Let's consider $\left| \Psi \right> $ some state of quantum system.

Let's also consider some Hermitian Operator $\hat{Q}$, with the discrete specrum: $$ \hat{Q}\left| Q_n \right> = Q_n\left| Q_n \right>. $$

Now, if one try to mearure some physical quantity $\hat{Q}$ in non eigenstate $\left| \Psi \right> $, the result should be one of the eigenstates of $\hat{Q}$, say $\left| Q_n \right> $. $$ \hat{Q}\left| \Psi \right> = Q_n\left| Q_n \right>. $$ Thus, we observe what called wave function collapse: $$ \left| \Psi\right> \xrightarrow{\text{collapse}} \left| Q_n \right>. $$

As described in literature, the definition of an operator is $$ \hat{Q}\left|\Psi\right> = \left|\Phi\right>, $$ the thing that convert one state vector $\left|\Psi\right>$ to another $\left|\Phi\right>$. I had never met that for the Hermitian Operator acting to non eigenstate gives the operator's eigenstate.

So, is the equation $\hat{Q}\left| \Psi \right> = Q_n\left| Q_n \right>$ mathematically correct?

Or, from another point of view, Does it correct to say The Hermitian Operator Reduce the Noneigenstate State to the Self Eigenstate?

  • 1
    $\begingroup$ There is no understanding of the phenomenon here, just a description. The proper English name is wave function collapse and its proper understanding (and even whether it happens at all) is tightly linked with quantum interpretations in general and the measurement problem specifically. I'm therefore not quite sure what you expect as an answer to this question. $\endgroup$
    – ACuriousMind
    Commented Dec 9, 2018 at 11:20

1 Answer 1


The measurement of an observable $\hat{Q}$ on a state $|\Psi\rangle$ is not represented by the equation $$ \hat{Q}|\Psi\rangle = Q_n |Q_n\rangle.\qquad \rm (wrong!)$$ This is a common misconception for those learning QM for the first time.

A Hermitian operator $\hat{Q}$ represents an observable in the sense that:

  1. The eigenvalues $Q_m$ of $\hat{Q}$ represent the physical outcomes of the measurement, e.g. different possible values of charge, momentum, energy, etc.
  2. The probability of obtaining outcome $Q_m$ is given by $p_m = |\langle Q_m|\Psi\rangle|^2$, where $|Q_m\rangle$ is the eigenvector of $\hat{Q}$ associated with eigenvalue $Q_m$, i.e. $\hat{Q}|Q_m\rangle = Q_m|Q_m\rangle$.
  3. The post-measurement state, conditioned on the fact that outcome $Q_m$ was obtained, is $|Q_m\rangle$.

These properties do not imply that measurement is represented by "applying operator $\hat{Q}$ to state $|\Psi\rangle$".

  • $\begingroup$ I agree with you the $Q_n \left| Q_n \right> = \hat{Q}\left| \Psi \right>$ wrong. But, for general case $\hat{Q}\left|\Psi\right> = \left|\Phi\right>$, from the acting $\hat{Q}$ to $\left|\Psi\right>$ we must get some vector. What physical meaning of this vector? $\endgroup$
    – Sergio
    Commented Dec 9, 2018 at 12:07
  • $\begingroup$ @Sergio Please do not edit your question to completely change its meaning and invalidate the given answers. It is not fair to the people who put time into answering. If you have a different question, you should first check to see if it has been asked already, and if not you should make a new post. In this case, your new question is essentially a duplicate of physics.stackexchange.com/questions/44366/… $\endgroup$ Commented Dec 9, 2018 at 14:19

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