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?

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    $\begingroup$ Using only Clifford gates, you will only obtain the six stabiliser states, i.e. the $X,Y,Z$ eigenstates. $\endgroup$ Commented Mar 3, 2021 at 8:55
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    $\begingroup$ there is something missing in the sentence. You want to know which states are reachable with those gates, or show whether any state is reachable with those gates, or something else? $\endgroup$
    – glS
    Commented Mar 3, 2021 at 13:30
  • $\begingroup$ @glS Yes, I mean, I wanted to know which states can be reached using those gates $\endgroup$
    – heromano
    Commented Mar 3, 2021 at 16:27

1 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.

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.

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 $\mathcal{S}$.

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.


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