Heisenberg's Uncertainty - ΔλΔx

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  • #1
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So we have a light bulb emitting through a small slit. I read somewhere that a blackbody shining through a small slit acts like a perfect emitter, and I wanted to know why. I figured it must have something to do with Heisenberg's Uncertainty. Here's how I went about it:

[tex]\Delta x\Delta p \geq \frac{\hbar}{2}[/tex]

The momentum of light is just E/c, so

[tex]p = \frac{E}{c} = \frac{h}{\lambda}[/tex]

and

[tex]\lambda^{2} = \frac{h^{2}c^{2}}{E^{2}} [/tex]

To find Δp, just take the derivative wrt λ

[tex]\frac{\Delta p}{\Delta \lambda}=\frac{-h}{\lambda^{2}}[/tex]

and

[tex]\Delta p = \frac{h}{\lambda^{2}}\Delta \lambda[/tex]

I ignore the negative because this is uncertainty in either direction. So..

[tex]\Delta x \Delta p = \Delta x\frac{h}{\lambda^{2}}\Delta\lambda[/tex]

Then I subbed back into the first eq and solved for ΔxΔλ

[tex]\Delta x \Delta \lambda \geq \frac{\lambda^{2}}{4\pi}[/tex]

thus

[tex]\Delta x \Delta \lambda \geq \frac{h^{2}c^{2}}{4\pi E^{2}}[/tex]

Is this why they say that the light shining through a small slit a near perfect emitter? Because since we don't know exactly where the light goes through the slit, it must have an uncertainty in the wavelength when it comes out. If so, what is that E for? Is it the energy of the photons before they enter the slit? Thanks.
 

Answers and Replies

  • #2
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The correct equation is [itex]\sigma_x\sigma_p \geq \frac{\hbar}{2}[/itex], where [itex]\sigma[/itex] denotes standard deviation. I'm not sure it's correct to treat them as [itex]\Delta[/itex]s the way you did, treating them as differentials. Maybe if I knew statistics better I would be able to tell you.
 
  • #3
Bill_K
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So we have a light bulb emitting through a small slit. I read somewhere that a blackbody shining through a small slit acts like a perfect emitter, and I wanted to know why. I figured it must have something to do with Heisenberg's Uncertainty.
This is known as cavity radiation. EM waves inside a hollow body interact with the walls. Given enough time they come to thermal equilibrium. The hole gives them a way to escape and be observed. The fact that the hole is "small" retains them long enough to equilibrate into black body radiation. Uncertainty principle not involved.
 
  • #4
jtbell
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The correct equation is [itex]\sigma_x\sigma_p \geq \frac{\hbar}{2}[/itex], where [itex]\sigma[/itex] denotes standard deviation. I'm not sure it's correct to treat them as [itex]\Delta[/itex]s the way you did, treating them as differentials. Maybe if I knew statistics better I would be able to tell you.
In physics, we traditionally indicate uncertainties with Δ (e.g. Δx, Δy, Δp), with the understanding that they are standard deviations. Usually we assume a normal (Gaussian) distribution unless explicitly specified otherwise.
 
  • #5
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In physics, we traditionally indicate uncertainties with Δ (e.g. Δx, Δy, Δp), with the understanding that they are standard deviations. Usually we assume a normal (Gaussian) distribution unless explicitly specified otherwise.
Well if you read the OP, (s)he also treated them as differentials e.g. [itex]\frac{\Delta f}{\Delta x} = \frac{d f}{d x}[/itex]. My comment was addressing this.
 
  • #6
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Well if you read the OP, (s)he...
It's he :p

So I'm not allowed to use them as differentials? My modern professor has done so on a few occasions to get at a couple more uncertainty relations.
 
  • #7
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It's he :p

So I'm not allowed to use them as differentials? My modern professor has done so on a few occasions to get at a couple more uncertainty relations.
I think there is the approximation

[itex]\frac{\sigma_{f(X)}}{\sigma_X} \approx f'(E[X])[/itex]

Where [itex]E[X][/itex] is the mean value of X. However I think it's important that you understand the distinction. The expression on the left is not a derivative.

This is a relation between the standard deviations of a random variable and a function of that random variable that comes from a first order Taylor series expansion. It will not necessarily always give decent results.

source
 
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  • #8
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Okay thanks a lot, I'll have to be more careful in the future. I'm going to start using the sigma notation instead of delta, just to remind myself.
 
  • #9
ZapperZ
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I'm scratching my head here because the actual, relevant response to the entire premise has been completely overlooked. Bill K's response addresses directly the starting scenario, especially the part where this has nothing to do with the HUP.

Somehow, the emphasis on the distraction, i.e. the notation or the validity of the derivation took center stage.

Zz.
 
  • #10
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Oh man I didn't even see his post. :uhh:

As it happens, after my original post, I read the Wiki page on black bodies, which has a section about cavity radiation. But I appreciate the answer and wish I'd have actually seen it.
 

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