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Difference/similarity between superposition and uncertainty

  1. Feb 3, 2016 #1
    What is the relation between "superposition" and "the Heisenberg uncertainty relation"?
     
    Last edited: Feb 3, 2016
  2. jcsd
  3. Feb 3, 2016 #2

    jfizzix

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    When the quantum state of a system [itex]|\psi\rangle[/itex] is expressed as a sum over multiple eigenstates of an observable [itex]\hat{Q}[/itex], we say that the system is in a superposition of these different eigenstates (where each eigenstate corresponds to a particular possible measurement outcome of [itex]\hat{Q}[/itex]). Depending on the particular superposition, you could be almost equally likely to measure any outcome of [itex]\hat{Q}[/itex] (maximum uncertainty), or have one outcome's probability be much larger than the others (minimum uncertainty).

    If [itex]|\psi\rangle[/itex] happens to be in a single eigenstate of [itex]\hat{Q}[/itex], then when you measure [itex]\hat{Q}[/itex], you will get exactly the outcome associated to that eigenstate with 100 percent probability. In this case, the uncertainty in [itex]\hat{Q}[/itex] is zero, since you know exactly what your measurement outcome would be.

    However, just as [itex]|\psi\rangle[/itex] can be expressed as a sum over the eigenstates of [itex]\hat{Q}[/itex], it can also be expressed as a different sum over the different eigenstates of another observable [itex]\hat{R}[/itex].

    The uncertainty principle comes in to play for observable pairs [itex]\hat{Q}[/itex] and [itex]\hat{R}[/itex] where there is no eigenstate of [itex]\hat{Q}[/itex] that is also an eigenstate of [itex]\hat{R}[/itex]. When that happens, there is no quantum state [itex]|\psi\rangle[/itex] where you will be able to predict the measurement outcome of both [itex]\hat{Q}[/itex] and [itex]\hat{R}[/itex] with 100 percent certainty.
     
  4. Feb 3, 2016 #3
    Thank you very much for the explanation! :smile: To get this clear: is the Heisenberg uncertainty relation equal to this general form, or is it specific to impulse and position?
     
  5. Feb 3, 2016 #4

    jfizzix

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    The Heisenberg relation was originally expressed in terms of position and momentum, but it wasn't long before it was defined for general pairs of observables (where the position-momentum pair is a special case).
     
  6. Feb 3, 2016 #5
    Is the superposition of eigenstates called a "mixed state"?
    Is that non-commutation?
     
  7. Feb 3, 2016 #6

    jfizzix

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    Yes; that and non-commutation are equivalent.

    Oops.. I answered the second question without looking at the first one. See radium's response for an answer to the first one.
     
    Last edited: Feb 5, 2016
  8. Feb 3, 2016 #7

    radium

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    No, a mixed state is a statistical ensemble of several quantum states. Like an ensemble of spin 1/2 particles. The density matrix would be diagonal
     
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