Squeezed-state quantum key distribution with a Rindler observer

Abstract

Lengthening the maximum transmission distance of quantum key distribution plays a vital role in quantum information processing. In this paper, we propose a directional squeezed-state protocol with signals detected by a Rindler observer in the relativistic quantum field framework. We derive an analytical solution to the transmission problem of squeezed states from the inertial sender to the accelerated receiver. The variance of the involved signal mode is closer to optimality than that of the coherent-state-based protocol. Simulation results show that the proposed protocol has better performance than the coherent-state counterpart especially in terms of the maximal transmission distance.

Appendix

The output of an energy scaled homodyne detector can be represented by the operator [26]

$$\begin{aligned} \hat{O}_J(\tau ,\phi )=\hat{b}_\mathrm{S}(\tau ){\hat{b}_\mathrm{L}}^\dag (\tau )e^{i\phi }+{\hat{b}_\mathrm{S}}^\dag (\tau )\hat{b}_\mathrm{L}(\tau )e^{-i\phi }, \end{aligned}$$

(32)

where $\hat{b}_\mathrm{S}$ and $\hat{b}_\mathrm{L}$ denote the boson annihilation field operators, while ${\hat{b}_\mathrm{S}}^\dag$ and ${\hat{b}_\mathrm{L}}^\dag$ represent the creation field operators, respectively. The coherent state can be obtained by applying a unitary displacement operation on a vacuum state. The displacement operator is generated by a linear Hamiltonian [8] which can be expressed as

$$\begin{aligned} D(\beta )=\exp {\left[ \beta \left( b^\dag _\mathrm{D}-b_\mathrm{D}\right) \right] }, \end{aligned}$$

(33)

where the subscript D refers to the mode on which the displacement is applied. The operator can be expressed as

$$\begin{aligned} \hat{b}_\mathrm{K}=\int \hbox {d}k_df_\mathrm{K}(k_d,\tau )\hat{b}_{k_d}, \end{aligned}$$

(34)

where $k_d=(k_{d1},k_{d_2},k_{d_3})$ refers to the detection wave vector and $k_{d_1}$ denotes the Rindler frequency while $k_{d_2},k_{d_3}$ being the Minkowski frequencies. The integral is over the whole wave-vector space. Due to the first component of this wave vector is a right Rindler mode which is strictly positive then we have $\int \hbox {d}k_d=\int ^\infty _0\int ^\infty _{-\infty }\int ^\infty _{-\infty } \hbox {d}k_{d_1}\hbox {d}k_{d_2}\hbox {d}k_{d_3}$. The operator $\hat{b}_{k_d}$ obeys the usual boson communication relation

$$\begin{aligned} \left[ \hat{b}_{k_d},\hat{b}^\dag _{k'_d}\right] =\delta (k_{d1}-k'_{d1})\delta \left( k_{d2}-k'_{d2}\right) \delta \left( k_{d3}-k'_{d3}\right) . \end{aligned}$$

(35)

The function $f_\mathrm{K}(k_d)$ localizes the yielded modes in a certain region of space-time, and hence, the mode operators $\hat{b}_\mathrm{K}$ denote modes detected by a local accelerating observer. The average value of the signal received by Bob is

$$\begin{aligned} X=\left\langle \int \hbox {d}\tau \hat{O}(\tau )\right\rangle . \end{aligned}$$

(36)

Similarly, we also define the variance of the integrated signal

$$\begin{aligned} V=\left\langle \left( \int \hbox {d}\tau \hat{O}(\tau )\right) ^2\right\rangle - \left\langle \int \hbox {d}\tau \hat{O}(\tau )\right\rangle ^2. \end{aligned}$$

(37)

The initial state of both the signal and the local oscillator is the Minkowski vacuum state, $|0\rangle _M$. Then X can be expressed as

$$\begin{aligned} X(\phi )=\int \hbox {d}\tau \langle 0|_M\hat{D}^\dag (\beta )\left( \hat{b}_\mathrm{S}\hat{b}^\dag _\mathrm{L}e^{i\phi }+\hat{b}^\dag _\mathrm{S}\hat{b}_\mathrm{L}e^{-i\phi }\right) \hat{D}(\beta )|0\rangle _M. \end{aligned}$$

(38)

Moreover, the displacement operation can be expressed as

$$\begin{aligned} \hat{D}^\dag (\beta )\hat{b}_\mathrm{L}\hat{D}(\beta )=\hat{b}_\mathrm{L}+\beta \int \hbox {d}k_df_\mathrm{L}(k_d,\tau )f^*_\mathrm{D}(k_\mathrm{D},\tau ). \end{aligned}$$

(39)

In this equation, the integral part means the overlap part between the displacement and the mode that is being displaced. Iterating the mode operator definitions into the integral, we have

$$\begin{aligned} \begin{aligned} X(\phi )&=\int \hbox {d}\tau \langle 0|_M\left( \hat{b}_\mathrm{S}^\dag e^{i\phi }\left( \hat{b}_\mathrm{L}+\beta \int \hbox {d}k_df_\mathrm{L}(k_d,\tau )\right) \right) \\&\quad +\,\hat{b}_\mathrm{S}e^{-i\phi }\left( \hat{b}^\dag _\mathrm{L}+\beta \int \hbox {d}k_df^*_\mathrm{L}(k_d,\tau )f_\mathrm{D}(k_d,\tau )\right) |0\rangle _M\\&\cong \beta \int \hbox {d}\tau \langle 0|_M\int \hbox {d}k_d\left( \hat{b}^\dag _\mathrm{S}e^{i\phi }f_\mathrm{L}(k_d,\tau )f^*_\mathrm{D}(k_d,\tau )\right. \\&\left. \quad +\,\hat{b}_\mathrm{S}e^{-i\phi }f^*_\mathrm{L}(k_d,\tau )f_\mathrm{D}(k_d,\tau )\right) |0\rangle _M. \end{aligned} \end{aligned}$$

(40)

To calculate out the expectation value of Eq. (40), Bob’s measurement operator is written in Minkowski modes using the transformation relation given in [27]. The Minkowski annihilation operator annihilates the Minkowski vacuum if we assume the transformation of Eq. (39) is linear in the creation and annihilation operators.

$$\begin{aligned} \hat{a}_{k_s}|0\rangle =0, \end{aligned}$$

(41)

and hence, we have equation $\langle 0|_M\hat{b}_\mathrm{K}|0\rangle _M=\langle 0|_M\hat{b}^\dag _\mathrm{K}|0\rangle _M=0$. We draw a conclusion that the average value of the homodyne signal is zero.

The signal variance can be obtained with the conditions,

$$\begin{aligned} V(\phi )=\langle 0|_M\left( \int \hbox {d}\tau \int \hbox {d}k_d\left[ \hat{b}^\dag _se^{i\phi }f^*_\mathrm{L}(\tau )f_\mathrm{D}(\tau )+\hat{b}_se^{-i\phi }f_\mathrm{L}(\tau )f^*_\mathrm{D}(\tau )\right] ^2\right) . \end{aligned}$$

(42)

The explicit form of the detector wave function is introduced in order to calculate the variance. It is assumed that the detection direction is same as the acceleration and the signal and the local oscillator modes are focused onto the detector. Then the detector wave function can be factored into its inverse and longitudinal components after making the paraxial approximation,

$$\begin{aligned} f_\mathrm{K}(k_d,\tau )=e^{-ik_{d1}a\tau }f_\mathrm{K}(k_{d1})g_\mathrm{K}({k_{d2}})h_\mathrm{K}(k_{d3}). \end{aligned}$$

(43)

It is important to well localize the longitudinal component of the detector wave function. Then we consider a very broad detector wave function in $k_{d1}$ which is localized spatiotemporally. Specially, we have the definition

$$\begin{aligned} f(x)=\left\{ \begin{array}{ll} \frac{1}{\sqrt{2\pi a}} , &{}\quad k_{d1} > 0 \\ 0 , &{}\quad k_{d1} \le 0\\ \end{array} \right. . \end{aligned}$$

(44)

We can further simplify the second part of Eq. (42)

$$\begin{aligned} \begin{aligned}&\int \hbox {d}\tau \hat{b}_\mathrm{S}e^{-i\phi }\int \hbox {d}k_d f^*_L(k_d)f_\mathrm{D}(k_d)\\&\quad =\int \hbox {d}\tau \int \hbox {d}k'_d\int \hbox {d}k_{d1}\frac{e^{-i(k'_{d1}-k_{d1})a\tau }}{2\pi a}\hat{b}_{k'_d}g_\mathrm{S}(\mathbf {k}'_d)e^{-i\phi }\\&\quad =\int \hbox {d}k_{d1}\hat{b}_{k_{d1},s}f_\mathrm{D}(k_{d1})e^{-i\phi }. \end{aligned} \end{aligned}$$

(45)

The new boson annihilation operator is defined as

$$\begin{aligned} \hat{b}_{k_{d1},\mathrm{K}}\equiv \int \hbox {d}\mathbf {k}_d g_\mathrm{K}(\mathbf {k}_d)\hat{b}_{k_d}. \end{aligned}$$

(46)

The vector form can be expanded as $g_\mathrm{K}({\mathbf {k}_d})=g_\mathrm{K}(k_{d2})h_\mathrm{K}(k_{d3})$, where $g_\mathrm{S}$ and $g_\mathrm{L}$ are assumed to be vertical and the new boson operators satisfy $[\hat{b}_{k_{d1},\mathrm{K}},\hat{b}^\dag _{k_{d1},\mathrm{K}}]=1$ and $[\hat{b}_{k_{d1},\mathrm{S}},\hat{b}^\dag _{k_{d1},\mathrm{L}}]=0$. Besides, it is also assumed that the integral over $\tau$ is sufficient long and equation $\int \hbox {d}\tau \frac{1}{2\pi a}e^{-i(k_{d1}-k'_{d1})a\tau }\approx \delta (k_{d1}-k'_{d1})$ holds.

Iterating Eq.(45) into Eq. (42), we have

$$\begin{aligned} \begin{aligned} V(\phi )&\approx \beta ^2\langle 0|_M\int \hbox {d}k_{d1}\int \hbox {d}k'_{d1}\bigg (f_\mathrm{D}(k_{d1})f^*_\mathrm{D}(k'_{d1})\hat{b}_{k_{d1,\mathrm{S}}}\hat{b}^\dag _{k'_{d1,\mathrm{S}}}\\&\quad +\,f^*_\mathrm{D}(k_{d1})f_\mathrm{D}(k'_{d1})\hat{b}^\dag _{k_{d1},\mathrm{S}}\hat{b}_{k'_{d1},\mathrm{S}}\\&\quad +\,f_\mathrm{D}(k_{d1})f_\mathrm{D}(k'_{d1})\hat{b}_{k_{d1},\mathrm{S}}\hat{b}_{k'_{d1},\mathrm{S}}e^{2i\phi }\\&\quad +f^*_\mathrm{D}(k_{d1})f^*_\mathrm{D}(k'_{d1})\hat{b}^\dag _{k_{d1},\mathrm{S}}\hat{b}^\dag _{k'_{d1},\mathrm{S}} e^{-2i\phi }\bigg ). \end{aligned} \end{aligned}$$

(47)

The operator $\hat{b}_{k_{d1},\mathrm{S}}$ is then expressed in Minkowski modes. Using Eq. (10), we have

$$\begin{aligned} \begin{aligned} \hat{b}_{k_{d1},\mathrm{S}}&=\int \hbox {d}\mathbf {k}_dg_\mathrm{S}(\mathbf {k}_d)\int \hbox {d}k_s(A_{k_dk_s}\hat{a}_{k_s}+B_{k_dk_s}\hat{a}^\dag _{k_s})\\&=\int \hbox {d}k_s\frac{1}{\sqrt{2\pi \omega _s(e^{2\pi k_{d1}}-1)}}\left( \frac{\omega _s+k_{s1}}{\omega _s-k_{s1}}\right) ^{i\frac{1}{2}k_{d1}}\\&\quad \times \left[ e^{\pi k_{d1}}\hat{a}_{k_s}g_\mathrm{S}(\mathbf {k}_s)+\hat{a}^\dag _{k_s}g_\mathrm{S}(-\mathbf {k}_s)\right] . \end{aligned} \end{aligned}$$

(48)

Combining with Eqs. (11), (41) and the identity

$$\begin{aligned} \int \hbox {d}k_s\frac{1}{2\pi \omega _s}\left( \frac{\omega _s+k_{s1}}{\omega _s-k_{s1}}\right) ^{i\frac{1}{2} (x-x')}=\delta (x-x'), \end{aligned}$$

(49)

we find

$$\begin{aligned} V(\phi )=\beta ^2\int \hbox {d}k_{d1}|f_\mathrm{D}(k_{d1})|^2\frac{e^{2\pi k_{d1}}+1}{e^{2\pi k_{d1}}-1}. \end{aligned}$$

(50)

In order to calculate the concrete value of V, we need know $f_\mathrm{D}(k_d,\tau )$,

$$\begin{aligned} \begin{aligned} f_\mathrm{D}(k_d,\tau )&=e^{-i\sqrt{a^2k^2_{d1}-k^2_{d2}-k^2_{d3}}\ \xi +ik_{d2}x_2+ik_{d3}x_3 -ik_{d1}a\tau }\\&\quad \times \,f_\mathrm{D}(k_{d1})g_\mathrm{D}(k_{d2})h_\mathrm{D}({k_{d3}}). \end{aligned} \end{aligned}$$

(51)

Moreover, we give some more assumptions, the propagation direction is $-\xi$, the average values of $k_{d2}$ and $k_{d3}$ are zero. Bob’s local oscillator mode function is a Gaussian and reaches the maximum around the frequency $k_{do}$ with respect to his proper time, $\tau$. Then it is suitable to make the approximation,

$$\begin{aligned} |f_\mathrm{D}(k_{d1})|^2\approx \delta (k_{d1}-k_{do}/a). \end{aligned}$$

(52)

Ultimately, the variance of the signal mode is,

$$\begin{aligned} \langle \Delta X_\mathrm{S}(0)^2\rangle =V/\beta ^2=\frac{e^{2\pi k_{do}/a}+1}{e^{2\pi k_{do}/a}-1}. \end{aligned}$$

(53)