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How to form a picture in mind of coherent state?

  1. Nov 17, 2011 #1
    Dear all:
    I knew the coherent state is the eigenstate of annihilation operator with a complex eigenvalue, which seriously disturbed me of thinking it. Can any one give me a simple and understandable picture of what is it, thanks a lot in advance.
     
  2. jcsd
  3. Nov 17, 2011 #2

    Cthugha

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    Science Advisor

    In simple terms:

    The photon annihilation operator corresponds to the destruction of a photon. So an eigenstate of the annihilation operator has the property of not being changed if a photon from the field is detected.

    Almost any common light field does not have a fixed photon number, but a fluctuating one. Let us now consider two extreme light fields having the same mean photon number. One sends out exactly one photon per second without any noise. The other sends out a packet of 10 photons randomly once inside every interval of ten seconds. Both have the same mean photon number, but show very different behavior with respect to photon detections. If you detect a photon from the source emitting one photon per second, the detection event changes the light field, so right at that moment there will be zero photons left and the probability to detect another photon has changed to 0. If you detect a photon from the noisy source sending out the packets, you know that you are right now at the moment where many photons are emitted. Therefore the probability to detect more photons at the same time will increase. Both states are obviously not eigenstates of the annihilation operator as the detection of a photon decreases or increases the probability to detect the next one which puts the light field into a different state.

    Coherent states are now found in the middle. When starting from the noiseless source and adding a bit of noise, the behavior of the field with respect to detection events will change from decreased probability to detect more photons to an increased probability to detect more photons as I explained above. At some intermediate amount of noise, this probability will neither be enhanced, nor decreased. This is exactly the amount of noise present in a coherent state and as the detection of one photon does not change the probability to detect more of them, this will be an eigenstate of the annihilation operator.

    The right amount of noise is by the way present if the photon number distribution is Poissonian.
     
  4. Nov 17, 2011 #3
    thank you very much indeed, and your answer is absolutely fantastic! I am very appreciated it, people like you make the forum a great place!
     
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