# Confused over a line in L&L Quantum Mechanics (non-relativistic) involving linear algebra

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.

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,
$$\hat{f}\Psi = \hat{f}\Bigg( \sum_n a_n \Psi_n \Bigg) = \sum_n a_n \hat{f}\Psi_n = \sum_n a_n f_n \Psi_n.$$
• 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! – maaarrr Jan 27 at 5:12