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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?


After the answers given below:

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

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  • $\begingroup$ 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. $\endgroup$ Commented Dec 7, 2021 at 13:09
  • $\begingroup$ Yes, to find a formula or an estimate on what $E_0E_{\rm top}$ is. Or, more generally, how I would estimate $E_{\rm top}$. $\endgroup$
    – Marion
    Commented Dec 7, 2021 at 13:24
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    $\begingroup$ 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. $\endgroup$ Commented Dec 7, 2021 at 13:28
  • $\begingroup$ 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. $\endgroup$
    – Marion
    Commented Dec 7, 2021 at 14:09

1 Answer 1

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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.

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  • $\begingroup$ Why $2^3 \times 2^3$? If the Hamiltonian was only the first term, for example, then would it not be $H = {\rm BlockDiag}(\sigma_z,\sigma_z,1)$? $\endgroup$
    – Marion
    Commented Dec 7, 2021 at 13:41
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    $\begingroup$ 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. $\endgroup$
    – Andrea
    Commented Dec 7, 2021 at 13:46
  • $\begingroup$ Indeed, I get two degenerate eigenvalues. $\endgroup$
    – Marion
    Commented Dec 7, 2021 at 14:10
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    $\begingroup$ Problem solved then! I bet there is a more clever way to go about it, but if you have access to mathematica this brute force approach works very well. $\endgroup$
    – Andrea
    Commented Dec 7, 2021 at 14:21
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    $\begingroup$ 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. $\endgroup$
    – Andrea
    Commented Dec 8, 2021 at 14:02

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