# How is eigendecomposition of a Hamiltonian equivalent to finding the minimum of an energy function?

This question is in regards to Dwave's quantum computer which is tailored to solve QUBO problems using quantum annealing.

QM tells us that the ground state of a quantum system is given by the eigenvector corresponding to lowest eigenvalue of the associated Hamiltonian $$H$$.

I would like to understand how does finding the minimum of $$x^T Q x$$ equate to finding (the lowest eigenvalued) eigenvector of a matrix $$H$$ and if so what is the relation between $$Q$$ and $$H$$?

• The problem with the example in your EDIT is that your $x$ vector is not normalized. DaftWullie is assuming $\lvert x \rangle$ to be normalized since its coordinates $\alpha_n$ in an orthonormal basis satisfy $\sum_n \lvert \alpha_n \rvert^2 = 1$. I think the concept of the "Rayleigh quotient" is the answer you're looking for. Commented Jul 30, 2021 at 20:27
• Does this answer your question? Standard to select base hamiltonaian for Adiabatic quantum computing Commented Jul 31, 2021 at 6:34
• If you have follow-up questions, please ask them separately. I was not one of the two close voters (in fact I was the only +1 voter), but in this case, I don't blame them if you are to keep perpetually editing your question to ask your follow-ups to what seems to be a perfectly good answer. Commented Jul 31, 2021 at 19:52
• A follow-up question is posted when the asker wants to ask another question that is a follow-up of the question. In my case I wasn't asking follow up questions or even changing my question. I was posting addendum to show why the posted answer cannot be accepted. Those addendums were detailed enough that they couldn't be posted as comments. Posting the comment as response to comment by user12... Commented Aug 1, 2021 at 14:11
• @morpheus note that you can revert to a previous version of your question if you think it more appropriate. As a general comment, yes, it is better to avoid long edits that seem to change what is being asked, although as you say, adding clarifications that do not change the question itself is fine. In this specific case, the edits seem more or less fine to me, albeit I understand why they might look like you keep asking for new questions.
– glS
Commented Aug 1, 2021 at 16:13

Consider $$Q=H$$. Let $$|\lambda_n\rangle$$ be the eigenbasis of $$H$$, i.e. $$H=\sum_n\lambda_n|\lambda_n\rangle\langle\lambda_n|.$$ Now consider the $$|x\rangle$$ that minimises $$\langle x|H|x\rangle$$. Since the eigenbasis is an orthonormal basis, we can choose to write $$|x\rangle=\sum_n\alpha_n|\lambda_n\rangle,\qquad\sum_n|\alpha_n|^2=1,$$ where it is now our job to determine the $$\{\alpha_n\}$$ the minimise $$\langle x|H|x\rangle=\sum_n|\alpha_n|^2\lambda_n.$$ Hopefully it is now obvious that you want to pick $$\alpha=1$$ for the smallest eigenvalue and $$\alpha_n=0$$ for all others, leaving you with $$|x\rangle$$ as the eigenvector with smallest eigenvalue.