Hadamard cascade

Question

Anyone of you can explain (with mathematical steps) me this circuit:

I do not understand why the first qubit phase (as show on IBM composer) is influenced by the second. More precisely: In circuit with single qubit set to zero, the two Hadamard gates at the end, return a quantum state with prob of $|1\rangle$ equal to 0% and phase 0. On circuit like that in the figure, the first qubit (q[0]), return a quantum state with prob of $|1\rangle$ equal to 0% but phase is $\pi$.

Thanks

Answer 1

You have a circuit with 2 qubits. But you only apply Hadamard gates on one of them. So q1 remains unchanged (think of it as if you apply identity to it), as for the q0, at the end it is also unchanged since the two Hadamard gates cancel each other.

First note that

$H|0\rangle = \dfrac{|0\rangle + |1\rangle}{\sqrt{2}}$ and $H|1\rangle = \dfrac{|0\rangle - |1\rangle}{\sqrt{2}}$

The math behind it:

Let is assume your initial state is $|0\rangle $

Apply Hadamard once, it $H|0\rangle$ becomes $$\frac{1}{\sqrt2}(|0\rangle+|1\rangle)= \frac{1}{\sqrt2}|0\rangle + \frac{1}{\sqrt2}|1\rangle $$

Now we apply Hadamard again to the result $H (\frac{1}{\sqrt2}|0\rangle + \frac{1}{\sqrt2}|1\rangle) $, I am gonna split it in 2 steps.

First, apply Hadamard to the first part, So $ H\frac{1}{\sqrt2}|0\rangle$ becomes $$ \frac{1}{\sqrt2}( \frac{1}{\sqrt2}|0\rangle + \frac{1}{\sqrt2}|1\rangle)=\frac{1}{2}|0\rangle + \frac{1}{2}|1\rangle $$

With the second part, $ H\frac{1}{\sqrt2}|1\rangle$ becomes $$ \frac{1}{\sqrt2}( \frac{1}{\sqrt2}|0\rangle - \frac{1}{\sqrt2}|1\rangle)=\frac{1}{2}|0\rangle - \frac{1}{2}|1\rangle $$

Put them back together:

$$ \frac{1}{2}|0\rangle + \frac{1}{2}|1\rangle + \frac{1}{2}|0\rangle - \frac{1}{2}|1\rangle\\ = (\frac{1}{2}+\frac{1}{2}) |0\rangle + (\frac{1}{2}- \frac{1}{2}) |1\rangle \\ = |0\rangle $$

You can do the same calculation if your initial state was $|1\rangle$ and you will similarly $|1\rangle$ .

Another way to think about it is in terms of matrices, you can see that HH=Identity matrix:

$H = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} $ H applied twice is
$HH = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} \times \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1\\ 1 & -1 \end{pmatrix} $ Now, I leave the remaining math to OP since it is an easy one. But by computing $HH$ you will find $\dfrac{1}{2}\begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix}$ which is the identity matrix