# Two-site fermion system

I've to study a two-site fermion system with hamiltonian $$H=\sum_{\sigma=\uparrow,\downarrow}[\epsilon_1 c^+_{1\sigma}c_{1\sigma}+\epsilon_2 c^+_{2\sigma}c_{2\sigma}+w(c^+_{1\sigma}c_{2\sigma}+c^+_{2\sigma}c_{1\sigma})].$$

I've proven that this hamiltonian preserves the number of particles. When $$N=0$$ or $$N=4$$ there is only possible state, respectively using occupation number rapresentation: $$|0\rangle$$ and $$|1111\rangle$$.

1. Are these state eigenstates? How can i calculate their energy?

I diagonalized the Hamiltonian. The new creation and annihilation operators that I found, which preserve the commutation rules being a canonical transformation, what do they represent? I write the Hamiltonian diagonalized $$H=\sum_{k,\sigma}\epsilon_ka^+_{k\sigma}a_{k\sigma}$$

1. If I apply $$a^+_{1\uparrow}$$ to the vacuum state, I actually get a fermion in state one with spin up or I get the fermions related to the various $$c^+_{i,\sigma}$$ associated with the creation operator $$a^+_{1\uparrow}=\sum_iU^+_{i1}c^+_{i\uparrow}$$ with $$U$$ the matrix of normalized eigenvectors of the basis change.

Sorry for maybe very stupid question

1. Are these state eigenstates? How can i calculate their energy?


Act Hamiltonian on the states $$\vert 00,00\rangle$$ and $$\vert 11,11\rangle$$. The 4 number denotes occupation for site 1 spin-up, site 1 spin-down, site 2 spin-up, and site 2 spin-down:

$$H=\sum_{\sigma=\uparrow,\downarrow}[\epsilon_1 c_{1\sigma}^\dagger\,c_{1\sigma} + \epsilon_2 c_{2\sigma}^\dagger\,c_{2\sigma} + w(c_{1\sigma}^\dagger\,c_{2\sigma}+c_{2\sigma}^\dagger\,c_{1\sigma})].$$

$$H \vert 00, 00\rangle = 0 \vert 00, 00\rangle.$$ The state $$\vert 00, 00\rangle$$ is a eigen state of $$H$$ with eigen value $$0$$.

Next, examine $$H \vert 11, 11\rangle$$:

\begin{align} \epsilon_1 c^\dagger_{1\uparrow}c_{1\uparrow} \vert 11, 11\rangle & = \epsilon_1 \vert 11, 11\rangle \\ \epsilon_2 c^\dagger_{2\uparrow}c_{2\uparrow} \vert 11, 11\rangle &= \epsilon_2 \vert 11, 11\rangle\\ w c^\dagger_{1\uparrow}c_{2\uparrow} \vert 11, 11\rangle &=w c^\dagger_{1\uparrow} \vert 11, 01\rangle = 0 \vert 11, 01\rangle = 0\vert 11, 11\rangle;\\ w c^\dagger_{2\uparrow}c_{1\uparrow} \vert 11, 11\rangle &=w c^\dagger_{2\uparrow} \vert 01, 11\rangle = 0 \vert 01, 11\rangle = 0\vert 11, 11\rangle; \end{align} The last two terms vanish, due to exclusion of fermion $$c^\dagger\vert 1\rangle = 0$$.

Adding the cases for spin-down, we have $$H \vert 11, 11\rangle = (\epsilon_1+\epsilon_1+ \epsilon_2+\epsilon_2+0+0+0+0) \vert 11, 11\rangle =2 (\epsilon_1+ \epsilon_2) \vert 11, 11\rangle.$$ The state $$\vert 11, 11\rangle$$ is eigen state of eigen value $$2 (\epsilon_1+ \epsilon_2)$$.

The basis changed


Writing the Hamitonian in matrix form:

$$\sum_{\sigma=\uparrow,\downarrow} \begin{bmatrix} c_{1\sigma}^\dagger & c_{2\sigma}^\dagger \\ \end{bmatrix} \begin{bmatrix} \epsilon_1 & w \\ w & \epsilon_2 \end{bmatrix} \begin{bmatrix} c_{1\sigma} \\ c_{2\sigma} \end{bmatrix}$$ To diagonlize the Hamitonian, by an unitary matrix $$U =\begin{bmatrix} \cos \frac{\theta}{2} & \sin \frac{\theta}{2} \\ -\sin \frac{\theta}{2} & \cos \frac{\theta}{2} \end{bmatrix};\,\,\, U^T =\begin{bmatrix} \cos \frac{\theta}{2} & -\sin \frac{\theta}{2} \\ \sin \frac{\theta}{2} & \cos \frac{\theta}{2} \end{bmatrix}$$ where $$\tan\theta = \frac{2 \,w}{\epsilon_2 - \epsilon_1}$$. The Hamiltonian will become diagonal by $$U^T H U$$. In the following process.

$$\sum_{\sigma=\uparrow,\downarrow} \begin{bmatrix} c_{1\sigma}^\dagger & c_{2\sigma}^\dagger \\ \end{bmatrix} \mathbf{U} \mathbf{U^T} \begin{bmatrix} \epsilon_1 & w \\ w & \epsilon_2 \end{bmatrix} \mathbf{U} \mathbf{U^T} \begin{bmatrix} c_{1\sigma} \\ c_{2\sigma} \end{bmatrix} =\sum_{\sigma=\uparrow,\downarrow} \begin{bmatrix} a_{1\sigma}^\dagger & a_{2\sigma}^\dagger \\ \end{bmatrix} \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \begin{bmatrix} a_{1\sigma} \\ a_{2\sigma} \end{bmatrix}$$ where $$\lambda_{1, 2} = \frac{\epsilon_2+\epsilon_1}{2}\mp \sqrt{\left(\frac{\epsilon_2-\epsilon_1}{2}\right)^2 + w^2}$$. The new bases: $$\tag{1} \begin{bmatrix} a_{1\sigma} \\ a_{2\sigma} \end{bmatrix} = \begin{bmatrix} \cos \frac{\theta}{2} & -\sin \frac{\theta}{2} \\ \sin \frac{\theta}{2} & \cos \frac{\theta}{2} \end{bmatrix} \begin{bmatrix} c_{1\sigma} \\ c_{2\sigma} \end{bmatrix}$$

The Hamiltonian with new bases is diagonal $$H=\sum_{\sigma=\uparrow,\downarrow}[\lambda_1 a_{1\sigma}^\dagger\,a_{1\sigma} + \lambda_2 a_{2\sigma}^\dagger\,a_{2\sigma}.$$

Show in the equation (1), the new basis $$a_1$$ and $$a_2$$ mixed the site $$1$$ and $$2$$, but didn't mixed the spin states. \begin{align} a_{1\sigma} & = \cos\frac{\theta}{2}\, c_{1\sigma} - \sin\frac{\theta}{2} \, c_{2\sigma};\\ a_{2\sigma} & = \sin\frac{\theta}{2}\, c_{1\sigma} + \cos\frac{\theta}{2} \, c_{2\sigma}; \end{align}

Appendix

1. Diagonal of 2x2 symmetry matrix, by finding its eigen vectors: $$\tag{A1} \mathbf{A} = \begin{bmatrix} a & c \\ c & b \end{bmatrix}; \,\,\,\text{ eigen vector }\,\, \begin{bmatrix} a & c \\ c & b \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \lambda \begin{bmatrix} x \\ y \end{bmatrix};$$ First find its eigen values, by solving determinant equation $$\Vert \mathbf{A} - \lambda \mathbf{I} \Vert = 0$$ The quadratic equation render two roots (eigenvalues) for $$\lambda$$ $$\tag{A2} \lambda_{+, -} = \frac{a+b}{2}\pm \sqrt{\left(\frac{b-a}{2}\right)^2 + c^2}$$ Substitue Eq. (A2) into Eq. (A1) to find the ratio between $$x:y$$

\begin{align} & (a-\lambda) x + c y = 0;\\ \text{Let } & \,\,\,x = c; \,\,\, y = \lambda -a; \\ \text{Normalization } &\,\,\, N = \sqrt{x^2 + y^2} = 4 \cos^2\left(\frac{\theta}{2}\right)\\ \text{Define } & \,\,\,\tan\theta = \frac{c}{(b-a)/2}. \end{align}

The eigen vector $$\begin{bmatrix} x \\ y \end{bmatrix}; = \begin{bmatrix} \cos \frac{\theta}{2} \\ -\sin \frac{\theta}{2} \end{bmatrix}; \,\,\,\text{ and }\,\,\, \begin{bmatrix} \sin \frac{\theta}{2} \\ \cos \frac{\theta}{2} \end{bmatrix};$$ The two eigen vectors corresponding to $$\lambda_{\pm}$$ are mutual orthogonal

These tqo eigen vectors form the column of the unitary matrix $$\mathbf{U}$$ $$\mathbf{U} = \begin{bmatrix} \cos \frac{\theta}{2} & \sin \frac{\theta}{2} \\ -\sin \frac{\theta}{2} & \cos \frac{\theta}{2} \end{bmatrix}$$

1. Non-eigen state

\begin{align} H &\vert 11,10\rangle \\ = & \sum_{\sigma=\uparrow,\downarrow}[\epsilon_1 c_{1\sigma}^\dagger\,c_{1\sigma} + \epsilon_2 c_{2\sigma}^\dagger\,c_{2\sigma} + w(c_{1\sigma}^\dagger\,c_{2\sigma}+c_{2\sigma}^\dagger\,c_{1\sigma})] \vert 11,10\rangle \end{align}

The terms: \begin{align} c_{1\sigma}^\dagger\,c_{1\sigma} \,\vert 11,10\rangle &= 1\vert 11,10\rangle;\text{ for both spins;} \\ c_{2\uparrow}^\dagger\,c_{2\uparrow}\, \vert 11,10\rangle &= 1\vert11,10\rangle;\\ c_{2\downarrow}^\dagger\,c_{2\downarrow}\, \vert 11,10\rangle &= 0\vert11,10\rangle;\\ c_{1\sigma}^\dagger\,c_{2\sigma}\, \vert 11,10\rangle &= 0; \text{ due to the term } c_{1\sigma}^\dagger\\ c_{2\downarrow}^\dagger\,c_{1\downarrow})\, \vert 11,10\rangle &= 1 \vert 10,11\rangle; \end{align}

Put the above results all together: \begin{align} H \vert 11,10\rangle =& \,\left( 2\epsilon_1 + \epsilon_2 +0+0+0+0\right)\,\vert 11,10\rangle + \,w \, \vert 10,11\rangle\\ = & \,\left( 2\epsilon_1 + \epsilon_2 \right)\,\vert 11,10\rangle + \,w \, \vert 10,11\rangle \end{align}

• Thankyou very much 1)Please can you tell the method you have used to get that $U$. Because i have the same eigenvalues. Tell me the method so i search for it and study So the action of the new operator $a$ is both in 1 and 2 site. 2) If I want to calculate the energy of a state with N=3, for example |11,10>, how can i do? This state is not an eigenstate of H Apr 5, 2021 at 21:45
• I edited the answer including your comment in the appendix. Let me know if you had further questions.
– ytlu
Apr 6, 2021 at 4:32