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By definition, the Hadamard gate can be written in the computational basis as:

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

Now, note that applying a Hadamard gate to the state, says, $|\psi \rangle = |0\rangle$, then it is equivalent to be doing the following rotation:

So what if you apply another Hadamard gate? What would happen? Well, firs note that he inverse of Hadamard gate is itself. That is:

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

Therefore, you apply another Hadamard gate then you will get back to the same spot as indicated by the picture below:

The takeaway is:

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

Hence, applying the Hadmard gate twice to the state $|\psi\rangle$ will keep the state as $|\psi\rangle$.

This doesn't change even if $|\psi \rangle$ is in some superposition state. That is, suppose, $|\psi \rangle = \sqrt{\dfrac{2}{3}}|0\rangle + \sqrt{\dfrac{2}{3}}|1\rangle $,

and if you apply a Hadamard gate, then geometrically, it is equivalent as doing:

and if you apply another Hadamard gate you will get back to the original vector/state $|\psi \rangle = \sqrt{\dfrac{2}{3}}|0\rangle + \sqrt{\dfrac{2}{3}}|1\rangle $

So if starting at the stage $A$, you have arbitrary state $|\psi_A \rangle = \begin{pmatrix} \cos \dfrac{\theta}{2} \\ e^{i \phi} \sin \dfrac{\theta}{2} \end{pmatrix}$ then by applying Hadamard gate you get:

$$ H |\psi \rangle = \dfrac{1}{\sqrt{2}} \begin{pmatrix} 1& 1\\ 1 & -1 \\ \end{pmatrix} \begin{pmatrix} \cos \dfrac{\theta}{2} \\ e^{i \phi} \sin \dfrac{\theta}{2} \end{pmatrix} = \dfrac{1}{\sqrt{2}} \begin{pmatrix} \cos \dfrac{\theta}{2} + e^{i \phi} \sin \dfrac{\theta}{2} \\ \cos \dfrac{\theta}{2} - e^{i \phi} \sin \dfrac{\theta}{2} \end{pmatrix} $$

So at stage $B$, your qubit is in the state $|\psi_B \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} \cos \dfrac{\theta}{2} + e^{i \phi} \sin \dfrac{\theta}{2} \\ \cos \dfrac{\theta}{2} - e^{i \phi} \sin \dfrac{\theta}{2} \end{pmatrix} $

and like what we discussed above, applying another Hadamard gate will get it back to the starting vector $|\psi_A\rangle$. That is, the qubit at stage $C$ have the state $|\psi_C \rangle = |\psi_A\rangle$

So what is the quit at B and C? – Oliver Custance – 2021-01-17T17:06:51.027

It depends on what $|\psi\rangle$ is, but the state of the qubit at $C$ is the same as the state of the qubit at $A$. – KAJ226 – 2021-01-17T17:21:08.693

1Perfect, thanks very much for answering this. – Oliver Custance – 2021-01-17T17:24:55.347

@Oliver You are welcome. – KAJ226 – 2021-01-17T17:35:23.320

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**Qubit states at points A, B and C**

Let the input pure state be $|\psi\rangle = a|0\rangle + b|1\rangle$. At point A the state is simply

$$ |\psi_A\rangle = |\psi\rangle = a|0\rangle + b|1\rangle $$

since there are no gates that change the input state before the point A. At point B the state is

$$ |\psi_B\rangle = H|\psi\rangle = \frac{a}{\sqrt{2}}(|0\rangle + |1\rangle) + \frac{b}{\sqrt{2}}(|0\rangle - |1\rangle) = \frac{a+b}{\sqrt{2}}|0\rangle + \frac{a-b}{\sqrt{2}}|1\rangle $$

due to the action of the first Hadamard

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

Finally, at point C the state is

$$ |\psi_C\rangle = HH|\psi\rangle = a|0\rangle + b|1\rangle $$

because Hadamard is self-inverse, i.e. $HH = I$, as can be checked by matrix multiplication.

**Why superpositions cannot be described as probabilistic mixtures?**

The behavior of the quantum circuit in the question rules out description of quantum superpositions as probabilistic mixtures (e.g. the qubit is in the $0$ state with probability 50% and in the $1$ state with probability 50%), because the process of forming a probabilistic mixture is not invertible. In particular, applying it twice does not return back to the initial state as it does in the case of the Hadamard gate. There are two compatible explanations of this fact: one from purely mathematical perspective and the other from the physics point of view.

Mathematically, a process of forming an equal probabilistic mixture is described by the stochastic matrix

$$ \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}. $$

This does not describe a quantum gate because - unlike the Hadamard above - it is not invertible. In particular, applying it twice to a vector does not return it to its original state. Note that by the postulates of quantum mechanics evolution of any closed quantum system is unitary and hence invertible.

Physically, the description of superpositions as probabilistic mixtures turns out to be insufficient because probabilities cannot interfere destructively whereas amplitudes of a quantum states can. Indeed, we can see destructive interference in action as it cancels out the amplitude of the $|1\rangle$ state when we compute $HH|0\rangle$ step by step

$$ \begin{align} HH|0\rangle &= \frac{1}{\sqrt{2}}H|0\rangle + \frac{1}{\sqrt{2}}H|1\rangle\\ &= \frac{1}{2}(|0\rangle +|1\rangle)+ \frac{1}{2}(|0\rangle -|1\rangle)\\ &= \left(\frac{1}{2} + \frac{1}{2}\right)|0\rangle + \left(\color{red}{\frac{1}{2} - \frac{1}{2}}\right)|1\rangle\\ &= 1|0\rangle + \color{red}{0}|1\rangle =\\ & = |0\rangle. \end{align} $$

Probabilities of terms in a mixture are never subtracted this way, because probabilities cannot be negative.

1Awesome, thanks very much for answering my question. – Oliver Custance – 2021-01-17T18:01:00.010

mainly to ask what's the state of the qubits at A, B and C? – Oliver Custance – 2021-01-17T16:32:42.430

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Does this answer your question? What happens if $|\psi\rangle$ = $|0\rangle$ or $|\psi\rangle$ = $|1\rangle$ is passed as an input to two Hadamard gates in sequence? (it's the same exact exercise)

– glS – 2021-01-18T13:32:47.067In simple words, since $HH=I$, it happens nothing. – Martin Vesely – 2021-01-19T12:32:01.193