# How is the ground state of a Hamiltonian defined?

I'm studying VQE, but there is something I don't get.

We know (I think) that for a given Hamiltonian the minimum eigenvalue is associated with the ground state. But if we take the Hamiltonian to be Pauli Z, then it has two eigenvalues: 1 associated with state $$|0 \rangle$$ and -1 associated with state $$|1 \rangle$$. Clearly the minimum eigenvalue is -1 so the ground state should be $$|1\rangle$$. But the ground sate for a qubit is $$|0\rangle$$

What am I getting wrong?

A couple of points:

1. The ground state is by definition the eigenvector associated with the minimum valued eigenvalue.
2. Lets consider the Pauli Z matrix as you have. First, \begin{align*} Z = \begin{pmatrix}1 & 0\\ 0 & -1 \end{pmatrix}. \end{align*} As this matrix is diagonal, we can immediately see that the eigenvalues are the values on the main diagonal (so 1 and -1), and they are associated with the standard basis vectors $$|0\rangle$$ and $$|1\rangle$$ respectively. Thus, since the eigenvector with the lowest associated eigenvalue is the $$|1\rangle$$ state, the $$|1\rangle$$ state is the ground state.

Your confusion may have simply been with the definition of the ground state -- it is not always the $$|0\rangle$$ state, although for some matrices (such as the identity matrix), it can be.

A potentially elucidating example may be found in considering the Hadmard matrix, \begin{align*} H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix}. \end{align*} The eigenvalues of this matrix are $$1, -1$$ with the associated eigenvectors $$\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$ and $$\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$$. Thus, we know that the ground state of the Hadamard matrix is the $$|-\rangle \equiv \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$$ state.

Edit: Upon reflection, I realize that you may be asking what the ground state of a qubit is. This question doesn't entirely make sense to me, as the state of a qubit is represented by a vector rather than a matrix (and thus, does not have eigenstates and eigenvalues on its own). If you could clarify your question, I would be happy to address it more directly.

• You did answer my question so thanks, but your update raises another one. It was my impression that that qubit has an excited state |1> and a relaxed state |0>. Is that not the case? Apr 14 '20 at 19:10
• No problem! And you may be thinking something along the line of "if a qubit is in one of the eigenstates of some Hamiltonian H, which eigenstate is it in?". An excited state is simply one of the eigenstates with an eigenvalue larger than the ground state energy. Apr 14 '20 at 20:46
• @Arthur-1 Indeed, it is common to label the ground state of the qubit as $|0\rangle$, and the excited state as $|1\rangle$. In that case, the Hamiltonian of the system is $H=-E Z/2$, where $E$ is the energy gap between the two levels. The key here is the negative sign which switches maximum and minimum eigenvectors. Apr 15 '20 at 8:16
• @DaftWullie I could be wrong, but I believe that is more common in physics. In quantum computing, I was under the impression we tend to label states in terms of the computational basis vectors, as those are the states we need to do everything in terms of while performing computations. Thanks for the correction, though. Apr 15 '20 at 14:24
• But you have to have some way of deciding what the computational basis is. That's generally guided by the physics of the device... Apr 15 '20 at 15:08

I think the confusion arises from the fact we usually measure the energy of the spins in a magnetic field. In that case, the energy is given by a Hamiltonian $$H=-\vec{\mu}\vec{B}$$ where $$\vec{\mu} = -1/2g\mu_B\vec{\sigma}\approx \mu_B\vec{\sigma}$$ and $$\mu_B$$ is a positive constant called Bohr magneton. When B is only in the $$z$$-direction, it becomes $$H=-\mu_zB_z \approx -\mu_BB_z\sigma_z$$ In such case, $$|0\rangle$$ is the ground state.