What is physical meaning of state getting by acting of Hermitian operator to non-eigenstate?

Question

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 $\left|\Phi\right>$.

For example, the one of such operators is time evolution operator $\hat{U}(t_1,t_2) = e^{-\frac{\hat{H}}{\hbar}(t_2 - t_1)}$: $$ \left|\Psi(t_2)\right> = \hat{U} \left|\Psi(t_1)\right>. $$

Here we know, the result of acting $\hat{U}$ to $\left|\Psi(t_1)\right>$ lead to some another state $\left|\Psi(t_2)\right>$.

Now, let's consider some Hermitian Operator $\hat{Q}$, with the discrete specrum: $$ \hat{Q}\left| Q_n \right> = Q_n\left| Q_n \right>, $$ acting on some non-eigenstate $\left| {\Psi} \right\rangle = \sum\limits_n \left| {Q_n}\right\rangle a_n = \sum\limits_n \left| {Q_n} \right\rangle \left\langle {Q_n} | {\Psi} \right\rangle $.

We get $$ \hat{Q}\left|\Psi\right> = \left|\Phi\right>,$$ where $$ \left|\Phi\right> = \sum\limits_n Q_n \left| {Q_n}\right\rangle a_n = \sum\limits_n Q_n\left| {Q_n} \right\rangle \left\langle {Q_n} | {\Psi} \right\rangle $$

Except for calculating averages, what physical meaning of $\left|\Phi\right>$? Is the real state, or some ghost? Because we know the measurement of physical quantity should collapsed $\left|\Psi\right>$ to some of $\{\left| Q_n \right>\}$.

Answer 1

$\def\ket#1{|#1\rangle}$

Except for calculating averages, what physical meaning of $\ket\Phi$? Is the real state, or some ghost? Because we know the measurement of physical quantity should collapsed $\ket\Psi$ to some of {$\ket{Q_n}$}.

I always wonder why it's so common for beginners in QM to confuse mathematical action of an operator on a ket vector with a physical measurement. As it always happens with mathematics applied to physics, there are some mathematical objects directly endowed of a physical significance, and others not. After all, in any mathematical sequence of passages there are intermediate steps whose only meaning is just of being temporary phases towards a meaningful result.

In case of QM we have a vector (Hilbert) space of kets, and operators defined thereon. This means that generally you are allowed to apply whichever operator to whichever ket (if it's in the operator's domain) and shouldn't expect to always get something meaningful.

In your example $\ket\Phi$ is no "ghost". It's a bona fide representative of a possible state, apart for not being normalized. This doesn't mean that you know how and when it can be reached starting from $\ket\Psi$. To know this you need more information on your system and its dynamics.

Answer 2

The physical action of a measurement of $Q$ is a projection operator which can be written $$ | {Q_n} \rangle \langle {Q_n} | $$

where $ \langle {Q_m} | {Q_n} \rangle = \delta_{mn}$, and $ \{Q_n \} $ runs over the possible values for the measurement. The probability of result $Q_n$ given the state $\left|\Psi\right>$ is

$$P(Q_n) = \langle\Psi| {Q_n} \rangle \langle {Q_n} \left|\Psi\right> $$

Then the expectation of the measurement in the state $\left|\Phi\right>$ is

$$ E(Q) =\sum\limits_n Q_n P(Q_n) = \sum\limits_n \langle\Psi| {Q_n} \rangle Q_n \left\langle {Q_n} | {\Psi} \right\rangle$$

which justifies the definition of the Hermitian operator

$$ Q = \sum\limits_n | Q_n \rangle {Q_n}\langle {Q_n} | $$

Then it should be clear that $Q$ only has meaning for calculating averages. The physical meaning concerns the projection operators leading to the particular results of the measurement.