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Why 2 $H$ gates in series create a probability of 100% for one value of the qubit and 0% of the second value of the qubit since an $H$ gate acts like a superposition generator?

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  • $\begingroup$ Consider using the title for the question itself, so it will be easier to find for others. $\endgroup$
    – luciano
    Nov 4 '20 at 18:35
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The reason for this is because the inverse of Hadamard gate is itself. That is, giving that

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

then

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

so therefore, $$HH|\psi\rangle = H H^{-1} |\psi \rangle = I |\psi \rangle = |\psi\rangle \ \ \ \textrm{where $I$ is the Identity operator.}$$

Now when $|\psi \rangle = |0 \rangle$, which is the starting state of your quantum circuit, then applying two consecutive Hadamard gates is the same as doing nothing... hence the qubit stayed at the state $|0\rangle$. Thus giving you a 100% probability is it the state $|0\rangle$.


To see this in the geometric sense, applying two consecutive Hadamard gates to the qubit in the state $|0\rangle$ is like doing the following rotations:

enter image description here

The first Hadamard gate rotate the qubit in the red trajectory to the state $|+\rangle = \dfrac{|0 \rangle + |1 \rangle}{\sqrt{2}} $. Then the second Hadamard gate rotate the qubit in the blue trajectory, back to the state $|0\rangle$ again.

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Simply speaking, $H^2 = HH = I$ which is identity (you can verify this by direct matrix multiplication). As a result, such operator does not change anything in a quantum state.

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The answers above noting that $H^2=I$ neatly capture this. If you want an alternative way of seeing this, you can look at what $H$ does to each of the computational basis states if applied twice. (Since any state will be a superposition of the basis states, understanding how the operator acts on each basis state will tell you what it does to a general state.)

  • When $H$ is applied to $|0 \rangle \equiv \begin{bmatrix}1\\0\end{bmatrix}$, matrix multiplication means you can read off the first column of $H$ to see what the coefficients of $|0 \rangle$ are in the standard basis after applying $H$ to it. Specifically, $H$ takes $|0 \rangle$ to $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$.
  • When $H$ is applied to $|1 \rangle \equiv \begin{bmatrix}0\\1\end{bmatrix}$, matrix multiplication means you can read off the second column of $H$ to see what the coefficients of $|1 \rangle$ are in the standard basis after applying $H$ to it. Specifically, $H$ takes $|1 \rangle$ to $\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$.

To apply $H$ to each of the computational basis states a second time, we can just re-use the results from above: $$HH |0 \rangle = H\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) \\= \frac{1}{2}((|0\rangle + |1\rangle)+(|0\rangle - |1\rangle)) \\=\frac{1}{2}(2|0\rangle)=|0\rangle,$$ $$HH |1 \rangle = H\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) \\= \frac{1}{2}((|0\rangle + |1\rangle)-(|0\rangle - |1\rangle)) \\=\frac{1}{2}(2|1\rangle)=|1\rangle.$$ Since applying $H$ twice to both computational basis states results in no change, applying $H$ to any state results in no change, i.e., $H^2=I$.

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