# Questions about expectation values and definite values (quantum physics)

1. Mar 11, 2009

### kehler

Is the expectation value of momentum/position/energy the value that we're most likely to measure? So suppose we measure 100 particles with the same wavefunction, would we expect most of them to have momentum/position/energy that's equal to the expectation value? And I was wondering, how do we even measure the momentum of a particle described by a wavefunction?

Also, suppose a wavefunction is found to have a definite value of momentum/position/energy, does this necessarily mean that the expectation value of its momentum/position/K.Energy will be equal to that definite value?
And lastly, is it possible to have a wavefunction that has both definite momentum and position?

I'm a bit confused by these concepts. Any help would be much appreciated :)

2. Mar 12, 2009

### lanedance

no if you made the same measurement on repeatedly on particles perpeared in the same state it is the average of all you measurements. Consider a particle in a even superposition of spin up or down states (z axis). The expectation value is 0 - If you measure spin in the z axis, you will only ever find it up or down, 50% of the time for each in average, never zero.
no you would expect the average of all your measurements to be the expectation value. Whether it is most of them will depend on the probaility distribution gerated from the wavefunction
not totally sure, there's probably many differnt methods & perhaps someone can correct me, but something like measuring the path in magnetic field(eg. bubble chamber) will give a charge to speed ratio and so yield momentum... (though that raises an interesting question as to measure path you are measureing position...)
theoretically yes though it would mean you other variable is totally unkown, which is not very physical, relations are
dp.dx and dE.dt >= hbar/2
no, the unceratinty princlple give the product of the unceratinties as:
dp.dx >= hbar/2
so if there is only dx uncertainty in x, there must be at least dp = hbar/(2dx) uncertainty in p
google could help as well there's quite a few discussions going round, or you could try the PF library...

3. Mar 12, 2009

### kehler

Thanks :). That helped.

4. Mar 12, 2009

### turin

If you really want it to be described by a wavefunction, then diffraction and scattering come to mind, such as sending the particle through a grating (or crystal lattice). deBroglie told us that the wavelength is inversely proportional to the momentum through the proportionality constant, h, so if you can determine the wavelength, then you can determine the momentum. Notice that diffraction and scattering obscure position, so there's your uncertainty principle at work.

The way momentum is measured in practice is basically what lanedance said. However, this relies on the particle nature rather than the wave nature, so it isn't really like measuring the momentum of the wavefunction. In the big colliders, they use silicon, wire mesh, and electronic channel readout instead of superheated liquid and photography, but the kinematical concept is the same: the momentum is proportional to the radius of curvature of the track.

I will also point out that the tracking measurement that lanedance and I described requires some finite length of track, which translates into an uncertainty in position (in the same direction as the momentum being measured, as it should). The longer the track the better the momentum measurement the more uncertain is the position. So there's your uncertainty principle at work again. You cannot escape from it even if the particle is acting like a particle. (Actually, that really isn't the uncertainty principle at work, it's detector resolution, which is much much worse than uncertainty-principle-limited. In order for tracking to be truly limited by the uncertainty principle, you would have to resolve the track down to the scale at which it receives random kicks from the magnetic field.)

Last edited: Mar 12, 2009
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