用LIGO观测排除标准$\hbarω$引力子
Exclusion of standard $\hbarω$ gravitons by LIGO observation
摘要：
戴森（Dyson，2013）认为，引力波中的引力子数量非常多，使得它们不可能被分解为单个粒子。虽然是真的，它是在本文中示出的LIGO干涉探测器也经历了频繁的和{ \it离散}量子相互作用与入射的引力波，在这样一种方式，以允许波和探测器之间的能量和动量的交换。这为寻找引力子的另一种方法打开了大门。相互作用的最基本形式是激光光子的一阶费米加速（减速），因为它被在引力波中振荡的测试质量镜反射，导致光子的频率蓝移（红移），这取决于反射前镜子是否朝向光子前进（后退）。如果例如发生蓝移，则波能量被吸收并且振荡将被阻尼。有人认为，这种能量交换相互作用是LIGO观测到的辐射反应噪声的原因（尽管计算这种噪声的相同幅度的更常见方法是基于动量考虑）。最重要的是，在每一次相互作用中，探测器吸收或发射的波能量远远小于标准引力子能量$\hbarω$，其中$ω$是引力波的角频率。这对波能量的量化设置了非常严格的上限，即它必须至少比$\hbarω$低$\approx 10^{11}$倍，而与$ω$本身的值无关。
英文摘要：
Dyson (2013) argued that the extraordinarily large number of gravitons in a gravitational wave makes them impossible to be resolved as individual particles. While true, it is shown in this paper that a LIGO interferometric detector also undergoes frequent and {\it discrete} quantum interactions with an incident gravitational wave, in such a way as to allow the exchange of energy and momentum between the wave and the detector. This opens the door to another way of finding gravitons. The most basic form of an interaction is the first order Fermi acceleration (deceleration) of a laser photon as it is reflected by a test mass mirror oscillating in the gravitational wave, resulting in a frequency blueshift (redshift) of the photon depending on whether the mirror is advancing towards (receding from) the photon before the reflection. If e.g. a blueshift occurred, wave energy is absorbed and the oscillation will be damped. It is suggested that such energy exchanging interactions are responsible for the observed radiation reaction noise of LIGO (although the more common way of calculating the same amplitude for this noise is based on momentum considerations). Most importantly, in each interaction the detector absorbs or emits wave energy in amounts far smaller than the standard graviton energy $\hbarω$ where $ω$ is the angular frequency of the gravitational wave. This sets a very tight upper limit on the quantization of the wave energy, viz. it must be at least $\approx 10^{11}$ times below $\hbarω$, independently of the value of $ω$ itself.
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