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Author Topic: How does one measure the momentum of a quantum particle?  (Read 2799 times)

Offline Pmb

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As I'm brushing up on my quantum mechanics it has occured to me that when its said that one measures the observable momentum etc. that I don't know how to measure momentum of quantum mechanical paricles.

Does anybody have any idea of how to measure the momentum of, say, a free electron or photon?


 

Offline syhprum

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Re: How does one measure the momentum of a quantum particle?
« Reply #1 on: 10/07/2012 11:04:13 »
The deflection of the path of an electron thru a magnetic field of known strength can be observed in a cloud chamber and its velocity calculated.
This was a very old technique and has probably long been superseded.
 

Offline JP

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Re: How does one measure the momentum of a quantum particle?
« Reply #2 on: 10/07/2012 18:49:21 »
They still use the same basic principle that syphrum mentioned: http://public.web.cern.ch/public/en/research/Detector-en.html#tracking

This only works for charged particles, of course, but most particles in the standard model are either photons (which are fairly easy to detect), charged, or very unstable (and generally quickly decay into charged products).  The big exception are neutrinos, which are very rarely directly detected.
 

Offline Pmb

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Re: How does one measure the momentum of a quantum particle?
« Reply #3 on: 11/07/2012 12:19:46 »
The deflection of the path of an electron thru a magnetic field of known strength can be observed in a cloud chamber and its velocity calculated.
This was a very old technique and has probably long been superseded.
I was thinking abstractly, more along the lines of measuring a quantum mechanical system. For example: If you have a particle in a box then cn you measure the partilces momentum with a single measurement? Moving through a magnetic field means that its position is being measured in a continuals sense. Is that neccesary? What about chargless particles like neutrinos
 

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Re: How does one measure the momentum of a quantum particle?
« Reply #3 on: 11/07/2012 12:19:46 »

 

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