Considering a spin-$\frac{1}{2}$ qubit, which is the ground state, $|0\rangle$ or $|1\rangle$?

Question

Considering a spin-$\frac{1}{2}$ qubit, which is the ground state, $|0\rangle$ or $|1\rangle$? I apologize for the simplicity of the question.

Answer 1

The question is not so simple I don't think - but just to expand a bit on @DaftWullie's answer:

• Informally, we can define a qubit $|\psi\rangle$ as a two-dimensional Hilbert space having two privileged orthonormal basis vectors $|0\rangle$ and $|1\rangle$, and
• Again informally, we can define the ground state of a Hamiltonian $\mathcal H$ acting on $|\psi\rangle$ as the eigenvector(s) that has the lowest energy eigenvalue.

Much of the rest of what's meant by $|0\rangle$ or $|1\rangle$ and how they relate to a given Hamiltonian are matters of historical convention. For example you may be familiar with defining a qubit with the two lowest orbitals of an electron around a nucleus. By convention we call the lowest orbital $|0\rangle$ and the second-lowest orbital $|1\rangle$, probably for much of the same reason that (classical) computer engineers identify logical 0 when a node is grounded (or at zero volts) and being logical 1 when the node is positive (or about 0.9 volts).

Furthermore from your question about spin, we tend to identify the spin-up vector $\mid\uparrow\rangle$ as $|1\rangle$ and the spin-down vector $\mid\downarrow\rangle$ as $|0\rangle$, but again that's a matter of convention, probably because 1 is above, or "up", from 0. We could just as easily equate $\mid\uparrow\rangle$ as $|0\rangle$.

Indeed I learn from @hft here that a common convention is for $\mid\uparrow\rangle$ to often be equated with $|0\rangle$ while $\mid\downarrow\rangle$ is often equated to $|1\rangle$.

Answer 2

Ground state always means ground state of a Hamiltonian. We proceed to precisely define the ground state of a Hamiltonian.

Let $\mathcal{H}$ be a Hilbert space representing the space of states of your system. Let $H: \mathcal{H} \rightarrow \mathcal{H}$ represent the Hamiltonian of your system. The Hamiltonian has a spectrum, i.e. the collection of its eigenvalues $\{E_n\}$ and corresponding eigenstates $\{ \lvert E_n \rangle\}$ such that $$H \lvert E_n \rangle = E_n \lvert E_n \rangle.$$ The ground state of the Hamiltonian is the eigenstate of the Hamiltonian with the smallest eigenvalue $E_{\min} = \min(\{E_n\})$, i.e. the ground state of the Hamiltonian is $\lvert E_{\min} \rangle$ such that $$H\lvert E_{\min} \rangle = E_{\min}\lvert E_{\min} \rangle.$$

Thus, it does not really make sense to ask your question as you wrote it. One would need to ask something like "what state is the ground state of the given Hamiltonian $H$".

Answer 3

The ground state is only defined if you have a Hamiltonian applied to the qubit. For example, if the Hamiltonian is $Z$, the ground state (the eigenvector with minimum eigenvalue) is $|1\rangle$, while if it's $-Z$, the ground state is $|0\rangle$. Other Hamiltonians, such as $X$, have a ground state that is neither $|0\rangle$ nor $|1\rangle$ (and is rather $(|0\rangle+|1\rangle)/\sqrt{2}$ in this case).