Alternating Tight Binding Hamiltonian

Question

The alternating Hamiltonian may be written as:

$$H = t \sum_{n} (-1)^{n} \left[c^{\dagger}_{n+1}c_{n} + c^{\dagger}_{n}c_{n+1} \right] \; \; .$$

I wanted to know the energy dispersion for this system, so I wrote in mommentum space; After some calculations, I got:

$$H = t\sum_{k} 2\cos\left( k \right)\, c^{\dagger}_{k} c_{k+\pi} \; \; .$$

However, I can't diagonalize this. What should I do? 

Answer 1

Due to the factor of $(-1)^n$, we can see that this lattice is actually periodic with a period equal to two sites. That is, we've got a lattice with a two-site basis. As such, when we define our momentum-space operators, we have to take this periodicity into account. To do this, we define two sets of operators via \begin{align} \hat{a}_k&=\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{-ika(2j)/2}\hat{c}_{2j}\,,\\ \hat{b}_k&=\frac{1}{\sqrt{N}}\sum_{j=0}^{N-1}e^{-ika(2j+1)/2}\hat{c}_{2j+1}\,, \end{align} where $a$ is the lattice spacing (i.e., the distance between every two sites), $N$ is the number of primitive unit cells (and so half the total number of sites), and we apply periodic boundary conditions, i.e., $\hat{c}_{N}=\hat{c}_0$, etc. The allowed values of $k$ are determined by assuming periodic boundary conditions, yielding $$ k\to k_m=m\frac{2\pi}{Na}\,,~~~~~~~m=0,1,\dots,N-1\,. $$ Accordingly, we have the orthogonality condition, $$ \frac{1}{N}\sum_ne^{i(k-k')an} = \delta_{kk'}\,, $$ which allows us to invert both of the previous definitions as \begin{align} \hat{c}_{2j}&=\frac{1}{\sqrt{N}}\sum_ke^{ika(2j)/2}\hat{a}_{k}\,,\\ \hat{c}_{2j+1}&=\frac{1}{\sqrt{N}}\sum_ke^{ika(2j+1)/2}\hat{b}_{k}\,. \end{align} Using these definitions, let's rewrite terms in the Hamiltonian as follows. First $$ \sum_n(-1)^n\hat{c}_{n+1}^{\dagger}\hat{c}_n = \sum_n\hat{c}_{2n+1}^{\dagger}\hat{c}_{2n} - \sum_n\hat{c}_{2n+2}^{\dagger}\hat{c}_{2n+1}\,. $$ Plugging in for the operators yields \begin{align} \sum_n\hat{c}_{2n+1}^{\dagger}\hat{c}_{2n} &= \sum_n\frac{1}{\sqrt{N}}\sum_ke^{-ika(2n+1)/2}\hat{b}_{k}^{\dagger} \frac{1}{\sqrt{N}}\sum_{k'}e^{ik'a(2n)/2}\hat{a}_{k}\\ &= \sum_{kk'} \hat{b}_{k}^{\dagger}\hat{a}_{k'} e^{-iak/2}\frac{1}{{N}}\sum_ne^{ia(k'-k)n}\\ &= \sum_{k} e^{-iak/2}\hat{b}_{k}^{\dagger}\hat{a}_{k}\,, \end{align} and \begin{align} \sum_n\hat{c}_{2n+2}^{\dagger}\hat{c}_{2n+1} &= \sum_n\frac{1}{\sqrt{N}}\sum_ke^{-ika(2n+2)/2}\hat{a}_{k}^{\dagger} \frac{1}{\sqrt{N}}\sum_{k'}e^{ik'a(2n+1)/2}\hat{b}_{k}\\ &= \sum_{kk'} \hat{a}_{k}^{\dagger}\hat{b}_{k'} e^{-ika}e^{ik'a/2}\frac{1}{{N}}\sum_ne^{ia(k'-k)n}\\ &= \sum_{k} e^{-iak/2}\hat{a}_{k}^{\dagger}\hat{b}_{k}\,, \end{align} where we have used the orthogonality relation to compute the sums over $n$ and then collapsed the double sum over $k$ and $k'$ using the resulting $\delta_{kk'}$.

The other two terms in the Hamiltonian are just the Hermitian conjugates of these ones, and so all told we have \begin{align} \hat{H} &= t\sum_{k} e^{-iak/2}\hat{b}_{k}^{\dagger}\hat{a}_{k} + t\sum_{k} e^{iak/2}\hat{a}_{k}^{\dagger}\hat{b}_{k} - t\sum_{k} e^{-iak/2}\hat{a}_{k}^{\dagger}\hat{b}_{k} - t\sum_{k} e^{iak/2}\hat{b}_{k}^{\dagger}\hat{a}_{k}\\ &= 2it\sum_{k}\sin\left(\frac{ka}{2}\right)\left( \hat{a}_{k}^{\dagger}\hat{b}_{k} -\hat{b}_{k}^{\dagger}\hat{a}_{k} \right)\,. \end{align} We can see that we have succeeded in diagonalizing the Hamiltonian in the quasi-momentum $k$. Now we just need to diagonalize the remaining piece.

To do this, wee define new operators as $$ c_{k,\pm} = \frac{\pm i\hat{a}+\hat{b}}{\sqrt{2}}\,, $$ in which case $$ \hat{a}_k = \frac{i(\hat{c}_{k,-} - \hat{c}_{k,+})}{\sqrt{2}}\,,~~~~~~~~\hat{b}_k = \frac{\hat{c}_{k,-} + \hat{c}_{k,+}}{\sqrt{2}}\,. $$ Then, plugging this into the terms above, we get $$ \hat{a}_k^{\dagger}\hat{b}_k = \frac{-i(\hat{c}_{k,-}^{\dagger} - \hat{c}_{k,+}^{\dagger})}{\sqrt{2}}\frac{\hat{c}_{k,-} + \hat{c}_{k,+}}{\sqrt{2}} =\frac{-i}{2}( \hat{c}_{k,-}^{\dagger}\hat{c}_{k,-} +\hat{c}_{k,-}^{\dagger}\hat{c}_{k,+} -\hat{c}_{k,+}^{\dagger}\hat{c}_{k,-} -\hat{c}_{k,+}^{\dagger}\hat{c}_{k,+} )\,, $$ which implies that $$ \hat{b}_k^{\dagger}\hat{a}_k=(\hat{a}_k^{\dagger}\hat{b}_k)^{\dagger} =\frac{i}{2}( \hat{c}_{k,-}^{\dagger}\hat{c}_{k,-} +\hat{c}_{k,+}^{\dagger}\hat{c}_{k,-} -\hat{c}_{k,-}^{\dagger}\hat{c}_{k,+} -\hat{c}_{k,+}^{\dagger}\hat{c}_{k,+} )\,. $$ Then, \begin{align} \hat{a}_k^{\dagger}\hat{b}_k - \hat{b}_k^{\dagger}\hat{a}_k &= \frac{-i}{2}( \hat{c}_{k,-}^{\dagger}\hat{c}_{k,-} +\hat{c}_{k,-}^{\dagger}\hat{c}_{k,+} -\hat{c}_{k,+}^{\dagger}\hat{c}_{k,-} -\hat{c}_{k,+}^{\dagger}\hat{c}_{k,+} ) - \frac{i}{2}( \hat{c}_{k,-}^{\dagger}\hat{c}_{k,-} +\hat{c}_{k,+}^{\dagger}\hat{c}_{k,-} -\hat{c}_{k,-}^{\dagger}\hat{c}_{k,+} -\hat{c}_{k,+}^{\dagger}\hat{c}_{k,+} ) \\ &= -\frac{i}{2}( 2\hat{c}_{k,-}^{\dagger}\hat{c}_{k,-} -2\hat{c}_{k,+}^{\dagger}\hat{c}_{k,+}) \end{align} Finally, plugging this back into the Hamiltonian, we get \begin{align} \hat{H} &= 2it\sum_{k}\sin\left(\frac{ka}{2}\right)\left( -\frac{i}{2}( 2\hat{c}_{k,-}^{\dagger}\hat{c}_{k,-} -2\hat{c}_{k,+}^{\dagger}\hat{c}_{k,+}) \right) = \sum_{k}2t\sin\left(\frac{ka}{2}\right)\left( \hat{c}_{k,-}^{\dagger}\hat{c}_{k,-} -\hat{c}_{k,+}^{\dagger}\hat{c}_{k,+} \right) \end{align}