also omit the number 2 in dp, since the uncertainity is defined by average distance from the center of distribution.

 

[peter0302]

Ok, so I just learned for the first time that in the familiar uncertainty relation
dX dP >= h_bar / 2

dX and dP are not literally the range of values X and P can have, but are the standard deviation in those values. Glad I know that now.

So the Fraunhoffer formula should provide the right wavefunction for the photon going through a single slit.

 

[lbrits]

Show me how to solve the Schrodinger equation for a photon.

Cute.

 

[lbrits]

Huygen's principle works just fine at all wave-lengths... it just needs to be applied properly.

 

[peter0302]

sin(theta) becomes greater than 1 when the slit width is less than the wavelength. So it simply doesn't work. Fortunately, we can make the slit width larger than a wavelength and still get uncertainty effects.