# Tag Info

## New answers tagged eigenvalue

1

In the proof for : If A and B are two commuting hermitian operators, there exists an orthonormal basis consisting of eigenvectors shared between the two operators. the following lemma is often used : Let $A,B$ be two commuting hermitian operators and $E_\lambda$ be an eigenspace of $A$. Then $E_\lambda$ is stable by $B$. The eigenspace $E_\lambda$ is the ...

4

TL;DR: Assuming that $A,B$ are self-adjoint, the product $AB$ does not need to be diagonalizable. And if $AB$ is diagonalizable, the eigenvalues need not be real or imaginary. Example 1: $AB$ is not diagonalizable: $$A~=~\begin{pmatrix} 0 & 1 \cr 1 & 0 \end{pmatrix} \quad\wedge\quad B~=~\begin{pmatrix} 1 & 0 \cr 0 & 0 \end{pmatrix}\quad\... 4 Suppose A and B are Hermitian operators: what is the nature of the eigenvalues of [A,B]+\{B,A\}=2AB? First, "A is Hermitian" from the practical point of view means that: A=A^\dagger, it is unitarily diagonalizable and has real eigenvalues. The same for B. Now, (AB)^\dagger = B^\dagger A^\dagger =BA: this tells us that, in general AB\... 5 In general, we can say that C=AB will have real, imaginary and complex eigenvalues (complex of the form z=a+ib where and \{a,b\in \mathbb{R}\mid a,b \ne 0\} as shown in the comments by Mark and Qmechanic's answer). For example, if$$A=\begin{bmatrix} 0 &1 \\ 1& 0 \end{bmatrix}\ \ \text{and}\ \ B=\begin{bmatrix} 1 & 0\\ 0& -1 \end{...

2

As I pointed in the comments, one possible application is calculations of the density-of-states (DOS), which determines many physical properties in various contexts: absorption of crystals (DOS of final states) thermodynamic properties of materials (e.g., via the DOS of phonons) electric conductance lifetimes of atomic levels critical parameters of phase ...

0

Another way to obtain the coherent state $|\alpha\rangle$ is by acting the displacement operator $$\hat{D}(\alpha) = \exp(\alpha \hat{a}^\dagger - \alpha^* \hat{a})$$ on the vacuum state $|0\rangle$. $$|\alpha\rangle = \hat{D}(\alpha) |0\rangle$$ You can verify that $$\hat{a} |\alpha\rangle = \alpha |\alpha\rangle$$ This looks deceptively simple, and you do ...

0

I am not familiar with your "para-unitary" scheme. As I understand it, the diagonalization of the general bosonic hamiltonian via Bogoliubov transformations relies on Williamson's theorem and the associated symplectic spectrum of the matrix. A link to the basic linear algebra is here.

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