Let's say I want to calculate the energy spectrum of Heisenberg model:

$ H = J \sum \limits_{\langle i,j \rangle} \vec S_i \cdot \vec S_j -g ~\mu_B \sum \limits_{i} \vec S_i \cdot \vec B$

but value of $J$ is given as $J/k_B= 20 ~K $ (kelvin). now If I put $J= 20 * 1.38* 10^{−23} $ joule in $H$ and find the energy eigenvalues numerically by diagonalizing , then I may not get correct results because of such a low $J$.

Can I do it in following way:

By Converting H in kelvin unit: Keeping J as 20 K and $ \mu_B = 0.671 ~ K ~T^{-1}$ and B in $T $ ($T$ =Tesla). then Calculated Energy will be in $K$ unit, next I can convert it back to joule or eV.

Am I correct or any better way ?


1 Answer 1


You're correct, but there's also an easier way. I tend to rephrase such problems in terms of an overall energy scale and a ratio, e.g. $$ H=J\left[ \sum \limits_{\langle i,j \rangle} \vec S_i \cdot \vec S_j -\frac{g ~\mu_B}{J} \sum \limits_{i} \vec S_i \cdot \vec B \right] = J\left[ \sum \limits_{\langle i,j \rangle} \vec S_i \cdot \vec S_j -\alpha \sum \limits_{i} \vec S_i \cdot \vec B \right] $$ Then I usually set $J=1$ in the numerical calculations (that is, solve the dimensionless Hamiltonian inside the square brackets), and only worry about converting back afterwards. And converting to whatever units of $J$ you prefer is easy, since it's just a multiplicative factor.

This approach also has the benefit of making it easy to directly compare one's results with already published literature. Many papers on model Hamiltonians are about the Hamiltonian itself, not some experimental system, and so they present data either in units where $J=1$ or in units of $\epsilon/J$.

  • $\begingroup$ Thank you very much. I have one more doubt, and I will get back to you. $\endgroup$ Commented Apr 16, 2021 at 13:10
  • 1
    $\begingroup$ @Quantum_Magnet I'm curious why almost a year after your last comment here, you un-accepted this answer, un-upvoted the same user's answer to your other question, and then downvoted that answer, all within 3 minutes? Is there something wrong with these answers? They seem perfectly fine to me! $\endgroup$ Commented Mar 12, 2023 at 22:31
  • $\begingroup$ @NikeDattani, Sorry, It could be due to a mistake. accepted again $\endgroup$ Commented Mar 20, 2023 at 11:11
  • $\begingroup$ @Magnet You unaccepted the answer again. It's of course within your rights to do so, but given the history here I find the messaging unclear. Should I infer anything about my answer's quality or suitability, or is this more likely to be a user interface issue? $\endgroup$
    – Anyon
    Commented Jul 4, 2023 at 17:05
  • $\begingroup$ @Anyon, Hi, I am sorry about that, appears to be user interface issue. Resolved it $\endgroup$ Commented Jul 5, 2023 at 5:17

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