In the article, Environment-assisted quantum transport, $\gamma$ is a constant equal to $2\pi kT/\hbar*E_{R}/(\hbar\omega_{c})$ where $T$ is the temperature, $k$ is the Boltzmann constant. Supposedly, this constant should be in units of $\rm cm^{-1}$, but when I do calculations with different values of temperature (in K), my units end up in $\rm cm/s^2$. How can the values end up in $\rm cm^{-1}$?

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    $\begingroup$ $\uparrow$ Link to article? $\endgroup$ – Qmechanic Jun 4 '15 at 19:55
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    $\begingroup$ kT has units of energy. $\hbar$ has units of energy*time. Assuming $E_R$ has energy units, and knowing $\hbar \omega$ has energy units, I get 1/time. Link/ref to the article, and your calculation would be appreciated. Edit: arxiv.org/abs/0807.0929 This seems to be the article. $\endgroup$ – Omry Jun 4 '15 at 20:09
  • $\begingroup$ What is $E_R$ ? $\endgroup$ – Joelafrite Jun 4 '15 at 21:48
  • $\begingroup$ $k=1.3806488(13)×10^{-23} J K^{-1} =8.6173324(78)×^{-5} eV K^{-1} =1.3806488(13)×^{-16} erg K^{-1}$ $\endgroup$ – user66432 Jun 4 '15 at 22:11
  • $\begingroup$ $E_R$ and $\omega_c$ are in units of $cm^{-1}$ $\endgroup$ – TanMath Jun 5 '15 at 4:37

Supposedly, this constant should be in units of cm^-1 but when I do calculations with different values of temperature (in K), my units end up in cm/s^2.

You did something wrong. $kT$ has units of energy. Assuming $E_R$ also has units of energy and $\omega_c$ has units of inverse time means $2\pi kT/\hbar\, E_R/(\hbar\omega_c)$ has units of inverse time -- in other words, frequency.

How can the values end up in cm-1?

Wavenumber (units of inverse length) is oftentimes used in lieu of frequency (units of inverse time) for convenience.

Edit in response to comments
The reason I said "for convenience" is multi-fold. One reason is that it is very convenient to drop those physical constants. For example, the article could have said

For the FMO complex, the reorganization energy is found to be $E_R = 35\,\text{cm}^{-1}\,\hbar c$ [29] and the cutoff $\omega_c = 150\,\text{cm}^{-1}\,c$, inferred from ...

There's no reason for being that pedantically correct. Dropping those conversion factors that involve $\hbar$, $c$, and $k$ is "convenient". The numerical value of each of those constants is one in natural units. Note that this system of natural units ($\hbar = c = k = 1$) leaves one free dimension-full quantity to be resolved. Whether one uses electron-volts or centimeters, the conversion factors between mass, length, time, and temperature, are obvious. Dropping those conversion factors is "convenient" in the sense that the paper is easier to write, and it's easier to read if you know what's going on. (Physics journal papers aren't written for lay audiences. They're written for other physicists to read.)

Another reason for my use of "convenient" is that the numbers that result are humanly convenient, at least in this case. An energy of 35 cm-1 is a much more "convenient" value than is its SI equivalent, 6.95×10-22 joules. The same goes with regard to frequency. A value of 105 cm-1 is a much more convenient value than is its SI equivalent, 3.15×1012 hertz.

As a sanity check that this is exactly what the paper is doing, consider the following in the cited paper:

For the above spectral density the rate turns out to be $γ_φ(T) = 2\pi \frac {kT}{\hbar}\frac{E_R}{\hbar \omega_c}$. This gives a rough estimate for the dephasing rate at room temperature of around 300 cm−1 ...

Room temperature in this system of units is about 205 cm−1 (or being pedantically correct, $205\,\text{cm}^{-1}\,\hbar c/k$). With this, $2\pi (205\,\text{cm}^{-1}) \frac {35}{105}$ is about 300 cm-1.

  • $\begingroup$ note that $E_R$ and $\omega_c$ are in units of cm^-1 as well but when I convert $E_R$ to J and $\omega_C$ to Hz I get gamma=12068 cm^-1 for 300 K but it should be close to 300 cm^-1 like it says in the paper... Any info on that? $\endgroup$ – TanMath Jun 5 '15 at 5:23
  • $\begingroup$ Why is it that you have in your wolfram alpha $hc/k$ while in the link title, it says $\hbar c/k$? Should you multiply by h bar in your wolfram alpha calculation? $\endgroup$ – TanMath Oct 7 '15 at 22:39

The author is using natural units in which both energy and temperature have units of $\mathrm{cm}^{-1}$, (and $ \hbar =k_b=c=1$)

  • $\begingroup$ are u sure? it isn't mentioned in a paper... note that natural units are used in cosmology and QCD, but this is the field of quantum mechanics and open quantum systems... $\endgroup$ – TanMath Jun 5 '15 at 5:26
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    $\begingroup$ Apparently I'm not supposed to say plus one or minus one anymore, but plus one. This answer is correct. Bruce, you missed one constant, which is $c=1$. Setting the speed of light to have a numerical value of one is what enables the authors of the paper to express frequency in units of cm$^{-1}$. $\endgroup$ – David Hammen Jun 5 '15 at 10:23

I believe the original article is http://research.smp.uq.edu.au/kassal/papers/kassal003.pdf

and the section concerned (clipped from the article) is

enter image description here

Note that the units of $\gamma$ can be derived by looking at the components whose dimensions are known (following @Omry's comment):

$$kT = \mathrm{energy}\\ E_R = \text{energy}\\ \hbar \omega = \text{energy}\\ \hbar = \mathrm{energy \cdot time}$$

Which leaves us with dimensions of $s^{-1}$ for $\gamma$.

I expect that dephasing can be expressed either in terms of "per second" or "per cm" - the two values being connected by the speed of light.


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