# What is physical meaning of an operator to applied on a wavefunction? [duplicate]

we know that an eigen operator give the eigen value and keep the wavefunction in corresponding position. but what is the simple physical meaning of this process because the wavefunction represent the matter waves.

• Possible duplicate of Physical meaning of quantum operators
– user191954
Jun 12 '18 at 8:25
• I would say that it depends on the operator. The eigenvalue corresponding to an hermitic operator is an observable, some quantity that is mesurable at the laboraty. And by the way the wavefunction has not physical meaning by itself, it's its squared moduli, $|\psi|^2$ what has a physical interpretation (it represents the probability of finding the particle at that point in space).
– Mat
Jun 12 '18 at 8:27

I'll answer with a question: What does the wave function $\psi$ itself mean? If you can't answer this concisely (and it's hard to!), you should not expect a short simple answer to the question what it means to apply an operator to the wave function.
Already the fact that a wave function multiplied by a phase factor $e^{i \varphi}$ describes the same physics should tell you that it's not simple.
You say $\psi$ represents the matter waves - yes, well, but that's a vague notion, in which way does it represent them? This depends on your preferred formulation of quantum mechanics. At least we can be sure that $|\psi|^2$ is the probability distribution for positions.
Usually, an operator $A$ applied to a wave function itself has no particular meaning. If you associate $A$ with an observable you are measuring, the important facts lie in the scalar products $\langle \phi, A \psi \rangle$, telling you the transition probabilities.
The Schrödinger momentum is $\vec p = \frac{\hbar}{i} \int {d^3x \psi \vec \nabla \psi }$. We can interpret this by defining the concept of the expectation value of an operator. Then we can do some interesting mathematics with operators. However all the physical meaning is represented by this expression.