What is going on inside quantum gates physically? What are the physical realities of the different quantum gates? I am particularly interested in how the CNOT gate works on the inside as well as how the Hadamard gate works. The Hadamard gate to me is strange because my assumption on superposition was that a particle left undisturbed would naturally assume a superposition, but that cannot be how a Hadamard gate functions because the same process that puts the qubit into superposition also has to be reversible to take that particle out of superposition and into the original state.
Any intuition or articles on the physics going on inside quantum gates would be appreciated.
 A: I do not know how it is implemented in a Josephson junction, but if we were working with photons, one way to realise the action of the Hadamard gate is by using a perfect beam splitter.
Operator
In particular, the Hadamard gate is given by:
$$H=\frac{1}{\sqrt{2}}\left[\begin{array}{cc}
1 & 1 \\
1 & -1
\end{array}\right]$$
In comparison, the action of a beam splitter with reflectivity $r$ and transmittivity $t$ is given by
$$H=\left[\begin{array}{cc}
r & t \\
t & -r
\end{array}\right]$$
with $r=t=1/\sqrt 2$ for a perfect beam splitter.

Physical intuition
Consider the following setup. There are two input modes. Namely $1$, corresponding to the state $|10\rangle_1$ and $2$, corresponding to the state $|01\rangle_2$. Similarly the output modes are $3$, corresponding to the state $|10\rangle_3$ and $4$, corresponding to the state $|01\rangle_4$.

If a photon is incoming in mode 1, then after passing through the beam splitter it is in a superposition of two output modes whose weights are given by the Hadamard gate. Similarly for a photon in mode 2.
$$|01\rangle_1 = \frac{1}{\sqrt 2}\big(|10\rangle_3 + |01\rangle_4 \big)$$
$$|10\rangle_2 = \frac{1}{\sqrt 2}\big(-|10\rangle_3 + |01\rangle_4 \big)$$
Now consider the input state in an equal superposition of the two modes. What will be the corresponding output mode?
$$\frac{1}{\sqrt 2} \big(|01\rangle_1+|10\rangle_2\big) = |01\rangle_4$$
As you can see, the output is now a single mode! How is that possible? Well, the amplitudes going to mode 3 from mode 1 and 2, destructively interfered. The reason they did so is because of the emergence of a phase difference of $\pi$ due to reflection through the dielectric in mode 3.
Since the system is symmetric with input and output modes, the same thing happens with a superposition of output modes giving you a single input mode. It’s the destructive interference of amplitudes!

And for an all-optical implementation of a CNOT gate, you may look at this paper.
