By Jordan-Wigner transform, we can tranfer spin-$1/2$ model into fermions, then how to choose the right hamiltonian so that we can solve the model?

Question

I know that by using Jordan-Wigner transform(JWT), we can transform spin-$1/2$ systems into fermions. My problem is, for example, after JWT, we have a hamiltonian of form $$\epsilon\left(c_{1}^{\dagger} c_{1}+c_{2}^{\dagger} c_{2}\right)+\lambda\left(c_{1}^{\dagger} c_{2}^{\dagger}+c_{2} c_{1}\right)\tag{1}$$ with $\lambda$ and $\epsilon\in \mathbb R$. If I directly write it as quadratic form $$\left( \begin{matrix} c_{1}^{\dagger}& c_2& c_{2}^{\dagger}& c_1\\ \end{matrix} \right) \left( \begin{matrix} \epsilon& \lambda& 0& 0\\ \lambda& 0& 0& 0\\ 0& 0& \epsilon& 0\\ 0& 0& 0& 0\\ \end{matrix} \right) \left( \begin{array}{c} c_1\\ c_{2}^{\dagger}\\ c_2\\ c_{1}^{\dagger}\\ \end{array} \right) $$ the eigenvalues of the middle hermitian matrix will be $$0,\epsilon ,\frac{\epsilon \pm \sqrt{\epsilon ^2+4\lambda ^2}}{2}$$

But if I change eq.(1) into another form with canonical commutation relation for fermions as follows: $$ \begin{align} &\epsilon \left( c_{1}^{\dagger}c_1+c_{2}^{\dagger}c_2 \right) +\lambda \left( c_{1}^{\dagger}c_{2}^{\dagger}+c_2c_1 \right) \\ &=\epsilon \left( c_{1}^{\dagger}c_1+c_{2}^{\dagger}c_2 \right) +\frac{\lambda}{2}\left( c_{1}^{\dagger}c_{2}^{\dagger}+c_2c_1 \right) -\frac{\lambda}{2}\left( c_{2}^{\dagger}c_{1}^{\dagger}+c_1c_2 \right) \\ &=\frac{\epsilon}{2}\left( c_{1}^{\dagger}c_1+c_{2}^{\dagger}c_2 \right) +\frac{\epsilon}{2}\left[ \left( 1-c_1c_{1}^{\dagger} \right) +\left( 1-c_2c_{2}^{\dagger} \right) \right] +\frac{\lambda}{2}\left( c_{1}^{\dagger}c_{2}^{\dagger}+c_2c_1 \right) -\frac{\lambda}{2}\left( c_{2}^{\dagger}c_{1}^{\dagger}+c_1c_2 \right) \\ &=\frac{\epsilon}{2}\left( c_{1}^{\dagger}c_1+c_{2}^{\dagger}c_2 \right) -\frac{\epsilon}{2}\left[ c_1c_{1}^{\dagger}+c_2c_{2}^{\dagger} \right] +\frac{\lambda}{2}\left( c_{1}^{\dagger}c_{2}^{\dagger}+c_2c_1 \right) -\frac{\lambda}{2}\left( c_{2}^{\dagger}c_{1}^{\dagger}+c_1c_2 \right) +\epsilon \end{align} $$

The same reason, we can write it as quadratic form $$\mathcal{H}=\frac{1}{2}\left(\begin{array}{cccc} c_{1}^{\dagger} & c_{2} & c_{2}^{\dagger} & c_{1} \end{array}\right)\left(\begin{array}{cccc} \epsilon & \lambda & 0 & 0 \\ \lambda & -\epsilon & 0 & 0 \\ 0 & 0 & \epsilon & -\lambda \\ 0 & 0 & -\lambda & -\epsilon \end{array}\right)\left(\begin{array}{c} c_{1} \\ c_{2}^{\dagger} \\ c_{2} \\ c_{1}^{\dagger} \end{array}\right)+\epsilon$$

But this time, eigenvalues become $$\pm \sqrt{\epsilon ^2+\lambda ^2}$$

I think this might be the reason that the unitary used to diagonalize the middle hermitian matrix cannot keep the canonical commutation relation for fermions.

So my problem is, after JWT, is there some routine that we can write the right hermitian matrix so the eigenvalue of the hermitian matrix is the answer we want?

Answer 1

The example you find has a mistake. When you diagonalize the matrix $\mathcal{H}$, you are applying some unitary transformation $U$, which is a $4 \times 4$ matrix, to the vector $(c_1,c^{\dagger}_2,c_2,c^{\dagger}_1)^T$. Let the transformed vector be $(\tilde{c}_1,\tilde{c}^{\dagger}_2,\tilde{c}_2,\tilde{c}^{\dagger}_1)^T$. We must have $\tilde{c}^{\dagger}_1=(\tilde{c}_1)^{\dagger}$ and $\tilde{c}^{\dagger}_2=(\tilde{c}_2)^{\dagger}$, but a $4 \times 4$ matrix $U$ gives extra degrees of freedoms to the transformation, and the relation generally no longer holds. Therefore, you will have different eigenvalues for the same Hamiltonian as invalid results.

In your previous edition of the post, you mentioned a link about Jordan-Wigner transformation and its use in $1$-dimensional quantum Ising model. The Hamiltonian after Jordan-Wigner transformation is $$H=\sum_{k}{\big(-c^{\dagger}_kc_k(2\cos{k})-(c^{\dagger}_kc^{\dagger}_{-k}e^{ik}+c_{-k}c_ke^{ik})+2h_zc^{\dagger}_kc_k\big)}$$ You argue that we can write the Hamiltonian as $$H=\sum_{k} \begin{pmatrix} c^{\dagger}_k & c_{-k} \end{pmatrix} \begin{pmatrix} -2\cos{k}+2h_z & -e^{ik} \\ -e^{-ik} & 0 \end{pmatrix} \begin{pmatrix} c_k \\ c^{\dagger}_{-k} \end{pmatrix}$$ This is true, but diagonalizing the $2 \times 2$ matrix in the above equation does not give you superposition of $c_k$ and $c^{\dagger}_{-k}$ which can diagonalize the Hamiltonian. There is the other $-k$ term as contribution in the sum, meaning \begin{align} H & = \sum_{k>0} \begin{pmatrix} c^{\dagger}_k & c_{-k} \end{pmatrix} \begin{pmatrix} -2\cos{k}+2h_z & -e^{ik} \\ -e^{-ik} & 0 \end{pmatrix} \begin{pmatrix} c_k \\ c^{\dagger}_{-k} \end{pmatrix} \\ & + \sum_{-k,k \geq 0} \begin{pmatrix} c^{\dagger}_{-k} & c_k \end{pmatrix} \begin{pmatrix} -2\cos{k}+2h_z & -e^{-ik} \\ -e^{ik} & 0 \end{pmatrix} \begin{pmatrix} c_{-k} \\ c^{\dagger}_k \end{pmatrix} \end{align} Therefore, you can see the diagonalization of $\begin{pmatrix} -2\cos{k}+2h_z & -e^{ik} \\ -e^{-ik} & 0 \end{pmatrix}$ only diagonalizes the first part of the Hamiltonian. As written in the link, the correct way to do this is to sum up the two parts, and have $$H=\sum_{k} \begin{pmatrix} c^{\dagger}_k & c_{-k} \end{pmatrix} \begin{pmatrix} -\cos{k}+h_z & -i\sin{k} \\ i\sin{k} & \cos{k}-h_z \end{pmatrix} \begin{pmatrix} c_k \\ c^{\dagger}_{-k} \end{pmatrix}$$ The diagonalization of the matrix gives you the superposition of $c_k$ and $c^{\dagger}$ which diagonalizes the Hamiltonian.