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%)Mach Zehnder Experiment, 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.

  • $\begingroup$ 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. $\endgroup$ – MayankB May 27 at 17:01

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