- #1
Mentz114
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When 2 identical photons are incident at the same time at the input channels of a 50:50 beam splitter, quantum interference causes the elimination of the both reflected and both transmitted outcomes ##\hat{a}_{1H}^{\dagger}\hat{a}_{2H}^{\dagger}|0\rangle_1|0\rangle_2 \ne |1\rangle_1|1\rangle_2##. It is straightforward to derive this by applying the BS transformation to the inputs to get the transformed operator ##\tfrac{1}{\sqrt{2}}(t\hat{a}_{1H}^{\dagger} + ir\hat{a}_{2H}^{\dagger})(ir\hat{a}_{1H}^{\dagger} + t\hat{a}_{1H}^{\dagger})## and on expanding we get ( with ##t=r=1/\sqrt{2}##) ##\tfrac{1}{\sqrt{2}} (i\hat{a}_{1H}^{\dagger}\hat{a}_{1H}^{\dagger} + i\hat{a}_{2H}^{\dagger}\hat{a}_{2H}^{\dagger} +[\hat{a}_{2H}^{\dagger},\hat{a}_{1H}^{\dagger}])##. For identical photons the commutator vanishes.
If the input photons are orthogonally polarised the transformed operator is
## \tfrac{1}{\sqrt{2}}(i\hat{a}_{1H}^{\dagger}\hat{a}_{1H}^{\dagger} + i\hat{a}_{2V}^{\dagger}\hat{a}_{2V}^{\dagger} +[\hat{a}_{2V}^{\dagger},\hat{a}_{1H}^{\dagger}])##. In a fair universe we could expect that ##\hat{a}_{2V}^{\dagger}\hat{a}_{1H}^{\dagger} = \pm e^{\pi/2}\hat{a}_{1V}^{\dagger},\hat{a}_{2H}^{\dagger}## (or something similar) which would give ##\tfrac{1}{2}(i\hat{a}_{1H}^{\dagger}\hat{a}_{1H}^{\dagger} + i\hat{a}_{2V}^{\dagger}\hat{a}_{2V}^{\dagger} -i\hat{a}_{1V}^{\dagger}\hat{a}_{2H}^{\dagger} + \hat{a}_{1V}^{\dagger}\hat{a}_{2H}^{\dagger})## and this operator applied to ##|0\rangle_1|0\rangle_2## gives four equal probabilty outcomes for independent photons.
I think this is what one would expect although not all possible combinations are present. The bad news is that the commutator always vanishes. So the last conclusion is ruled out. Apart from being an obvious fault in the universe the problem is - how does one calculate the BS output for distinguishable photons ?
In the notation used here there is no difference between the order we write the operators because they only operate on their own channel. This reflects the fact that physically the photons arrive at same time and any time lag will make them distinguishable. Unless the different polarizations cause a time lag, there is no such thing as the commutator above.
(My source for the theory is chapters 6 and 9 in Gerry and Knight "Introductory Modern Optics".)
If the input photons are orthogonally polarised the transformed operator is
## \tfrac{1}{\sqrt{2}}(i\hat{a}_{1H}^{\dagger}\hat{a}_{1H}^{\dagger} + i\hat{a}_{2V}^{\dagger}\hat{a}_{2V}^{\dagger} +[\hat{a}_{2V}^{\dagger},\hat{a}_{1H}^{\dagger}])##. In a fair universe we could expect that ##\hat{a}_{2V}^{\dagger}\hat{a}_{1H}^{\dagger} = \pm e^{\pi/2}\hat{a}_{1V}^{\dagger},\hat{a}_{2H}^{\dagger}## (or something similar) which would give ##\tfrac{1}{2}(i\hat{a}_{1H}^{\dagger}\hat{a}_{1H}^{\dagger} + i\hat{a}_{2V}^{\dagger}\hat{a}_{2V}^{\dagger} -i\hat{a}_{1V}^{\dagger}\hat{a}_{2H}^{\dagger} + \hat{a}_{1V}^{\dagger}\hat{a}_{2H}^{\dagger})## and this operator applied to ##|0\rangle_1|0\rangle_2## gives four equal probabilty outcomes for independent photons.
I think this is what one would expect although not all possible combinations are present. The bad news is that the commutator always vanishes. So the last conclusion is ruled out. Apart from being an obvious fault in the universe the problem is - how does one calculate the BS output for distinguishable photons ?
In the notation used here there is no difference between the order we write the operators because they only operate on their own channel. This reflects the fact that physically the photons arrive at same time and any time lag will make them distinguishable. Unless the different polarizations cause a time lag, there is no such thing as the commutator above.
(My source for the theory is chapters 6 and 9 in Gerry and Knight "Introductory Modern Optics".)
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