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Quantum Thermometry

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Abstract

We show how Berry phase can be used to construct a precision quantum thermometer. An important advantage of our scheme is that there is no need for the thermometer to acquire thermal equilibrium with the sample. This reduces measurement times and avoids precision limitations. We also discuss how such methods can be used to detect the Unruh effect.

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Notes

  1. Suddenly switching on the coupling is known to be problematic since it can give rise to divergent results. However, in this case such problems are avoided because we are considering an effective \((1+1)\) dimensional setting. In \((3+1)\) dimensions these divergences can be treated by introducing a continuous switching function [29]; the results are qualitatively the same.

  2. Laserscale, http://www.gebotech.de/pdf/LaserscaleGeneralCatalog_en_2010_04.

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Acknowledgments

This work was supported in part by the Natural Sciences and Engineering Research Council of Canada. R.B.M. is grateful to Fabio Scardigli and the organizers of the Horizons of Quantum Physics conference for their invitation to speak at this meeting. E. M-M. gratefully acknowledges the funding of the Banting Postdoctoral Fellowship Programme.

Author information

Authors and Affiliations

  1. Department of Physics and Astronomy, Institute for Quantum Computing, University of Waterloo, Waterloo, ON, Canada

    Robert B. Mann

  2. Department of Applied Mathematics, Institute for Quantum Computing, University of Waterloo, Waterloo, ON, Canada

    Eduardo Martín-Martínez

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  1. Robert B. Mann
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  2. Eduardo Martín-Martínez
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Correspondence to Robert B. Mann.

Appendix: Diagonalization of the Hamiltonian

Appendix: Diagonalization of the Hamiltonian

Consider a point-like detector, endowed with an internal structure, which couples linearly to a scalar field \(\phi (x(t))\) at a point \(x(t)\) corresponding to the world-line of the detector. The interaction Hamiltonian is of the form \(H_I\propto \hat{X} \hat{\phi }(x(t))\) where we have chosen the detector to be modeled by a harmonic oscillator with frequency \(\Omega _b\). In this case the operator \(\hat{X}\propto (b^\dagger + b)\) corresponds to the detector’s position where \(b^{\dagger }\) and \(b\) are creation and anihilation operators.

Suppose that the detector couples only to a single mode of the field with frequency \(|k|=\Omega _a\). The field operator takes the form

$$\begin{aligned} \hat{\phi }(x(t))\approx \hat{\phi }_k(x(t))\propto \left[ a\, e^{i(kx-\Omega _a t)}+a^\dagger \, e^{-i(kx-\Omega _a t)}\right] , \end{aligned}$$

where \(a^{\dagger }\) and \(a\) are creation and annihilation operators associated with the field mode \(k\). The Hamiltonian is therefore given by Eq. (1), which is

$$\begin{aligned} H_T\!=\!\Omega _a a^\dagger a+\Omega _b b^\dagger b + \lambda (b+b^\dagger )[a^\dagger e^{i(kx-\Omega _a t)}+ a e^{-i(kx-\Omega _a t)}], \end{aligned}$$
(20)

where \(\lambda \) is the coupling frequency, and resembles an Unruh-DeWitt detector in the case where the atom interacts with a single mode of the field. In what follows we employ a mixed picture, in which the detector’s operators are time independent, in contrast to standard approaches that employ the interaction picture. The latter is the most convenient picture for computing transition probabilities, whereas we find the former mathematically more convenient for Berry phase calculations.

To diagonalize the Hamiltonian (20) we begin with a diagonal Hamiltonian of the form

$$\begin{aligned} H_0={\omega _a} a^\dagger a + {\omega _b} b^\dagger b \end{aligned}$$
(21)

Our objective is to obtain the unitary transformation that diagonalises (20). We shall do this by finding the unitary transformation that transforms the Hamiltonian (21) into (20); the inverse operator is then the operator that diagonalizes (20). Once we obtain its eigenstates and eigenvalues we will be able to compute the geometrical phase acquired after cyclic evolution. Throughout we shall make use of the relation

$$\begin{aligned} e^{B}A\, e^{-B} = \exp \left( ad_B \right) A = A + [B,A] + \frac{1}{2}[B,[B,A]] + \cdots , \end{aligned}$$
(22)

where \(ad_B(A) \equiv [B,A] \).

Let us introduce the single mode squeeze operator

$$\begin{aligned} S_{a}=\exp \left( \alpha ^* {a^\dagger }^2 - \alpha a^2\right) \end{aligned}$$

whose action on the creation/annihilation operators

$$\begin{aligned} \nonumber S^\dagger _{a}\,a\, S_{a}&= a\, \cosh t + a^\dagger e^{-i\theta }\sinh t\nonumber \\ S^\dagger _{a}a^\dagger S_{a}&= a^\dagger \, \cosh t + a e^{i\theta }\sinh t \end{aligned}$$
(23)

is straightforward to show upon setting \(\alpha = \frac{t}{2} e^{i\theta }\).

We first apply a 2 single mode squeeze to the Hamiltonian \(H_0\) via

$$\begin{aligned} H_{1s}=S^\dagger _a(u,\theta _a) S^\dagger _b(v,\theta _b) H_0 S_b(v,\theta _b)S_a(u,\theta _a) \end{aligned}$$

obtaining

$$\begin{aligned} H_{1s}&={\omega _a} \left[ a^\dagger a\, \cosh 2u +\frac{1}{2}\sinh 2u \left( a^\dagger a^\dagger e^{-i\theta _a} + a a\, e^{i\theta _a} \right) \right] \nonumber \\&\quad +{\omega _b} \left[ b^\dagger b\cosh 2 v +\frac{1}{2}\sinh 2v \left( b^\dagger b^\dagger e^{-i\theta _b} + bb e^{i\theta _b} \right) \right] , \end{aligned}$$
(24)

where we have removed the constant term \(\sinh ^2 u+\sinh ^2 v\).

The 2-mode displacement operator is

$$\begin{aligned} D(\chi )=\exp \left[ \chi a^\dagger b- \chi ^* a b^\dagger \right] \end{aligned}$$
(25)

and its action of (25) on the creation/annihilation operators is

$$\begin{aligned} \nonumber D^\dagger (s,\phi )\,a\, D(s,\phi )&= a\, \cos s + b e^{i\phi }\sin s\nonumber \\ \nonumber D^\dagger (s,\phi )a^\dagger D(s,\phi )&= a^\dagger \, \cos s + b^\dagger e^{-i\phi }\sin s\nonumber \\ \nonumber D^\dagger (s,\phi )\,b\, D(s,\phi )&= b\, \cos s - a e^{-i\phi }\sin s\nonumber \\ D^\dagger (s,\phi )b^\dagger D(s,\phi )&= b^\dagger \, \cos s - a^\dagger e^{i\phi }\sin s, \end{aligned}$$
(26)

where we have defined \(\chi \equiv s e^{i\phi }\).

Computing the effect of the displacement on each of the 6 different operators in (24) we obtain

$$\begin{aligned} D^\dagger (s,\phi )\,a^\dagger a\, D(s,\phi )&= a^\dagger a\, \cos ^2 s \!+\! b^\dagger b\,\sin ^2 s\!+\! (1/2)\sin 2s \big (a^\dagger b\, e^{i\phi }+b^\dagger a\, e^{-i\phi }\big )\nonumber \\ D^\dagger (s,\phi )\,b^\dagger b\, D(s,\phi )&= a^\dagger a\, \sin ^2 s \!+\! b^\dagger b\,\cos ^2 s\!-\! (1/2)\sin 2s\, \big (a^\dagger b\, e^{i\phi }+b^\dagger a\, e^{-i\phi }\big )\nonumber \\ D^\dagger (s,\phi )\,a^\dagger a^\dagger \, D(s,\phi )&=a^\dagger a^\dagger \, \cos ^2 s + b^\dagger b^\dagger \,e^{-2i\phi }\sin ^2 s+ a^\dagger b^\dagger \, e^{-i\phi }\sin 2s \nonumber \\ D^\dagger (s,\phi )\,a\, a\, D(s,\phi )&= a a\, \cos ^2 s + bb\,e^{2i\phi }\sin ^2 s+ ab\,e^{i\phi }\sin 2s \nonumber \\ D^\dagger (s,\phi )\,b^\dagger b^\dagger \, D(s,\phi )&=b^\dagger b^\dagger \, \cos ^2 s + a^\dagger a^\dagger \,e^{2i\phi }\sin ^2 s- a^\dagger b^\dagger \, e^{i\phi }\sin 2s \nonumber \\ D^\dagger (s,\phi )\,b\, b\, D(s,\phi )&= bb\, \cos ^2 s + a a\,e^{-2i\phi }\sin ^2 s- a b\, e^{-i\phi }\sin 2s \end{aligned}$$
(27)

Next we compute \(H_{1s,2d}=D^\dagger (s,\phi )H_{1s}D(s,\phi )\). Using (27) we find

$$\begin{aligned} H_{1s,2d}&= {\omega _a} \left\{ \left[ a^\dagger a\, \cos ^2 s + b^\dagger b\,\sin ^2 s+ \frac{1}{2}\sin 2s \left( a^\dagger b\, e^{i\phi }+b^\dagger a\, e^{-i\phi }\right) \right] \cosh 2t_a \right. \nonumber \\&+\frac{1}{2}\sinh 2t_a \left[ \left( a^\dagger a^\dagger \, \cos ^2 s + b^\dagger b^\dagger \,e^{-2i\phi }\sin ^2 s+ a^\dagger b^\dagger \, e^{-i\phi }\sin 2s\right) e^{-i\theta _a}\right. \nonumber \\&\left. \left. + \left( a a\, \cos ^2 s + bb\,e^{2i\phi }\sin ^2 s+ ba\,e^{i\phi }\sin 2s\right) e^{i\theta _a} \right] \right\} \nonumber \\&+{\omega _b} \left\{ \left[ a^\dagger a\, \sin ^2 s + b^\dagger b\,\cos ^2 s- \frac{1}{2}\sin 2s \left( a^\dagger b\, e^{i\phi }+b^\dagger a\, e^{-i\phi }\right) \right] \cosh 2 t_b \right. \nonumber \\&+\frac{1}{2}\sinh 2t_b \left[ \left( b^\dagger b^\dagger \, \cos ^2 s + a^\dagger a^\dagger \,e^{2i\phi }\sin ^2 s- a^\dagger b^\dagger \, e^{i\phi }\sin 2s\right) e^{-i\theta _b} \right. \nonumber \\&\left. \left. +\left( bb\, \cos ^2 s + a a\,e^{-2i\phi }\sin ^2 s- a b\, e^{-i\phi }\sin 2s \right) e^{i\theta _b} \right] \right\} \end{aligned}$$

Regrouping terms we get

$$\begin{aligned} H_{1s,2d}&=g_{1} a^\dagger a+g_{2} b^\dagger b+g_{3} a^\dagger b+g^*_{3} b^\dagger a+g_{4} a^\dagger a^\dagger +g^*_{4} a a\nonumber \\&\quad +g_{5} b^\dagger b^\dagger +g^*_{5} bb+g_{6} a^\dagger b^\dagger +g^*_{6} b a, \end{aligned}$$
(28)

where

$$\begin{aligned} g_{1}&= {\omega _a}\,\cos ^2 s\,\cosh 2u + {\omega _b}\,\sin ^2s\, \cosh 2v,\\ g_{2}&= {\omega _a}\,\sin ^2 s\, \cosh 2u + {\omega _b}\,\cos ^2s\, \cosh 2v\\ g_{3}&= \frac{1}{2}\sin 2s\,e^{i\phi }\left( {\omega _a}\,\cosh 2u - {\omega _b}\cosh 2v\right) \\ g_{4}&= \frac{1}{2}\left( {\omega _a}\,e^{-i\theta _a}\sinh 2u\,\cos ^2 s + {\omega _b}\, e^{-i\theta _b}e^{2i\phi }\sinh 2v\,\sin ^2 s\right) \\ g_{5}&= \frac{1}{2}\left( {\omega _a}\,e^{-i\theta _a}e^{-2i\phi }\sinh 2u\,\sin ^2 s + {\omega _b}\, e^{-i\theta _b} \sinh 2v\,\cos ^2 s\right) \\ g_{6}&= \frac{1}{2}\sin 2s\left( {\omega _a}\,e^{-i\theta _a}e^{-i\phi }\sinh 2u - {\omega _b}\, e^{-i\theta _b}e^{i\phi } \sinh 2v\right) \end{aligned}$$

Applying a one mode rotation of the \(a\) operators

$$\begin{aligned} R_{a}=\exp \left( -i\varphi \, {a^\dagger a}\right) \end{aligned}$$

we find

$$\begin{aligned} R_a a R^\dagger _a&= e^{i\varphi }a \quad R_a a^\dagger R^\dagger _a =e^{-i\varphi }a^\dagger \end{aligned}$$
(29)
$$\begin{aligned} R^\dagger _a a R_a&= e^{-i\varphi }a \quad R^\dagger _a a^\dagger R_a =e^{i\varphi }a^\dagger \end{aligned}$$
(30)

yielding

$$\begin{aligned} H_{T}&=g_{1} a^\dagger a+g_{2} b^\dagger b+e^{i\varphi }g_{3} a^\dagger b+e^{-i\varphi }g^*_{3} b^\dagger a+e^{2i\varphi }g_{4} a^\dagger a^\dagger \nonumber \\&\quad +e^{-2i\varphi }g^*_{4} a a+g_{5} b^\dagger b^\dagger +g^*_{5} bb+e^{i\varphi }g_{6} a^\dagger b^\dagger +e^{-i\varphi }g^*_{6} b a \end{aligned}$$
(31)

for the resultant Hamiltonian \(H_{T}=R^\dagger _a\, H_{1s,2d}\, R_a\).

Next we demand two conditions in order to reproduce the interaction Hamiltonian (20). First we remove the squeezing terms \(a^\dagger a^\dagger \) of the field Hamiltonian. To do so, we fix

$$\begin{aligned} \tan ^2 s=\frac{{\omega _a}\,\sinh 2u}{{\omega _b}\,\sinh 2v} \end{aligned}$$

implying

$$\begin{aligned} g_{1}&={\omega _a}\,\cos ^2 s\left[ \cosh 2u +\frac{\sinh 2u}{\tanh 2v}\right] \\ g_{2}&=\cos ^2 s\left[ \frac{{\omega _a}^2}{2{\omega _b}}\, \frac{\sinh 4u}{\sinh 2v}+{\omega _b}\,\cos 2v\right] \\ g_{3}&=\frac{1}{2}\sin 2s\,e^{i\phi }\left( {\omega _a}\,\cosh 2u - {\omega _b}\cosh 2v\right) \\ g_{4}&=\frac{1}{2}{\omega _a}\,\cos ^2 s\,\sinh 2u\left( e^{-i\theta _a} + e^{-i\theta _b}e^{2i\phi }\right) \\ g_{5}&=\frac{1}{2}\cos ^2s\left( \frac{{\omega _b}^2_a}{{\omega _b}}\frac{\sinh ^2 2u}{\sinh 2v}e^{-i\theta _a}e^{-2i\phi } + {\omega _b} \sinh 2v\, e^{-i\theta _b}\right) \\ g_{6}&=\frac{1}{2}\sin 2s\left( {\omega _a}\,e^{-i\theta _a}e^{-i\phi }\sinh 2u - {\omega _b}\, e^{-i\theta _b}e^{i\phi } \sinh 2v\right) \end{aligned}$$

Setting \(\theta _b=2\phi +\theta _a-\pi \) yields

$$\begin{aligned} g_{1}&={\omega _a}\,\cos ^2 s\left[ \cosh 2u +\frac{\sinh 2u}{\tanh 2v}\right] \\ g_{2}&=\cos ^2 s\left[ \frac{{\omega _a}^2}{2{\omega _b}}\, \frac{\sinh 4u}{\sinh 2v}+{\omega _b}\,\cos 2v\right] \\ g_{3}&=\frac{1}{2}\sin 2s\,e^{i\phi }\left( {\omega _a}\,\cosh 2u - {\omega _b}\cosh 2v\right) \\ g_{4}&=0\\ g_{5}&=\frac{1}{2}e^{-i(2\phi +\theta _a)}\cos ^2s\left( \frac{{\omega _b}^2_a}{{\omega _b}}\frac{\sinh ^2 2u}{\sinh 2v}- {\omega _b} \sinh 2v\right) \\ g_{6}&=\frac{1}{2}e^{-i(\theta _a+\phi )}\sin 2s\left( {\omega _a}\,\sinh 2u + {\omega _b}\, \sinh 2v\right) \end{aligned}$$

and so the term corresponding to a squeezing of the field has been eliminated.

To reproduce the interaction part we require \(g_3=g_6\), implying

$$\begin{aligned} e^{i(2\phi +\theta _a)}\left( {\omega _a}\,\cosh 2u - {\omega _b}\cosh 2v\right) =\left( {\omega _a}\,\sinh 2u + {\omega _b}\, \sinh 2v\right) \end{aligned}$$

Setting \(\theta _a= 2n\pi -2\phi \) gives

$$\begin{aligned} \frac{{\omega _a}}{{\omega _b}}=\frac{\cosh 2v+\sinh 2v}{\cosh 2u-\sinh 2u}=\frac{e^{2v}}{e^{-2u}} \end{aligned}$$

and as a consequence

$$\begin{aligned} u=\frac{1}{2}\ln \left( \frac{{\omega _a}}{{\omega _b}}\right) -v \end{aligned}$$

Finally we need to demand that

$$\begin{aligned} \frac{{\omega _a}}{{\omega _b}}>e^{2v} \end{aligned}$$
(32)

to ensure that \(u>0\).

Recapitulating, we started from the Hamiltonian \(H_0\) and applied two 1-mode squeezing operators, a 1-mode displacement operator and a 1-mode rotation on the field operators

$$\begin{aligned} H_T = R_a^\dagger (\varphi )D^\dagger (s,\phi )S^\dagger _a(u,\theta _a)S^\dagger _b(v,\theta _b) H_0S_a(u,\theta _a)S_b(v,\theta _b)D(s,\phi )R_a(\varphi )\qquad \end{aligned}$$
(33)

yielding a Hamiltonian depending on 6 parameters. By fixing 4 of them

$$\begin{aligned} s&=\arctan \sqrt{\frac{{\omega _a}\,\sinh 2u}{{\omega _b}\,\sinh 2v}},\quad \theta _a= 2n\pi -2\phi \end{aligned}$$
(34)
$$\begin{aligned} \theta _b&= 2\phi +\theta _a-\pi ,\quad u=\frac{1}{2}\ln \left( \frac{{\omega _a}}{{\omega _b}}\right) -v \end{aligned}$$
(35)

with the extra requirement for \(v\) given by (32), we obtain the hamiltonian \(H_T\)

$$\begin{aligned} H_T =\Omega _a a^\dagger a+\hat{\Omega }_b b^\dagger b + \hat{\lambda }(b+b^\dagger )(a^\dagger e^{i(\phi +\varphi )}+ a\, e^{-i(\phi +\varphi )}) + Z\left( b^\dagger b^\dagger + bb\right) ,\nonumber \\ \end{aligned}$$
(36)

where

$$\begin{aligned} \Omega _a&=\frac{\sinh 2v\left[ \cosh \left[ 2(C-v)\right] +\frac{\sinh \left[ 2(C-v)\right] }{\tanh 2v}\right] }{{\omega _a^{-1}}\sinh 2v+\omega _b^{-1}\,\sinh \left[ 2(C-v)\right] }\nonumber \\ \hat{\Omega }_b&= \frac{\sinh 2v\left[ \omega _a^2\, \frac{\sinh \left[ 4\left( C-v\right) \right] }{2\sinh 2v}+\omega _b^2\,\cosh 2v\right] }{\omega _b\,\sinh 2v+\omega _a\,\sinh [2(C-v)]}\nonumber \\ \hat{\lambda }&=\frac{\sqrt{\omega _a\omega _b\,\sinh [2(C-v)]\,\sinh 2v}}{\omega _b\,\sinh 2v+\omega _a\,\sinh [2(C-v)]} \left[ \omega _a\,\cosh \left[ 2(C-v)\right] - \omega _b\cosh 2v\right] \nonumber \\ Z&=\frac{1}{2} \frac{\sinh 2v\left( \omega ^2_a\frac{\sinh ^2 \left[ 2(C-v)\right] }{\sinh 2v}- \omega _b^2 \sinh 2v\right) }{\omega _b\,\sinh 2v+\omega _a\,\sinh [2(C-v)]}\nonumber \\ \varphi&= kx-\Omega _a t \end{aligned}$$
(37)

with \(C=\frac{1}{2}\ln \left( \frac{{\omega _a}}{{\omega _b}}\right) \) and where \(2p= {\tanh }^{-1}\big [-2Z/\hat{\Omega }_b\big ]\) .

The rotation is necessary to account for the time evolution on a given trajectory as it is completely decoupled from the rest of parameters. Actually for a particular choice of the displacement parameter phase \(\phi \) (for example \(\phi =0\)) we trivially get

$$\begin{aligned} \hat{H}_T=\Omega _a\, a^\dagger a+\hat{\Omega }_b\, b^\dagger b + \hat{\lambda }(b+b^\dagger )(a^\dagger \, e^{i\varphi }+ a\, e^{-i\varphi })+ Z\left( b^\dagger b^\dagger + bb\right) \end{aligned}$$
(38)

Applying another squeezing operator \(S_{b}(p)\) (where \(p\) is real) yields

$$\begin{aligned} S^\dagger _a S^\dagger _bb^\dagger b S_b S_a&=b^\dagger b \, \cosh 2p + \frac{1}{2}\sinh 2p\left( b^\dagger b^\dagger + bb \right) + \sinh ^2 p\nonumber \\ S^\dagger _a S^\dagger _b bb S_b S_a&= bb \cosh ^2 p+b^\dagger b^\dagger \,\sinh ^2 p + b^\dagger b\, \sinh 2p +\frac{1}{2}\sinh 2p \nonumber \\ S^\dagger _a S^\dagger _bb^\dagger b^\dagger S_b S_a&= b^\dagger b^\dagger \cosh ^2 p+bb\,\sinh ^2 p + b^\dagger b\, \sinh 2p +\frac{1}{2}\sinh 2p\qquad \end{aligned}$$
(39)

and so the interaction Hamiltonian \(H_T=S^\dagger _b(p)\hat{H}_T S_b(p)\), after eliminating constant terms, is

$$\begin{aligned} H_T&=\Omega _a a^\dagger a+\hat{\Omega }_b\Big [b^\dagger b \cosh 2p + \frac{1}{2}\sinh 2p\left( b^\dagger b^\dagger + b b \right) + \sinh ^2 p\Big ]\nonumber \\&\quad + \hat{\lambda }(\sinh q + \cosh q) (b+b^\dagger )(a^\dagger e^{i\varphi }+ a e^{-i\varphi })\nonumber \\&\quad + Z\big (f b \cosh ^2 p+b^\dagger b^\dagger \sinh ^2 p + b^\dagger b\sinh 2p \nonumber \\&\quad +b^\dagger b^\dagger \cosh ^2 p+bb\,\sinh ^2 p + b^\dagger b\, \sinh 2p\big ) \end{aligned}$$
(40)

which can be rewritten as

$$\begin{aligned} H_T&=\Omega _a\, a^\dagger a+\left( \hat{\Omega }_b \cosh 2p + 2Z \sinh 2p\right) b^\dagger b\nonumber \\&\quad + e^{q}\hat{\lambda }(b+b^\dagger )(a^\dagger \, e^{i\varphi }+ a\, e^{-i\varphi })\end{aligned}$$
(41)
$$\begin{aligned}&\quad +(b^\dagger b^\dagger + bb)\Big (Z\cosh 2p + \frac{\hat{\omega }}{2}\sinh 2p\Big ) \end{aligned}$$
(42)

Fixing a value of \(p\) such that

$$\begin{aligned} 2p=\tanh ^{-1}\left( \frac{-2Z}{\hat{\omega }}\right) \end{aligned}$$

yields the Hamiltonian

$$\begin{aligned} H_T=\Omega _a a^\dagger a\!+\!\sqrt{\hat{\Omega }_b^2\!-4Z^2}\,b^\dagger b + e^{q}\hat{\lambda }(b+b^\dagger )(a^\dagger e^{i\varphi }\!+\! a e^{-i\varphi }) \end{aligned}$$

We can rewrite this as an Unruh DeWitt hamiltonian

$$\begin{aligned} H_T = \Omega _a a^\dagger a+{\Omega _b}\,b^\dagger b + \lambda (b+b^\dagger )(a^\dagger e^{i\varphi }+ a e^{-i\varphi }), \end{aligned}$$
(43)

where

$$\begin{aligned} \lambda =e^{p}\hat{\lambda }\quad \Omega _b = \sqrt{\hat{\Omega }_b^2-4Z^2} \end{aligned}$$
(44)

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Mann, R.B., Martín-Martínez, E. Quantum Thermometry. Found Phys 44, 492–511 (2014). https://doi.org/10.1007/s10701-014-9797-x

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