Testing a Quantum Inequality with a Meta-analysis of Data for Squeezed Light

Abstract

In quantum field theory, coherent states can be created that have negative energy density, meaning it is below that of empty space, the free quantum vacuum. If no restrictions existed regarding the concentration and permanence of negative energy regions, it might, for example, be possible to produce exotic phenomena such as Lorentzian traversable wormholes, warp drives, time machines, violations of the second law of thermodynamics, and naked singularities. Quantum Inequalities (QIs) have been proposed that restrict the size and duration of the regions of negative quantum vacuum energy that can be accessed by observers. However, QIs generally are derived for situations in cosmology and are very difficult to test. Direct measurement of vacuum energy is difficult and to date no QI has been tested experimentally. We test a proposed QI for squeezed light by a meta-analysis of published data obtained from experiments with optical parametric amplifiers and balanced homodyne detection. Over the last three decades, researchers in quantum optics have been trying to maximize the squeezing of the quantum vacuum and have succeeded in reducing the variance in the quantum vacuum fluctuations to $-\,15$ dB. To apply the QI, a time sampling function is required. In our meta-analysis different time sampling functions for the QI were examined, but in all physically reasonable cases the QI is violated by much or all of the measured data. This brings into question the basis for QI. Possible explanations are given for this surprising result.

Appendix A: Derivation of Marecki’s Quantum Inequality

Marecki [9] derives a Quantum Inequality for squeezed light and squeezed vacuum following the general approach of Fewster and Teo [27] and Pfenning [13]. We briefly describe his derivation to clarify the comparisons to the OPA data. Marecki defines the operator variance of the normally ordered electric field$\ \varDelta E^{2}(x,t)$:

$$\begin{aligned} \varDelta E^{2}(x,t)=E^{2}(x,t)-\langle E^{2}(x,t)\rangle _{vac} \end{aligned}$$

(A.1)

and considers a time sampling of the field squared

$$\begin{aligned} \varDelta =\int _{-\infty }^{+\infty }dtf(t)\varDelta E^{2}(x,t) \end{aligned}$$

(A.2)

where

$$\begin{aligned} 1=\int _{-\infty }^{+\infty }dtf(t) \end{aligned}$$

(A.3)

He also mentions the possibility of including a frequency sampling function $\mu _{p}=$$\mu (\omega _{p}-\omega _{0})$, peaked at $\omega _{0}$, that reflects the frequency response of the apparatus measuring the variance. Since it is necessary to use a frequency sampling function to get finite results for $\varDelta ,$ we will include it in our derivations. However, we note that there is a potential consistency issue using an independently selected frequency sampling function since the time sampling function f(t) implies a frequency selection determined by its Fourier transform. Using the Coulomb gauge, the vector potential is

$$\begin{aligned} A_{i}({\mathbf {x}},t)= & {} \frac{1}{\sqrt{(2\pi )^{3}}}\int \frac{d^{3}k}{\sqrt{2\omega _{k}}}\nonumber \\&\times \sum _{\alpha =1,2}{\mathbf {e}}_{i}^{\alpha }({\mathbf {k}})\{a_{\alpha }^{\dag }({\mathbf {k}})e^{ikx}+a_{\alpha }({\mathbf {k}})e^{-ikx}\} \end{aligned}$$

(A.4)

where $\omega _{k}=|k|$ and $\alpha$ denotes the two polarization states which are normalized and orthogonal to ${\mathbf {k}}$. In the exponentials, $kx=-{\mathbf {k}}\cdot {\mathbf {x}}+\omega t$ represents the scalar product. The electric field operator is

$$\begin{aligned} E_{i}=-\frac{\partial A_{i}}{\partial t} \end{aligned}$$

(A.5)

The expectation value of the time sampled free vacuum field squared is

$$\begin{aligned} \langle E^{2}\rangle _{vac}= & {} \int _{-\infty }^{+\infty }dtf(t)\langle E^{2}({\mathbf {x}},t)\rangle _{vac} \end{aligned}$$

(A.6)

$$\begin{aligned}= & {} \frac{1}{2(2\pi )^{2}}\int \mu _{k}d^{3}k\mu _{p}d^{3}p\sqrt{\omega _{k} \omega _{p}}2\pi f_{FT}(\omega _{p}-\omega _{k})\nonumber \\&\times \sum _{\alpha ,\beta =1,2}{\mathbf {e}}_{i}^{\alpha }({\mathbf {k}}){\mathbf {e}}_{i}^{\beta }({\mathbf {p}} )\delta _{\alpha \beta }\delta ({\mathbf {p}}-{\mathbf {k}}) \end{aligned}$$

(A.7)

where we have used the commutator $[a_{\alpha }({\mathbf {p}}),a_{\beta }^{\dag }({\mathbf {k}})]=\delta _{\alpha \beta }\delta ({\mathbf {p}}-{\mathbf {k}})$ and included the frequency function. The Fourier transform of the time sampling function f(t) is defined as

$$\begin{aligned} f_{FT}(\omega )=\frac{1}{2\pi }\int _{-\infty }^{+\infty }dtf(t)e^{-i\omega t} \end{aligned}$$

(A.8)

Integrating Eq. A.7 over k, using the unity normalization of the polarization vectors ${\displaystyle \sum \limits _{i}} {\mathbf {e}}_{i}^{\alpha }({\mathbf {k}}){\mathbf {e}}_{i}^{\alpha }({\mathbf {p}})=1$ , and that $f_{FT}(0)=1/2\pi$ because of the f(t) nomalization, gives

$$\begin{aligned} \langle E^{2}\rangle _{vac}=\frac{1}{(2\pi )^{3}}\int (\mu _{p}^{2}\omega _{p})d^{3}p \end{aligned}$$

(A.9)

Substituting this result into the expression for the variance $\varDelta$ gives, after integration over time,

$$\begin{aligned} \varDelta= & {} \frac{1}{2(2\pi )^{2}}\int d^{3}kd^{3}p\mu _{k}\mu _{p}\sqrt{\omega _{k}\omega _{p}}\nonumber \\&\times \sum _{\alpha ,\beta =1,2}{\mathbf {e}}_{i}^{\alpha } ({\mathbf {k}}){\mathbf {e}}_{i}^{\beta }({\mathbf {p}})\{a_{\alpha }^{\dag } ({\mathbf {k}})a_{\beta }({\mathbf {p}})e^{i(-{\mathbf {k}}+{\mathbf {p}}){\mathbf {x}}} f_{FT}(\omega _{p}-\omega _{k})\nonumber \\&-a_{\alpha }({\mathbf {k}})a_{\beta }({\mathbf {p}})e^{i({\mathbf {k}}+{\mathbf {p}} ){\mathbf {x}}}f_{FT}(\omega _{p}+\omega _{k})+HC\} \end{aligned}$$

(A.10)

where HC is the Hermitian conjugate. To derive a quantum inequality, Marecki defines a vector operator $B_{i}(\omega )$ and computes the integral over frequency of the norm of ${\mathbf {B}}$ which has to be positive

$$\begin{aligned} \int _{0}^{\infty }d\omega B_{i}^{\dag }(\omega )B_{i}(\omega )>0 \end{aligned}$$

(A.11)

We choose

$$\begin{aligned} B_{i}(\omega )= & {} \frac{1}{\sqrt{2\pi ^{2}}}\int d^{3}p\sqrt{\omega _{p}} \nonumber \\&\times \sum _{\alpha ,\beta =1,2}{\mathbf {e}}_{i}^{\alpha }({\mathbf {p}})\{a_{\alpha }(p)(f^{1/2})_{FT}^{*}(\omega -\omega _{p})e^{i{{\mathbf {p}}}{{\mathbf {x}}}}\nonumber \\&- a_{\alpha }^{\dag }(p)(f^{1/2})_{FT}^{*}(\omega +\omega _{p})e^{-i{{\mathbf {p}}}{{\mathbf {x}}}}\} \end{aligned}$$

(A.12)

and substitute this into Eq. A.11, and use the result in Eq. A.10. After taking the expectation value with respect to state A, we obtain Eq. 4 in Sect. 2. Note that in Eq. A.12, $(f^{1/2})_{FT}^{*}(\omega - \omega _{p})$ means the complex conjugate of the Fourier transform of the square root of f(t).