I am studying second quantization for first time. Let we have a a Hamiltonian (corresponding to a chain of particles with nearest neighbor hopping 't', ignore spin, on-site potential and interaction potential between two particles) that is given in first quantization as:
$$H_{1st}=t\sum_{m=1}^N|m><m+1|+h.c.$$ $$H_{1st}=t(|1><2|+|2><3|+|3><4|+|4><1|+h.c.)$$ It can be written in matrix form as following (N=4 and with periodic boundary condition): $$H_{1st}=t(|1><2|+|2><3|+|3><4|+|4><1|+h.c.)$$ using $$|1>=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}; |2>=\begin{bmatrix}0\\1\\0\\0\end{bmatrix}...$$ and so on... we can write our matrix as: $$H_{1st}=t\begin{bmatrix}0&1&0&1\\1&0&1&0\\0&1&0&1\\1&0&1&0\end{bmatrix}$$ In second quantization this Hamiltonian can be written as: $$H_{2nd}=t\sum_{m=1}^Nc_m^\dagger c_{m+1}+h.c.$$ that can be written as for N=4 $$H_{2nd}=t(c_1^\dagger c_{2}+c_2^\dagger c_{3}+c_3^\dagger c_{4}+c_4^\dagger c_{1}+h.c.)$$

How do we write matrix from this last equation of $H_{2nd}?$

  • $\begingroup$ Given a linear transformation $T$, the matrix element $T_{ij}$ means that you start with basis vector $\left \lvert j \right \rangle$, act transformation $T$, and then ask what is the component along basis element $\left \lvert i \right \rangle$ of the result. In other words, $T_{ij} \equiv \langle i | T | j \rangle$. So really you just need to think about what your basis vectors are in 2nd quantization. $\endgroup$ – DanielSank Sep 20 '17 at 23:08
  • $\begingroup$ in first quantization our basis can be $|m>$ states as I used. Can you tell me how can I choose basis in 2nd quantization for this example? $\endgroup$ – Sana Ullah Sep 20 '17 at 23:18
  • $\begingroup$ I'm pretty sure you already are using second quatization. Take a look at this question. $\endgroup$ – DanielSank Sep 20 '17 at 23:20
  • $\begingroup$ I'm glad I could help you. If you liked the answer to that post on second quantization, you can vote up. If something is not clear, you can leave a comment asking for more information. $\endgroup$ – DanielSank Sep 21 '17 at 0:00
  • $\begingroup$ I don't have enough reputation points to give a Up vote right now. That answer was really very very helpful. I will give a up vote to that answer right after getting enough reputation points for up-vote. $\endgroup$ – Sana Ullah Sep 21 '17 at 0:17

Taking single particle as basis we can write our basis set as: (for N=4) 0001, 0010, 0100, 1000.

Matrix $H_{2nd}$ becomes $$H_2=\begin{bmatrix}<0001|H|0001>&<0001|H|0010>& ...& <0001|H|1000>\\.&.&.&.\\.&.&.&.\\.&.&.&.\\.&.&.&.\\<1000|H|0001>&<1000|H|0010>&.&<1000|H|1000>\end{bmatrix}$$ $$H_2=t\begin{bmatrix}0&1&0&1\\1&0&1&0\\0&1&0&1\\1&0&1&0\end{bmatrix}$$

  • $\begingroup$ So as you can see, you were already using second quantization! $\endgroup$ – DanielSank Sep 20 '17 at 23:59

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