Can an operator change a wave equation without changing the system?

[ShayanJ]

I read somewhere that the meaning of applying an operator to a wave equation is measuring the quantity associated with that operator.And because the result is a function different than the wave function,the system is changed because of the measurment.
But there is a problem here.If the above meaning of an operator is correct,the result function should be different because its a function giving us a different quantity than the one given by wave equation so the fact that the operator changes the wave function doesn't mean the system is changed.I'm sure either my explanation or that book's is wrong because uncertainty principle is a result of this fact.
thanks

 

[tom.stoer]

What you are saying is not correct.

Assume we have a momentum eigenfunction in position space

\psi_k(x) = e^{ikx}

and we apply the momentum operator; what we get is just

\hat{p}\psi_k(x) = k\psi_k(x)

So we just get the eigenfunction back.

Now we do the same with the position operator; what we get is simply

\hat{x}\psi_k(x) = x\psi_k(x)

The right hand side is neither a momentum eigenfunction (there is the additional x) nor a position eigenfunction - this would be

\psi_{x_0}(x) = \delta(x-x_0)

So applying an operator does NOT mean to measure the observable related to the operator. It just means that IF you have measured the observable you KNOW that the state is in an eigenstate and that applying the operator gives you the eigenvalue.

The operator (or observable) does not describe the process of measurement.

 

[Fredrik]

I read somewhere that the meaning of applying an operator to a wave equation is measuring the quantity associated with that operator.

This isn't true. Self-adjoint operators are the mathematical objects that represent measuring devices in QM, but having the operator act on a state vector doesn't represent the act of measurement.

<f,Af> is the expected average result in a long series of measurements, using the device that A represents, on systems all prepared in the same state f. But the state after each (idealized) measurement will be Pf, where P is the projection operator for the eigenspace of A that's associated with the eigenvalue of A that was the result of the measurement. If that eigenspace is 1-dimensional, and g is a normalized state vector in it, then the state will be <g,f>g. In bra-ket notation, we would write |a> instead of g, and |a><a|f> instead of <g,f>g. So the operator that takes the "before" state to the "after" state in this case is |a><a|. Of course, you can't know what state |a> is until you have actually performed the measurement.

 

[tom.stoer]

So the operator that takes the "before" state to the "after" state in this case is |a><a|. Of course, you can't know what state |a> is until you have actually performed the measurement.

This is a rather formal statement as it is not embedded in the dynamical framework.

 

[ShayanJ]

Strange to see such a mistake in a formal book.It was not my interpretation of the text but sth mentioned explicitly.
And thanks fredrik but from operator algebra I just know what are operators,eigenfunctions and eigenvalues.
Any way.So you mean all functions describing properties of a quantum system are eigen functions?
And should we apply all operators just to wave function?is that \Psi _{k} the wave function?what is that "k"?
I heard the amplitude of the matter wave is given by wave function.so it should be in meters.then the position will be in squared meters.So wave function should give sth else.what is that?
thanks

 

[tom.stoer]

So you mean all functions describing properties of a quantum system are eigen functions?

No. All formally allowed wave functions describe properties of the quantum system. One can extract them e.g. by calculating expectation values, e.g.

\langle p \rangle_\psi = \langle\psi|p|\psi\rangle = \int dx\,\psi^\ast(x)\left(-i\hbar\partial_x\right)\psi(x)

is that \Psi _{k} the wave function?what is that "k"?

Yes, it is the wave function in position space (that's the variable x).The k indicates that it is a momentum eigenfunction where the k is just the eigenvalue. Of course other wave functions are possible.

 

[ShayanJ]

And all formally allowed wave functions give eigen functions when applied by operators?

 

[tom.stoer]

No.

Do you know what the defintion of an eigenfunction really means?

An eigenfunction f(x) of an operator A is a function f(x) that - when multiplied by A - gives you a (= the constant eigenvalue) times the function f(x).

A f(x) = a * f(x)

So the exp(ikx) is an eigenfunction of the momentum operator -id/dx. When acting with -id/dx on this function it gives you k times the function.

-id/dx exp(ikx) = k * exp(ikx)

But of course there are a lot of functions and not all of them are eigenfunctions of reasonable operators.

 

[gerrardz]

No.

Do you know what the defintion of an eigenfunction really means?

An eigenfunction f(x) of an operator A is a function f(x) that - when multiplied by A - gives you a (= the constant eigenvalue) times the function f(x).

A f(x) = a * f(x)

So the exp(ikx) is an eigenfunction of the momentum operator -id/dx. When acting with -id/dx on this function it gives you k times the function.

-id/dx exp(ikx) = k * exp(ikx)

But of course there are a lot of functions and not all of them are eigenfunctions of reasonable operators.

May i try to explain the above facts in simple english ? Kindly confirm if i am right to understand all this eigen entities.

For a given allowed wavefunction, there is a specific operator when applied on it. yield a series of unique constants k (or may be just a single value). And these k's (eigenvalues) are the only possible values of the observable.

In other words, when the wavefunction is eigenfunction, when a specific operator is applied on it, eigenvalues are produced. Then the equation is eigen equation.

Possible to explain all this to someone layman ?

 

[A. Neumaier]

This is a rather formal statement as it is not embedded in the dynamical framework.

The right dynamical framework to make this more precise is that of a quantum dynamical semigroup. See http://en.wikipedia.org/wiki/Lindblad_equation