# Difference/similarity between superposition and uncertainty

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1. Feb 3, 2016

### entropy1

What is the relation between "superposition" and "the Heisenberg uncertainty relation"?

Last edited: Feb 3, 2016
2. Feb 3, 2016

### jfizzix

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

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

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

The uncertainty principle comes in to play for observable pairs $\hat{Q}$ and $\hat{R}$ where there is no eigenstate of $\hat{Q}$ that is also an eigenstate of $\hat{R}$. When that happens, there is no quantum state $|\psi\rangle$ where you will be able to predict the measurement outcome of both $\hat{Q}$ and $\hat{R}$ with 100 percent certainty.

3. Feb 3, 2016

### entropy1

Thank you very much for the explanation! To get this clear: is the Heisenberg uncertainty relation equal to this general form, or is it specific to impulse and position?

4. Feb 3, 2016

### jfizzix

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).

5. Feb 3, 2016

### entropy1

Is the superposition of eigenstates called a "mixed state"?
Is that non-commutation?

6. Feb 3, 2016

### jfizzix

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
7. Feb 3, 2016