- #1

bhthiang

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At the same time, for photons we take lamda= h/p as the actual/physical wave length of the light.

Can someone shed some light on this apparent inconsistency?

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- Thread starter bhthiang
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- #1

bhthiang

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At the same time, for photons we take lamda= h/p as the actual/physical wave length of the light.

Can someone shed some light on this apparent inconsistency?

- #2

futz

- 80

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Originally posted by bhthiang

At the same time, for photons we take lamda= h/p as the actual/physical wave length of the light.

Can someone shed some light on this apparent inconsistency?

I don't see any inconsistency. If you perform an electron diffraction experiment (for example, Bragg reflection in a crystal lattice) diffraction will occur for specific values of d (lattice spacing), theta (angle of incidence) and lambda. If you can get values for d and theta, and calculate the wavelength, it should correspond (within experimental error) to the theoretical value given by the de Broglie relation.

- #3

bdkeenan00

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- #4

futz

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Originally posted by bdkeenan00

The HUP says that dx*dp = h. That is, the product of the uncertainty in x (position) and p (momentum) is equal to a constant h (Planck's constant). So, the smaller one of these gets, the larger the other one does to keep this relation. Or, the more precisely we know one value, the less precisely we know the other.

So, if we could measure the momentum of a particle to be exactly zero, then technically it would have infinite wavelength, which is of course not physically meaningful. The HUP prevents this from happening. If we measured exactly 0 for p, then dp would also be 0, meaning dx would have to be infinite -- clearly, we could not tell anything useful about its position with an infinite uncertainty (not to mention the above equality would be violated).

- #5

Sonty

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- #6

bdkeenan00

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- #7

futz

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- #8

bhthiang

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Thanks for your reply.

So what you are saying is that the "physical" wavelength is also the wavelength of the wave function, whose modulus calculated at a location represents the probability of finding the electron at that position?

Can we apply this to photons and say that their wavelengths are the same as the wavelengths of their wave function?

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