Hermitian Operators: Finding Psi(p) from Psi(x)

[khemist]

I recently thought of this, please excuse me if it is way off the mark!

If I act on a state with a hermitian operator, am I able to find the psi(p) (momentum), where I had psi(x) (position) before (and wise versa)? Or does the operator do what it appears to do, and that is find the derivative of a particular state (if it does this, what does a derivative of a state mean?)?

Thanks

 

[eaglelake]

If we solve the eigenvalue equation for the momentum operator we get the momentum eigenvalues and momentum eigenfunctions. The eigenvalues of momentum are the possible values obtained when we measure momentum. If we are measuring momentum, then the wavefunction (obtained from the experimental configuration) must be expanded in terms of momentum eigenfunctions. The momentum eigenfunctions are the basis functions in this expansion and the expansion coefficients are the probability amplitudes.

Operating on any wavefunction with the momentum operator has no special significance, unless that wavefunction is a momentum eigenfunction. Operating on a momentum eigenfunction gives the momentum eigenvalue corresponding to that eigenfunction.
Best wishes

 

[khemist]

The eigenfunction/ value equation being $$\hat{A}$$x = $$\lambda$$x

In this case, x is the eigenfunction, and lambda is the eigenvalue. Now I assume there is a method to find these functions, rather than simply plugging in particular ones and hoping that when we apply the momentum operator to it that it yields an eigenvalue. Now if that function is in fact an eigenfunction, we are able to use that function as a basis vector for the vector space (is it an arbitrary space, or a particular space?), and the scalar lambda will be the probability amplitude for the particular eigenfunction.

Does this make sense and sound reasonably correct? Thanks for the help!

 

[eaglelake]

The eigenfunction/ value equation being $$\hat{A}$$x = $$\lambda$$x

In this case, x is the eigenfunction, and lambda is the eigenvalue. Now I assume there is a method to find these functions, rather than simply plugging in particular ones and hoping that when we apply the momentum operator to it that it yields an eigenvalue. Now if that function is in fact an eigenfunction, we are able to use that function as a basis vector for the vector space (is it an arbitrary space, or a particular space?), and the scalar lambda will be the probability amplitude for the particular eigenfunction.

Does this make sense and sound reasonably correct? Thanks for the help!

For the momentum operator $$\hat p_x = - i\hbar \partial /\partial x$$, the eigenvalue equation is $$- i\hbar \partial \varphi /\partial x = \lambda \varphi$$. You must solve this differential equation subject to the appropriate boundary conditions for eigenfunctions $$\varphi$$ and eigenvalues $$\lambda$$. $$\lambda$$ is not the probability amplitude. If we measure momentum for arbitrary state function
$$\psi$$, the probability amplitude is $$\left\langle {\varphi }<br /> \mathrel{\left | {\vphantom {\varphi \psi }}<br /> \right. \kern-\nulldelimiterspace}<br /> {\psi } \right\rangle = 1/\sqrt {2\pi \hbar } \int\limits_{ - \infty }^\infty {\varphi ^* \psi dx}$$. You may recognize this as the Fourier Transform of $$\psi$$. The mathematical methods are well known.

 

[khemist]

So is the eigenvalue not a necessarily important number, rather the important part is the fact that there are eigenvalues. The eigenfunction represents the basis for a particular space that one is solving for.

 

[dextercioby]

So is the eigenvalue not a necessarily important number, rather the important part is the fact that there are eigenvalues

Eigenvalues (or generally spectral values) of operators describing observables are very important numbers, they are the only possible values of the observables, if one measures these physical properties/quantities.

From the mathematical perspective, not choosing properly the operators accounting for the description of the observables in the mathematical formalism will generally not give you any eigenvalue.