# How are Hermitian operators derived

The von Neumann-Dirac Theory postulates that physical observables are represented by Hermitian operators. For example, let's assume the physical observable I am measuring is the spin of an electron, and my orthonormal basis is composed of the eigenvectors $$|\uparrow\rangle=\left(\begin{array}{c} 1 \\ 0 \\ \end{array}\right)$$ and $$|\downarrow\rangle = \left(\begin{array}{c} 0 \\ 1 \\ \end{array}\right) .$$

The Hermitian operator for these eigenvectors is $$\hat{A} = \left( \begin{array}{c} 1 & 0 \\ 0 & -1 \\ \end{array}\right) ,$$ because each eigenvector corresponds to an eigenvalue. In the case above, $$|\uparrow\rangle$$ has an eigenvalue of $$1$$ and $$|\downarrow\rangle$$ has an eigenvalue of $$-1$$. However, if I wasn't given the Hermitian operator $$\hat{A}$$ how would I derive it? Would it even be logical to derive it only knowing the eigenvectors?

• There are deductions you can do in specific cases, but at some point you just have to postulate that some Hermitian operator corresponds to your observable and see if the observations match the predictions. – Javier Oct 9 '20 at 0:02
• It's pretty easy to deduce that $\hat A$ is the Hermitian operator for the two basis vectors that I gave. However, there are an infinite number of orthonormal basis vectors in $\Bbb R^2$. Is there some prosses using matrix algebra that I can use to derive a Hermitian operator from basis vectors? – mrhumanzee Oct 9 '20 at 0:11
• If the eigenvectors are orthogonal and you are given their eigenvalues, you can trivially do this, as here. You may then unitarily transform your diagonal operator so it does not look diagonal. – Cosmas Zachos Oct 9 '20 at 0:30

Since a Hermitian matrix can be diagonalized as $$A = U^{\dagger} D U ,$$ where $$U$$ is composed of the eigenvectors and $$D$$ is a diagonal matrix with the eigenvalues on the diagonal, one can use it to compose $$A$$. For the two-dimensional case it would be represented by $$A = |a\rangle \lambda_a \langle a| + |b\rangle \lambda_b \langle b| ,$$ where $$|a\rangle$$ and $$|b\rangle$$ are the two eigenvectors. If you don't know the eigenvalues, it is reasonable to pick $$1$$ and $$-1$$ as their eigenvalues, because they are real valued (as required for a Hermitian operator), and would give convenient measurement values.