Polarization and the beam splitter

[Mentz114]

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 probability 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".)

 

[.Scott]

I wouldn't normally touch this question because there are way better people on this forum to respond.
But it's been almost a day now - so let me make some comments.

Right away, two different photons cannot interfere with each other. I believe the ket notation you are using is appropriate for two different paths for the same photon. If there are two photons whether "distinguishable" or not, they will each make their own independent path through the BS. So one output will have a 25% chance of yielding both photons, 50% chance of yielding one, and a 25% chance of remaining dark. The other output will yield whatever the first did not.

But I can't really tell if this is a full answer to your questions. If not, take another shot at it in your reply.

 

[PeterDonis]

Right away, two different photons cannot interfere with each other.

Yes, they can. But they have to be produced in such a way that they are not distinguishable.

 

[Mentz114]

[..]
Right away, two different photons cannot interfere with each other. I believe the ket notation you are using is appropriate for two different paths for the same photon. If there are two photons whether "distinguishable" or not, they will each make their own independent path through the BS. So one output will have a 25% chance of yielding both photons, 50% chance of yielding one, and a 25% chance of remaining dark. The other output will yield whatever the first did not.
[..]
.

I agree with that. It the same as saying that independent photons reflect or transmit in the BS with equal probability 1/2.
I think what is required is to combine the input state operator $(\hat{a}_1^{\dagger}, \hat{a}_2^{\dagger})$ and the polarization and work with vector $(\hat{a}_1^{\dagger}, \hat{a}_1^{\dagger})\otimes (\cos(\theta),\sin(\theta))$ and similarly transform the BS transformation. It seems to be working but I can only get one good prediction at present.

I hope this has been done and I might get a reference.

Yes, they can. But they have to be produced in such a way that they are not distinguishable.

Orthogonally polarised photons do not interfere which is why the BS equation must include polarization.

 

[Cthugha]

You need to be careful with the BS transformation relation here.

The basic idea is always that you define new operators for the modes at the output ports \hat{a}_{o,1} and \hat{a}_{o,2} such that they are a linear combination of the modes \hat{a}_{i,1} and \hat{a}_{i,2} of the input ports, such as:
\hat{a}_{o,1}=r\hat{a}_{i,1}+t'\hat{a}_{i,2} and \hat{a}_{o,2}=t\hat{a}_{i,1}+r'\hat{a}_{i,2}.

However, the BS is in principle a multimode device, which acts on every single mode of the full light field. For most of the modes, both inputs will of course simply be vacuum, so these processes can be simply neglected, but still, what the BS really gives you is a sum over terms that look like the one above for every possible mode of the light field. Accordingly, the situation is quite clear. For two indistinguishable photons coming in at the input ports, you get exactly one mode, where you have one photon at each input and an incredible number of modes-1, where only vacuum arrives. For two distinguishable photons you get one mode, where one photon arrives at the first port and vacuum arrives at the second, one mode where vacuum arrives at the first port and one photon arrives at the second and an incredible number of modes-2, where only vacuum arrives. If you do the math for this scenario, you will find that the interference is gone.

 

[Mentz114]

You need to be careful with the BS transformation relation here.

The basic idea is always that you define new operators for the modes at the output ports \hat{a}_{o,1} and \hat{a}_{o,2} such that they are a linear combination of the modes \hat{a}_{i,1} and \hat{a}_{i,2} of the input ports, such as:
\hat{a}_{o,1}=r\hat{a}_{i,1}+t'\hat{a}_{i,2} and \hat{a}_{o,2}=t\hat{a}_{i,1}+r'\hat{a}_{i,2}.

However, the BS is in principle a multimode device, which acts on every single mode of the full light field. For most of the modes, both inputs will of course simply be vacuum, so these processes can be simply neglected, but still, what the BS really gives you is a sum over terms that look like the one above for every possible mode of the light field. Accordingly, the situation is quite clear. For two indistinguishable photons coming in at the input ports, you get exactly one mode, where you have one photon at each input and an incredible number of modes-1, where only vacuum arrives. For two distinguishable photons you get one mode, where one photon arrives at the first port and vacuum arrives at the second, one mode where vacuum arrives at the first port and one photon arrives at the second and an incredible number of modes-2, where only vacuum arrives. If you do the math for this scenario, you will find that the interference is gone.

If the BS works by reflection then there is a phase change for the refelected light and a factor of i is given to the second term in that transformation as I say in my first post. But polarization is the issue and the result that is stated without calculation in the textbook I cite is wrong. I can't find any other treatment of this calculation.

 

[Cthugha]

If the BS works by reflection then there is a phase change for the refelected light and a factor of i is given to the second term in that transformation as I say in my first post. But polarization is the issue and the result that is stated without calculation in the textbook I cite is wrong. I can't find any other treatment of this calculation.

I am not sure how this is relevant to what I wrote. The phase upon reflection and any complex phase factor enter trivially via the difference in t and t' (or r and r' respectively).

I also do not see how Gerry/Knight gets this wrong. They do not state what you claim they do. Horizontal and vertical polarization are two different modes. You cannot apply the BS transformation to two different modes. They give a single mode description. For example their equation (6.10)
$$\hat{a}_2=\frac{1}{\sqrt{2}}(\hat{a}_0+i\hat{a}_1), \hat{a}_3=\frac{1}{\sqrt{2}}(i\hat{a}_0+\hat{a}_1)$$
is valid for input modes 0 and 1 being the same single mode except for the position. They need to have the same energy, polarization and be the same in any other distinguishable degree of freedom. Otherwise the BS will not mix them. If you want a multi-mode treatment, you need to set up the BS transformation for every mode separately. For example, for horizontal and vertical modes, you will get:
$$\hat{a}_{2,h}=\frac{1}{\sqrt{2}}(\hat{a}_{0,h}+i\hat{a}_{1,h}), \hat{a}_{3,h}=\frac{1}{\sqrt{2}}(i\hat{a}_{0,h}+\hat{a}_{1,h}), \hat{a}_{2,v}=\frac{1}{\sqrt{2}}(\hat{a}_{0,v}+i\hat{a}_{1,v}), \hat{a}_{3,v}=\frac{1}{\sqrt{2}}(i\hat{a}_{0,v}+\hat{a}_{1,v})$$

For two photons with orthogonal polarizations, you have four input modes and four output modes and two of the input modes are vacuum. And to correctly transform this, you need to set up the BS transform e.g. using one horizontal photon at port 0 and vacuum for the horizontal mode at port 1 and transform this and then do the same again for a vertical photon at port 1 and vacuum for the vertical mode at port 0, transform this again and sum up the results.

There is no transform that goes as
$$\hat{a}_2=\frac{1}{\sqrt{2}}(\hat{a}_{0,v}+i\hat{a}_{1,h})$$,
which is what you would need to get to your scenario.

 

[Mentz114]

I am not sure how this is relevant to what I wrote. The phase upon reflection and any complex phase factor enter trivially via the difference in t and t' (or r and r' respectively).

I also do not see how Gerry/Knight gets this wrong. They do not state what you claim they do. Horizontal and vertical polarization are two different modes. You cannot apply the BS transformation to two different modes. They give a single mode description. For example their equation (6.10)
$$\hat{a}_2=\frac{1}{\sqrt{2}}(\hat{a}_0+i\hat{a}_1), \hat{a}_3=\frac{1}{\sqrt{2}}(i\hat{a}_0+\hat{a}_1)$$
is valid for input modes 0 and 1 being the same single mode except for the position. They need to have the same energy, polarization and be the same in any other distinguishable degree of freedom. Otherwise the BS will not mix them. If you want a multi-mode treatment, you need to set up the BS transformation for every mode separately. For example, for horizontal and vertical modes, you will get:
$$\hat{a}_{2,h}=\frac{1}{\sqrt{2}}(\hat{a}_{0,h}+i\hat{a}_{1,h}), \hat{a}_{3,h}=\frac{1}{\sqrt{2}}(i\hat{a}_{0,h}+\hat{a}_{1,h}), \hat{a}_{2,v}=\frac{1}{\sqrt{2}}(\hat{a}_{0,v}+i\hat{a}_{1,v}), \hat{a}_{3,v}=\frac{1}{\sqrt{2}}(i\hat{a}_{0,v}+\hat{a}_{1,v})$$

For two photons with orthogonal polarizations, you have four input modes and four output modes and two of the input modes are vacuum. And to correctly transform this, you need to set up the BS transform e.g. using one horizontal photon at port 0 and vacuum for the horizontal mode at port 1 and transform this and then do the same again for a vertical photon at port 1 and vacuum for the vertical mode at port 0, transform this again and sum up the results.

There is no transform that goes as
$$\hat{a}_2=\frac{1}{\sqrt{2}}(\hat{a}_{0,v}+i\hat{a}_{1,h})$$,
which is what you would need to get to your scenario.

OK, I've tried that with initial condition 1) one photon in each channel transformed separately and summing, and 2) one photon in each channel and the results depend only on whether we use
$BS_1=\left[ \begin{array}{cc} t & ir \\\ ir & t \end{array} \right]$ or $BS_2=\left[ \begin{array}{cc} t & r \\\ r & t \end{array} \right]$
If the BS transformation is altered to what you are using it becomes $\tfrac{1}{\sqrt{2}}(\hat{a}_{1H}^{\dagger} + \hat{a}_{2H}^{\dagger})(\hat{a}_{1H}^{\dagger} + \hat{a}_{1H}^{\dagger})$ and there is no interference - ever. Gerry&Knight use $BS_1$ in the polarized case and this is probably why their result is wrong.

What works is if we use $BS_1$ for one case and $BS_2$ in the other !

Thanks, that was helpful.