0
$\begingroup$

I'm struggling to find a hermitian operator whose eigenstate is a gaussian function in $|\psi(x)|^2$. How do i do this?

Just to be clear, this is in order to realistically model the wavefunction collapse when 'position' is measured, so the eigenvalue must be the mean of the gaussian

$\endgroup$
3
  • $\begingroup$ So you are not after coherent states? $\endgroup$ Commented Mar 13, 2021 at 12:19
  • $\begingroup$ @CosmasZachos physics.stackexchange.com/q/620582 I'm after an answer to this question, but I think this will help me get there $\endgroup$
    – user86425
    Commented Mar 13, 2021 at 12:51
  • $\begingroup$ It should. Coherent states are Gaussian states serving as eigenfunctions of annihilation operators. $\endgroup$ Commented Mar 13, 2021 at 14:48

1 Answer 1

3
$\begingroup$

The ground state wavefunction of the Harmonic oscillator is given by a gaussian function $$\psi_0(x)=Ce^{-m\omega x^2/2\hbar}$$ The Hamiltonian of the harmonic oscillator looks like $$\mathcal{H}=\frac{P^2}{2m}+\frac{1}{2}m\omega X^2$$ Or on a position basis $$\mathcal{H}=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega x^2$$

$\endgroup$
2
  • $\begingroup$ this is true but if applied to a general wavefunction, it does not produce a meaningful measurement operator $\endgroup$
    – user86425
    Commented Mar 13, 2021 at 8:10
  • $\begingroup$ The eigenstate with $n>1$ have nodes so it's not possible for them to be Gaussian function. $\endgroup$ Commented Mar 13, 2021 at 8:28