enter image description hereIn measurement based quantum computing, according to R. Raussendorf, Hadamard Gate can be realized by five qubit cluster state, where the first qubit is measured in X basis and the following three are measured in Y basis.

However, even though I tried to calculate the outcome of the following measurements, I did not find Hadamard gate acting on initial state.

I assumed, to measure the qubit in X basis, we need to apply H gate and measure in Z basis:

\begin{equation} \mathbf{H} = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix} \end{equation}

for measurements in Y basis, we apply a U gate and measure in the z basis, the required U gate is as follows:

\begin{equation} \mathbf{U} =\frac{1}{\sqrt{2}} \begin{bmatrix} 1 & -i\\ 1 & i \end{bmatrix} \end{equation}

Now, measuring the state in those basis simply means applying those gates consecutively, except for additional X gates due to measurements, which we are not dealing right now. So, the result of this operation:

\begin{equation} \mathbf{X^{m_4}U X^{m_3}U X^{m_2}U X^{m_1}H}|\psi\rangle \end{equation}

which is not equal to:

\begin{equation} \mathbf{H}|\psi\rangle \end{equation}

However, if we only had 2 qubit cluster state and measure the first qubit in X basis, we would get:

\begin{equation} \mathbf{X^{m}H}|\psi\rangle \end{equation}

which is exactly Hadamard gate applied on initial state up to correction.

What am I missing here? Why don't we use two qubit cluster state in representation of Hadamard gate? How does this five qubit cluster state represents Hadamard gate?

  • $\begingroup$ Could you show your full calculation? You don't show (at the moment) the effect of the measurements (e.g. if you assume a particular outcome) - it's not the same as taking the observable and applying it as if it were a unitary - and especially not its effect on the entangled state. (The net effect of an X measurement should be $H$, up to correction, and Y measurement should give $HS$.) $\endgroup$
    – DaftWullie
    Nov 20, 2023 at 11:51

1 Answer 1


Your calculation shows that you are doing the right thing for a single step. You just don't seem to be carrying it through for a set of 4 steps. Overall, you should be getting $$ (X^{m_4}HS)(X^{m_3}HS)(X^{m_2}HS)(X^{m_1}H)|\psi\rangle $$

For simplicity, assume that all the $m_i=0$, so you have the sequence $$ HSHSHSH. $$ Now there's a neat identity: $HSHSHS=I$ (up to a global phase), so you're left with just $H$.

Now, it's a fair question which you're faffing about with this sequence rather than just using a single step. Yes, you can get Hadamard from a single step, but the sequence of 4 things is a building block. By changing some of the measurement bases, you can get an arbitrary single-qubit rotation. But the Hadamard is the natural starting point for building up to that.

  • $\begingroup$ Thank you for your answer, I now realized my mistake when you said "up to a global phase", where I was considering my result as wrong because of the global phase factor which I did not realize that it has no effect. $\endgroup$ Nov 20, 2023 at 13:44

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