Generation of two-mode photon-added displaced squeezed states

Beam-splitter (BS), a device that transforms two input modes to two new output modes, is described by the unitary operator

$$\begin{eqnarray}&&{{B}_{ab}}\left( t \right)={\rm exp} \left[ \theta \left( {{a}^{\dagger }}b-{{b}^{\dagger }}a \right) \right],\end{eqnarray} \tag{ 10 }$$

with $t={\rm cos} \theta$ the transmittivity and $r={\rm sin} \theta$ the reflectivity. Figure 1 sketches the arrangement of the devices necessary for the generation process. DC denotes a nondegenerate parametric downconverter described by the two-mode squeezer ${{S}_{ab}}\left( s \right)$. In the downconversion process, a mother-photon from the beam that pumps a ${{\chi }^{(2)}}$ nonlinear medium gives birth to two daughter-photons belonging to two different modes labeled $a$ and $b$. Mode $a\ \;\left( b \right)$ is then displaced by ${{D}_{a}}\left( \alpha \right)\ \;\ \;\left( {{D}_{b}}\left( \beta \right) \right)$ which can be implemented by means of a low-reflectivity (low-$r$) and an intense coherent beam with amplitude ${{\alpha }^{\prime }}\left( {{\beta }^{\prime }} \right)$ such that $r{{\alpha }^{\prime }}=\alpha \left( r{{\beta }^{\prime }}=\beta \right)$. As a result, the state ${{\left| \alpha ,\beta ,s \right\rangle }_{ab}}$ of the form (1) appears after such operations. To simulate the action of ${{a}^{\dagger m}}{{b}^{\dagger n}},$ let mode $a$ of state ${{\left| \alpha ,\beta ,s \right\rangle }_{ab}}$ be an input to a beam-splitter (BS1) having transmittivity $t$ and, at the same time, mode $b$ be an input to another beam-splitter (BS2) having the same transmittivity. As for the other input modes ${{a}^{\prime }}$ and ${{b}^{\prime }}$ of BS1 and BS2, they are Fock states ${{\left| m \right\rangle }_{{{a}^{\prime }}}}$ and ${{\left| n \right\rangle }_{{{b}^{\prime }}}},$ respectively. Behind the beam-splitters there are two photo-detectors, PD1 and PD2, to detect photons of outgoing modes ${{a}^{\prime }}$ and ${{b}^{\prime }}$, respectively. We are interested in the case when neither detectors click, projecting the two modes $a$ and $b$ onto

$$\begin{eqnarray}&&{{\left| {{\psi }_{BS}} \right\rangle }_{ab}}=\frac{{{\left| \psi _{BS}^{\prime } \right\rangle }_{ab}}}{\sqrt{{{P}_{BS}}}},\end{eqnarray} \tag{ 11 }$$

with

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{\left| \psi _{BS}^{\prime } \right\rangle }_{ab}} & = & {}_{{{a}^{\prime }}}^{{}}\left\langle 0 \right|_{{}}^{{}}{{B}_{a{{a}^{\prime }}}}\left( t \right){{\left| m \right\rangle }_{{{a}^{\prime }}}} \\ {} & {} & \times {}_{{{b}^{\prime }}}^{{}}\left\langle 0 \right|_{{}}^{{}}{{B}_{b{{b}^{\prime }}}}\left( t \right){{\left| n \right\rangle }_{{{b}^{\prime }}}}{{\left| \alpha ,\beta ,s \right\rangle }_{ab}} \\ \end{array}\end{eqnarray} \tag{ 12 }$$

and

$$\begin{eqnarray}&&{{P}_{BS}}={}_{ab}^{{}}\left\langle \psi _{BS}^{\prime }|\psi _{BS}^{\prime } \right\rangle _{ab}^{{}}\end{eqnarray} \tag{ 13 }$$

being the success probability (i.e., the probability that there are no clicks at all). Using the decomposition of the beam-splitter [19]

$$\begin{eqnarray}&&{{B}_{a{{a}^{\prime }}}}\left( t \right)={\rm exp} \left( -\frac{r}{t}{{a}^{\prime }}^{\dagger }a \right){{t}^{{{a}^{\prime }}^{\dagger }{{a}^{\prime }}-{{a}^{\dagger }}a}}{\rm exp} \left( \frac{r}{t}{{a}^{\dagger }}{{a}^{\prime }} \right),\end{eqnarray} \tag{ 14 }$$

we have

$$\begin{eqnarray}&&{}_{{{a}^{\prime }}}^{{}}\left\langle 0 \right|_{{}}^{{}}{{B}_{a{{a}^{\prime }}}}\left( t \right){{\left| m \right\rangle }_{{{a}^{\prime }}}}=\frac{{{r}^{m}}}{{{t}^{m}}\sqrt{m!}}{{t}^{{{a}^{\dagger }}a}}{{a}^{\dagger m}}\end{eqnarray} \tag{ 15 }$$

and

$$\begin{eqnarray}&&{}_{{{b}^{\prime }}}^{{}}\left\langle 0 \right|_{{}}^{{}}{{B}_{b{{b}^{\prime }}}}\left( t \right){{\left| n \right\rangle }_{{{b}^{\prime }}}}=\frac{{{r}^{n}}}{{{t}^{n}}\sqrt{n!}}{{t}^{{{b}^{\dagger }}b}}{{b}^{\dagger n}}.\end{eqnarray} \tag{ 16 }$$

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Substituting equations (15) and (16) into equation (12) yields

$$\begin{eqnarray}&&{{\left| \psi _{BS}^{\prime } \right\rangle }_{ab}}=\frac{{{r}^{m+n}}}{{{t}^{m+n}}\sqrt{m!n!}}{{t}^{{{a}^{\dagger }}a}}{{a}^{\dagger m}}{{t}^{{{b}^{\dagger }}b}}{{b}^{\dagger n}}{{\left| \alpha ,\beta ,s \right\rangle }_{ab}}.\end{eqnarray} \tag{ 17 }$$

Putting ${{t}^{{{a}^{\dagger }}a}}$ and ${{t}^{{{b}^{\dagger }}b}}$ in equation (17) in the form ${\rm exp} \left( {{a}^{\dagger }}a{\rm ln} t \right)$ and ${\rm exp} \left( {{b}^{\dagger }}b{\rm ln} t \right)$ then using the equalities [20]

$$\begin{eqnarray}&&{\rm exp} \left( {{a}^{\dagger }}a{\rm ln} t \right)=\frac{1}{t}\sum \limits_{j=0}^{\infty } \frac{{{\left( 1-{{t}^{-1}} \right)}^{j}}}{j!}{{a}^{j}}{{a}^{\dagger j}},\end{eqnarray} \tag{ 18 }$$

$$\begin{eqnarray}&&{\rm exp} \left( {{b}^{\dagger }}b{\rm ln} t \right)=\frac{1}{t}\sum \limits_{{{j}^{\prime }}=0}^{\infty } \frac{{{\left( 1-{{t}^{-1}} \right)}^{{{j}^{\prime }}}}}{{{j}^{\prime }}!}{{b}^{{{j}^{\prime }}}}{{b}^{\dagger {{j}^{\prime }}}},\end{eqnarray} \tag{ 19 }$$

we can write ${{P}_{BS}}$, equation (13), in the explicit form ready for numerical calculations as

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{P}_{BS}} & = & \frac{{{\left( 1-{{t}^{2}} \right)}^{m+n}}}{m!n!{{t}^{2\left( m+n+2 \right)}}} \\ {} & {} & \times \sum \limits_{j=0}^{\infty } \sum \limits_{{{j}^{\prime }}=0}^{\infty } \frac{{{\left( 1-{{t}^{-2}} \right)}^{j+{{j}^{\prime }}}}}{j!{{j}^{\prime }}!}{{C}_{m+j,n+{{j}^{\prime }}}}\left( \alpha ,\beta ,s \right). \\ \end{array}\end{eqnarray} \tag{ 20 }$$

As for the fidelity ${{F}_{BS}}={{\left| {}_{ab}^{{}}\left\langle {{\psi }_{BS}}|\alpha ,\beta ,s,m,n \right\rangle _{ab}^{{}} \right|}^{2}}$ of the generated state (11) with respect to the target state (2), its explicit form reads

$$\begin{eqnarray}&&\begin{array}{ccccccccccccccc} {{F}_{BS}} \\ =\;\frac{|\sum \limits_{j=0}^{\infty } \sum \limits_{{{j}^{\prime }}=0}^{\infty } \frac{{{(1-{{t}^{-1}})}^{j+{{j}^{\prime }}}}}{j!{{j}^{\prime }}!}{{C}_{m+j,n+{{j}^{\prime }}}}(\alpha ,\beta ,s){{|}^{2}}}{{{C}_{m,n}}(\alpha ,\beta ,s)\sum \limits_{j=0}^{\infty } \sum \limits_{{{j}^{\prime }}=0}^{\infty } \frac{{{(1-{{t}^{-2}})}^{j+{{j}^{\prime }}}}}{j!{{j}^{\prime }}!}{{C}_{m+j,n+{{j}^{\prime }}}}(\alpha ,\beta ,s)} \\ \end{array}.\end{eqnarray} \tag{ 21 }$$

Precisely speaking, from equation (17), the net effect of the beam-splitters BS1 and BS2 in figure 1, conditioned on the outcome that no photons at all are detected by both detectors, is the action of ${{t}^{{{a}^{\dagger }}a}}{{a}^{\dagger m}}{{t}^{{{b}^{\dagger }}b}}{{b}^{\dagger n}}$ on ${{\left| \alpha ,\beta ,s \right\rangle }_{ab}}$. So, a naïve inference would tell us that the desired addition of photons (i.e., the action of ${{a}^{\dagger m}}{{b}^{\dagger n}}$) would be achieved if $t=1$. However, such a mathematical 'result' cannot be accepted since it is unphysical: in fact, a beam-splitter with $t=1$ is tantamount to nothing and thus the state ${{\left| \alpha ,\beta ,s \right\rangle }_{ab}}$ remains itself without any photons added. What can be expected is that the generated state would more resemble the intended one if $t$ is getting closer to 1. To consolidate this we plot in figures 2 and 3 the fidelity (21) as a function of $t$ for several sets of the other parameters. As expected, the figures indicate that, though the fidelity can never be one, it always increases with $t$ and asymptotically tends to 1 in the limit $t\to 1$. The bad thing however is reduction of the corresponding success probability with increasing $t$, as seen from their curves (see the bottom parts of figures 2 and 3), which also agrees with physical intuition. Namely, the greater $t$ becomes, the easier the photons of mode ${{a}^{\prime }}\ \;\left( {{b}^{\prime }} \right)$ get through the beam-splitter BS1 (BS2) towards the detector PD1 (PD2) and, as a consequence, detection of no photons is less possible. Furthermore, as follows from figure 2, for given $\alpha ,\ \;\beta ,\ \;s$ and $t$, both the fidelity and the corresponding generation probability decrease with increasing $m$ or/and $n$. Such a behavior of probability is due to the fact that the chance for a detector to click is proportional to the number of incoming photons. Practically, this implies that adding more photons is more challenging and, even when it succeeded, the price to pay is the reduction of fidelity. Finally, as seen from figure 3, for given $m,\ \;n$ and $t$, the fidelity decreases but the corresponding success probability increases with increasing $\alpha ,\ \;\beta ,\ \;s$. Physically, these properties can be interpreted as follows. Here the state to which we add photons is ${{\left| \alpha ,\beta ,s \right\rangle }_{ab}}$ defined by equation (1). Its total averaged number of photons in both modes can be calculated to be $\left\langle {{{\hat{n}}}_{{\rm total}}} \right\rangle$ = ${}_{ab}^{{}}\left\langle \alpha ,\beta ,s \right|_{{}}^{{}}\left( {{a}^{\dagger }}a+{{b}^{\dagger }}b \right){{\left| \alpha ,\beta ,s \right\rangle }_{ab}}$ = $2{{{\rm sinh} }^{2}}s+{{\alpha }^{2}}+{{\beta }^{2}}$. Then, an increase in any of $\alpha ,\ \;\beta$ or/and $s$ certainly raises $\left\langle {{{\hat{n}}}_{{\rm total}}} \right\rangle$, thus triggering a decrease in the fidelity $\propto \;{{t}^{\left\langle {{{\hat{n}}}_{{\rm total}}} \right\rangle }}$ since $t\lt 1$. As for the success probability behavior, when $t$ is high enough, a larger number of photons in mode $a\ \;\left( b \right)$ may cause fewer photons from both modes $a$ and ${{a}^{\prime }}$ ($b$ and ${{b}^{\prime }}$) to come into the photo-detector PD1 (PD2), thus increasing the possibility of no clicks. As a concrete situation, consider two states ${{\left| 01 \right\rangle }_{a{{a}^{\prime }}}}$ and ${{\left| 11 \right\rangle }_{a{{a}^{\prime }}}}$ as inputs to BS1. The corresponding outputs will be $t{{\left| 01 \right\rangle }_{a{{a}^{\prime }}}}+{\rm i}r{{\left| 10 \right\rangle }_{a{{a}^{\prime }}}}$ and ${\rm i}tr\sqrt{2}\left( {{\left| 02 \right\rangle }_{a{{a}^{\prime }}}}+{{\left| 20 \right\rangle }_{a{{a}^{\prime }}}} \right)+\left( {{t}^{2}}-{{r}^{2}} \right){{\left| 11 \right\rangle }_{a{{a}^{\prime }}}}$. For the first input state the probability of no clicks is $P\left( {{\left| 01 \right\rangle }_{a{{a}^{\prime }}}} \right)={{r}^{2}}$, while for the second input state the probability of no clicks is $P\left( {{\left| 11 \right\rangle }_{a{{a}^{\prime }}}} \right)=2{{t}^{2}}{{r}^{2}}.$ Clearly, $P\left( {{\left| 11 \right\rangle }_{a{{a}^{\prime }}}} \right)\gt P\left( {{\left| 01 \right\rangle }_{a{{a}^{\prime }}}} \right)$ if $t\gt 1/\sqrt{2}\approx 0.707$.

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The Fock states ${{\left| m \right\rangle }_{{{a}^{\prime }}}}$, ${{\left| n \right\rangle }_{{{b}^{\prime }}}}$ in the previous scheme are difficult to prepare and detections of no photons are also difficult to observe due to the always present vacuum noise. So, in this subsection we present an alternative way to add photons. Namely, nondegenerate parametric downconverters will be used instead of beam-splitters. The associated schematic setup is sketched in figure 4. Let the squeezing degree of both the downconverters DC2 and DC3 be $z$ which is also assumed real for simplicity. The generation is regarded to be successful when at the same time PD1 registers $m$ photons and PD2 $n$ photons. If this event happens, it projects the states of modes $a$ and $b$ onto

$$\begin{eqnarray}&&{{\left| {{\psi }_{DC}} \right\rangle }_{ab}}=\frac{{{\left| \psi _{DC}^{\prime } \right\rangle }_{ab}}}{\sqrt{{{P}_{DC}}}},\end{eqnarray} \tag{ 22 }$$

with

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{\left| \psi _{DC}^{\prime } \right\rangle }_{ab}} & = & {}_{{{a}^{\prime }}}^{{}}\left\langle m \right|_{{}}^{{}}{{S}_{a{{a}^{\prime }}}}\left( z \right){{\left| 0 \right\rangle }_{{{a}^{\prime }}}} \\ {} & {} & \times {}_{{{b}^{\prime }}}^{{}}\left\langle n \right|_{{}}^{{}}{{S}_{b{{b}^{\prime }}}}\left( z \right){{\left| 0 \right\rangle }_{{{b}^{\prime }}}}{{\left| \alpha ,\beta ,s \right\rangle }_{ab}} \\ \end{array}\end{eqnarray} \tag{ 23 }$$

and

$$\begin{eqnarray}&&{{P}_{DC}}={}_{ab}^{{}}\left\langle \psi _{DC}^{\prime }|\psi _{DC}^{\prime } \right\rangle _{ab}^{{}}\end{eqnarray} \tag{ 24 }$$

being the probability of successful generation. Using the decomposition of the two-mode squeezers [19]

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{S}_{a{{a}^{\prime }}}}\left( z \right) & = & {\rm exp} \left[ -{\rm tanh} (z){{a}^{\dagger }}{{a}^{\prime }}^{\dagger } \right] \\ {} & {} & \times {\rm exp} \left[ -{\rm ln} ({\rm cosh} \;z)\left( {{a}^{\dagger }}a+{{a}^{\prime }}{{a}^{\prime }}^{\dagger } \right) \right] \\ {} & {} & \times {\rm exp} \left[ {\rm tanh} (z)a{{a}^{\prime }} \right], \\ \end{array}\end{eqnarray} \tag{ 25 }$$

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{S}_{b{{b}^{\prime }}}}\left( z \right) & = & {\rm exp} \left[ -{\rm tanh} (z){{b}^{\dagger }}{{b}^{\prime }}^{\dagger } \right] \\ {} & {} & \times {\rm exp} \left[ -{\rm ln} ({\rm cosh} \;z)\left( {{b}^{\dagger }}b+{{b}^{\prime }}{{b}^{\prime }}^{\dagger } \right) \right] \\ {} & {} & \times {\rm exp} \left[ {\rm tanh} (z)b{{b}^{\prime }} \right], \\ \end{array}\end{eqnarray} \tag{ 26 }$$

we can calculate the matrix elements in equation (23) and obtain

$$\begin{eqnarray}&&{}_{{{a}^{\prime }}}^{{}}\left\langle m \right|_{{}}^{{}}{{S}_{a{{a}^{\prime }}}}\left( z \right){{\left| 0 \right\rangle }_{{{a}^{\prime }}}}=\frac{{{\left( -{\rm tanh} \;z \right)}^{m}}}{{\rm cosh} \;z\sqrt{m!}}{{a}^{\dagger m}}{{\left( {\rm cosh} \;z \right)}^{-{{a}^{\dagger }}a}}\end{eqnarray} \tag{ 27 }$$

and

$$\begin{eqnarray}&&{}_{{{b}^{\prime }}}^{{}}\left\langle n \right|_{{}}^{{}}{{S}_{b{{b}^{\prime }}}}\left( z \right){{\left| 0 \right\rangle }_{{{b}^{\prime }}}}=\frac{{{\left( -{\rm tanh} \;z \right)}^{n}}}{{\rm cosh} \;z\sqrt{n!}}{{b}^{\dagger n}}{{\left( {\rm cosh} \;z \right)}^{-{{b}^{\dagger }}b}}.\end{eqnarray} \tag{ 28 }$$

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Substituting equations (27) and (28) into equation (23) yields

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{\left| \psi _{DC}^{\prime } \right\rangle }_{ab}} & = & \frac{{{\left( -{\rm tanh} \;z \right)}^{m+n}}}{{{\left( {\rm cosh} \;z \right)}^{2}}\sqrt{m!n!}}{{a}^{\dagger m}}{{\left( {\rm cosh} \;z \right)}^{-{{a}^{\dagger }}a}} \\ {} & {} & \times {{b}^{\dagger n}}{{\left( {\rm cosh} \;z \right)}^{-{{b}^{\dagger }}b}}{{\left| \alpha ,\beta ,s \right\rangle }_{ab}}. \\ \end{array}\end{eqnarray} \tag{ 29 }$$

Using again the equalities (18) and (19) for ${{\left( {\rm cosh} z \right)}^{-{{a}^{\dagger }}a}}$ and ${{\left( {\rm cosh} z \right)}^{-{{b}^{\dagger }}b}}$ we can write ${{P}_{DC}}$, equation (24), explicitly as

$$\begin{eqnarray}\begin{array}{lllllllllllllll} {{P}_{DC}} & = & \frac{{{\left( {\rm sinh} \;z \right)}^{2(m+n)}}}{m!n!} \\ {} & {} & \times \sum \limits_{j=0}^{\infty } \sum \limits_{{{j}^{\prime }}=0}^{\infty } \frac{{{\left( -{{{\rm sinh} }^{2}}z \right)}^{j+{{j}^{\prime }}}}}{j!{{j}^{\prime }}!}{{C}_{m+j,n+{{j}^{\prime }}}}\left( \alpha ,\beta ,s \right). \\ \end{array}\end{eqnarray} \tag{ 30 }$$

and the fidelity ${{F}_{DC}}={{\left| {}_{ab}^{{}}\left\langle {{\psi }_{DC}}|\alpha ,\beta ,s,m,n \right\rangle _{ab}^{{}} \right|}^{2}}$ of the generated state (22) with respect to the target state (2) as

$$\begin{eqnarray}&&\begin{array}{ccccccccccccccc} {{F}_{DC}} \\ =\;\frac{{{\left| \sum \limits_{j=0}^{\infty } \sum \limits_{{{j}^{\prime }}=0}^{\infty } \frac{{{\left( 1-\;{\rm cosh} \;z \right)}^{j+{{j}^{\prime }}}}}{j!{{j}^{\prime }}!}{{C}_{m+j,n+{{j}^{\prime }}}}\left( \alpha ,\beta ,s \right) \right|}^{2}}}{{{C}_{m,n}}\left( \alpha ,\beta ,s \right)\sum \limits_{j=0}^{\infty } \sum \limits_{{{j}^{\prime }}=0}^{\infty } \frac{{{\left( -{{{\rm sinh} }^{2}}z \right)}^{j+{{j}^{\prime }}}}}{j!{{j}^{\prime }}!}{{C}_{m+j,n+{{j}^{\prime }}}}\left( \alpha ,\beta ,s \right)} \\ \end{array}.\end{eqnarray} \tag{ 31 }$$

As seen from equation (29), the net effect of the downconverter DC2 (DC3) in figure 4, conditioned on the outcome that $m\ \;(n)$ photons are detected by the photo-detector PD1 (PD2), is the implementation of the action of ${{a}^{\dagger m}}{{\left( {\rm cosh} z \right)}^{-{{a}^{\dagger }}a}}\ \;\ \;({{b}^{\dagger m}}{{\left( {\rm cosh} z \right)}^{-{{b}^{\dagger }}b}})$ on mode $a\ \;(b)$ of the state ${{\left| \alpha ,\beta ,s \right\rangle }_{ab}}$. So, the smaller the value of $z$ is the better ${{a}^{\dagger m}}{{\left( {\rm cosh} z \right)}^{-{{a}^{\dagger }}a}}\ \;\ \;({{b}^{\dagger n}}{{\left( {\rm cosh} z \right)}^{-{{b}^{\dagger }}b}})$ simulates ${{a}^{\dagger m}}({{b}^{\dagger n}})$. This explains why in figures 5 and 6 the fidelity always decreases with increasing $z$. For $z\to 0$ the fidelity is approaching but never reaches $1$ because when $z=0$ nothing will occur. An increase in the fidelity when $z$ is decreasing is, however, accompanied with a decrease in the corresponding success probability, as shown by the bottom parts of figures 5 and 6. This is due to the properties of two-mode squeezers which are reflected by equations (27) and (28). Furthermore, figure 5 shows that, for given $\alpha ,\beta ,s$ and $z$, both the fidelity and generation probability decrease with increasing $m$ or/and $n$. This agrees with the case of using beam-splitters in the previous subsection that adding more photons is both worse in quality and more difficult in realization. Finally, similar to the beam-splitter-based scheme, the fidelity decreases but the corresponding success probability increases with increasing $\alpha ,\beta ,s$ for given $m,n$ and $z,$ as observed from figure 6.

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