# Product of ground eigen-energy with highest eigen-energy of 3 qubits

Consider a 3-qubit quantum system with ground state $$|\psi_0\rangle$$ and highest energy state (for the problem at hand, in general there might be higher) $$|\psi_{\rm top}\rangle$$. The corresponding eigen-energies are $$E_0$$ and $$E_{\rm top}$$.

Given that these are eigenstates of the 3-local Hamiltonian: $$\hat{H} = \sigma_z \otimes \sigma_z \otimes \mathbb{1} - \mathbb{1}\otimes \sigma_y \otimes \sigma_y - \sigma_x \otimes \mathbb{1} \otimes \sigma_x$$ what is the product $$E_0 E_{\rm top}$$?

Ideas: I am not so sure how to compute this. I can write $$|\psi_0\rangle$$ as follows: $$|\psi_0\rangle = a|000\rangle + b|010\rangle + c|100\rangle + d|110\rangle + e|001\rangle + f|011\rangle + g|101\rangle + h|111\rangle.$$ Then I can apply the Hamiltonian $$\hat H$$ and transform the kets. What guarantees that in the end I will get an eigen-energy? If I do, so if in the end I fing $$\hat{H}|\psi_0\rangle= c|\psi_0\rangle$$ then $$c=E_0$$.

But how will I approach the same thing for the unknown $$|\psi_{\rm top}\rangle$$?

Also, does this problem have some sort of a name?

$$\hat{H} = \begin{pmatrix} 1 & 0 & 0 & 1 & 0 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 & -1 & 0 & 0 & 0 \\ 0 & -1 & -1 & 0 & 0 & 0 & 0 -1 \\ 1 & 0 & 0 &-1 & 0 & 0 & -1 & 0 \\ 0 & -1 & 0 & 0 & -1 & 0 & 0 & 1 \\ -1 & 0 & 0 & 0 & 0 & -1 &-1 & 0 \\ 0 & 0 & 0 &-1 & 0 & -1 & 1 & 0 \\ 0 & 0 & -1 & 0 & 1 & 0 & 0 & 1 \\ \end{pmatrix}$$ and one can plug this into Matlab or such to find the two eigenvalues.

• Could you elaborate on what you exactly are asking for? what is the product $E_0E_{\mathrm{top}}$? does not make much sense to me. Dec 7, 2021 at 13:09
• Yes, to find a formula or an estimate on what $E_0E_{\rm top}$ is. Or, more generally, how I would estimate $E_{\rm top}$. Dec 7, 2021 at 13:24
• Okay, so your question is how to find an (presumably) approximation of the product $E_0 \, E_{\mathrm{top}}$, where $E_0$ is the ground state energy and $E_{\mathrm{top}}$ the highest energy eigenvalue? If so, it could make sense to edit the question and not simply ask what is ... - because it is a product of two real numbers and in principle you can obtain each number by diagonalizing $H$ (i.e. find its eigenvalues) and then multiplying them. Dec 7, 2021 at 13:28
• Thanks. Yes, I got it now. I thought the question was tricky, this is why I got confused. I edited the question with the answer. Dec 7, 2021 at 14:09

You are looking for the energy eigenvalues. Write $$H$$ as a $$2^3$$ by $$2^3$$ matrix and compute its eigenvalues (some of them will be degenerate). Take the highest and the lowest and multiply them together.
• Why $2^3 \times 2^3$? If the Hamiltonian was only the first term, for example, then would it not be $H = {\rm BlockDiag}(\sigma_z,\sigma_z,1)$? Dec 7, 2021 at 13:41
• That is not how tensor products work. The matrix you wrote is 6x6, but you have 8 basis vectors in $\mathbb C^2 \otimes \mathbb C^2 \otimes \mathbb C^2$, as you write correctly in your second formula. Dec 7, 2021 at 13:46
• At second order perturbation theory you get expressions of the form $(E_n-E_0)^2$... but yes, probably they are just testing you can compute eigenvalues. Dec 8, 2021 at 14:02