Recently I've been trying to understand why the scattering matrices that describe an interferometer should be $SU(2)$ matrices rather than $U(2)$. The condition of unitarity is undiscussed as it follows from energy conservation. But why should the determinant be especially 1?

I understand that $$\textrm{U(2)}=\left\{ \left(\begin{array}{cc} A & B\\ -B^{*}e^{i\theta} & A^{*}e^{i\theta} \end{array}\right)\left|A,B\in\mathbb{C},\theta\in\mathbb{R},\left|A\right|^{2}+\left|B\right|^{2}=1\right.\right\} ,$$ while $$\textrm{SU(2)}=\left\{ \left(\begin{array}{cc} A & B\\ -B^{*} & A^{*} \end{array}\right)\left|A,B\in\mathbb{C},\left|A\right|^{2}+\left|B\right|^{2}=1\right.\right\} .$$ If we have input annihilation operators $\hat{a}_{1/2}$ and output $\hat{b}_{1/2}$, than a matrix in $U(2)$ gives $$\left(\begin{array}{c} \hat{b}_{1}\\ \hat{b}_{2} \end{array}\right)=\left(\begin{array}{cc} A & B\\ -B^{*}e^{i\theta} & A^{*}e^{i\theta} \end{array}\right)\left(\begin{array}{c} \hat{a}_{1}\\ \hat{a}_{2} \end{array}\right)=\left(\begin{array}{cc} A \hat{a}_{1}+ B\hat{a}_{2}\\ e^{i\theta}\left(-B^{*}\hat{a}_{1} + A^{*}\hat{a}_{2}\right) \end{array}\right).$$

It's not clear to me, how physically this phase indeterminacy $e^{i\theta}$ in one of the two outputs should be discarded. Nevertheless, if I compute quantities like the number of photons $\hat{b}_{2}^\dagger\hat{b}_{2}$, this phase disappears!

The same situation happens with the Jones formalism, which has always $SU(2)$ matrices.

  • $\begingroup$ Phase indeterminacy is important; look up The Aharanov-Bohm experiment and also the gauge principle in QED, QCD, QFT and even in gravity... $\endgroup$ Jun 19 '18 at 8:19
  • $\begingroup$ Which kind of makes me think that when you've made the phase indeterminacy disappear it was merely as a formal exercise and not really due to physical thinking/reasoning. $\endgroup$ Jun 19 '18 at 8:20
  • $\begingroup$ It is important, but then why everybody talks about SU(2) interferometry rather than U(2) interferometry? See SU(2) and SU(1,1) interferometers $\endgroup$
    – QuOpt
    Jun 19 '18 at 8:22
  • $\begingroup$ Seems like a waste of time chasing up those links ... why? See my comment above. $\endgroup$ Jun 19 '18 at 8:24

Phase is a funny thing. It only has meaning as a relative concept. For this reason, if I have a state $|\psi\rangle$ and I multiply it by some phase factor $|\psi\rangle\exp(i\theta)$ I still have exactly the same state. This phase is not observable. If on the other hand, I have a superposition $$|\psi\rangle+|\phi\rangle$$ then another superposition given by $$|\psi\rangle\exp(i\theta)+|\phi\rangle$$ would not be the same state again, because it contains a relative phase. This is observable.

So the reason why the phase factor you got is discarded, is because it does not produce a relative phase. That is why you cannot observe it with you number operator.

  • $\begingroup$ I agree: the number operator won't make me observe any relative phase. And since in general, we're not superposing the two output ports of the interferometer, even if there's a relative phase it won't matter. But definitely, some relative phase is produced. Thank you for your contribution! @flippiefanus $\endgroup$
    – QuOpt
    Jun 22 '18 at 7:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.