In an open quantum system setup where the system is coupled to infinite harmonic oscillators as bath (as in the Caldeira leggett model, for example), through the Hamiltonian,

$$H = \frac{\hat{p}^2}{2m} + V(\hat{x}) + \sum_n \frac{\hat{p}^2_n}{2m_n}+ \frac{1}{2}m_n \omega_n ^2 \hat{x}^2 - \hat{x} \sum_n g_n \hat{x}_n$$

Here, $\hat{p}, \hat{x}, \hat{p_n}, \hat{x_n}$ are, respectively, system momentum and position operator and nth bath oscillator momentum and position operator and $g_n$ is the coupling strength between the system and the nth bath oscillator.

One then defines the spectral density as following,

$$j(w)=\frac{1}{m}\sum_i \frac{g_i ^2}{\omega_i ^2}\delta(\omega-\omega_i)$$

The dimension of $j(\omega)$ can be calculated to be $\frac{mass}{time}$.

But in the same paper as linked above (in the supplementary info), the spectral density is later expressed in terms of ${time}^{-1}$. (please refer to the following image). What happened to the mass dimension? Are they now calling $\frac{j(\omega)}{m}$ spectral density?

enter image description here

On the other hand, the following paper expresses the weighted spectral density $\rho(\omega)\omega^2$ as $cm^{-1}$. (please refer to the following image).It appears that in the notation of this paper, $\rho(\omega)$ is the spectral density. If so, then the correct dimension of weighted spectral density should be $\frac{mass}{{time}^3}$. How is it $cm^{-1}$ instead?

enter image description here

I know that sometimes, in experimental literature, frequency is expressed as wavenumber and one needs to multiply the term with speed of light expressed in cms ($c=3 \times 10^{10}$)in order to get the frequency. But in the second paper I have linked, one would need to multiply the expression with $c^3$ instead, because the original expression is in terms of $\frac{mass}{{time}^3}$.

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    $\begingroup$ I believe the main problem is the pigeon hole effect with this terminology. Too many people use the term "spectral density" and it is a different version for each in each paper. It is a loosely defined term in my opinion. $\endgroup$
    – ACR
    Commented Nov 17, 2022 at 2:41
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    $\begingroup$ You will literally have to follow each equation in a given paper and try to recreate their figure if possible. $\endgroup$
    – ACR
    Commented Nov 17, 2022 at 2:43


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