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Quantum Mechanics Operator

  • Thread starter astrozilla
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  • #1
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Homework Statement


How can we compute the error or uncertainty in measuring an operator O ?


Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
dextercioby
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Technically, we don't measure the operator, but the observable. We measure whatever we've got instruments for, depending on the particularly chosen experimental set up. We can measure energy, wavelength, spin component, etc. As for the errors, this is statistical analysis. You record the numbers and then play with them.
 
  • #3
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This is definately a very stupid exam question.
I just thought that there is some mathematical formula or that ΔO (uncertainty) is related somehow to Werner Heisenberg's uncertainty principle.
 
  • #4
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This is probably what you're looking for:

The uncertainty of an observable Q is:

[tex](\Delta Q)^2_{\psi}=\langle Q^2 \rangle_{\psi}-\langle Q \rangle^2_{\psi}[/tex]
 
  • #5
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o:)

...
 
  • #6
dextercioby
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This is definately a very stupid exam question.
I just thought that there is some mathematical formula or that ΔO (uncertainty) is related somehow to Werner Heisenberg's uncertainty principle.
It is stupid, because it's first of all poorly worded and then wrongly formulated. As i said above, we measure obervables, but we can't compute the errors, nor the any uncertainty. We can compute some statistical quantities, like mean square deviation, dispersion, mean, probability of an outcome, etc.
 
  • #7
vela
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If you know what the state of the system is, you can calculate the uncertainty in an observable without making any measurements. The uncertainty is inherent to the state. In the case of the observable [itex]\hat{x}[/itex], for example, the uncertainty Δx simply characterizes the spatial extent of the wave function ψ(x).

You could, of course, take a bunch of identically prepared systems and perform the same measurement on each, and you would find that the spread in the results reveals this inherent uncertainty in the state.
 

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