# Diagonalizing Hamiltonian in Second Quantization

I have a fermionic system with states 1,2. They are coupled by a harmonic oscillator. The Hamiltonian of the system should then be

$$H=\left[\gamma(a^\dagger+a)-\delta\right]\left( c^\dagger_1 c_2+c^\dagger_2 c_1 \right) + \xi \left ( c^\dagger_1 c_1-c^\dagger_2 c_2 \right) + \hbar\omega\left(a^\dagger a + \frac{1}{2} \right)$$

where $$\gamma$$ is the coupling constant, $$\delta$$ some initial overlap integral and $$\xi$$ the energy difference between the uncoupled states.

In order to get the eigenvalues of this system I would like to diagonalize $$H$$. However I fail to see how to approach this because of the terms with three operators in them such as $$a^\dagger c^\dagger_1 c_2$$, which make me wonder how to even display $$H$$ as a matrix.

• I assume that should that be $c_1^\dagger c_2 + c_2^\dagger c_1$, right? – J. Murray May 2 '20 at 21:09
• @J.Murray right, thanks. Changed it – Liberty May 2 '20 at 21:17

Actually, after reading my reply I understood that all of this is by no means necessary. The solution is extremely simple. \begin{align} \left( \hat{H}_0 + \hat{H}_{int} \right) \vert \Psi_k \rangle = E_k \vert \Psi_k \rangle \end{align} The most general state $$\vert \Psi_k \rangle$$ is in the tensor space therefore it is of the form \begin{align} \vert \Psi_k \rangle = \sum_{n_1,f_1,f_2} c_{1,2,3} \;\vert n_1 \rangle \times \vert f_1\rangle \times \vert f_2 \rangle \end{align} The fermion product $$\vert f_1 \rangle \times \vert f_2 \rangle$$ is one of the 4 possible candidates 1) $$\vert0,0\rangle$$ 2) $$\vert0,1\rangle$$ 3) $$\vert1,0\rangle$$ 4) $$\vert1,1\rangle$$

The free hamiltonian has eigenvectors $$\vert n_b, f_1, f_2 \rangle$$ with eigenvalues $$E(n_b,f_1,f_2) = \hbar\left(n_b + \frac{1}{2}\right) + \xi(f_1-f_2)$$. Here the term proportional to $$\xi$$ contributes only if $$\vert f_1 \neq f_2 \vert$$. However, the states 1-4 are not eigenstates of $$H_{int}$$. A linear combination of these eigenstates is still an eigenstate of $$H_0$$. Therefore I seek for eigenstates of $$H_{int}$$ in the form of such linear combination.

Let us analyse the interaction term $$\hat{H}_{int}$$. It is the sum of two term, the first in which there are three operators ($$a c^\dagger c$$ etc.) and the last with two operator. \begin{align} \hat{H}_{int} &= \hat{H}_{int,1} + \hat{H}_{int,2}\\ &=\gamma \Big( \hat{a}^\dagger + \hat{a} \Big)\left(\hat{c}_1^\dagger \hat{c}_2 + \hat{c}_2^\dagger \hat{c}_1\right) - \delta \left(\hat{c}_1^\dagger \hat{c}_2 + \hat{c}_2^\dagger \hat{c}_1\right) \end{align} Let us start from the latter - which is easier. The eigenvectors of $$H_{int2}$$ are the rays $$c_1\Big( \vert \alpha ,1,0\rangle + \vert \alpha ,0,1 \rangle\Big)$$ for any $$\vert\alpha \rangle \equiv \sum_{n_b=0}^{\infty} c_{n_b} \vert n_b\rangle$$. Also, we have the linear combinations $$c_1 \vert \alpha ,0,0\rangle + c_2 \vert \alpha ,1,1\rangle$$.

Finally, let us turn to $$H_{int1}$$. The fermionic part is analogue, however this time we have the bosonic creation/annihilation operators. Therefore any linear combination $$c_1 \vert \alpha ,0,0\rangle + c_2 \vert \alpha ,1,1\rangle$$ is eigenstate (with zero eigenvalue). The other case is where I use the state $$c_1\Big( \vert \alpha ,1,0\rangle + \vert \alpha ,0,1 \rangle\Big)$$. Here $$\vert \alpha\rangle$$ is a generic linear combination of number-eigenstates of the bosonic operator $$n_b$$. The result is \begin{align} \hat{H}_{int1} \vert \alpha\rangle \times \vert(...)\rangle &=\gamma \Big(\hat{a}^\dagger + \hat{a} \Big) \sum_{n_b=0}^{\infty} c_{n_b} \vert n_b\rangle \times \vert(...)\rangle \\ &=\gamma \Big(\sum_{n_b=1}^{\infty} \sqrt{n_b}\,c_{n_b-1} \vert n_b\rangle + \sum_{n_b=0}^{\infty} \sqrt{n_b+1}\,c_{n_b+1} \vert n_b\rangle \Big)\times \vert(...) \\ &=\gamma \Big(\sum_{n_b=1}^{\infty} (\sqrt{n_b}\,c_{n_b-1}+ \sqrt{n_b+1}\,c_{n_b+1}) \vert n_b\rangle\Big)\times \vert(...) + c_1 \vert0\rangle\times \vert(...) \end{align}

Comparing the lhs with rhs I've $$c_1=c_0$$ and $$c_{n_b} = (\sqrt{n_b}\,c_{n_b-1}+ \sqrt{n_b+1}\,c_{n_b+1})$$ therefore $$c_{n_b + 1} = \frac{c_{n_b} - \sqrt{n_b}\,c_{n_b-1}}{\sqrt{n_b+1}}$$

Modulo few errors I've made in ladder operators factors etc., you can find the state $$\vert \alpha \rangle$$ and the eigenvectors of the full $$\hat{H}$$ taking the intersection of the three sets.

The hamiltonian operator we have is that of three harmonic oscillator (two fermionic and one bosonic) in the presence of a interaction. If we assume of treating this term as a perturbation over the free theory. We can find the spectrum of the full hamiltonian by using time-independent perturbation theory. If $${\rm O}\left(H_{int}\right)$$ where $$\lambda$$ is the perturbative parameter, the eigenvalue problem reads $$\begin{gather} \left( \hat{H}_0 + \hat{H}_{int} \right)\left\vert n\right\rangle = E_n \left\vert n \right\rangle \\ {\rm where}\\ H_0 = \xi\left( \hat{n}_{f1} - \hat{n}_{f2} \right) + \hbar\omega\left( \hat{n}_b + \frac{1}{2}\right) \\ H_{int} = \left[ \gamma\left(\hat{a}^\dagger + \hat{a} \right) - \delta \right] \Big( \hat{c}_1^\dagger \hat{c}_2 + \hat{c}_2^\dagger \hat{c}_1 \Big) \end{gather}$$ here $$n_i$$ is a shorthand for the oscillator's occupation number, while $$n$$ is the label of the eigenstate.

If we schematically indicate the eigenstates of the free theory with $$\vert n^{(0)} \rangle=\left\vert n_b,n_{f1},n_{f2} \right\rangle$$, that is \begin{align} \hat{H}_0 \vert n^{(0)} \rangle = E^{(0)}_n \vert n^{(0)} \rangle \end{align} the full theory eigenstates and eigenvalues, in time-independent perturbation theory, are given in terms of the exact free-theory solutions as eq.(1.44) of Sakurai,Napolitano - Modern quantum mechanics \begin{align} \vert n \rangle &= \vert n^{(0)} \rangle + \lambda \sum_{k\neq n} \vert k^{(0)} \rangle \frac{\langle k^{(0)} \vert H^{(int)} \vert n^{(0)} \rangle}{E^{(0)}_n - E^{(0)}_k} + \lambda^2... \end{align}

Before you apply to the problem you should notice that the fermionic creation operator gives zero if the corresponding state is occupied. There are few possible states

1. $$\vert n_b,0,0 \rangle$$
2. $$\vert n_b,0,1 \rangle$$
3. $$\vert n_b,1,0 \rangle$$
4. $$\vert n_b,1,1 \rangle$$

And $$H_{int}$$ conserves the sum of occupation numbers of fermionic oscillators, that is $$n_{f1} + n_{f2}$$ of the ket must be the same of the bra here $$\langle k^{(0)} \vert H^{(int)} \vert n^{(0)} \rangle = 0$$ otherwise it gives zero. The energy spectrum you're looking for is eq. (1.53b) of Sakurai, Napolitano.

Therefore, I think that the problem may further reduced to finding only the state $$\vert\alpha\rangle$$ since the hamiltonian eigenstates can be written as the tensor product $$\vert \alpha \rangle \times \Big( \vert 0,1 \rangle + \vert 1,0 \rangle \Big)$$

• So some of the matrix elements $H_{ij}$ can also contain ladder operators? How does that make sense? Also what about the term $\frac{1}{2}\hbar\omega$ that isn't linked with any operators? – Liberty May 2 '20 at 21:54