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So I was reading Quantum Mechanics: Non-relativistic Theory by Landau and Lifshitz when I came across this line (Edition 2, page 10):

$$\hat{f} \Psi = \sum_n{a_n f_n \Psi_n},$$

where $\Psi = \sum_n a_n \Psi_n$ and $\hat{f} \Psi_n = f_n \Psi_n$.

I tried understanding this line a bit better by making an analogy with $\Psi$ as a vector in Euclidean space and $\hat{f}$ a matrix, etc., but can't recover this relation. By the way, I can follow the derivation they use in the book, but was trying to understand this on a more intuitive level.

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The notation might be a little confusing: the vectors $\Psi_n$ are the eigenvectors of the operator $\hat{f}$, as can be seen by the relation: $$\hat{f}\Psi_n = f_n \Psi_n.$$ The numbers $f_n$ are the eigenvalues associated with the eigenvectors $\Psi_n$.

If $\hat{f}$ is a Hermitian operator (which are used to represent physical observables in Quantum Mechanics), then we are guaranteed that its eigenvectors form a complete basis, meaning that any arbitrary vector in the space can be written as a linear combination of these vectors. Thus, an arbitrary vector $\Psi$ can be written as: $$\Psi = a_0 \Psi_0 + a_1 \Psi_1 + a_2 \Psi_2 + ... = \sum_n a_n \Psi_n.$$

We can now combine the above two relations using the fact that the operator $\hat{f}$ is linear. Therefore,

$$\hat{f}\Psi = \hat{f}\Bigg( \sum_n a_n \Psi_n \Bigg) = \sum_n a_n \hat{f}\Psi_n = \sum_n a_n f_n \Psi_n.$$

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  • $\begingroup$ Oh wow, that was actually really simple. I don't know why I didn't think to use the linearity of $\hat{f}$. Thank you so much! $\endgroup$ – maaarrr Jan 27 at 5:12

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