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

5

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.

6

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:

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

since

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

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:

1

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

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

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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.

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