Algebraic treatment of non-Hermitian quadratic Hamiltonians

Abstract

We generalize a recently proposed algebraic method in order to treat non-Hermitian Hamiltonians. The approach is applied to several quadratic Hamiltonians studied earlier by other authors. Instead of solving the Schrödinger equation we simply obtain the eigenvalues of a suitable matrix representation of the operator. We take into account the existence of unitary and antiunitary symmetries in the quantum-mechanical problem.

Appendix

In this “Appendix” we derive some additional results that may be useful for future applications of the algebraic method.

For every eigenvalue $\lambda _{i}$ we construct the operator

$$\begin{aligned} Z_{i}=\sum _{j=1}^{2K}c_{ij}O_{j}. \end{aligned}$$

(71)

For convenience we label the eigenvalues in such a way that $\lambda _{j}=-\lambda _{2K-j+1}$, $j=1,2,\ldots ,K$, and when they are real we organize them in the following way:

$$\begin{aligned} \lambda _{1}<\lambda _{2}<\cdots<\lambda _{K}<0<\lambda _{K+1}<\cdots <\lambda _{2K}. \end{aligned}$$

(72)

If we take into account that $[H,Z_{i}Z_{j}]=\left( \lambda _{i}+\lambda _{j}\right) Z_{i}Z_{j}$ then we conclude that

$$\begin{aligned}{}[H,[Z_{i},Z_{j}]]=\left( \lambda _{i}+\lambda _{j}\right) [Z_{i},Z_{j}]=0, \end{aligned}$$

(73)

which tells us that $Z_{i}$ and $Z_{j}$ commute when $\lambda _{i}+\lambda _{j}\ne 0$. If $[Z_{j},Z_{2K-j+1}]=\sigma _{j}\ne 0$ for all $j=1,2,\ldots ,K$ then we can write H in the following way

$$\begin{aligned} H=-\sum _{j=1}^{K}\frac{\lambda _{j}}{\sigma _{j}}Z_{2K-j+1}Z_{j}+E_{0}. \end{aligned}$$

(74)

If $\psi _{0}$ is a vector in the Hilbert space where H is defined that satisfies

$$\begin{aligned} Z_{j}\psi _{0}=0,\;j=1,2,\ldots ,K, \end{aligned}$$

(75)

then $H\psi _{0}=E_{0}\psi _{0}.$

Consider the time-evolution of the dynamical variables

$$\begin{aligned} O_{j}(t)=e^{itH}O_{j}e^{-itH}, \end{aligned}$$

(76)

so that

$$\begin{aligned} \dot{O}_{j}(t)=ie^{itH}[H,O_{j}]e^{-itH}=i\sum _{k=1}^{2K}H_{kj}O_{k}(t). \end{aligned}$$

(77)

If we define the row vector $\mathbf {O}(t)=\left( O_{1}(t)\,O_{2}(t)\,\ldots \,O_{2K}(t)\right)$ then we have the matrix differential equation $\dot{\mathbf {O}}(t)=i\mathbf {O}(t)\mathbf {H}$ with the following solution:

$$\begin{aligned} \mathbf {O}(t)=\mathbf {O}e^{it\mathbf {H}},\;\mathbf {O}=\mathbf {O}(0). \end{aligned}$$

(78)

Since $P(\mathbf {H})=0$ then

$$\begin{aligned} P\left( -i\frac{d}{dt}\right) \mathbf {O}(t)=\mathbf {O}P(\mathbf {H})e^{it \mathbf {H}}=0, \end{aligned}$$

(79)

gives us a differential equation of order 2K for the dynamical variables. Obviously, $Z_{j}(t)=e^{it\lambda _{j}}Z_{j}$, $j=1,2,\ldots ,2K$, satisfies this equation.