Atomic Clock theory

  • #1

Summary:

Can Quantum mechanics calculate the 9192631770 Hz frequency from first principles.

Main Question or Discussion Point

I have searched in vain for a detailed calculation of the frequency 9192631770 Hz from basic physical constants using Quantum Mechanics.
Can anyone help with this please
 

Answers and Replies

  • #2
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Summary:: Can Quantum mechanics calculate the 9192631770 Hz frequency from first principles.

I have searched in vain for a detailed calculation of the frequency 9192631770 Hz from basic physical constants using Quantum Mechanics.
Can anyone help with this please
There is no such calculation. It is a definition. The second is a man-made quantity and could be any number of oscillations of the Caesium hyperfine transition. The BIPM picked a number and defined it thus.
 
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  • #3
Vanadium 50
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Dale is right, it's a definition.

However, what you may have meant is if 9,192,631,770 was calculable using the pre-1967 definitions. The answer is still no. This is a 56-body problem and there is not even a classical solution for 3 bodies. It would have to be calculated numerically and that is well beyond today's generation of computers, and the next, and the one after that.

More fundamentally, the mass of the electron is known 10x worse than the number you are trying to calculate, so you will have an input uncertainty an order of magnitude too large. (If you could do the calculation, you could turn it around and use this to measure the electron's mass more precisely)
 
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  • #4
It is more than a number, it is a frequency corresponding to a coupling energy between the nuclear magnetic moment and the electron. Can QM calculate the energy levels of the doublet?
 
  • #5
DrClaude
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It is more than a number, it is a frequency corresponding to a coupling energy between the nuclear magnetic moment and the electron.
I can quite confidently say that everyone who replied in this thread understands this quite well.

Can QM calculate the energy levels of the doublet?
Yes, but not at the level of precision needed to match exactly the number used in the definition of the second. See @Vanadium 50's answer above for the explanation as to why this is the case.
 
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