2 The two golden rules

A physical system which is determined by a Hamiltonian is a Hilbert space (a complex vector space with inner product), and the Hamiltonian corresponds to a linear operator. Suppose that $H \in B(\mathbb {C}^{d})$ is a Hamiltonian, $H^{\dagger }$ is the adjoint of H, $H^{T}$ is the transpose of H, $|\psi \rangle $ is a unit column vector, $\langle \psi |=(|\psi \rangle )^{\dagger }$ is a unit row vector.

We note from the Definition 2.1 that, in particular, when $W=I$ , the pseudo-Hermitian Hamiltonian coincides with a Hermitian Hamiltonian. Therefore, all Hermitian Hamiltonians form a subset of the set of pseudo-Hermitian Hamiltonians.

We now present some results about the pseudo-Hermitian Hamiltonian and its metric operators.

In order to understand the metric operators for a given pseudo-Hermitian Hamiltonian operator, we introduce the operator–vector correspondence as follows. We know that

$$ \begin{align*}\mathrm{vec}: L(\mathbb{X},\mathbb{Y})\rightarrow \mathbb{Y}\otimes \mathbb{X}\end{align*} $$

is the linear bijection and isometry, mapping an operator to a vector [Reference Bellman1]. For $u\in \mathbb {X}$ and $v\in \mathbb {Y}$ , we have

$$ \begin{align*} \mathrm{vec}(uv^{\dagger})=u\otimes \bar{v}, \end{align*} $$

where the bar denotes the complex conjugate.

Theorem 2.3. Let $H\in B(\mathbb {C}^{d})$ be a pseudo-Hermitian Hamiltonian and $W\in B(\mathbb {C}^{d})$ be invertible. Then W is a metric operator for H if and only if

$$ \begin{align*}(I\otimes H^{T}-H^{\dagger}\otimes I)\mathrm{vec}(W)=0.\end{align*} $$

Proof. Necessity. Let W be a metric operator for H. Then $H^{\dagger }=WHW^{-1}$ and

$$ \begin{align*}\mathrm{vec}(H^{\dagger}W) =\mathrm{vec}(WH).\end{align*} $$

Using the property of the vec mapping,

$$ \begin{align*}\mathrm{vec}(AXB^{T}) = (A\otimes B)\mathrm{vec}(X),\end{align*} $$

we have

$$ \begin{align*}(H^{\dagger}\otimes I)\mathrm{vec}(W)=(I\otimes H^{T})\mathrm{vec}(W).\end{align*} $$

Thus,

$$ \begin{align*} (I\otimes H^{T}-H^{\dagger}\otimes I)\mathrm{vec}(W)=0. \end{align*} $$

Sufficiency. Let W satisfy $(I\otimes H^{T}-H^{\dagger }\otimes I)\mathrm {vec}(W)=0$ . Then

$$ \begin{align*}(I\otimes H^{T})\mathrm{vec}(W)=(H^{\dagger}\otimes I)\mathrm{vec}(W) \end{align*} $$

and

$$ \begin{align*}\mathrm{vec}(WH)=\mathrm{vec}(H^{\dagger}W).\end{align*} $$

Since the vec mapping is a linear bijection, we have $WH=H^{\dagger }W$ . Since W is invertible, it is a metric operator for H. This completes the proof.

For example, when $H=(\begin {smallmatrix} \mathrm {i} & 1 \\ 2 & -\mathrm {i}\end {smallmatrix})$ , we have

$$ \begin{align*}I\otimes H^{T}-H^{\dagger}\otimes I=\left(\!\!\! \begin{array}{cccc} 2\mathrm{i} & 2 & -2 & 0 \\ 1 & 0 & 0 & -2 \\ -1 & 0 & 0 & 2 \\ 0 & -1& 1 & -2\mathrm{i} \\ \end{array}\!\!\! \right).\end{align*} $$

By calculation, the eigenvalue zero of the operator $(I\otimes H^{T}-H^{\dagger }\otimes I)$ has multiplicity 2, and $\alpha _{1}=(0,1,1,0)^{T}$ and $\alpha _{2}=(2,-2\mathrm {i},0,1)^{T}$ are two linear independent eigenvectors of $I\otimes H^{T}-H^{\dagger }\otimes I$ . Then we obtain two metric operators for H:

$$ \begin{align*}W_{1}=\mathrm{vec}^{-1}(\alpha_1)=\left(\!\!\! \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}\!\!\! \right), \quad W_{2}=\mathrm{vec}^{-1}(\alpha_2)=\left(\!\!\! \begin{array}{cc} 2 & -2\mathrm{i} \\ 0 & 1 \\ \end{array}\!\!\! \right).\end{align*} $$

For any nonzero linear combination $\beta $ of $\alpha _1$ and $\alpha _2$ , if $\mathrm {vec}^{-1}(\beta )$ is invertible, then $\mathrm {vec}^{-1}(\beta )$ is also a metric operator for H. We see that the metric operator $W_{2}$ as above is not Hermitian. From the proof of Theorem 2.2(ii) a new Hermitian metric operator,

$$ \begin{align*}\tilde{W_{2}}=e^{\mathrm{i}\pi/3}W_{2}+e^{\mathrm{-i}\pi/3}W_{2}^{\dagger}=\left(\!\!\! \begin{array}{cc} 2 & \sqrt{3}-\mathrm{i} \\ \sqrt{3}+\mathrm{i} & 1 \\ \end{array}\!\!\! \right)\end{align*} $$

as $\theta ={\pi }/{3}$ , is obtained for H.

In the following section, for a pseudo-Hermitian Hamiltonian H, the corresponding metric operator W may always be chosen to be Hermitian, that is, $W=W^{\dagger }$ .

3 The results of metric operators for pseudo-Hermitian Hamiltonian

In this section we assume that the Hamiltonian $H\in B(\mathbb {C}^{d})$ has different eigenvalues. Let H admit a complete biorthonormal set of eigenvectors $\{|E_{n}\rangle , |\hat {E}_{n}\rangle \}$ . Then it satisfies the following defining relations:

(3.1) $$ \begin{align} H|E_{n}\rangle=\gamma_n|E_{n}\rangle, \quad H^{\dagger}|\hat{E}_{n}\rangle=\overline{\gamma_n}|\hat{E}_{n}\rangle, \end{align} $$

(3.2) $$ \begin{align} \langle \hat{E}_n|E_m\rangle=\delta_{nm}, \end{align} $$

(3.3) $$ \begin{align} \sum_n|\hat{E}_n\rangle\langle E_n|=\sum_n|E_n\rangle\langle \hat{E}_n|=I. \end{align} $$

Here, $\delta _{nm}$ stands for the Kronecker delta function, and I is the identity operator. In view of equations (3.1)–(3.3), we have

(3.4) $$ \begin{align} H=\sum_n\gamma_n|E_n\rangle\langle \hat{E}_n|, \quad H^{\dagger}=\sum_n\overline{\gamma_n}|\hat{E}_n\rangle\langle E_n|. \end{align} $$

In addition, if H is a pseudo-Hermitian Hamiltonian with metric operator W, that is, $H^{\dagger }=WHW^{-1}$ , put

$$ \begin{align*} \langle x|y\rangle_{W} =\langle x|W |y\rangle, \quad \text{for all}\, \, |x\rangle, |y\rangle \in \mathbb{K}. \end{align*} $$

Then, for any m,n,

$$ \begin{align*}\gamma_n\langle E_m|E_n\rangle_{W}=\langle E_m|WH| E_n\rangle=\langle E_m|H^{\dagger}W| E_n\rangle=\overline{\gamma_m}\langle E_m| E_n\rangle_{W}, \end{align*} $$

and we can write

(3.5) $$ \begin{align} (\gamma_n-\overline{\gamma_m})\langle E_m|E_n\rangle_{W}=0. \end{align} $$

Therefore, this determines

(3.6) $$ \begin{align} \langle E_m|E_n\rangle_{W}=\rho_{m,n} e^{\mathrm{i}\theta{(m,n)}}\delta_{\gamma_n, \overline{\gamma_m}}, \end{align} $$

where $\rho _{m,n}=|\langle E_m|E_n\rangle _{W}|\geq 0$ and $\theta {(m,n)}=\text {Arg}(\langle E_m|E_n\rangle _{W})$ is the argument of $\langle E_m|E_n\rangle _{W}$ .

If there exists an eigenvalue $\gamma _j\notin \mathbb {R}$ , then $\langle E_j|E_j\rangle _{W}=0$ . In this case, it is impossible to define an inner product with respect to W. If all eigenvalues are real, the product in equation (3.6) which can be become negative due to this exponential factor (for example, W has negative eigenvalues), cannot define an inner product. In fact, the quadratic form in equation (3.6) satisfies the inner product conditions except positive-definiteness, because the metric operator W is not necessarily positive, and H is Hermitian with respect to this quadratic form since

$$ \begin{align*}\langle x|H| y\rangle_{W}=\langle x|WH| y\rangle=\langle x|H^{\dagger}W| y\rangle=\langle H x|W| y\rangle=\langle H x|y\rangle_{W}.\end{align*} $$

Thus, when the spectra of pseudo-Hermitian Hamiltonian are all real, it is significant to find a positive metric operator such that equation (3.6) is a well-defined inner product, and the non-Hermitian Hamiltonian can be exchanged for the Hermitian case. Besides, the existence of this positive metric operator was shown by Mostafazadeh [Reference Mostafazadeh13], who proved that the spectra of H are real if and only if there is a positive invertible linear operator $\eta $ such that $H^{\dagger }=\eta H\eta ^{-1}$ .

Hence, for a pseudo-Hermitian Hamiltonian H with real spectra, we provide in the following a method to change the given nonpositive metric operator ( $W\ngtr 0$ ) into a positive one ( $\eta>0$ ).

Proof. Let the eigenvalues of H be all real, that is, $\gamma _n\in \mathbb {R}$ for all n. By equation (3.4),

(3.7) $$ \begin{align} H=\sum_n\gamma_n|E_n\rangle\langle \hat{E}_n|, \quad H^{\dagger}=\sum_n\gamma_n|\hat{E}_n\rangle\langle E_n|. \end{align} $$

Put

(3.8) $$ \begin{align} P_{n}=|E_n\rangle\langle \hat{E}_n| \end{align} $$

and

(3.9) $$ \begin{align} A=\sum_n a_{n}P_{n}, \quad a_{n}\in \mathbb{R}\backslash\{0\}. \end{align} $$

We have

$$ \begin{align*}A^{-1}=\sum_n \frac{1}{a_{n}}P_{n},\end{align*} $$

$$ \begin{align*}AH=\sum_n a_{n}|E_n\rangle\langle \hat{E}_n|H=H\sum_n a_{n}|E_n\rangle\langle \hat{E}_n|=HA\end{align*} $$

and

$$ \begin{align*}A|E_n\rangle= a_{n}|E_n\rangle.\end{align*} $$

So $|E_n\rangle $ are simultaneous eigenstates of A and H, and the coefficients of expansion $a_{n}$ in equation (3.9) are indeed the eigenvalues of A. Take

$$ \begin{align*} \eta=WA. \end{align*} $$

Then

$$ \begin{align*} \eta^{\dagger}=&A^{\dagger}W=WW^{-1}\sum_n \bar{a}_{n}P_{n}^{\dagger}W=WA=\eta. \end{align*} $$

Since A commutes with H, we get that $\eta $ is also a metric operator for H by Theorem 2.2(v). Next define

$$ \begin{align*} \langle\phi|\psi\rangle_{\eta}=\langle\phi|\eta|\psi\rangle. \end{align*} $$

We have

$$ \begin{align*}\langle E_m|E_n\rangle_{\eta}= a_{n}\langle E_m|E_n\rangle_{W}.\end{align*} $$

Since W is invertible and Hermitian, by equation (3.6), we have

$$ \begin{align*}\langle E_m|E_n\rangle_{W}=\begin{cases} 0, & m\neq n, \\ \langle E_m|W|E_m\rangle\in \mathbb{R}\backslash 0, & m= n. \end{cases} \end{align*} $$

Then

$$ \begin{align*} \langle E_m|E_n\rangle_{W}=\tilde{\rho}_{n}\delta_{m,n}, \end{align*} $$

where $\tilde {\rho }_{n}$ are positive or negative real numbers. Thus,

$$ \begin{align*} \langle E_m|E_n\rangle_{\eta}= a_{n}\tilde{\rho}_{n}\delta_{m,n}. \end{align*} $$

If we choose the coefficients of A as

$$ \begin{align*}a_{n}=\frac{1}{\tilde{\rho}_{n}},\end{align*} $$

then

(3.10) $$ \begin{align} \langle E_m|E_n\rangle_{\eta}=\delta_{m,n}. \end{align} $$

By this method of constructing $\eta $ , we have

$$ \begin{align*} \eta=WA=W\sum_j a_{j}P_{j}=\sum_j \frac{W}{\langle E_j|E_j\rangle_{W}}|E_j\rangle\langle \hat{E}_j|. \end{align*} $$

According to equation (3.3),

$$ \begin{align*} W\sum_n|E_n\rangle\langle \hat{E}_n|=\sum_n|\hat{E}_n\rangle\langle E_n|W, \end{align*} $$

and multiplying $|E_j\rangle $ yields

$$ \begin{align*} W|E_j\rangle=|\hat{E}_j\rangle\langle E_j|W|E_j\rangle. \end{align*} $$

Thus,

(3.11) $$ \begin{align} \eta=\sum_j |\hat{E}_j\rangle\langle \hat{E}_j|. \end{align} $$

This completes the proof.

We see that $\eta $ is positive-definite, $\langle \cdot |\cdot \rangle _{\eta }=\langle \cdot |\eta |\cdot \rangle $ is a well-defined inner product, and the eigenstates of H form an orthonormal basis by equation (3.10). Indeed, the positive-definite metric operator $\eta $ we obtained in equation (3.11) which was shown by Mostafazadeh [Reference Mostafazadeh13], satisfies our Theorem 2.3. By equation (3.7) and vec mapping,

$$ \begin{align*}I\otimes H^{T}-H^{\dagger}\otimes I=I\otimes\sum_n\gamma_n\overline{|\hat{E_{n}}\rangle\langle E_{n}|}-\sum_n\gamma_n|\hat{E}_n\rangle\langle E_n|\otimes I,\end{align*} $$

By equation (3.7) and vec mapping,

$$ \begin{align*}I\otimes H^{T}-H^{\dagger}\otimes I=I\otimes\sum_n\gamma_n\overline{|\hat{E_{n}}\rangle\langle E_{n}|}-\sum_n\gamma_n|\hat{E}_n\rangle\langle E_n|\otimes I,\end{align*} $$

$$ \begin{align*}\mathrm{vec}(\eta)=\sum_j |\hat{E}_j\rangle\otimes\overline{|\hat{E}_{j}\rangle},\end{align*} $$

followed by biorthogonality relation equation (3.2), we have

$$ \begin{align*}(I\otimes H^{T}-H^{\dagger}\otimes I)(\mathrm{vec}(\eta))=\sum_n\gamma_n|\hat{E_{n}}\rangle\otimes\overline{|\hat{E_{n}}\rangle}-\sum_n\gamma_n|\hat{E_{n}}\rangle\otimes\overline{|\hat{E_{n}}\rangle}=0.\end{align*} $$

Moreover, since $\eta $ is positive-definite, there exists an operator $\rho $ such that $\eta =\rho ^{\dagger }\rho $ . Put

$$ \begin{align*} \rho=\sum_j |e_j\rangle\langle \hat{E}_j|, \end{align*} $$

where $\{|e_j\rangle \}$ is the standard basis of $\mathbb {C}^{d}$ . Then

$$ \begin{align*} \rho^{-1}=\sum_j |E_j\rangle\langle e_j| \end{align*} $$

and

$$ \begin{align*} (\rho H\rho^{-1})^{\dagger}=\sum_j r_j|e_j\rangle\langle e_j|=\rho H\rho^{-1}. \end{align*} $$

Hence, H is similar to a Hermitian operator.

As we know, the eigenvalues of pseudo-Hermitian Hamiltonians are all real or complex conjugate pairs. When there exists a pair of nonreal eigenvalues $\gamma _{j}=\overline {\gamma _{k}}$ such that

$$ \begin{align*} W|E_j\rangle=\langle E_k|W|E_j\rangle|\hat{E}_k\rangle, \end{align*} $$

we get

(3.12) $$ \begin{align} \eta=\sum_{t\neq j,k}|\hat{E}_t\rangle\langle \hat{E}_t| +\delta_{\gamma_{j},\overline{\gamma_{k}}}\,\big(|\hat{E}_k\rangle\langle \hat{E}_j|+|\hat{E}_j\rangle\langle \hat{E}_k|\big), \end{align} $$

which is not positive-definite. In this case, $\langle E_j|E_j\rangle _{\eta }=\langle E_k|E_k\rangle _{\eta }=0$ . Thus, it is impossible to define an orthonormal basis with respect to the metric operator $\eta $ .

Let us next apply our construction to a simple $2\times 2$ matrix Hamiltonian which is pseudo-Hermitian. Consider the Hamiltonian

$$ \begin{align*} H=\left(\!\!\! \begin{array}{cc} re^{\mathrm{i}\theta} & se^{\mathrm{i}\phi} \\ te^{-\mathrm{i}\phi} & re^{-\mathrm{i}\theta} \\ \end{array}\!\!\! \right)\neq H^{\dagger}, \quad r,s,t,\theta,\phi \in \mathbb{R}, \end{align*} $$

and the given metric operator

$$ \begin{align*} W=\left(\!\!\! \begin{array}{cc} 0 & e^{\mathrm{i}\phi} \\ e^{-\mathrm{i}\phi} & 0 \\ \end{array}\!\!\! \right)= W^{\dagger}. \end{align*} $$

Clearly, $WH=H^{\dagger }W$ . The eigenvalues of W are $\lambda _{W\pm }=\pm 1 $ which means $W\ngtr 0$ , so the inner product $\langle \cdot |\cdot \rangle _{W}=\langle \cdot |W|\cdot \rangle $ is not well defined. The eigenvalues of H are

$$ \begin{align*} \gamma_{\pm}=r\cos\theta\pm\sqrt{st-r^{2}\sin^{2}\theta}. \end{align*} $$

Therefore, for $st>r^{2}\sin ^{2}\theta $ , the eigenvalues are real, while for $st<r^{2}\sin ^{2}\theta $ , the eigenvalues are complex conjugates of each other (for $st=r^{2}\sin ^{2}\theta $ , the Hamiltonian cannot be diagonalized).

Case 1. For $st>r^{2}\sin ^{2}\theta $ , the eigenvalues of Hamiltonian H are

$$ \begin{align*} \gamma_{\pm}=r\cos\theta\pm\sqrt{st-r^{2}\sin^{2}\theta} \in \mathbb{R}. \end{align*} $$

Then H is quasi-Hermitian, and there must exist a positive-definite metric operator. The eigenstates of H corresponding to the two eigenvalues are

$$ \begin{align*}|E_{\pm}\rangle=\frac{1}{\sqrt{2(st-r^{2}\sin^{2}\theta)}}\left(\!\!\! \begin{array}{c} s e^{\mathrm{i}\phi/2}\\[.3cm] \Big(-\mathrm{i}r\sin\theta\pm\sqrt{st-r^{2}\sin^{2}\theta}\Big)e^{-\mathrm{i}\phi/2}\\ \end{array}\!\!\! \right), \end{align*} $$

the eigenstates of $H^{\dagger }$ corresponding to the two eigenvalues are

$$ \begin{align*}|\hat{E}_\pm\rangle=\frac{1}{\sqrt{2}\Big(\sqrt{st-r^{2}\sin^{2}\theta}\pm\mathrm{i}r\sin\theta \Big)}\left(\!\!\! \begin{array}{c} te^{\mathrm{i}\phi/2} \\[.3cm] \Big(\mathrm{i}r\sin\theta\pm\sqrt{st-r^{2}\sin^{2}\theta}\Big)e^{-\mathrm{i}\phi/2}\\ \end{array}\!\!\! \right), \end{align*} $$

and

$$ \begin{align*}\langle\hat{E}_i|E_j\rangle=\delta_{ij},\quad i,j\in\{+,-\}.\end{align*} $$

The operator P in this case can be determined according to

$$ \begin{align*} P_{+}=|E_+\rangle\langle\hat{E}_+| &=\frac{1}{2\sqrt{st-r^{2}\sin^{2}\theta}}\\ &\quad \times \left(\!\!\! \begin{array}{cc} \sqrt{st-r^{2}\sin^{2}\theta}+\mathrm{i}r\sin\theta & se^{\mathrm{i}\phi} \\[.3cm] te^{-\mathrm{i}\phi} & \sqrt{st-r^{2}\sin^{2}\theta}-\mathrm{i}r\sin\theta\\ \end{array}\!\!\! \right),\\ P_{-}=|E_-\rangle\langle\hat{E}_-| &=\frac{1}{2\sqrt{st-r^{2}\sin^{2}\theta}} \\ &\quad \times \left(\!\!\! \begin{array}{cc} \sqrt{st-r^{2}\sin^{2}\theta}-\mathrm{i}r\sin\theta & -se^{\mathrm{i}\phi} \\[.3cm] -te^{-\mathrm{i}\phi} & \sqrt{st-r^{2}\sin^{2}\theta}+\mathrm{i}r\sin\theta\\ \end{array}\!\!\! \right). \end{align*} $$

The coefficients $a_{\pm }$ of A are

$$ \begin{align*} a_{+}=\frac{1}{\langle E_+|E_+\rangle_{W}}&=\frac{\sqrt{st-r^{2}\sin^{2}\theta}}{s},\\ a_{-}=\frac{1}{\langle E_-|E_-\rangle_{W}}&=\frac{-\sqrt{st-r^{2}\sin^{2}\theta}}{s} \end{align*} $$

and

$$ \begin{align*} A=\frac{\sqrt{st-r^{2}\sin^{2}\theta}}{s}(P_{+}-P_{-})=\frac{1}{s} \left(\!\!\! \begin{array}{cc} \mathrm{i}r\sin\theta & se^{\mathrm{i}\phi} \\ te^{-\mathrm{i}\phi} & -\mathrm{i}r\sin\theta\\ \end{array}\!\!\! \right). \end{align*} $$

Thus,

$$ \begin{align*} \eta=WA=\frac{1}{s}\left(\!\!\! \begin{array}{cc} t & -\mathrm{i}r\sin\theta e^{\mathrm{i}\phi} \\ \mathrm{i}r\sin\theta e^{-\mathrm{i}\phi} & s \\ \end{array}\!\!\! \right). \end{align*} $$

It can now be checked that

$$ \begin{align*} \langle E_m|E_n\rangle_{\eta}=\delta_{n,m}, \quad m,n=\pm. \end{align*} $$

Therefore, $\{|E_\pm \rangle \}$ is an orthonormal basis with respect to new inner product $\langle \cdot |\cdot \rangle _{\eta }$ in $\mathbb {C}^{2}$ . We see that

$$ \begin{align*} \eta=|\hat{E}_{+}\rangle\langle\hat{E}_{+}|+|\hat{E}_{-}\rangle\langle\hat{E}_{-}|. \end{align*} $$

Case 2. For $st<r^{2}\sin ^{2}\theta $ , the eigenvalues of H are

$$ \begin{align*} \gamma_{\pm}=r\cos\theta\pm \mathrm{i}\sqrt{r^{2}\sin^{2}\theta-st}\notin \mathbb{R}, \end{align*} $$

the eigenstates corresponding to two eigenvalues are

$$ \begin{align*}|E_{\pm}\rangle=\frac{1}{\sqrt{2(r^{2}\sin^{2}\theta-st)}}\left(\!\!\! \begin{array}{c} s e^{\mathrm{i}\phi/2}\\[.3cm] -\mathrm{i}\Big(r\sin\theta\mp\sqrt{r^{2}\sin^{2}\theta-st}\Big)e^{-\mathrm{i}\phi/2}\\ \end{array}\!\!\! \right), \end{align*} $$

the eigenstates of $H^{\dagger }$ corresponding to the two eigenvalues are

$$ \begin{align*}|\hat{E}_\pm\rangle=\frac{-1}{\sqrt{2}\Big(\sqrt{r^{2}\sin^{2}\theta-st}\pm r\sin\theta \Big)}\left(\!\!\! \begin{array}{c} te^{\mathrm{i}\phi/2} \\[.3cm] \mathrm{i}\Big(r\sin\theta\mp\sqrt{r^{2}\sin^{2}\theta-st}\Big)e^{-\mathrm{i}\phi/2}\\ \end{array}\!\!\! \right), \end{align*} $$

and

$$ \begin{align*}\langle\hat{E}_\pm|E_\pm\rangle=\delta_{ij},\quad i,j\in\{+,-\}.\end{align*} $$

The operator P in this case can be determined according to

$$ \begin{align*} P_{+}=|E_+\rangle\langle\hat{E}_+| &=\frac{1}{2\sqrt{r^{2}\sin^{2}\theta-st}} \\ &\quad \times \left(\!\!\! \begin{array}{cc} \sqrt{r^{2}\sin^{2}\theta-st}+r\sin\theta & -\mathrm{i}se^{\mathrm{i}\phi} \\[.3cm] -\mathrm{i}te^{-\mathrm{i}\phi} & \sqrt{r^{2}\sin^{2}\theta-st}-r\sin\theta\\ \end{array}\!\!\! \right),\\ P_{-}=|E_-\rangle\langle\hat{E}_-| &=\frac{1}{2\sqrt{r^{2}\sin^{2}\theta-st}} \\ &\quad \times \left(\!\!\! \begin{array}{cc} \sqrt{r^{2}\sin^{2}\theta-st}-r\sin\theta & \mathrm{i}se^{\mathrm{i}\phi} \\[.3cm] \mathrm{i}te^{-\mathrm{i}\phi} & \sqrt{r^{2}\sin^{2}\theta-st}+r\sin\theta\\ \end{array}\!\!\! \right). \end{align*} $$

According to equation (3.5), we have

$$ \begin{align*}(\gamma_{-}-\overline{\gamma_+})\langle E_+|E_-\rangle_{W}=0, \quad (\gamma_{+}-\overline{\gamma_-})\langle E_-|E_+\rangle_{W}=0.\end{align*} $$

Since $\gamma _{-}=\overline {\gamma }_{+}$ , we have

$$ \begin{align*}\langle E_+|E_+\rangle_{W}=0, \quad \langle E_-|E_-\rangle_{W}=0.\end{align*} $$

The coefficients $a_{\pm }$ of A are

$$ \begin{align*} a_{+}=\frac{1}{\langle E_-|E_+\rangle_{W}}&=\frac{-\mathrm{i}\sqrt{r^{2}\sin^{2}\theta-st}}{s},\\ a_{-}=\frac{1}{\langle E_+|E_-\rangle_{W}}&=\frac{\mathrm{i}\sqrt{r^{2}\sin^{2}\theta-st}}{s} \end{align*} $$

and

$$ \begin{align*} A=\frac{-\mathrm{i}\sqrt{st-r^{2}\sin^{2}\theta}}{s}(P_{+}-P_{-})=\frac{-1}{s}\left(\!\!\! \begin{array}{cc} \mathrm{i}r\sin\theta & se^{\mathrm{i}\phi} \\ te^{-\mathrm{i}\phi} & -\mathrm{i}r\sin\theta\\ \end{array}\!\!\! \right). \end{align*} $$

Thus,

$$ \begin{align*} \eta=WA=\frac{-1}{s}\left(\!\!\! \begin{array}{cc} t & -\mathrm{i}r\sin\theta e^{\mathrm{i}\phi} \\ \mathrm{i}r\sin\theta e^{-\mathrm{i}\phi} & s \\ \end{array}\!\!\! \right), \end{align*} $$

which is also not positive-definite. Furthermore,

$$ \begin{align*} \langle E_+|E_+\rangle_{\eta}=0,\quad \langle E_-|E_-\rangle_{\eta}=0,\quad \langle E_+|E_-\rangle_{\eta}=1, \end{align*} $$

and

$$ \begin{align*} \eta=|\hat{E}_{+}\rangle\langle\hat{E}_{-}|+|\hat{E}_{-}\rangle\langle\hat{E}_{+}|. \end{align*} $$