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Hong-Ou-Mandel effect

  1. Feb 15, 2012 #1
    I've been reading about this two-photon interference experiment


    But the wikipedia article does not explain the origin of the "-" sign before states 3 and 4. Just because one photon reflecting off the bottom side changes its phase, how does it follow that states 2 and 3 cancel each other? The transmitted photon in state 2 still has the same phase as the photon reflected off the upper side in state 3, so why would the states cancel completely? And why does it take only one photon from the pair changing its phase for the "-" sign to appear in front of the entire 2-photon state?
    (As a side note, isn't it true that if the number of photons is known, 2 photons in this case, then phase is indeterminate anyway?)
  2. jcsd
  3. Feb 16, 2012 #2


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    Let me start at the end.

    Well, the phase is indeterminate, but the phase difference between indeterminate and indeterminate - pi/2 will still always be - pi/2.

    For an ideal beam splitter the transmitted and the reflected beam will always be 90 degrees out of phase. This is true for any entrance port and a necessary result to ensure conservation of energy. This is just the same as in a RF coupler.

    One can just represent this by assuming a complex factor t for a transmission process and a complex amplitude i*r for a reflection process. The imaginary part ensures the 90 degree phase shift. For a 50-50 beam splitter these are normalized such that |t|^2 =0.5 and |r|^2=0.5, so that a single photon has a 50/50 probability of being transmitted or reflected.

    Assuming one photon arriving at each entrance port of the beam splitter, we have 4 possibilities: Both are transmitted, both are reflected, photon 1 is transmitted, while photon 2 is reflected and photon 2 is transmitted, while photon 1 is reflected, giving us the 3 possible output states |1,1>, |0,2> and |2,0>.

    Now one can calculate the probability amplitudes for each of these three outcomes.

    Considering two transmission processes, the outcome will be |t|^2 |1,1>. Considering two reflection processes, the outcome will be i |r| i |r| |1,1>=-|r|^2 |1,1>. The cases involving one reflection and one transmission event are i |r t| |0,2> and i |r,t| |2,0>, respectively. The output state is now just the sum all of these parts:
    |out>=(|t|^2 -|r|^2) |1,1> +i |r t| (|2,0>+|0,2>).

    You see that for a 50/50 beam splitter where |t|^2=|r|^2=0.5 the difference in the first term amounts to zero and can be interpreted as destructive interference. So the remaining output state is:
    |out>= sqrt(2)*i |r t| (|2,0>+|0,2>)
    where the sqrt(2) ensures normalization to unity.
  4. Feb 16, 2012 #3
    Thanks, Cthugha, the math works just fine. Should we use a similar equation if both photons arrive from the SAME direction? Then two transmissions or two reflections will mean that both photons will emerge together from the same port, and if they emerge from different ports, it means one was reflected and the other transmitted. When I tried to add amplitudes for this case, I got

    (-1) |r|^2 |2,0> + |t|^2 |0,2> + 2 i |r t| |1,1>

    Is that correct? It seems to imply that 2 times out of 3 the photon pair will split, and 1 time out of 3 the photon will emerge together?
  5. Feb 16, 2012 #4


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    You have a factor of two in the final term which means that 50% (2 out of 4 times) of the time the pair will split and 25% of the time both photons will leave via exit port A and 25% of the time both photons will leave via exit port B. This is pretty much what one would expect naively.

    If you have access to the Mandel/Wolf (Optical coherence and quantum optics), you will find that it devotes a whole subchapter to this kind of problem. It is called "Effect of an attenuator or beam splitter on the quantum field". If you have access to this book and are interested in that topic, it is really worth reading.
  6. Feb 16, 2012 #5
    But from normalization, |-1|^2 + |1|^2 + |2i|^2 = 6, so after dividing by sqrt 6, the amplitude of |1,1> term becomes 2i/sqrt 6 and the respective probability 4/6 = 2/3?
  7. Feb 16, 2012 #6


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    Oh, sorry. I misread. I thought you had already written the squared values last time.

    You have to square the two possibilities to arrive at the |1,1> situation sperately instead of adding them up before squaring. You only add up the probability amplitudes for indistinguishable events.
  8. Feb 16, 2012 #7
    That's my understanding as well. But since the two possibilities r,t and t,r both result in the same |1,1> state, how can we distinguish them? One photon exits port 1, and another identical photon exits port 2. What will change if we swap the photons?
    So why cannot we add these two amplitudes, just like we did for the original HOM effect?
  9. Feb 16, 2012 #8


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    You have no indistinguishable events because loosely speaking you just have one initial state to start from.

    You just have one two-photon state arriving at the beam splitter. To get indistinguishable events you would need to have two different states at some point. You cannot break down the initial two-photon state into two seperate photons. This is why swapping the photons does not make any sense as it does not create a different event.

    I must admit that it was maybe irritating to say "You have to square the two possibilities to arrive at the |1,1> situation sperately instead of adding them up before squaring." as this is purely mathematical and maybe misleading. If you do the whole math, you will find that any Fock state arriving at a single port of the beam splitter behaves as expected classically if you just have the vacuum field at the other entrance port.
  10. Feb 16, 2012 #9
    Your point is well taken. But how should I then write the output state? If I count the two |1,1> outcomes as distinguishable events, and write them separately,

    (-1) |r|^2 |2,0> + |t|^2 |0,2> + i |r t| |1,1> + i |r t| |1,1> (1)

    then elementary algebra will make that form identical to

    (-1) |r|^2 |2,0> + |t|^2 |0,2> + 2 i |r t| |1,1> (2)

    and we still don't have the correct expression. Does your answer imply that rather than two distinguishable events, I should instead count them as one and the same event? But then I must drop the last term in (1) and we still don't have anything resembling the classical outcome. What we really need to get the "classical" statistics of outcomes is sqrt 2 coefficient so that

    (-1) |r|^2 |2,0> + |t|^2 |0,2> + sqrt 2 i |r t| |1,1> (3)

    But is there a way to get this by adding amplitudes, or does one have to derive it in terms of creation operators (photon algebra)? Can you suggest any short primer on this topic aside from the Mandel-Wolf textbook?

    Also, isn't a Fock state |2> equivalent to the product state |1>|1>?
  11. Feb 16, 2012 #10


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    The best way is always doing this in operator algebra. This will also give the correct amplitudes for any case. I think I can give you some more details tomorrow as it is already quite late and I do not have any good reference at hand right now.

    What makes you think that |2> is identical to |1>|1>? The latter is a product state and this notation usually refers to a product state of two states having occupation n=1 each.
  12. Feb 16, 2012 #11
    Thanks, I will be looking forward to more details tomorrow. Also I wondered, what would happen if these two input photons were not identical (for example, had different frequency or polarization). Would this input Fock state |1v,1h> also behave "classically" at the beam splitter?
  13. Feb 17, 2012 #12


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    If these photons are distinguishable, you will not see an interference effect at the beam splitter. This is why the HOM-dip can be used for example to check whether single photon sources produce indistinguishable photons.

    Returning to the original question. Let us assume input modes 0 and 1 and output modes 2 and 3. Assuming the general case of any beam splitter we have reflectivity r' and t' for mode 0 and r and t for mode 1. Now we also know the simple commutator relations:
    [tex][a_0,a_0^\dagger]=[a_1,a_1^\dagger]=1[/tex] and

    One can show that the relations also hold for the output modes for which the operators are given by:

    [tex]a_2=t' a_0+r a_1[/tex] and
    [tex]a_3=r' a_0 + t a_1[/tex].

    Now one can calculate the average photon numbers and the correlators of the output photon numbers. The photon numbers are straightforward:

    [tex]n_2=a_2^\dagger a_2=(t^{'*} a_0^\dagger + r^* a_1^\dagger)(t^' a_0 + r a_1)= |t|^2 n_0 + |r|^2 n_1 + t^{'*}r a_0^\dagger a_1 + r^*t^' a_1^\dagger a_0[/tex]

    For output mode 3 just exchange any r and t in the above expression.
    The next thing is the probability to detect a photon at both output ports simultaneously. This is:
    [tex]P_{23}=\langle a_2^\dagger a_3^\dagger a_3 a_2\rangle=\langle n_2 n_3\rangle[/tex]. From here on things become quite lengthy, but you just need to insert the right expressions. If input 0 is the vacuum state, things become a bit easier. Inserting all the stuff just gives:
    [tex]\langle n_2 n_3 \rangle= |r|^2 |t|^2 \langle n_1^2\rangle - |r|^2 |t|^2 \langle a_1^\dagger a_0 a_0^\dagger a_1 \rangle[/tex] which is equivalent to
    [tex]|r|^2 |t|^2 (\langle n_1^2\rangle - \langle n_1 (a_0^\dagger a_0 +1) \rangle)= |r|^2 |t|^2 (\langle n^2\rangle - \langle n_1 \rangle)[/tex]
    That is 0.5 for a n=2 Fock state, just as one would expect classically. One can now also evaluate the correlation
    [tex]\langle \Delta n_2 \Delta n_3 \rangle=\langle n_2 n_3 \rangle-\langle n_2\rangle \langle n_3 \rangle[/tex]
    as this quantity tells as bit more about the correlations present.
  14. Feb 19, 2012 #13

    Here is why I think they mean the same thing. The way I understand the definition, a multi-particle state of the form |n1,n2,n3...> denotes a state where there are n1 particles in one state, n2 in another etc. etc. Once we define |n> as the state where there are n particles, each in that same single particle state, it follows that |n> is the product state |1>|1>....|1>. If it is not, then of course we should be able to detect correlations between particles, but then it will no longer be a simple Fock state |n>, but rather a general Fock state of the form sum over k1, k2 of c_{k1,k2}|k1>|k2>. Such a state will be entangled since we cannot factorize it, but it would be wrong to write it simply as |n> because then we don't have n particles is the SAME one-particle state (entangled particles don't know what their state is!). So we must just leave it in that form, sum over k1, k2 of c_{k1,k2}|k1>|k2>.

    So my understanding was that there are two types of Fock states: simple Fock state |n> which is a product |1>...|1>, and a general Fock state which is entangled, written as a superposition of product terms, and applies only when the particles are known to differ by some parameter (polarization, momentum, etc.), and that would also make such particles distinguishable.

    Can you tell me if my understanding is wrong?
  15. Feb 20, 2012 #14


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    I am still not sure I get your point, but let me try to get it.

    To have a state of the form |1>|1>....|1> you would really need many different modes with occupation of one each. The Fock state is rather a |n>|0>....|0> -like state. Maybe discussion gets easier if you tell me where you got your information from. Is it by chance the wikipedia entry on Fock states or quantum field theory?

    Well, states resulting from SPDC for example are indeed a bit different as they offer phase-sensitive correlation functions if that is what you mean. I would not describe these states really as Fock states. However, they are expressed best in a Fock state basis.
  16. Feb 20, 2012 #15
    This makes sense. Actually, I just realized that the wikipedia article implies your definition, once we define a Fock state as the eigenstate of the number operator. So it makes sense that |1>|1>...|1> (or more generally, |n1>|n2>...|nk>) would just be a multimode Fock state, and then SPDC processes would be described in terms of Fock states would not themselves be called Fock states. Anyway, this looks like a terminological difference. But this still doesn't explain to me what, if anything the "bona fide" Fock state such as |2> tell us about the individual photons in this state.

    If they are identical, and there are certainly two of them, then each of them should know its state? (Will the state should be defined by the mode associated with the creation operator, (a+)^2 that created the |2> state?) On the other hand, when each particle possesses its own wave function, this should give us grounds to write |2> as a product state? I don't see how one can avoid that conclusion (unless one claims that the two photons in |2> form an entangled pair, and writes the corresponding combination of operators such as
    [(a+)(b+) - (b+)(a+)] acting on |0,0>).
  17. Feb 20, 2012 #16


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    Yes, basically a |2> state is what comes out if you apply the same creation operator of some mode twice to the vacuum.

    Wave functions for photons are a major problem. For example the mere fact that detection destroys a photon makes the definition of position eigenstates problematic to impossible. While there are some approaches, there is no generally accepted mainstream way to formulate wave functions for photons. Accordingly, I do not think that it is appropriate to think of a n=2 Fock state as just some sum of two independent particles. It is somewhat like "more is different" holds true for photons, too. Also, things like bosonic final state stimulation and similar effects are quite hard to explain in terms of particles keeping some individual nature.
  18. Feb 20, 2012 #17
    But in most standard situations a wave function is defined in terms of amplitudes of possible experimental outcomes. So shouldn't this be true for multi-photon states as well?

    Suppose we have a bi-photon |2V,0H> i.e. both photons are vertically polarized, and none is polarized horizontally. Can Quantum Optics predict the amplitude of all 4 possible pairs of outcomes RR, LL, RL, LR if one measures both photons in the same (e.g. circular) basis?
  19. Feb 20, 2012 #18


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    Sure, probability amplitudes just work fine. However, these just give you probabilities for processes and outcomes of measurements, but no eigenstates.

    Yes, sure. Outcomes of measurements are no problems.

    Just one more issue I just noticed where I am not sure whether the point is clear or not. The photons in a Fock state are not entangled, but that does not mean that they are statistically independent. Even photons in classical states are not necessarily statistically independent.
  20. Feb 20, 2012 #19
    Just for curiosity, what are those probabilities equal to? Are they 25% each as one would expect if the photons were totally independent?

    I think I can understand the bunching of photons (at least in a thermal source) and antibunching (e.g. resonance fluorescence) and can see that it does not involve any entanglement. I just can't see what would cause such behavior in a Fock state (whether single-mode or multimode).
  21. Feb 20, 2012 #20


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    That may depend on how you measure it. However, a "simple" and straightforward measurement would typically give classically expected values.

    Hmm, but the case of a general Fock state is not that different from resonance fluorescence. Resonance fluorescence gives you the n=1 Fock state and g2=g3=...=gn=0. The n=2 Fock state will give you g2=0.5 and g3=...=gn=0. The n=3 Fock state will give you g2=0.6666666, g3=2/9, g4=0,...gn=0 and so on.

    There is not really a conceptual difference between resonance fluorescence and Fock states of higher photon number.
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