# What are the units pm/K?

I can only think of picometres, but it doesn't seem to make sense. Here is the context, from the paper 'Towards Reproducible Ring Resonator Based Temperature Sensors', Klimov et al., Sensors & Transducers 191, 63 (2015):

Our results suggest that consistently high performance temperature sensors are obtained from the zone of stability (waveguide width > 600 nm, air gap ≈ 130 nm and ring radius >10 µm) such that quality factors are consistent ≈ 104 and the temperature sensitivity is in the 70 pm/K to 80 pm/K range.

• Sounds like picometers to me, but I can't make sense of the noise floor... should that be $80\mu K/\sqrt{Hz}$? The noise should still be dependent on the integration time... or is that with 1/f noise? – CuriousOne Apr 28 '16 at 8:47
• Further in the paper, "temperature dependent shifts in resonant wavelength of 10 pm/K", so yes, picometres per Kelvin. – lemon Apr 28 '16 at 8:59
• @lemon Yes. But not a particularly useful measurement when I was looking for the resolution in milliKelvin. Noise floor suggests it's not good enough for what I want. – user56903 Apr 28 '16 at 9:10

it is fairly clear that the unit is picometer per kelvin. That is, you have some device with a resonance wavelength $\lambda_\mathrm{R}$ which depends on temperature, $$\lambda_\mathrm{R}=\lambda_\mathrm{R}(T)=\lambda_\mathrm{R,0}+\alpha (T-T_0),$$ where I've expanded linearly around $\lambda_\mathrm{R,0}= \lambda_\mathrm{R} (T_0)$. The sensitivity then has dimensions of length over temperature (consistent with $\mathrm{pm/K}$), and the range seems about right - a few parts per thousand increase in wavelength in the visible range for a $1\:\mathrm{K}$ temperature increase.