JohnGringo
- 1
- 0
Hey
Given an anisotropic hamiltonian
[itex] \mathcal{H} = -\sum_{j,\rho} \left( J_\rho^z s_j^z s_{j+\rho}^z + \frac{J_\rho^{xy}}{2}\left( s_j^+ s_{j+\rho}^- + s_j^- s_{j+\rho}^+ \right)\right) - g\mu_B H\sum_j s_j^z[/itex]
Here [itex]\rho[/itex] is a vector connecting the neighbouring sites.
How do I show that the state
[itex] |k> = \frac{1}{\sqrt{2S}}s_{k}^{-}|0>[/itex]
where
[itex] s_{k}^-=\frac{1}{\sqrt{N}}\sum_j\exp(ik\cdot r_j)s_j^-[/itex]
is an eigenstate of the hamiltonian?
So the plan is the use the Fourier transform some how, but I am kind of lost with this. What do I substitute where and why?
Thanks!
Given an anisotropic hamiltonian
[itex] \mathcal{H} = -\sum_{j,\rho} \left( J_\rho^z s_j^z s_{j+\rho}^z + \frac{J_\rho^{xy}}{2}\left( s_j^+ s_{j+\rho}^- + s_j^- s_{j+\rho}^+ \right)\right) - g\mu_B H\sum_j s_j^z[/itex]
Here [itex]\rho[/itex] is a vector connecting the neighbouring sites.
How do I show that the state
[itex] |k> = \frac{1}{\sqrt{2S}}s_{k}^{-}|0>[/itex]
where
[itex] s_{k}^-=\frac{1}{\sqrt{N}}\sum_j\exp(ik\cdot r_j)s_j^-[/itex]
is an eigenstate of the hamiltonian?
So the plan is the use the Fourier transform some how, but I am kind of lost with this. What do I substitute where and why?
Thanks!
_{ij} = -J_z^z s_i^z s_j^z -J_{xy}^{xy} (s_i^{+}s_j^{-} + s_i^{-}s_j^{+}) -g\mu_B H\delta_{ij}s_i^zwhere J_z^z and J_{xy}^{xy} are the exchange parameters and H is the applied magnetic field.Now we can use the Fourier transform to express the state |k> in terms of the basis states. Recall that the Fourier transform is defined as:s_k^- = \frac{1}{\sqrt{N}} \sum_j \exp(ik\cdot r_j) s_j^-where r_j is the position vector of site j. Plugging this into the definition of |k>, we have:|k> = \frac{1}{\sqrt{2S}} \frac{1}{\sqrt{N}} \sum_j \exp(ik\cdot r_j) s_j^- |0>Now we can calculate the action of the Hamiltonian on the state |k>. Using the definition of the Hamiltonian matrix above, we have