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 [tex]A[/tex] sublattice, and the spins up the [tex]B[/tex] sublattice. Perform a canonical transformation on the [tex]B[/tex] (but not on the [tex]A[/tex] spins: rotate them by [tex]180^{\circ}[/tex] about the [tex]\hat{S}^x[/tex] axis,

[tex]\hat{S}_j^{\pm}\rightarrow +\hat{S}_j^{\mp}[/tex]


[tex]\hat{S}_j^z \rightarrow -\hat{S}_j^z[/tex] [tex](j in B)[/tex]

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


[tex]\hat{S}_j^z \rightarrow -\hat{S}_j^z[/tex] [tex](j in B)[/tex]

are satisfied?

 

[DrDu]

A rotation by 180 deg around x is [tex] \exp(i\sigma_x \pi/2)=\cos(\pi/2) +i\sigma_x \sin(\pi/2)=\sigma_x [/tex].
Hence [tex] \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 [/tex], 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]

[tex]
\exp(i\sigma_x \pi/2)=\cos(\pi/2) +i\sigma_x \sin(\pi/2)=i \sigma_x
[/tex]

How you get this?

[tex]
\exp(i\sigma_x \pi/2)=\cos(\sigma_x \pi/2) +isin(\sigma_x \pi/2)[/tex]


from this?

 

[DrDu]

use [tex]\sigma_x^2=1[/tex] and the Taylor representation of the sin.

 

[Petar Mali]

Thanks!

[tex]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}=
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)!}
=i\sigma_x\sum^{\infty}_{n=0}\frac{{(\frac{\pi}{2})}^{2n+1}}{(2n+1)!}=i\sigma_x[/tex]

I have one more question?

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

 

[Petar Mali]

You want to say that

[tex]e^{\frac{i\sigma_x\pi}{2}}\hat{S}_j^z=-\hat{S}_j^z[/tex]?

and

[tex]e^{\frac{i\sigma_x\pi}{2}}\hat{S}_j^{\pm}=\hat{S}_j^{\mp}[/tex]?

 

[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

[tex] e^{\frac{i\sigma_x\pi}{2}}\hat{S}_j^z e^{-\frac{i\sigma_x\pi}{2}}=-\hat{S}_j^z [/tex]

and

[tex] e^{\frac{i\sigma_x\pi}{2}}\hat{S}_j^{\pm} e^{-\frac{i\sigma_x\pi}{2}}=\hat{S}_ j^{\mp} [/tex]

 

[Petar Mali]

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

[tex]\vec{s}=\frac{1}{2}\vec{\sigma}[/tex]

for [tex]\hbar=1[/tex].

So

[tex]
e^{i s_x\pi}\hat{S}_j^z e^{-i s_x\pi}=-\hat{S}_j^z
[/tex]

[tex]
e^{i s_x\pi}\hat{S}_j^{\pm} e^{-i s_x\pi}=\hat{S}_ j^{\mp}
[/tex]

for any [tex]s_x[/tex]. 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.

[tex]
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}=
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)!}
=i\sigma_x\sum^{\infty}_{n=0}\frac{{(\frac{\pi}{2} )}^{2n+1}}{(2n+1)!}=i\sigma_x
[/tex]

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

[tex]
e^{\frac{i\sum_k\sigma_k^x\pi}{2}}=\prod_ke^{\frac{i\sigma_k^x\pi}{2}}[/tex]

Am I right?

 

[DrDu]

Yes, you have to take this product.