Operators 'act on' the wavefunction

[says]

The wavefunction describes the state of a system. When an operator 'acts on' the wavefunction are we saying, in layman's terms, that the operator is changing the state of the system?

 

[DrClaude]

Not necessarily. If the state is an eigenstate of the operator, then the operator obviously will not change the state.

Also, it depends on what is meant by "acting on." It can be simply a mathematical operation, like if we operate on a wave function with the time evolution operator, which will tell us how the system will evolve naturally, or it can correspond to a physical process, such as a measurement.

Personally, I see "acting on" a wave function as a mathematical operation, whose physical significance depends on why we are acting on the wave function.

 

[kuruman]

Yes, unless the the wavefunction is a "proper" state (eigenstate) of the operator in which case the state of the system after the application of the operator is the same as before. In other words, if the state of the system is a mixture of more than one proper states, when the operator acts on the system, the wavefunction of the system is changed to one of these proper states. Any subsequent "action" of the operator on the ensuing proper state leaves it unchanged.

 

[hilbert2]

Yes, unless the the wavefunction is a "proper" state (eigenstate) of the operator in which case the state of the system after the application of the operator is the same as before. In other words, if the state of the system is a mixture of more than one proper states, when the operator acts on the system, the wavefunction of the system is changed to one of these proper states. Any subsequent "action" of the operator on the ensuing proper state leaves it unchanged.

This sounds like you're saying that the operator causes a quantum collapse of the state, which isn't true... Isn't that what happens only in a measurement? A measurement is an irreversible process, while acting with a hermitian operator causes no loss of information unless some of its eigenvalues are zero.

 

[secur]

@says, did you mean to ask about observables, not operators (as @kuruman assumed)? Generally we assume an operator is bounded and linear, but that's all. It's a linear function "on" a vector space (in QM, that would be a Hilbert Space), meaning, its domain is the whole space and its range is a subspace.

Since the identity matrix is (can be considered to be) an operator then clearly, an operator doesn't necessarily change the vector it's acting on (in QM, that would be the wavefunction). Also there's no change if the wavefunction is an eigenvector of the operator. More generally: the operator will leave it unchanged iff the eigenvalues associated with all the wavefunction's nonzero eigenstates are equal (degenerate), as for instance with identity operator.

The above are mathematical facts (unless I made a mistake) about all operators. They're true in physics also of course, except for one thing, often referred to as the "collapse".

If the Hermitian operator is applied as a measurement, and it has no degenerate eigenvalues, and if we suppose the usual collapse interpretation, then the wavefunction must become a single one of its eigenvectors. Only in the special case that the wavefunction already was an eigenvector of that operator, will there be no change. If there's degeneracy, then there's no change if the wave is in the degenerate subspace - except maybe in the phase. I'm not quite sure ... @DrClaude knows, ask him.