Can Quantum Mechanics Calculate the Frequency of an Atomic Clock?

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Discussion Overview

The discussion centers around whether quantum mechanics can calculate the frequency of 9,192,631,770 Hz, which is defined as the frequency of the cesium hyperfine transition used in atomic clocks. Participants explore the theoretical and practical implications of this frequency in relation to quantum mechanics and physical constants.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant requests assistance in calculating the frequency 9,192,631,770 Hz from basic physical constants using quantum mechanics.
  • Another participant asserts that this frequency is a definition rather than a calculable quantity, indicating it is a man-made standard chosen by the BIPM.
  • A different participant agrees that it is a definition and adds that calculating this frequency using pre-1967 definitions is also not feasible due to its complexity, likening it to a 56-body problem.
  • Concerns are raised about the precision of the mass of the electron, suggesting that the uncertainty in this measurement would hinder any calculation of the frequency.
  • One participant emphasizes that the frequency corresponds to a coupling energy between the nuclear magnetic moment and the electron, questioning whether quantum mechanics can calculate the energy levels of the doublet.
  • Another participant reiterates the importance of the frequency as more than just a number and confirms that while quantum mechanics can calculate energy levels, it may not achieve the precision required to match the defined frequency.

Areas of Agreement / Disagreement

Participants generally agree that the frequency is a defined quantity rather than a calculable one. However, there is disagreement regarding the feasibility of calculating related energy levels and the implications of uncertainty in measurements.

Contextual Notes

Participants note limitations related to the precision of physical constants, particularly the mass of the electron, which introduces significant uncertainty into any potential calculations.

Bruce Harvey
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TL;DR
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
 
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Bruce Harvey said:
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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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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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?
 
Bruce Harvey said:
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.

Bruce Harvey said:
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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