How to create the state $\vert 0 \rangle+i \vert 1 \rangle$ using elementary gates?


I am trying to write $|0\rangle+i|1\rangle$ in terms of elementary gates like H, CNOT, Pauli Y, using the IBM QE circuit composer.

I was thinking some kind of combination of H and Y since $Y|0\rangle=i|1\rangle$, so it is close but not quite.

Please help! Thanks


Posted 2021-01-18T16:15:06.850

Reputation: 101



Starting with the state $|\psi_0 \rangle = |0\rangle$, and we want to get to the state $|\psi_f \rangle = \dfrac{|0\rangle + i|1\rangle}{\sqrt{2}}$ then we must realize that we need to create some sort of a superposition between the state $|0\rangle$ and the state $|1\rangle$. This is where the Hadamard gate will come into play. The Hadamard gate which defined in the computational basis $\bigg\{ |0\rangle = \begin{pmatrix} 1\\ 0 \\ \end{pmatrix} , |1\rangle = \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix} \bigg\}$ as: $$ H = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix}$$ and it takes the state $|0\rangle$ to the state $\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$. This can be workout explicitly as through Matrix algebra as follow:

$$ H|0\rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix}\begin{pmatrix} 1\\ 0 \\ \end{pmatrix} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 1 \\ \end{pmatrix} = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}} $$

Now, we need to change the relative phase from of the above state. That is we want to change the state from $\dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$ to the state $\dfrac{|0\rangle + i|1\rangle}{\sqrt{2}} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1\\ i \\ \end{pmatrix} $. If you look at it closely, you will see that the unitary matrix (quantum gate) you want to apply should take the form as:

$$ \begin{pmatrix} 1& 0\\ 0 & i \\ \end{pmatrix}$$


$$ \begin{pmatrix} 1& 0\\ 0 & i \\ \end{pmatrix}\dfrac{1}{\sqrt{2}}\begin{pmatrix} 1\\ 1 \\ \end{pmatrix} = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1\\ i \\ \end{pmatrix} = \dfrac{|0\rangle + i|1\rangle}{\sqrt{2}} $$

The matrix $ \begin{pmatrix} 1& 0\\ 0 & i \\ \end{pmatrix}$ has a special name, it is called the Phase gate (or S gate) in quantum computing.

So to summarize, we first apply the Hadamard gate follows by the Phase gate (As Lena pointed out in her answer as well) to get from the state $|\psi_0 \rangle = |0\rangle$ to the state $|\psi_f \rangle = \dfrac{|0\rangle + i|1\rangle}{\sqrt{2}}$.

The quantum circuit looks like:

enter image description here


Posted 2021-01-18T16:15:06.850

Reputation: 6 322


H then S do the trick, it gives me this : Statevector


Posted 2021-01-18T16:15:06.850

Reputation: 1 188

Okay, I'm not sure how you can apply S to a superposition like H|0> though. Since S is a 2x2 matrix, I;m not sure how to write 1/sqrt(2)(|0>+|1>) so that you can apply S to it – Natasha – 2021-01-18T16:34:18.493

1As you can see in the other responses, S can be applied to a superposed state, it will just apply to |0> and |1> easy thanks to the linearity of the operations in quantum computation ;) – Lena – 2021-01-18T16:50:51.437


It helps to know the action of various gates on the computational basis states. For example, the S gate

$$ S = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix} $$

keeps $|0\rangle$ fixed while phasing $|1\rangle$ by $i$. This means that if we had the state $|0\rangle+|1\rangle$ then $S(|0\rangle+|1\rangle)$ is $|0\rangle+ i|1\rangle$, i.e. the state we need. We can obtain $|0\rangle+|1\rangle$ by applying Hadamard

$$ H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} $$

to the $|0\rangle$ state.

In summary, you can apply Hadamard and the S gate to the $|0\rangle$ state

$$ SH|0\rangle = S\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) = \frac{1}{\sqrt{2}}(S|0\rangle + S|1\rangle) = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)\tag1 $$

where we included normalization factors.

Note that $(1)$ is not the only way to obtain $|0\rangle + i|1\rangle$. For example, the $\frac{\pi}{2}$ Y rotation

$$ R_y\left(\frac{\pi}{2}\right) = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} $$

can be used in place of the Hadamard. This is useful on hardware platforms that may not implement Hadamard natively.

Adam Zalcman

Posted 2021-01-18T16:15:06.850

Reputation: 4 307

Hi, thanks for your explanation! For the purpose of implementing quantum algorithms (I'm trying to implement phase estimation), does the normalisation constant not make much of a difference? It was trying to "get rid" of this that was confusing me. – Natasha – 2021-01-18T16:47:37.823

It only matters at the end if/when you compute output probabilities. However, you can still keep ignoring the normalization factors as long as you remember to renormalize before measurement. – Adam Zalcman – 2021-01-18T16:52:14.637

Ah okay, thank you! – Natasha – 2021-01-18T17:03:22.240