2
$\begingroup$

A spin model i.e. $H_s = \sum_i^{L-1} S_i^x\cdot S_{i+1}^x$ can be written in matrix form as following

$$H_s = \big(S_1^x \otimes S_2^x \otimes I_3^2 \otimes I_4^2\otimes \cdots\otimes I_{L-1}^2\big) + \big( I_1^2 \otimes S_2^x \otimes S_3^x \otimes I_4^2 \otimes \cdots I_{L-1}^2\big) + \cdots +\big( I_1^2 \otimes I_2^2 \otimes I_3^2 \cdots \otimes S_{L-2}^x \otimes S_{L-1}^x \big) $$ where $S_i^x =\begin{bmatrix} 0&0.5\\0.5&0\end{bmatrix} $ and $I_i^2 =\begin{bmatrix} 1&0\\0&1\end{bmatrix}$

On the other hand, let we have a fermionic model $H_f = \sum_i^{L-1} c_i^\dagger c_{i+1}$. For spinless fermions, one can represent creation operator $c_i^\dagger$ and annihilation operator $c_i$ in matrix form as $c_i^\dagger =\begin{bmatrix} 0&1\\0&0\end{bmatrix} $ and $c_i =\begin{bmatrix} 0&0\\1&0\end{bmatrix}$.

So if we can write fermionic operators in 2-by-2 matrix just like spin operators, we can write complete Hamiltonian matrix just like above equation i.e.

$$H_f = \big(c_1^\dagger \otimes c_2 \otimes I_3^2 \otimes I_4^2\otimes \cdots\otimes I_{L-1}^2\big) + \big( I_1^2 \otimes c_2^\dagger \otimes c_3 \otimes I_4^2 \otimes \cdots I_{L-1}^2\big) + \cdots +\big( I_1^2 \otimes I_2^2 \otimes I_3^2 \cdots \otimes c_{L-2}^\dagger \otimes c_{L-1} \big) $$

Question: If my approach is correct then what is the difference between spins models and fermionic models? Spin models obey commutation relations $[S_i^a,S_j^b] = i\hbar \epsilon_{abc} S_c \delta_{ij}$ while fermions obey anti-commutation relations $\{c_i^\dagger, c_j\} = \delta_{ij}$. Where do these relations go when we write Hamiltonian in matrix form?

$\endgroup$
  • $\begingroup$ Spin is an intrinsic property of fermions and bosons, they are totally different things. Fermions have half integer spin, and bosons have integer spin. $\endgroup$ – WarreG Jan 29 at 9:59
  • $\begingroup$ @WarreG yes you are absolutely right. I have rephrased my question, I think I didn't explain my question clearly in previous version. $\endgroup$ – Sana Ullah Jan 29 at 10:11
  • $\begingroup$ You are making a model for spinless fermions, that seems just weird to me because fermions always have spin or they wouldn't be fermions. Electrons for example are fermions and they do obey the first commutation relation you stated. $\endgroup$ – WarreG Jan 29 at 10:16
  • $\begingroup$ Actually, I did not want to deal with spins explicitly that's why I took spinless model (by spinless I mean only one type of spin particles which obey fermionic statistics). The same model can be taken for spinful fermions with creation and annihilation operator given by 4-by-4 matrix. $\endgroup$ – Sana Ullah Jan 29 at 10:25
  • 2
    $\begingroup$ @WarreG The notion of "spinless fermions" is standard: what it really means is "spin-polarised fermions", i.e. fermions in a high magnetic field $B \gg k_B T/\mu$, with $\mu$ the magnetic moment of 1 fermion. $\endgroup$ – Mark Mitchison Jan 29 at 10:57
4
$\begingroup$

Your matrix expressions for the fermionic operators are wrong because they do not obey the anti-commutation relations. More precisely, they are correct if you have only a single fermionic mode, but are wrong for $L>1$. If you want to get a matrix representation of fermionic operators you need to use the Jordan-Wigner transformation: \begin{align}\hat{c}_k & = \hat{\sigma}_k^- \prod_{j<k}\hat{\sigma}^z_j \\ & = \bigotimes_{j=1}^{k-1}\hat{\sigma}^z_j \otimes \hat{\sigma}^-_k \bigotimes_{l = k+1}^L \hat{I}^2_l, \end{align} where the second line is the tedious full expression making the tensor product structure explicit. Here, $\hat{\sigma}_k^- = (\hat{\sigma}_k^x - {\rm i}\hat{\sigma}^y_k)/2$ is the spin lowering operator on site $k$. From this you can construct any fermionic operator you want as strings of $\hat{c}_k$, $\hat{c}_l^\dagger$, and linear combinations thereof.

As a simple example of how this works, consider the simplest non-trivial case of $L=2$ sites. If one (wrongly!) postulates that $\hat{c}_{1} = \hat{\sigma}^-_1\otimes\hat{I}^2_2$ and $\hat{c}_{2} = \hat{I}^2_1\otimes\hat{\sigma}^-_2$, the anti-commutation relation would be $\{\hat{c}_1,\hat{c}_2\} = 2\hat{\sigma}^-\otimes\hat{\sigma}^-\neq 0$, so that's no good. Using the correctly defined operators instead, we have \begin{align} \{\hat{c}_1,\hat{c}_2\} &= \hat{\sigma}^-_1 \hat{\sigma}^-_2 \hat{\sigma}^z_1 + \hat{\sigma}^-_2\hat{\sigma}^z_1\hat{\sigma}^-_1\\ & = \hat{\sigma}^- \hat{\sigma}^z \otimes \hat{\sigma}^- + \hat{\sigma}^z\hat{\sigma}^- \otimes \hat{\sigma}^- \\ & = \{\hat{\sigma}^-,\hat{\sigma}^z\} \otimes \hat{\sigma}^- \\ & = 0, \end{align} since $\{\hat{\sigma}^-,\hat{\sigma}^z\} = 0$.

$\endgroup$
  • $\begingroup$ I am very sorry for my naive understanding. Can you please elaborate it little bit more. How operators in question are not obeying anti-commutation relation while operators constructed by your formula obey? $\endgroup$ – Sana Ullah Jan 29 at 13:46
  • 1
    $\begingroup$ I've updated the answer with an example. I've written the tensor products out in painstaking detail starting on the second line, but I think this example aptly demonstrates how that notation is bulky and not very helpful. The whole point of adding subscripts indexing the site numbers to your operators is to avoid having to do that, i.e. $\hat{\sigma}^z_2 = \hat{I}\otimes \hat{\sigma}^z$ etc. by definition $\endgroup$ – Mark Mitchison Jan 29 at 23:15
  • $\begingroup$ Thank you very much Sir. It is very helpful. thank you again for explaining it with an example. $\endgroup$ – Sana Ullah Feb 1 at 11:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.