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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

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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 $$\mathcal{H}=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+\frac{1}{2}m\omega x^2$$

Young Kindaichi
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  • this is true but if applied to a general wavefunction, it does not produce a meaningful measurement operator –  Mar 13 '21 at 08:10
  • The eigenstate with $n>1$ have nodes so it's not possible for them to be Gaussian function. – Young Kindaichi Mar 13 '21 at 08:28