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Suppose we have normalised and orthogonal wavefunctions $\psi_1$ and $\psi_2$ as different solutions to the same equations with eigenvalues of $\lambda_1$ and $\lambda_2$ for the momentum operator $\hat p$. Then, I construct a new wavefunction as $\psi = sin \theta \cdot\psi_1 + cos \theta \cdot \psi_2$.

Now $\psi$ is normalised. On applying the momentum operator, we have $$\hat p\psi = sin \theta \cdot \lambda_1\psi_1 + cos \theta \cdot \lambda_2\psi_2$$ As far as I can tell, this $\psi$ doesn't turn out to be an eigenfunction of $\hat p$. What is happening here? It should have an eigenvalue, right?

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closed as unclear what you're asking by Norbert Schuch, user36790, ACuriousMind, Jon Custer, Cosmas Zachos Sep 21 '16 at 21:30

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    $\begingroup$ Nope, the superposition of two eigenvectors is not necessarily an eigenvector. It is only an eigenvector if the original ones were degenerate. $\endgroup$ – Jannick Sep 20 '16 at 8:05
  • $\begingroup$ It's not clear why you think this should be an eigenfunction. $\endgroup$ – ACuriousMind Sep 20 '16 at 13:20
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As far as math is concerned, I cannot put it better then @Jannick did in the comment (which I've upvoted): no, a sum of eigenvectors with different eigenvalues is not an eigenvector.

As far as physics is concerned, $\psi_{1,2}$ admit the following interpretation: whenever you measure $p$ and the system is at one of these states, you will definitely get $\lambda_{1,2}$ as a result. The superposition $\psi$ is interpreted as follows: when measuring $p$ you could get either $\lambda_1$ or $\lambda_2$ with probabilities given by $\sin^2 \theta$ and $\cos^2 \theta$ respectively. Thus $p$ does not have a definite value when your system is at $\psi$, so $\psi$ it is not an eigenstate of $p$.

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  • $\begingroup$ As far as the math is concerned, the superposition of eigenvectors is almost never an eigenvector. In fact the linear span of eigenvectors will be the set of all quantum states. $\endgroup$ – Neal Sep 20 '16 at 10:19
  • $\begingroup$ So what would be the equation for the value of $p$? $\endgroup$ – Akshit Sep 21 '16 at 4:37
  • $\begingroup$ And do we now calculate $<p>$ the same as before, $<\psi \mid \hat{p} \mid \psi>$? $\endgroup$ – Akshit Sep 21 '16 at 4:39
  • $\begingroup$ @Akshit sure. Its easy to show that it is equal to $\sin^2 \theta \cdot \lambda_1 + \cos^2 \theta \cdot \lambda_2$, so the probabilistic interpretation is correct. $\endgroup$ – Prof. Legolasov Sep 21 '16 at 4:50

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