# How does uncertainty apply to an electron's magnetic moment measurement?

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1. Mar 8, 2015

### nortonian

The electron paths in the path integral formulation of qm are in ordinary space-time.

The end-points of the paths in the steady states must be in ordinary space-time as well.

The electron's potential is exact or nearly exact either by measurement or by calculation.

The electron's locations when it leaves the excited state and arrives at the ground state must be exact or nearly exact as well.

Is it possible to say therefore that uncertainty does not apply to space-time as a background for natural phenomena?

2. Mar 8, 2015

### jfizzix

The uncertainty principle does apply, for in the path integral formulation, the calculation of the probability amplitude of getting from A to B is a sum over all paths taken.

Plus, the uncertainty principle gives you a limit on how well you can predict the outcome of a pair of measurements (like position and momentum).

Using multiple path integrals, you can find the probability distribution of different final states for a given initial state.

What the uncertainty principle tells us here is that there is no interaction where the final states will be well localized in both position and momentum.

3. Mar 9, 2015

### nortonian

Sorry, I posed the question wrong.
As I understand the determination of an electron's magnetic moment a magnetic field is applied to a hydrogen atom and the change in frequency of the emission is observed. The energy of the emitted photons, whether measured or predicted, is nearly exact so for the energy-time uncertainty we have a very large spread for the time, true or false? In the case of the path integral formulation of qm all times, past and future, are considered so time of emission is indeterminate and this interpretation of uncertainty seems to be accurate.
I assume in the case of the experimental measurement the same is true: change in potential is precise but time is not.

4. Mar 9, 2015

### jfizzix

If a transition has a very well-defined frequency, the resulting transition lifetime will be broad in accordance with the energy time uncertainty principle.

In the path integral formulation (in my limited understanding), you calculate the transition amplitude from a given initial state to a given final state, with given initial and final times. This does not include all times, past present and future, but only those between and including the specified initial and final times.

If you calculate the transition amplitude from a given initial state to a given final state at different times, and re-normalize, you can get the probability density for the transition to occur within a given time, cinditioned on it occurring at all.

The energy-time uncertainty relation tells us that if this transition occurs with a very precise frequency, the uncertainty in the time (or decay lifetime) of the transition will be broad. What this means is that the the temporal probability density for the transition will be broad.

5. Mar 9, 2015

### naima

It was seen in another thread that Feynman never said that!

6. Mar 9, 2015

### jfizzix

I don't know enough about Feynman one way or the other to comment. Is what I've said wrong, though?

Last edited: Mar 9, 2015
7. Mar 10, 2015

### nortonian

You are using a statistical distribution typical of spectroscopic measurements. The actual experiment
ElectronMagneticMoment2006.pdf
uses a single electron in a Penning trap with classically induced one-quantum transitions. Spontaneous emission is inhibited so energy-time uncertainty cannot be applied as usually described in the literature
The path-integral formulation (also limited understanding) uses a Lagrangian where space and time coordinates are treated symmetrically as opposed to the Hamiltonian where they are not. So why is e-t uncertainty applied the same way in both formulations? It seems that there should be a different uncertainty relation for each method.
The transition occurs when there is classical resonance with the “wobble” frequency of the electron which is not the same as decay lifetime. Normally there is a statistical ensemble of atoms emitting radiation for e-t uncertainty derivations. In this case there is only one. I am not disputing that there is e-t uncertainty, but I'm trying to figure out how it is applied.

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8. Mar 15, 2015

### nortonian

There seems to be some interest in the experiment, but none in trying to resolve the problem; that is, to interpret it with the energy-time uncertainty relation.
My own opinion is that quantum mechanics, though very effective at analyzing statistical ensembles, has no good method of analyzing single events or processes on the microscopic level. As a result they end up as curiosities that cannot be described by either classical or quantum theory.
Are there any other ideas out there?

9. Mar 17, 2015

### nortonian

Here's another article of interest to the topic.

10. Mar 19, 2015

### nortonian

When we want to imagine how uncertainty applies to position-momentum we think of trying to localize a single particle with a single photon using a microscope. A more highly defined position leads to a less defined momentum and vice versa. However, if we use the experiment described in the above article which describes how a single electron changes energy levels, to imagine how energy-time uncertainty works the idea becomes more confusing. Instead of an indefinite measurement of energy leading to a more precise measurement of time we get a statistically smaller chance of the transition occurring at all. This is because the microwave stimulation is a classically varied parameter which causes a gradual increase in transitions as it approaches the actual transition value. On the other hand, if we consider the meaning of time in the interaction it is even more confusing. How does time uncertainty figure into it? Is it the actual clock time when the transition occurs or is it the time it takes the electron to change energy levels?

11. Mar 19, 2015

### Khashishi

How do you exactly measure the electric potential?

12. Mar 19, 2015

### nortonian

Check this link for a layman's description.
In my own words, energy is supplied to the oscillating electron with microwaves and when a transition occurs it drains energy from a resonant circuit without disturbing the motion of the electron itself. Thus the transition is observed in the resonant circuit. This is referred to as a Quantum Non-demolition measurement, meaning it occurs outside of the uncertainty principle. If a single electron can be observed undergoing a QND transition, why can't we assume that all transitions are outside the influence of uncertainty? In other words, the statistical ensemble used to define energy-time uncertainty in spectroscopic studies is actually a summation of exact individual transitions which are smeared out by thermal effects.