Which states can reached using single-qubit Clifford gates?

Question

Starting with the qubit state $|0\rangle$, which single-qubit states can be obtained by applying single-qubit Clifford gates, i.e. Pauli + Hadamard + $S$ gates?

Answer 1

The states reachable from $|0\rangle$ using Clifford gates are precisely the six eigenstates of the single-qubit Pauli operators, i.e.

$$\mathcal{S} = \{|0\rangle, |1\rangle, |+\rangle, |-\rangle, |{+i}\rangle,|{-i}\rangle\}\tag1$$

which are also known as the single-qubit stabilizer states.

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Remark on generators It is easy to express Pauli operators $X$, $Y$ and $Z$ in terms of $H$ and $S$, so

$$\langle H, S\rangle = \langle H, S, X, Y, Z\rangle$$

and we do not need to include Pauli operators among the generators of the Clifford group.

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Elementary proof of $(1)$ First note that all states in $(1)$ are reachable from $|0\rangle$ using $H$ and $S$ gates

$$ \begin{array}{ccc} |0\rangle = I|0\rangle && |1\rangle = HS^2H|0\rangle \\ |+\rangle = H|0\rangle && |-\rangle = S^2H|0\rangle \\ |{+i}\rangle = SH|0\rangle && |{-i}\rangle = S^3 H|0\rangle. \\ \end{array}\tag2 $$

Next, compute the action of $H$ and of $S$ on all six states in $\mathcal{S}$

$$ \begin{array}{ccc} H|0\rangle = |+\rangle && S|0\rangle = |0\rangle \\ H|1\rangle = |-\rangle && S|1\rangle = i|1\rangle \equiv |1\rangle \\ H|+\rangle = |0\rangle && S|+\rangle = |{+i}\rangle \\ H|-\rangle = |1\rangle && S|-\rangle = |{-i}\rangle \\ H|{+i}\rangle = e^{i\pi/4}|{-i}\rangle \equiv |{-i}\rangle && S|{+i}\rangle = |-\rangle \\ H|{-i}\rangle = e^{-i\pi/4}|{+i}\rangle \equiv |{+i}\rangle && S|{-i}\rangle = |+\rangle \\ \end{array} $$

and conclude that no combination of $H$ and $S$ gates maps a state in $\mathcal{S}$ to a state outside of $\mathcal{S}$.

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Proof of $(1)$ using stabilizer formalism Suppose $|\psi\rangle$ is a $+1$ eigenstate of an operator $A$. Then for every unitary operator $U$, the state $U|\psi\rangle$ is a $+1$ eigenstate of $UAU^\dagger$.

Now, it is easy to check that

$$ \begin{array}{ccc} HXH^\dagger=Z && SXS^\dagger=Y \\ HYH^\dagger=-Y && SYS^\dagger=-X \\ HZH^\dagger=X && SZS^\dagger = Z \end{array} $$

and we see that conjugation by $H$ and conjugation by $S$ permute the single-qubit Pauli group $G_1=\langle X, Y, Z\rangle$. Therefore, if $|\psi\rangle$ is a $+1$ eigenstate of a non-identity Pauli operator then for every $V\in\langle H, S\rangle$ the state $V|\psi\rangle$ is a $+1$ eigenstate of a non-identity Pauli operator.

Finally, $|0\rangle$ is a $+1$ eigenstate of $Z \in G_1$. Therefore, every state reachable from $|0\rangle$ using $H$ and $S$ is $+1$ eigenstate of a non-identity operator in $G_1$. There are fifteen non-identity operators in $G_1$, but only six of them have $+1$ as an eigenvalue. These are: $X$, $-X$, $Y$, $-Y$, $Z$ and $-Z$. The corresponding $+1$ eigenstates are: $|+\rangle$, $|-\rangle$, $|{+i}\rangle$, $|{-i}\rangle$, $|0\rangle$ and $|1\rangle$, respectively.

For the converse, see $(2)$ above.