2020-02-16 Kitaev model note 2 :

2020-02-16 本文已影响0人低维量子系统

Fourier transformations

$d_\bf{r} = \frac{1} {\sqrt{\Omega}} \sum_{\bf{q}} e^{ i \bf{qr}} d_{\bf{q} }$

$d^{\dagger}_\bf{r} = \frac{1} {\sqrt{\Omega}} \sum_{\bf{q}} e^{ -i \bf{qr}} d^{\dagger}_{\bf{q} } = \frac{1} {\sqrt{\Omega}} \sum_{\bf{q}} e^{ i \bf{qr}} d^{\dagger}_{\bf{-q} }$

The first term

$\sum_{\bf{r}} ( d^{\dagger}_{\bf{r}} + d_{\bf{r}} )( d^{\dagger}_{\bf{r+n_x}} - d_{\bf{r+n_x}} )$

$=\sum_{\bf{r}} \frac{1} {\sqrt{\Omega}} \sum_{\bf{q}} e^{ i \bf{qr}} ( d^{\dagger}_{\bf{-q}} + d_{\bf{q} })\cdot \frac{1} {\sqrt{\Omega}} \sum_{\bf{q'}} e^{ i \bf{q'(r+n_x)}} ( d^{\dagger}_{\bf{-q'}} - d_{\bf{q'} })$

$= \sum_{\bf{q}} \sum_{\bf{q'}} [ \frac{1} {{\Omega}} \sum_{\bf{r}} e^{ i \bf{(q+q')r}} ] \cdot e^{ i \bf{q'n_x}} ( d^{\dagger}_{\bf{-q}} + d_{\bf{q} })( d^{\dagger}_{\bf{-q'}} - d_{\bf{q'} })$

$= \sum_{\bf{q}} \sum_{\bf{q'}} \delta_{\bf{q'},\bf{-q}} \cdot e^{ i \bf{q'n_x}} ( d^{\dagger}_{\bf{-q}} + d_{\bf{q} })( d^{\dagger}_{\bf{-q'}} - d_{\bf{q'} })$

$= \sum_{\bf{q}} e^{- i \bf{qn_x}} ( d^{\dagger}_{\bf{-q}} + d_{\bf{q} })( d^{\dagger}_{\bf{q}} - d_{\bf{-q} })$

$= \sum_{\bf{q}} e^{ -i \bf{qn_x}} ( d^{\dagger}_{\bf{-q}}d^{\dagger}_{\bf{q}} - d^{\dagger}_{\bf{-q}} d_{\bf{-q} } + d_{\bf{q} } d^{\dagger}_{\bf{q}} - d_{\bf{q} } d_{\bf{-q} })$

$P_1 = \sum_{\bf{q}} e^{ -i \bf{qn_x}} ( - d^{\dagger}_{\bf{-q}} d_{\bf{-q} } + d_{\bf{q} } d^{\dagger}_{\bf{q}})$

= $\sum_{\bf{q}} e^{ -i \bf{qn_x}} ( - d^{\dagger}_{\bf{-q}} d_{\bf{-q} } + 1- d^{\dagger}_{\bf{q} } d_{\bf{q}})$

= $-\sum_{\bf{q}} ( e^{ -i \bf{qn_x}} d^{\dagger}_{\bf{-q}} d_{\bf{-q} } + e^{ -i \bf{qn_x}} d^{\dagger}_{\bf{q} } d_{\bf{q}})$

= $-\sum_{\bf{q}} ( e^{ i \bf{qn_x}} d^{\dagger}_{\bf{q}} d_{\bf{q} } + e^{ -i \bf{qn_x}} d^{\dagger}_{\bf{q} } d_{\bf{q}})$

= $-\sum_{\bf{q}} ( e^{ i \bf{qn_x}} + e^{- i \bf{qn_x}} ) d^{\dagger}_{\bf{q} } d_{\bf{q}}$

= $-\sum_{\bf{q}} 2\cos( \bf{qn_x} ) d^{\dagger}_{\bf{q}} d_{\bf{q}}$

$P_2 = \sum_{\bf{q}} e^{ -i \bf{qn_x}} d^{\dagger}_{\bf{-q}}d^{\dagger}_{\bf{q}}$

= $\sum_{\bf{q^+}} e^{ -i \bf{q^+n_x}} d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}}+ \sum_{\bf{-q^+}} e^{ i \bf{q^+n_x}} d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}}$

= $\sum_{\bf{q^+}} [ \cos(q^+n_x) - i\sin(q^+n_x) ]d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}}+ \sum_{-\bf{q^+}} [\cos(q^+n_x) +i\sin(q^+n_x) ]d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}}$

$= \sum_{\bf{q^+}} \cos(q^+n_x)d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}} - i\sin(q^+n_x)d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}} +\cos(q^+n_x)d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}} +i\sin(q^+n_x) d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}}$

$= \sum_{\bf{q^+}} \cos(q^+n_x)d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}} +\cos(q^+n_x)d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}} - i\sin(q^+n_x)d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}} +i\sin(q^+n_x) d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}}$

$= \sum_{\bf{q^+}} [ \cos(q^+n_x)d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}} +\cos(q^+n_x)d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}} ]- i\sin(q^+n_x)d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}} +i\sin(q^+n_x) d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}}$

$= \sum_{\bf{q^+}} \cos(q^+n_x)(d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}} +d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}} )- i\sin(q^+n_x)d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}} +i\sin(q^+n_x) d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}}$

$= \sum_{\bf{q^+}} 0 - i\sin(q^+n_x)d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}} +i\sin(q^+n_x) d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}}$

$= \sum_{\bf{q^+}} i\sin(-q^+n_x)d^{\dagger}_{\bf{-q^+}}d^{\dagger}_{\bf{q^+}} +i\sin(q^+n_x) d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}}$

$= \sum_{\bf{q^-}} i\sin(q^-n_x)d^{\dagger}_{\bf{q^-}}d^{\dagger}_{\bf{-q^-}} +\sum_{\bf{q^+}} i\sin(q^+n_x) d^{\dagger}_{\bf{q^+}}d^{\dagger}_{\bf{-q^+}}$

$= \sum_{\bf{q}} i\sin(qn_x)d^{\dagger}_{\bf{q}}d^{\dagger}_{\bf{-q}}$

Similarly,

$P_3 = \sum_{\bf{q}} e^{ -i \bf{qn_x}} ( - d_{\bf{q} } d_{\bf{-q} })$

$=-e^{ -i \bf{3n_x}} d_{\bf{3} } d_{\bf{-3} }-e^{ i \bf{3n_x}} d_{\bf{-3} } d_{\bf{3} }$

$=-[\cos 3\bf{n_x} -i\sin3\bf{n_x}] d_{\bf{3} } d_{\bf{-3} }-[\cos 3\bf{n_x} +i\sin3\bf{n_x}] d_{\bf{-3} } d_{\bf{3} }$

$=-[\cos 3\bf{n_x}d_{\bf{3} } d_{\bf{-3} } -i\sin3\bf{n_x}d_{\bf{3} } d_{\bf{-3} }] -[\cos 3\bf{n_x} d_{\bf{-3} } d_{\bf{3} }+i\sin3\bf{n_x}d_{\bf{-3} } d_{\bf{3} }]$

$=-[\cos 3\bf{n_x}d_{\bf{3} } d_{\bf{-3} } + \cos 3\bf{n_x} d_{\bf{-3} } d_{\bf{3} } ] -[-i\sin3\bf{n_x}d_{\bf{3} } d_{\bf{-3} }+i\sin3\bf{n_x}d_{\bf{-3} } d_{\bf{3} }]$

$=-[0 ] -[-i\sin3\bf{n_x}d_{\bf{3} } d_{\bf{-3} }+i\sin3\bf{n_x}d_{\bf{-3} } d_{\bf{3} }]$

$=-[i\sin(-3\bf{n_x})d_{\bf{3} } d_{\bf{-3} }+i\sin3\bf{n_x}d_{\bf{-3} } d_{\bf{3} }]$

$=-\sum_{\bf{q}} i\sin \bf{q} \bf{n_x} d_{\bf{-q}} d_{\bf{q}}$

$=\sum_{\bf{q}} i\sin (\bf{-q} \bf{n_x}) d_{\bf{-q}} d_{\bf{q}}$

$=\sum_{\bf{q}} i\sin (\bf{q} \bf{n_x}) d_{\bf{q}} d_{\bf{-q}}$

In summary,

$\sum_{\bf{r}} ( d^{\dagger}_{\bf{r}} + d_{\bf{r}} )( d^{\dagger}_{\bf{r+n_x}} - d_{\bf{r+n_x}} )$

$=-\sum_{\bf{q}} 2\cos( \bf{qn_x} ) d^{\dagger}_{\bf{q}} d_{\bf{q}}+\sum_{\bf{q}} i\sin(qn_x)d^{\dagger}_{\bf{q}}d^{\dagger}_{\bf{-q}} -\sum_{\bf{q}} i\sin (\bf{q} \bf{n_x}) d_{\bf{-q}} d_{\bf{q}}$

The second term

$\sum_{\bf{r}} ( d^{\dagger}_{\bf{r}} + d_{\bf{r}} )( d^{\dagger}_{\bf{r+n_y}} - d_{\bf{r+n_y}} )$

$=-\sum_{\bf{q}} 2\cos( \bf{qn_y} ) d^{\dagger}_{\bf{q}} d_{\bf{q}}+\sum_{\bf{q}} i\sin(qn_y)d^{\dagger}_{\bf{q}}d^{\dagger}_{\bf{-q}} -\sum_{\bf{q}} i\sin (\bf{q} \bf{n_y}) d_{\bf{-q}} d_{\bf{q}}$

The third term

$\sum_{\bf{r}}2d_r^{\dagger}d_r$

$=\sum_r 2\frac{1} {\sqrt{\Omega}} \sum_{\bf{q}} e^{ -i \bf{qr}} d^{\dagger}_{\bf{q} }\cdot \frac{1} {\sqrt{\Omega}} \sum_{\bf{q'}} e^{ i \bf{q'r}} d_{\bf{q'} }$

$= 2 \sum_{\bf{q}}\sum_{\bf{q'}} \frac{1} {\Omega} \sum_r e^{ -i \bf{(q'-q)r}} d^{\dagger}_{\bf{q} } d_{\bf{q'} }$

$= 2 \sum_{\bf{q}}\sum_{\bf{q'}} \delta_{\bf{q,q'}} d^{\dagger}_{\bf{q} } d_{\bf{q'} }$

$= 2 \sum_{\bf{q}} d^{\dagger}_{\bf{q} } d_{\bf{q} }$

Finally, we shall have

$H=-J_x\sum_{\bf{q}} 2\cos( \bf{qn_x} ) d^{\dagger}_{\bf{q}} d_{\bf{q}}+\sum_{\bf{q}} i\sin(qn_x)d^{\dagger}_{\bf{q}}d^{\dagger}_{\bf{-q}} -\sum_{\bf{q}} i\sin (\bf{q} \bf{n_x}) d_{\bf{-q}} d_{\bf{q}}$

$-J_y\sum_{\bf{q}} 2\cos( \bf{qn_y} ) d^{\dagger}_{\bf{q}} d_{\bf{q}}+\sum_{\bf{q}} i\sin(qn_y)d^{\dagger}_{\bf{q}}d^{\dagger}_{\bf{-q}} -\sum_{\bf{q}} i\sin (\bf{q} \bf{n_y}) d_{\bf{-q}} d_{\bf{q}}$

$+ J_z\sum_{\bf{q}} 2 d^{\dagger}_{\bf{q} } d_{\bf{q} }$

$H=\sum_{\bf{q}} (2J_z - 2 J_x - 2 J_y) d^{\dagger}_{\bf{q} } d_{\bf{q} }+[ ( iJ_x\sin q_x + iJ_y\sin q_y) d^{\dagger}_{\bf{q}} d^{\dagger}_{\bf{-q}} + h.c.]$

Let's define

$\epsilon_q = 2J_z - 2 J_x - 2 J_y$ ,

$\Delta=J_x\sin q_x + J_y\sin q_y$

Then the Hamiltonian is finally expressed as the standard quadratic form

$H=\sum_{\bf{q}} \epsilon_q d^{\dagger}_{\bf{q} } d_{\bf{q} }+i\Delta d^{\dagger}_{\bf{q}} d^{\dagger}_{\bf{-q}} - i\Delta d_{\bf{-q}} d_{\bf{q}}$

2020-02-16 Kitaev model note 2 :

Fourier transformations

The first term

The second term

The third term

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