How Does This Quantum Spin Equation Simplify?

  • Thread starter Thread starter Petar Mali
  • Start date Start date
Petar Mali
Messages
283
Reaction score
0
\frac{1}{2}S\sum_{\vec{n},\vec{m}}I_{\vec{n}-\vec{m}}[(S-S_{\vec{n}}^z)+(S-S_{\vec{m}}^z)]=SI(0)\sum_{\vec{m}}(S-S_{\vec{m}}^z)

How can I get this result?
 
Physics news on Phys.org
Separate the terms into two different sums, then you will have one sum that looks like \sum_{n,m} I_{n-m} (S-S^z_n) and another with S^z_m. For the former, the sum over m can be carried out.
 
Petar Mali said:
\frac{1}{2}S\sum_{\vec{n},\vec{m}}I_{\vec{n}-\vec{m}}[(S-S_{\vec{n}}^z)+(S-S_{\vec{m}}^z)]=SI(0)\sum_{\vec{m}}(S-S_{\vec{m}}^z)

What does this equation describe (just curious)?
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top