Planck's assumption and Uncertainty principle

  • Thread starter kevin0960
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Hi guys,

I earned that

[tex] \Delta E \Delta t \gtrapprox \frac{\hbar}{2} [/tex]

But one thing that's really strange is, according to Planck's assumption on black-body radiation problem, it just say that energy of each osciliator can be written as

[tex]E = nhf[/tex]

where n is an integer. (I hope this would be true)

So, my question is that if we can measure the energy of an osciliator precisly enough so that Energy of a certain osciliator can meet this inequality.

[tex](n - 1)hf < E < (n+1)hf[/tex]

At the same time, we can measure [tex] \Delta t[/tex] in some way but it won't be [tex]\infty[/tex] cause [tex] \Delta E[/tex] is obviously not 0.
But here is the problem, we can actually know the energy of osciliator because the energy of it have to be multiple of n. So, E is nhf, and we can get the [tex] \Delta t[/tex] not an infinity value which is not reasonable. Because if we measure the energy of an object, the uncertainty of t have to be infinite.

So, what is the problem?

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I'm not good at eng though :(
 

DrClaude

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For starters, a caveat: the energy-time uncertainty relation is not a case of the Heisenberg uncertainty principle, since time is not an operator in QM.

If an eigenstate has a precisely defined energy, then indeed ##\Delta E = 0##, which would mean ##\Delta t = \infty##, which is not absurd in that it means that that state has an infinite lifetime. A system in that state will remain in such a state forever.

This is why states with finite lifetimes (for example, atomic states that can decay by emission of a photon) don't have a precise energy, but a certain width in energy.
 

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