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.
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3 Answers
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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).
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$\begingroup$ Do you have some ressources for beginners about that ? I'm still confused about the idea of "applying a Hamiltonian" $\endgroup$– DuenJun 8 at 12:30
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1$\begingroup$ @Duen The best thing is to take a course on quantum mechanics. $\endgroup$ Jun 9 at 5:41
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$\begingroup$ I concur with @AbdullahKhalid. But to summarise the main concept... the physics of any device is described by a Hamiltonian (a Hermitian operator), which specifies how it evolves in time. Basis states, unitary gates etc are an abstraction that sit on the top of that. Basis states generally coincide with the eigenstates of the operator. $\endgroup$ Jun 9 at 6:29
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$\begingroup$ @DaftWullie If we use $H = \omega_0 S_z$, then |1⟩ is the ground state, but if we use $H = -\omega_0 S_z$, then |0⟩ is the ground state, correct? When I do the calculations, I find that $H = \omega_0 S_z$, where $\omega_0 = -\gamma B_0$. Would there be any difference between using $H = \omega_0 S_z$ or $H = -\omega_0 S_z$? Note: $\gamma$ represents the gyromagnetic ratio. $\endgroup$– StudentJun 9 at 19:29
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$\begingroup$ @Student You want to be careful of signs. a $\pm$ sign on the front of a Hamiltonian completely switches which eigenvectors have the smallest/largest eigenvalue. So, assuming $\gamma$ and $B_0$ are positive, $\omega_0$ is negative as you write it. The $H=\omega_0 S_z$ would have $|1\rangle $as the ground state (assuming you define $S_Z$ like the Pauli $Z$ matrix). If you flip the sign, then $|0\rangle$ is the ground state. $\endgroup$ Jun 12 at 5:56
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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 $|1\rangle$.
answered Jun 8 at 12:51
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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 $H$".
