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my friend and i have been debating the answer to this question could someone help me with this by explaining their answer please :)

Is it possible to measure the energy of the particle if the wave function psi is not an eigenfunction of the hamiltonian?

Lalo
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    Of course you can measure it, it’s just in a superposition of one or more eigenstates. But you will measure an energy. – Jon Custer Oct 21 '23 at 16:00

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You can always measure the energy of the particle. When the particle's wavefunction $\psi(x)$ is not an eigenfunction $\psi_i(x)$ of the Hamiltonian, it can still be written as a linear combination of such eigenfunctions. $$\psi(x) = \sum_i c_i \psi_i(x)$$

When you now measure the energy of this state, the result is not unique. Instead, you can get any energy eigenvalue $E_i$ as the result, each with probability $|c_i|^2$. This is called Born's rule.

This rule contains as a special case the behavior when the wave function $\psi(x)$ happens to be an eigenfunction $\psi_i(x)$. Then one of the $|c_i|^2$ is $1$ and all the others are $0$. This means you get one eigenvalue $E_i$ with certainty (i.e. with $100$% probability), while you do not (i.e. with $0$ probability) get any of the other eigenvalues.

Thomas Fritsch
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