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I've heard an analogy on the news regarding the Webb telescope. It said Webb's resolution is such that it would be able to locate from Earth a bumble bee on the moon. I understand that it will be a spaced based telescope, and it will never view the moon from Earth. My question is: will the Uncertainty Principle be violated with this telescope? I must confess I can't do the math, and I do not expect anyone else to do it for me. But what are your thoughts regarding the tremendous resolution of the telescope and the Uncertainty Principal. Am I way off-base on this?

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No. The uncertainty principle is a result that can be derived from physical optics. The calculation for the diffraction limit of a telescope is mathematically equivalent to the uncertainty principle.

In more detail, the Heisenberg uncertainty principle is nothing more than a relationship between the central second moments of distributions that are Fourier duals of each other. Take the Heisenberg uncertainty principle: $$\sigma_p \sigma_x \ge \frac{\hbar}{2}$$ now square both sides and divide by $\hbar^2$ to get: $$\sigma_k^2 \sigma_x^2 \ge \frac{1}{4}, $$ where the wave number is defined as $\mathbf{k} = \mathbf{p}/\hbar$. Go read the proof of the Heisenberg uncertainty principle on Wikipedia - it's derived entirely using the properties of Fourier transforms. The only difference is whether you introduce an unnecessary scaling of $\hbar$ to the wave number.

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  • $\begingroup$ Very interesting. The diffraction limit = the Uncertainty Principle. Almost makes me think the universe has a pixel size. Thanks $\endgroup$ – Lambda Nov 5 '16 at 4:34
  • $\begingroup$ I've never seen that shown or spoken. And don't see how it can be true. The diffraction limit is derived from physical optics, using CLASSICAL electromagnetism. There is no hint of quantum effects there which is where the uncertainty principle comes from, that's the diffraction limit and the uncertainty principle are the same for light. @Sean, can you prove what you said? $\endgroup$ – Bob Bee Nov 5 '16 at 5:02
  • $\begingroup$ @BobBee Really? I haven't taken a quantum mechanics class where it wasn't mentioned that the HUP is just a relationship that applies to variables that are Fourier duals of each other. $\endgroup$ – Sean E. Lake Nov 5 '16 at 5:30
  • $\begingroup$ Yes, but for the quantum effect you need to introduce h. In Fourier analysis I think the constant is something other than h. $\endgroup$ – Bob Bee Nov 5 '16 at 5:51

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