This is more a check to ensure I know the physical meaning of eigenvectors and eigenvalues in quantum mechanics, and to ask the general community if this is wrong:

On some observable, represented by the operator $\hat{Q}$ the eigenfunction of this operator is as follows: $$\hat{Q} f = qf$$

Where $q$ is the eigenvalue of the operator. In this case, $q$ represents values that one could get while attempting to measure $\hat{Q}$, and the eigenvector produced by $q$ is the state that the system is in at the time of the measurement that produces $q$. Is this right?


1 Answer 1


..the eigenvector produced by 𝑞 is the state that the system is in at the time of the measurement that produces 𝑞.

That's not correct. Eigenvectors aren't "produced" by an eigenvalue: eigenvectors are "produced" by the operator $\hat Q$, inasmuch as "produced" is understood to mean "vectors (or functions) on which the operator acts by multiplying the original vectors by numbers". Synonyms for "eigenvectors" are proper vectors (in French vecteurs propres) or sometimes characteristic vectors. Characteristic vectors is good because it implies that the eigenvectors are enough to characterize the operator.

If the system is prepared in an eigenstate or eigenfunction $f$ of $\hat Q$, then measuring the observable $Q$ will always yield the outcome $q$, and $q$ is the only outcome from a measurement of $\hat Q$ with the system prepared in the state $f$.

If you make a measurement of $Q$ and obtain the result $q$, then yes the system will "collapse" to an eigenstate of $Q$, and evolve from that collapsed state thereafter. If there is more than one eigenstate with this eigenvalue, things are a little more complicated as you can get linear combinations of those eigenstates with eigenvalue $q$.


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