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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.

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  • $\begingroup$ Phase indeterminacy is important; look up The Aharanov-Bohm experiment and also the gauge principle in QED, QCD, QFT and even in gravity... $\endgroup$ – Mozibur Ullah 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. $\endgroup$ – Mozibur Ullah 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. $\endgroup$ – Mozibur Ullah Jun 19 '18 at 8:24
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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.

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  • $\begingroup$ 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

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