Mach-Zehnder interferometry wave functions

Consider the set up below:

I have read that in the apparatus the wavefunction is given by: $$|\psi \rangle=e^{i\theta}|c \rangle +i |b \rangle$$ where $\theta$ is the phase added by the phase adder. My question is what do the states $|b \rangle$ and $|c \rangle$ represent and why are we timesing them by the phases $i$ and $e^{i\theta}$ instead of just having the phases built into the wave functions (which I think would give different results) ?

• $|b>$ and $|c>$ are the wave functions corresponding to spatial output modes of the first beam splitter. The reflected spatial modes will only have phase $i$ attached to it after the reflection. journals.aps.org/pra/abstract/10.1103/PhysRevA.58.4904 this explains how they come about. – WInterfell May 29 '15 at 22:11

According to the diagram you provided, $|a\rangle$ is injected into a beam-splitter, resulting in the out-put state, $$|\psi_{o}\rangle =B|a\rangle= e^{\frac{i\pi}{2}}|b\rangle +|c\rangle.$$ Here the beam-splitter unitary is denoted by $B$. I have expressed $i$ as $e^{\frac{i\pi}{2}}$, to highlight the fact that $|b\rangle$ picks up a phase $\pi/2$, relative to the original state $|a\rangle$. Therefore, the $e^{\frac{i\pi}{2}}$ term (or $i$ as you have it) in front of $|b\rangle$ describes the fact that the reflected part of the state acquire a $\pi/2$ phase, whereas the transmitted part $|c\rangle$ has the same phase as $|a\rangle$.
Finally, the phase-adder describes the fact that a phase $\theta$ is acquired by $|c\rangle$ relative to $|b\rangle$. Physically, this can be accomplished by placing a physical media (e.g. dielectric) through which the state must propagate: if this medium is placed only in the path corresponding to $|c\rangle$, then there will be a phase acquired by $|c\rangle$. In other words, nothing happens to $|b\rangle$, but $|c\rangle$ acquires a phase $\theta$; which can be described mathematically by the unitary transformation $$U(\theta)= e^{i\theta |c\rangle \langle c|}.$$ Therefore, as you have it, the final state is $$U(\theta)|\psi_{o}\rangle=|\psi\rangle= i|b\rangle+e^{i\theta}|b\rangle.$$
• I think it should be |c>, not |b> at the end: $$U(\theta)|\psi_{o}\rangle=|\psi\rangle= i|b\rangle+e^{i\theta}|c\rangle.$$ – KIAaze Aug 1 at 11:46