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\(\mathcal {P}\mathcal {T}\)-symmetry in Compact Phase Space for a Linear Hamiltonian

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Abstract

We study the time evolution of a \(\mathcal {P}\mathcal {T}\)-symmetric, non-Hermitian quantum system for which the associated phase space is compact. We focus on the simplest non-trivial example of such a Hamiltonian, which is linear in the angular momentum operators. In order to describe the evolution of the system, we use a particular disentangling decomposition of the evolution operator, which remains numerically accurate even in the vicinity of the Exceptional Point. We then analyze how the non-Hermitian part of the Hamiltonian affects the time evolution of two archetypical quantum states, coherent and Dicke states. For that purpose we calculate the Husimi distribution or Q function and study its evolution in phase space. For coherent states, the characteristics of the evolution equation of the Husimi function agree with the trajectories of the corresponding angular momentum expectation values. This allows to consider these curves as the trajectories of a classical system. For other types of quantum states, e.g. Dicke states, the equivalence of characteristics and trajectories of expectation values is lost.

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Acknowledgements

The authors are grateful to A. B. Klimov for many fruitful discussions. T.G. received financial support from CONACyT through the grant “Ciencia de Frontera 2019”, No. 10872.

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CONACyT “Ciencia de Frontera 2019”, Number 10872.

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Appendices

Appendix A: Properties of coherent and Dicke states

We are using as initial states two states with different properties of their distributions: one that is localized and one that is not. In fact, it is well known [42] that for a coherent state \(\left | \theta _{0} , \phi _{0}\right \rangle \), centered at \(\left (\left \langle S_{x} \right \rangle , \left \langle S_{y} \right \rangle , \left \langle S_{z} \right \rangle \right )\) with

$$ \begin{array}{@{}rcl@{}} \left\langle S_{x} \right\rangle & = & S \sin \theta_{0} \cos \phi_{0}, \\ \left\langle S_{y} \right\rangle & = & S \sin \theta_{0} \sin \phi_{0}, \\ \left\langle S_{z} \right\rangle & = & S \cos \theta_{0}, \end{array} $$
(44)

its variances are,

$$ \begin{array}{@{}rcl@{}} {{\varDelta}}^{2} S_{x} & = & \frac{S}{2} \left( 1- \sin^{2} \theta_{0} \cos^{2} \phi_{0} \right), \\ {{\varDelta}}^{2} S_{y} & = & \frac{S}{2} \left( 1- \sin^{2} \theta_{0} \sin^{2} \phi_{0} \right), \\ {{\varDelta}}^{2} S_{z} & = & \frac{S}{2} \left( 1- \cos^{2} \theta_{0} \right), \end{array} $$
(45)

with the property

$$ {{\varDelta}}^{2} S_{x} + {{\varDelta}}^{2} S_{y} + {{\varDelta}}^{2} S_{z} = S. $$
(46)

So the state is localized because its variances are \(\sim S\).

The Dicke state \(\left | S, m\right \rangle \), fulfills

$$ \left\langle S_{x} \right\rangle = 0, \left\langle S_{y} \right\rangle = 0, \left\langle S_{z} \right\rangle = m, $$
(47)

with variances

$$ \begin{array}{@{}rcl@{}} {{\varDelta}}^{2} S_{x} & = & \frac{1}{2} \left( S^{2} + S - m^{2} \right), \\ {{\varDelta}}^{2} S_{y} & = & \frac{1}{2} \left( S^{2} + S - m^{2} \right), \\ {{\varDelta}}^{2} S_{z} & = & 0, \end{array} $$
(48)

one can see that the state is not localized, i.e. its variances are \(\sim S^{2}\), when \(m^{2} \sim S\).

Furthermore, consider the functional,

$$ \mathcal{L} \quad :\quad |{{\varPsi}}\rangle \to \langle S_{x}\rangle^{2} + \langle S_{x}\rangle^{2} + \langle S_{x}\rangle^{2} , $$

then we want to show two things: (i) In any case \({\mathscr{L}}[{{\varPsi }}] \le S\); (ii) \({\mathscr{L}}[{{\varPsi }}] = S\) if and only if Ψ is a coherent state. Both statements follow rather immediately from the invariance of \({\mathscr{L}}\) under rotations. For instance to prove (i) assume there exits a Ψ with \({\mathscr{L}}[{{\varPsi }}] > S\) then we can find a rotation which makes 〈Sx〉 = 〈Sy〉 = 0. Hence, we find 〈Sz〉> S which is impossible because of the eigenvalues of Sz.

To prove (ii) we note that one direction of the “if and only if” is clear: If Ψ is a coherent state then \({\mathscr{L}}[{{\varPsi }}] = S\). To show the other case, assume \({\mathscr{L}}[{{\varPsi }}] < S\) and do the rotation just as in the previous case. Then we find 〈Sz〉< S, which means that the state in question cannot be the eigenstate |S, S〉 which it should be (if Ψ really were a coherent state, according to (10)).

Appendix B: Evolution Equation on Phase Space

The von Neumann equation can be written as

$$ i\partial_{t}{\varrho} (t) = [H_{0},{\varrho}(t)]+ i [{{\varGamma}},{\varrho}(t)]_{+}. $$
(49)

To recast this equation on phase space, we multiply by the kernel and take the trace

$$ \begin{array}{@{}rcl@{}} i\partial_{t}\text{tr}\left( {\varrho} (t) \hat{\omega}_{Q}(\theta,\phi)\right) & = & \text{tr}\left( [H_{0},{\varrho}(t)]\hat{\omega}_{Q}(\theta,\phi)\right) \\ & & +i \text{tr}\left( [{{\varGamma}},{\varrho}(t)]_{+}\hat{\omega}_{Q}(\theta,\phi)\right). \end{array} $$
(50)

The correspondence rules or Bopp operators are very useful [40,41,42]:

$$ \left. \begin{array}{c}\hat{{\varrho}}\hat{S}_{z} \\ \hat{S}_{z}\hat{{\varrho}} \end{array}\right\}\longleftrightarrow \left\{ \mp \frac{1}{2}l_{z} + {{\varLambda}}_{0}(\theta,\phi), \right. $$
(51)
$$ \left. \begin{array}{c}\hat{{\varrho}}\hat{S}_{\pm} \\ \hat{S}_{\pm}\hat{{\varrho}} \end{array}\right\}\longleftrightarrow \left\{ \mp \frac{1}{2}l_{\pm} + {{\varLambda}}_{\pm}(\theta,\phi), \right. $$
(52)

where l±,z are the first order differential operators,

$$ l_{\pm}= e^{\pm i \phi}\left( \pm \partial_{\theta}+ i\cot\theta \partial_{\phi} \right), \quad l_{z}=-i\partial_{\phi}. $$
(53)

The operators Λ0,±(𝜃, ϕ) are

$$ {{\varLambda}}_{0}(\theta,\phi) = \frac{1}{2} \left( \frac{1}{\epsilon}\cos\theta -\cos\theta -\sin\theta \partial_{\theta} \right), $$
(54)
$$ {{\varLambda}}_{\pm}(\theta,\phi)= \frac{e^{\pm i\phi}}{2\epsilon}\pm \frac{1}{2}\Big[\cos\theta l_{\pm}-e^{\pm i\phi}\sin\theta (l_{z} \pm 1) \Big], $$
(55)

with 𝜖 = (2S + 1)− 1.

Using (51) and (52), (50) takes the following form

$$ i\partial_{t}Q(\theta,\phi) = -2\left( v l_{x} +2i\gamma S \cos\theta - i\gamma \sin\theta \partial_{\theta}\right)Q(\theta,\phi). $$
(56)

Another way to obtain (56) is through [39]. Equation (50) can be recasted in the following form

$$ i\partial_{t} Q(\theta,\phi) = \frac{i}{S} \{Q(\theta,\phi),Q_{H_{0}}(\theta,\phi)\}_{p} + i \ \hat{\Xi} (\theta,\phi)Q(\theta,\phi). $$
(57)

Here

$$ \{F,G\}_{p} = \frac{1}{\sin\theta}\left( \partial_{\phi}F \ \partial_{\theta}G - \partial_{\theta}F \ \partial_{\phi}G \right) $$
(58)

are the Poisson brackets on the sphere and

$$ \hat{\Xi} (\theta,\phi)Q(\theta,\phi) = \operatorname{tr}([{{\varGamma}}, {\varrho}(t)]_{+} \hat{\omega}_{Q} \left( \theta, \phi\right)) $$
(59)

is the corresponding action of the anticommutator of Γ and the density matrix on Q.

The structure of (57) comprises a symplectic structure given by the Possion brackets {,}p plus a term which originates from the \(\mathcal {P}\mathcal {T}\)-symmetric structure of the Hamiltonian. This structure is similar to the one found by Graefe and coworkers in [20].

Appendix C: Analytical solution of the Ehrenfest Equations for coherent states

Let x, y and z be 〈Sx〉/S, 〈Sy〉/S and 〈Sz〉/S, respectively. Then we get the “normalized” system of equations (c.f. 35),

$$ \begin{array}{@{}rcl@{}} \dot{x} & = & 2\gamma x z \\ \dot{y} & = & -2v z + 2\gamma y z \\ \dot{z} & = & 2v y + 2\gamma z^{2}- 2\gamma, \end{array} $$
(60)

valid for the evolution of coherent states. Note that if \(x=\sin \limits \theta \cos \limits \phi , y=\sin \limits \theta \sin \limits \phi \) and \(z=\cos \limits \theta \) we recover the first two equations in (28). Our aim is it to find the trajectories on the sphere. First, we divide the second equation by the first and obtain

$$ \frac{dy}{dx} = y^{\prime}(x) = \left( y-\frac{v}{\gamma} \right) \frac{1}{x} \ \Longrightarrow \ y(x) = C_{1} x + \frac{v}{\gamma}. $$
(61)

Here C1 is a first integration constant, which is determined from the initial conditions and yields

$$ C_{1} = -\frac{v/\gamma - y_{0}}{x_0} = -\frac{v/\gamma - \sin \theta_{0} \sin \phi_{0}}{\sin \theta_{0} \cos \phi_{0}}, $$
(62)

where 𝜃0 and ϕ0 are the parameters of the initial coherent state. Now we simply use the fact that ((36))

$$ x^{2} + y^{2} + z^{2} = 1. $$
(63)

This means thatt the trajectory is just the intersection of the unit sphere with the plane. This always gives a circle, with its center in the xy-plane. Let (xc,yc) be the center of that circle and (Δx, Δy) be such that

$$ \vec{r}(\alpha) = \left( \begin{array}{c} x_{c} \\ y_{c} \\ 0 \end{array} \right) + \left( \begin{array}{c} {{\varDelta}} x \\ {{\varDelta}} y \\ 0 \end{array} \right) \cos \alpha + \left( \begin{array}{c} 0 \\ 0 \\ \sqrt{{{\varDelta}} x^{2} + {{\varDelta}} y^{2}} \end{array} \right) \sin \alpha. $$
(64)

Then we can calculate all the unknowns xc, yc, Δx and Δy from the two intersection points, x1,2, of the line y = C1x + v/γ and the circle x2 + y2 = 1 in the xy-plane:

$$ x^{2} + \left( C_{1} x +v/\gamma \right)^{2} = 1, $$
(65)

then

$$ \begin{array}{@{}rcl@{}} x_{1,2} & = & -p \pm \sqrt{p^{2} - q}, \\ p & = & \frac{v}{\gamma} \frac{C_{1}}{1+{C_{1}^{2}}}, \\ q & = & \frac{v^{2}/\gamma^{2} - 1}{1 + {C_{1}^{2}}}. \end{array} $$
(66)

With this we obtain

$$ \begin{array}{@{}rcl@{}} x_{c} & = & \frac{x_{1} + x_{2}}{2} = -p, \\ y_{c} & = & C_{1} x_{c} +v/\gamma = -C_{1} p +v/\gamma, \end{array} $$
(67)

and also

$$ \begin{array}{@{}rcl@{}} {{\varDelta}} x & = & \frac{x_{1} - x_{2}}{2} = \sqrt{p^{2} - q}, \\ {{\varDelta}} y & = & C_{1} {{\varDelta}} x = C_{1} \sqrt{p^{2} - q}. \end{array} $$
(68)

Now, as we observed some trajectories that degenerated to a point, let us find a solution for

$$ \begin{array}{@{}rcl@{}} {{\varDelta}} x & = & {{\varDelta}} y = 0 \\ {{\varDelta}} z & = & \sqrt{{{\varDelta}} x^{2} + {{\varDelta}} y^{2}} = 0. \end{array} $$
(69)

It is clear from (68) that this is fulfilled when p2 = q, which yields

$$ \gamma = \frac{\pm i \cos \theta_{0} \cos \phi_{0} + \sin \phi_{0}}{\sin \theta_{0}} v . $$
(70)

We see that real solutions exist only for 𝜃0 = π/2. Only the trajectory of the evolution of an initial coherent state located on the x, y-plane degenerates to a point. The value of γ that leads to this is (v = 1)

$$ \gamma = \sin \phi_{0}. $$
(71)

Furthermore, in this case (𝜃0 = π/2, v = 1) we have

$$ \begin{array}{@{}rcl@{}} x_{c} - x_{0} & \sim & \sin \phi - \gamma , \\ y_{c} - y_{0} & \sim & \sin \phi - \gamma , \end{array} $$
(72)

so both, xcx0 and ycy0, suffer a change of sign before and after (71). This means that trajectories for \(\gamma < {\sin \limits } \phi _{0}\) and for \(\gamma > {\sin \limits } \phi _{0}\) are centered in different sides of the common point (𝜃0,ϕ0). All this is exemplified in Fig. 8.

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Valtierra, I.F., Gaeta, M.B., Ortega, A. et al. \(\mathcal {P}\mathcal {T}\)-symmetry in Compact Phase Space for a Linear Hamiltonian. Int J Theor Phys 60, 3286–3305 (2021). https://doi.org/10.1007/s10773-021-04905-x

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