Difference/similarity between superposition and uncertainty

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entropy1
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What is the relation between "superposition" and "the Heisenberg uncertainty relation"?
 
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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 into 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.
 
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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?
 
entropy1 said:
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?

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).
 
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Is the superposition of eigenstates called a "mixed state"?
jfizzix said:
The uncertainty principle comes into 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].

Is that non-commutation?
 
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.
 
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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