Classical ground state is Ne\'{e}l state:

[Petar Mali]

Tha classical ground state is Ne\'{e}l state: every spin up is surrounded by nearest neighbours which are down, and vice versa. To give them a name, denote the spins down the $$A$$ sublattice, and the spins up the $$B$$ sublattice. Perform a canonical transformation on the $$B$$ (but not on the $$A$$ spins: rotate them by $$180^{\circ}$$ about the $$\hat{S}^x$$ axis,

$$\hat{S}_j^{\pm}\rightarrow +\hat{S}_j^{\mp}$$


$$\hat{S}_j^z \rightarrow -\hat{S}_j^z$$ $$(j in B)$$

Can you explain me this transformation with more details? I can't see why relations
$$\hat{S}_j^{\pm}\rightarrow +\hat{S}_j^{\mp}$$


$$\hat{S}_j^z \rightarrow -\hat{S}_j^z$$ $$(j in B)$$

are satisfied?

 

[DrDu]

A rotation by 180 deg around x is $$\exp(i\sigma_x \pi/2)=\cos(\pi/2) +i\sigma_x \sin(\pi/2)=\sigma_x$$.
Hence $$\mathbf{\sigma}->\exp(i\sigma_x \pi/2)\mathbf{\sigma}\exp(-i \sigma_x \pi/2)=i\sigma_x \mathbf{\sigma} (-i \sigma_x)=(\sigma_x, -\sigma_y, -\sigma_z)^T$$, as sigma_x commutes with sigma_x and anti-commutes with sigma_y and sigma_z.
That's for spin 1/2 only. You will have to check yourself that it holds also for higher spins.

 

[Petar Mali]

$$\exp(i\sigma_x \pi/2)=\cos(\pi/2) +i\sigma_x \sin(\pi/2)=i \sigma_x$$

How you get this?

$$\exp(i\sigma_x \pi/2)=\cos(\sigma_x \pi/2) +isin(\sigma_x \pi/2)$$


from this?

 

[DrDu]

use $$\sigma_x^2=1$$ and the Taylor representation of the sin.

 

[Petar Mali]

Thanks!

$$e^{\frac{i\sigma_x\pi}{2}}=cos\frac{\sigma_x\pi}{2}+isin\frac{\sigma_x\pi}{2}=isin\frac{\sigma_x\pi}{2}=<br /> i\sum^{\infty}_{n=0}(-1)^n\frac{{(\frac{\sigma_x\pi}{2})}^{2n+1}}{(2n+1)!}=i\sum^{\infty}_{n=0}(-1)^n\frac{{(\frac{\sigma_x\pi}{2})}^{2n}\frac{\sigma_x\pi}{2}}{(2n+1)!}<br /> =i\sigma_x\sum^{\infty}_{n=0}\frac{{(\frac{\pi}{2})}^{2n+1}}{(2n+1)!}=i\sigma_x$$

I have one more question?

I'm rotate them by $$180^{\circ}$$. Why I use $$e^{\frac{i\sigma_x\pi}{2}}$$?
Why not $$e^{{i\sigma_x\pi}}$$

 

[Petar Mali]

You want to say that

$$e^{\frac{i\sigma_x\pi}{2}}\hat{S}_j^z=-\hat{S}_j^z$$?

and

$$e^{\frac{i\sigma_x\pi}{2}}\hat{S}_j^{\pm}=\hat{S}_j^{\mp}$$?

 

[DrDu]

To question #5:Because the spin 1/2 operator s_iis sigma_i /2 (setting hbar to 1).
To question #6:I want to say that

LaTeX Code: e^{\\frac{i\\sigma_x\\pi}{2}}\\hat{S}_j^z e^{-\\frac{i\\sigma_x\\pi}{2}}=-\\hat{S}_j^z

and

LaTeX Code: e^{\\frac{i\\sigma_x\\pi}{2}}\\hat{S}_j^{\\pm} e^{-\\frac{i\\sigma_x\\pi}{2}}=\\hat{S}_ j^{\\mp}

 

[Petar Mali]

I can't read your post!

 

[DrDu]

To question #5:Because the spin 1/2 operator s_iis sigma_i /2 (setting hbar to 1).
To question #6:I want to say that

$$e^{\frac{i\sigma_x\pi}{2}}\hat{S}_j^z e^{-\frac{i\sigma_x\pi}{2}}=-\hat{S}_j^z$$

and

$$e^{\frac{i\sigma_x\pi}{2}}\hat{S}_j^{\pm} e^{-\frac{i\sigma_x\pi}{2}}=\hat{S}_ j^{\mp}$$

 

[Petar Mali]

To question #5:Because the spin 1/2 operator s_iis sigma_i /2 (setting hbar to 1).

$$\vec{s}=\frac{1}{2}\vec{\sigma}$$

for $$\hbar=1$$.

So

$$e^{i s_x\pi}\hat{S}_j^z e^{-i s_x\pi}=-\hat{S}_j^z$$

$$e^{i s_x\pi}\hat{S}_j^{\pm} e^{-i s_x\pi}=\hat{S}_ j^{\mp}$$

for any $$s_x$$. Correct?

 

[DrDu]

Yes, in fact, the relations of #1 hold for any vector, not only for spin for a 180 deg rotation.

 

[DrDu]

First there is no spin 2/3. Second, the operator for higher spins is represented by matrices of dimension (2s+1)x(2s+1). E.g. for spin 1 you get the usual 3x3 rotation matrices.

 

[Petar Mali]

Hm. I know it past a little time but I'm now confused a bit.

$$e^{\frac{i\sigma_x\pi}{2}}=cos\frac{\sigma_x\pi}{2 }+isin\frac{\sigma_x\pi}{2}=isin\frac{\sigma_x\pi} {2}=<br /> i\sum^{\infty}_{n=0}(-1)^n\frac{{(\frac{\sigma_x\pi}{2})}^{2n+1}}{(2n+1) !}=i\sum^{\infty}_{n=0}(-1)^n\frac{{(\frac{\sigma_x\pi}{2})}^{2n}\frac{\sig ma_x\pi}{2}}{(2n+1)!}<br /> =i\sigma_x\sum^{\infty}_{n=0}\frac{{(\frac{\pi}{2} )}^{2n+1}}{(2n+1)!}=i\sigma_x$$

I agree that this is rotation for angle $$\pi$$. But I want to rotate whole B sublattice. I think that in this case I must use somethink like

$$e^{\frac{i\sum_k\sigma_k^x\pi}{2}}=\prod_ke^{\frac{i\sigma_k^x\pi}{2}}$$

Am I right?

 

[DrDu]

Yes, you have to take this product.