Exceptional points of the eigenvalues of parameter-dependent Hamiltonian operators

Abstract

We calculate the exceptional points of the eigenvalues of several parameter-dependent Hamiltonian operators of mathematical and physical interest. We show that the calculation is greatly facilitated by the application of the discriminant to the secular determinant. In this way, the problem reduces to finding the roots of a polynomial function of just one variable, the parameter in the Hamiltonian operator. As illustrative examples, we consider a particle in a one-dimensional box with a polynomial potential, the periodic Mathieu equation, the Stark effect in a polar rigid rotor and in a polar symmetric top.

Appendices

Resultant and discriminant

In this section, we summarize those properties of the discriminant of a polynomial that are relevant for present paper. All the expressions shown here are available in the literature mentioned earlier [14,15,16, 26] and also in the web (https://en.wikipedia.org/wiki/Resultant, https://en.wikipedia.org/wiki/Discriminant). The resultant of two polynomials

$$\begin{aligned} f(x)= & {} \sum _{j=0}^{m}a_{m-j}x^{j}, \nonumber \\ g(x)= & {} \sum _{j=0}^{n}b_{m-j}x^{j}, \end{aligned}$$

(A.1)

is given by the determinant

$$\begin{aligned} Res_{x}(f,g)=\left| \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} a_{0} &{} 0 &{} \cdots &{} 0 &{} b_{0} &{} 0 &{} \cdots &{} 0 \\ a_{1} &{} a_{0} &{} \cdots &{} 0 &{} b_{1} &{} b_{0} &{} \cdots &{} 0 \\ a_{2} &{} a_{1} &{} \ddots &{} 0 &{} b_{2} &{} b_{1} &{} \ddots &{} \vdots \\ \vdots &{} \vdots &{} \ddots &{} a_{0} &{} \vdots &{} \vdots &{} \ddots &{} b_{0} \\ a_{m} &{} a_{m-1} &{} \cdots &{} \vdots &{} b_{n} &{} b_{n-1} &{} \cdots &{} \vdots \\ 0 &{} a_{m} &{} \ddots &{} \vdots &{} 0 &{} b_{n} &{} \ddots &{} \vdots \\ \vdots &{} \vdots &{} \ddots &{} a_{m-1} &{} \vdots &{} \vdots &{} \ddots &{} b_{n-1} \\ 0 &{} 0 &{} \cdots &{} a_{m} &{} 0 &{} 0 &{} \cdots &{} b_{n} \end{array} \right| . \end{aligned}$$

(A.2)

It can be proved that

$$\begin{aligned} Res_{x}(f,g)=a_{0}^{m}b_{0}^{n}\prod _{i=1}^{m}\prod _{j=1}^{n}\left( \xi _{i}-\mu _{j}\right) , \end{aligned}$$

(A.3)

where $\xi _{i}$ and $\mu _{j}$ are the roots of the polynomials f and g , respectively. The discriminant of f(x) is defined as

$$\begin{aligned} Disc_{x}(f)=\frac{(-1)^{m(m-1)/2}}{a_{m}}Res_{x}(f,f^{\prime }), \end{aligned}$$

(A.4)

and in this case, we have

$$\begin{aligned} Disc_{x}(f)=a^{2m-2}\prod _{i<j}\left( \xi _{i}-\xi _{j}\right) ^{2}. \end{aligned}$$

(A.5)

Suppose that the nonlinear equation $Q(E,\lambda )=0$ gives us the eigenvalues $E(\lambda )$ of a quantum-mechanical system. If this equation is a polynomial function of E, then the roots of $F(\lambda )=Disc_{E}(Q(E,\lambda ))$ are the exceptional points $\lambda _{EP}$ in the complex $\lambda$ plane where at least two eigenvalues coalesce. We appreciate that the advantage of resorting to the discriminant is that we only have to search for the roots of a nonlinear function of just one variable. In all the examples studied here, the nonlinear equation $Q(E,\lambda )$ is a polynomial function of both E and $\lambda$ so that $F(\lambda )$ is a polynomial function of $\lambda$ (see “Appendix B,” and the examples). Consequently, the calculation is particularly simple because there are efficient algorithms for finding the roots of polynomials. Besides, most computer-algebra software enable one to obtain analytical expressions for $F(\lambda )$ because the discriminant is given by a determinant. Thus, the only numerical step of the calculation reduces to finding the roots of the polynomial $F(\lambda )$.

As an illustrative example, we consider a trivial toy problem that we deem to be quite interesting: the Hamiltonian operator in matrix form

$$\begin{aligned} {\mathbf {H}}(\beta ,\lambda )=\left( \begin{array}{ccc} 3-\lambda &{} \beta &{} 0 \\ \beta &{} 2 &{} \beta \\ 0 &{} \beta &{} 1+\lambda \end{array} \right) . \end{aligned}$$

(A.6)

When $\beta =0$ the three eigenvalues cross at $\lambda =1$ and the three eigenvectors are degenerate (they are obviously linearly independent). However, when $\beta \ne 0$ the eigenvalues do not cross for real values of $\lambda$ and exhibit avoided crossings as shown in Fig. 8 for $\beta =0.1$.

Fig. 8

Avoided crossings for the three-level model

The characteristic polynomial is

$$\begin{aligned} Q(E,\lambda )=\frac{\left( 2-E\right) \left( 50E^{2}-200E-50\lambda ^{2}+100\lambda +149\right) }{50}, \end{aligned}$$

(A.7)

so that

$$\begin{aligned} Disc_{E}(Q(E,\lambda ))=\frac{\left( 50\lambda ^{2}-100\lambda +51\right) ^{3}}{31250}. \end{aligned}$$

(A.8)

We appreciate that the three eigenvalues coalesce at any of the two EPs $\lambda _{EP}=1+\sqrt{2}i/10$ and $\lambda _{EP}^{*}$ that are branch points of order two. The structure of the avoided crossings in this toy model is similar to that of the modified Lipkin model for $N=3$ [21].

At $\lambda _{EP}$ (a similar analysis can be carried out at $\lambda _{EP}^{*}$) the matrix ${\mathbf {H}}$ has only one eigenvalue $E\left( \lambda _{EP}\right) =2$ and only one eigenvector

$$\begin{aligned} {\mathbf {v}}_{1}=\frac{1}{2}\left( \begin{array}{c} -i \\ \sqrt{2} \\ i \end{array} \right) , \end{aligned}$$

(A.9)

so that ${\mathbf {H}}$ is defective. By means of the Jordan chain

$$\begin{aligned} \left( {\mathbf {H}}-2{\mathbf {I}}_{3}\right) {\mathbf {v}}_{2}= & {} {\mathbf {v}}_{1}, \nonumber \\ \left( {\mathbf {H}}-2{\mathbf {I}}_{3}\right) {\mathbf {v}}_{3}= & {} {\mathbf {v}}_{2}, \end{aligned}$$

(A.10)

where ${\mathbf {I}}_{3}$ is the $3\times 3$ identity matrix, we obtain two additional vectors ${\mathbf {v}}_{2}$ and ${\mathbf {v}}_{3}$ and the matrix

$$\begin{aligned} {\mathbf {U}}=\left( \begin{array}{ccc} {\mathbf {v}}_{1}&{\mathbf {v}}_{2}&{\mathbf {v}}_{3} \end{array} \right) =\left( \begin{array}{c@{\quad }c@{\quad }c} -i/2 &{} 5\sqrt{2} &{} 0 \\ 1/\sqrt{2} &{} 5i &{} 50\sqrt{2} \\ i/2 &{} 0 &{} 50i \end{array} \right) , \end{aligned}$$

(A.11)

that converts ${\mathbf {H}}$ into a Jordan canonical form

$$\begin{aligned} {\mathbf {U}}^{-1}\mathbf {HU}=\left( \begin{array}{c@{\quad }c@{\quad }c} 2 &{} 1 &{} 0 \\ 0 &{} 2 &{} 1 \\ 0 &{} 0 &{} 2 \end{array} \right) . \end{aligned}$$

(A.12)

In this simple case, we can easily obtain the EPs directly from the eigenvalues but in most nontrivial problems the use of the discriminant leads to far simpler expressions.

Three-term recurrence relations

Suppose that there is an orthonormal basis set $\left\{ \left| i\right\rangle ,i=0,1,\ldots \right\}$ such that

$$\begin{aligned} H\left| i\right\rangle =H_{i-1,i}\left| i-1\right\rangle +H_{i,i}\left| i\right\rangle +H_{i+1,i}\left| i+1\right\rangle , \end{aligned}$$

(B.1)

where $H_{i,j}=H_{j,i}$. Therefore, if we expand

$$\begin{aligned} \psi =\sum _{i}c_{i}\left| i\right\rangle , \end{aligned}$$

the Schrödinger equation $H\psi =E\psi$ becomes a three-term recurrence relation for the coefficients $c_{j}$:

$$\begin{aligned} A_{i}c_{i-1}+B_{i}c_{i}+A_{i+1}c_{i+1}= & {} 0,\;A_{i}=H_{i,i-1},\;B_{i}=H_{ii}-E, \nonumber \\&i=0,1,2\ldots ,\;c_{-1}=0. \end{aligned}$$

(B.2)

One commonly obtains approximate energies by means of the truncation condition $c_{j}=0$, $j>N$, so that the roots of the characteristic polynomial given by the secular determinant

$$\begin{aligned} D_{N}=\left| \begin{array}{c@{\quad }c@{\quad }c@{\quad }c@{\quad }c@{\quad }c} B_{0} &{} A_{1} &{} 0 &{} \cdots &{} \cdots &{} 0 \\ A_{1} &{} B_{1} &{} A_{2} &{} 0 &{} \cdots &{} 0 \\ \vdots &{} \vdots &{} \vdots &{} \ddots &{} \cdots &{} \vdots \\ 0 &{} 0 &{} \cdots &{} A_{N-1} &{} B_{N-1} &{} A_{N} \\ 0 &{} 0 &{} \cdots &{} 0 &{} A_{N} &{} B_{N} \end{array} \right| , \end{aligned}$$

(B.3)

converge from above toward the actual energies of the physical problem when $N\rightarrow \infty$. These determinants can be efficiently generated by means of the three-term recurrence relation [3, 28, 29]

$$\begin{aligned} D_{N}=B_{N}D_{N-1}-A_{N}^{2}D_{N-2},\;N=0,1,\ldots , \end{aligned}$$

(B.4)

with the initial conditions $D_{-1}=1$, $D_{j}=0$ for $j<-1$. Notice that the dimension of the determinant $D_{N}$ is $N+1$.