How is the ground state of a Hamiltonian defined?

Question

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?

Answer 1

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.

Answer 2

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.