# Is there a way to create a superposition of all the possible states?

If there're four qubits in my circuit, how can I arrange my gates so that the final output state is a superposition of all the possible states of 4 qubits? (there're 16 of them in total). I've tried some 2-qubits circuits to generate the superposition states like $$|\Psi^+\rangle$$ and $$|\Phi^-\rangle$$, but I'm not exactly sure if there's a way to create the superposition of all the possible states.

• Initialize them all to $\vert 0\rangle$, and Hadamard them all individually. Dec 3, 2020 at 1:17
• @Mark S Thanks!!!
– ZR-
Dec 3, 2020 at 1:21

Do you mean mapping the state $$|0\rangle^{\otimes n} \to \dfrac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1} |i\rangle$$ ?

If that is the case then you can just apply $$H^{\otimes n}$$ to the state $$|0\rangle^{\otimes n}$$. That is, you apply a Hadamard gate to each of the qubit.

The reason for this is $$H |0\rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$$ and so

\begin{align} \overbrace{H|0\rangle \otimes H|0\rangle \otimes \cdots \otimes H|0\rangle}^{n \ \textrm{times}} &= \overbrace{ \bigg( \dfrac{|0\rangle +|1\rangle}{\sqrt{2} } \bigg)\otimes \bigg( \dfrac{|0\rangle +|1\rangle}{\sqrt{2} } \bigg) \otimes \cdots \otimes \bigg( \dfrac{|0\rangle +|1\rangle}{\sqrt{2} } \bigg) }^{n \ \textrm{times}} \\ &= \dfrac{1}{\sqrt{2^{n}}}\big( \overbrace{ |00\cdots0\rangle + |00\cdots1\rangle + \cdots + |11\cdots 1\rangle }^{2^n \ \textrm{terms} } \big)\\ &= \dfrac{1}{\sqrt{2^n}}\sum_{i=0}^{2^n-1} |i\rangle \end{align}

• For the sake of pedantry - this is not a superposition of all possible states because there are states that are orthogonal to this. It is a superposition of all possible computational basis states. Dec 3, 2020 at 8:24
• A superposition of all states would be the Haar average -- which is zero. Dec 3, 2020 at 8:35
• @DaftWullie If we're being pedantic, every state is a superposition of all possible states. OP's question is nontrivial only if we interpret "superposition of A and B" as being a state in which both A and B have a nonzero contribution, but it's not possible for all states to have a nonzero contribution, so we must interpret this to mean that all basis states have nonzero contribution (which of course means that this is defined only with respect to a particular basis). Dec 3, 2020 at 21:41
• @Zhengrong What I meant is an expression of the form $\int_{\mathbb{S}^{d-1}} \mathrm{d}\psi f(\psi) |\psi\rangle$ where $\mathrm{d}\psi$ is the (normalised) Haar measure on the unit sphere and $f$ is a measurable function. For $f=1$, the result is unitarily invariant, and thus has to be zero. It is BTW possible to write something more meaningful such as $|\varphi\rangle = d \int_{\mathbb{S}^{d-1}} \mathrm{d}\psi \langle \psi | \varphi \rangle|\psi\rangle$. But this is just fancy rewriting ;) Dec 4, 2020 at 9:19
• @MarkusHeinrich What do you mean by "finite linear combinations"? What are saying is not basis dependent? Whether a state has a nonzero contribution from every basis vector is basis dependent. "Superposition" is often to mean "state with nonzero contribution from at least two basis vectors", and that is also basis dependent. Dec 4, 2020 at 21:11