I have a question about the eigenvalue decomposition of an operator, more specifically about the matrix with the eigenvectors as columns. If i have an operator that i decompose as follows: $$ \hat{A} = U\Lambda U^\dagger $$ with $\psi_i$ being the eigenvectors and $a_i$ the according eigenvalues. I know that if i multiply $U^\dagger$ with any of the eigenvectors (for A being 4x4) i get $|00\rangle$, $|01\rangle$, $|10\rangle$ or $|11\rangle$. How do i prove this ?

  • 2
    $\begingroup$ Your last sentence implies that you are speaking of a specific case rather than general properties of (Hermitian?) operators. I suggest clarifying/expanding your question. $\endgroup$
    – Roger V.
    Commented Apr 18, 2023 at 9:49
  • $\begingroup$ I don't understand the question. What are $\psi_i$, what is $a_i$ (within your equation $\hat{A}=U\Lambda U^\dagger$). What is this basis $|00\rangle$... seems like some kind of quantum computing application but I can't really tell. Although in total it seems like you just want to know the spectral theorem. I'd recommend googling that, which at least is the proof of the existence of that equation for any hermetian/self adjoint operator, and kind of the entire basis of quantum mechanics. $\endgroup$
    – AXensen
    Commented Apr 18, 2023 at 10:10

1 Answer 1


$\newcommand{\ket}[1]{\left|#1\right\rangle}$ I assume you write your matrix $\hat{A}$ with respect to the basis $(\ket{00},\ket{01},\ket{10},\ket{11})$. Say it has orthormal eigenvectors $v_{00},v_{01},v_{10},v_{11}\in\mathbb{C}^4$ with eigenvalues $\lambda_{00},\lambda_{01},\lambda_{10},\lambda_{11}\in\mathbb{C}$. Diagonalization then is given by setting $\Lambda := \operatorname{diag}(\lambda_{01},\lambda_{10},\lambda_{11})$ and $$ U := \begin{pmatrix} | & | & | & | \\ v_{00}&v_{01}&v_{10}&v_{11} \\ | & | & | & | \end{pmatrix} $$ such that $U\ket{ij} = v_{ij}$ for $i,j\in\{0,1\}$ and $$\hat{A} = U^\dagger \Lambda U.$$ Note this is not what you have written down but this is the more common convention. Since $U$ is unitary (follows from the $v_{ij}$ being orthonormal), we have $U^\dagger v_{ij} = U^{-1} v_{ij} = \ket{ij}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.