# How did the book derive the following Unitary Operator expression in Mach Zehnder experiment?

I'm reading Renato Portugal's "Quantum Walks and Search Algorithms". In The Postulates of Quantum Mechanics, under the heading 'Evolution Postulate', there is the Mach Zehnder experiment, with the following image:- (detector 1 detects nothing, and detector 2 detects 100%)

The author then explains the experiment in the classical viewpoint where a wave gives the above result (in the image), followed by the quantum viewpoint wherein a single photon gives the same result.

From the book:-
In quantum mechanics, if the set of mirrors is isolated from the environment, the two possible paths are represented by two orthonormal vectors $$|0\rangle$$ and $$|1\rangle$$, which generate the state space that describes the possible paths to reach the photon detector. Therefore, a photon can be in superposition of “path A,” described by $$|0\rangle$$, together with “path B,” described by $$|1\rangle$$. The action of the half-silvered mirrors on the photon must be described by a unitary operator U. This operator must be chosen so that the two possible paths are created in a balanced way, i.e. $$U|0\rangle=\frac{|0\rangle + e^{i\phi}|1\rangle}{\sqrt{2}}$$

My first problem is that this equation isn't making total sense to me.
What I mean by that is if the equation had been something like this, $$\hat{a}^\dagger=\frac{1}{\sqrt{2}}(\hat{b}^\dagger+i\hat{c}^\dagger)$$ Where,

$$\hat{a}^\dagger$$ is the channel between $$source$$ and $$splitter 1$$
$$\hat{b}^\dagger$$ is from $$splitter 1$$ to $$splitter 2$$ through $$A$$ and
$$\hat{c}^\dagger$$ is from $$splitter 1$$ to $$splitter 2$$ through $$B$$
it would have made more sense to me. I would have understood it as, "the operation of creating a photon on channel $$a$$ is exactly identical to the superposition of creating photons on the $$b$$ and $$c$$ channels."

My second problem is what will the action of operator $$U$$ on state $$|1\rangle$$ be? And how do you explain it?

What I know about Unitary Operator is that:-
If the state of the quantum system at time $$t_1$$ is described by vector $$|\psi_1\rangle$$, the system state $$|\psi_1\rangle$$ at time $$t_2$$ is obtained from $$|\psi_2\rangle$$ by a unitary transformation $$U$$, which depends only on $$t_1$$ and $$t_2$$, as follows: $$|\psi_2\rangle = U|\psi_1\rangle.$$

What that equation says is that the silver mirror creates two paths with equal probability. One way of deriving that is as follows, let $$U|0\rangle=\alpha|0\rangle+\beta|1\rangle$$. Since we know the probability of photon ending up in either path is equal, $$\alpha^2=\beta^2$$. We also know that $$\alpha^2+\beta^2=1\implies\alpha^2=\beta^2=1/2$$. If we choose the convention that $$\alpha=1/\sqrt{2}$$, then a general value of $$\beta$$ is $$e^{i\theta}/\sqrt{2}$$. The reason we can choose $$\alpha=1$$ is that the overall phase of the wavefunction has no physical significance.
To find $$U|1\rangle$$, we can again use the conditions
1. $$U|0\rangle=\gamma|0\rangle+\delta|1\rangle$$
2. $$\gamma^2=\delta^2=1/2$$
3. $$U|0\rangle$$ is orthogonal to $$U|1\rangle$$
Picking $$\gamma=1/\sqrt{2}$$, we get $$\delta=-e^{i\theta}/\sqrt{2}$$. Note that this again is a convention and we could have chosen other values of $$\gamma$$ and $$\delta$$ that satisfy the above conditions.
• Thank you for your answer but this I am aware of. What I meant was, what do you mean by $U|0\rangle=\alpha|0\rangle+\beta|1\rangle$? What does $U|0\rangle$ signify? Like when I wrote :- $$\hat{a}^\dagger=\frac{1}{\sqrt{2}}(\hat{b}^\dagger+i\hat{c}^\dagger)$$ What I meant was, "the operation of creating a photon on channel a is exactly identical to the superposition of creating photons on the b and c channels." Similarly what does it mean in the $U|0\rangle$ context? Thank you. – MayankB May 27 at 17:01