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Coherence in spontaneous radiation processes

  1. Mar 11, 2015 #1
    Hi, I am trying to understand what keeps getting referred to as "super radiance".
    So this has led me to R. H. Dickes 1954 paper "Coherence in spontaneous radiation processes". http://dx.doi.org/10.1103/PhysRev.93.99 .

    On the first page Dicke presents a "simple example", that is not simple at all.
    I will go through and break this down, and present my questions where they occur, and I hope that someone here may be able to assist in working out what it means.

    "Assume that a neutron is placed in a uniform magnetic field in the higher energy of the two spin states. In due course the neutron will spontaneously radiate a photon via a magnetic dipole transition and drop to the lower energy state. The probability of finding the neutron in its upper energy state falls exponentially to zero. "

    So far this makes sense.


    "If, now, a neutron in its ground state is placed near the first excited neutron (a distance small compared with a radiation wavelength but large compared with a particle wavelength and such that the dipole-dipole interaction is negligible), the radiation process would, according to the above hypothesis of independence, be unaffected. Actually, the radiation process would be strongly affected. The initial transition probability would be the same as before but the probability of finding an excited neutron would fall exponentially to one-half rather than to zero."

    OK what the..?A half you say?Things start to take a confusing turn now.

    "The justification for these assertions is the following: The initial state of the neutron system finds neutron 1 excited and neutron 2 unexcited. (It is assumed that the particles have nonoverlapping space functions, so that particle symmetry plays no role.) This initial state may be considered to be a superposition of the triplet and singlet states of the particles. "

    QUESTION 1: How is this a superposition of the singlet and triplet states? The set up of the situation explicitly tells us that we have neutron 1 in the excited state and neutron 2 in the ground state. Thus don't they have opposite spins and thus occupy the singlet state?

    QUESTION 2: Would a triplet state not require both neutrons to be in the same state, either ground or excited? Or have I been misunderstanding spin states?


    "The triplet state is capable of radiating to the ground state (triplet) but the singlet state will not couple with the triplet system."

    QUESTION 3: Is this saying that if both neutrons are excited, then they have the same spin and are thus in a total spin of 1 system and thus are in an excited triplet state, then they can both radiate and drop into a ground state where they are again both having the same spin and thus in the ground triplet?

    QUESTION 4: How can the singlet state not couple with the triplet system?
    Am I correct in interpreting this as saying that if one neutron is excited, then it CANNOT drop into a lower energy state as that would require going from a singlet( one excited and one in ground) to a triplet ( both in ground state)?

    "Consequently, only the triplet part is modified by the coupling with the field. After a long time there is still a probability of one-half that a photon has not been emitted. "

    QUESTION 5: Where does this 1/2 probability come from?

    "If, after a long period of time, no photon has been emitted, the neutrons are in a singlet state and it is impossible to predict which neutron is the excited one."

    QUESTION 6: If the nuetrons are in a singlet state, are they not both excited or in the ground state?

    " On the other hand, if the initial state of the two neutrons were triplet with s= I, m, =0 namely a state with one excited neutron, a photon would be certain to be emitted and the transition probability would be just double that for a lone excited neutron. Thus, the presence of the unexcited neutron in this case doubles the radiation rate."


    QUESTION 6: How can a state with one excited neutron be the triplet? It explicitly has the spin orientation of the two neutrons being opposite, -1/2 + 1/2 = 0 as far as I recall, which means they are in a singlet state. But they are apparently not in a singlet state.

    QUESTION 7: Why would the transition probability be just double that for a lone excited neutron?


    If anyone can please help me understand this I would be extremely grateful. I am so very very confused now.
     
  2. jcsd
  3. Mar 12, 2015 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Following the Pauli principle, the wave function for two identical fermions must be anti-symmetric upon interchange. The spin wave function must itself have a definite parity, meaning that there are two possible situations:

    symmetric
    $$
    | \uparrow \uparrow \rangle \\
    | \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle \\
    | \downarrow \downarrow \rangle
    $$
    or anti-symmetric
    $$
    | \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle
    $$
    So you see that the state ##| \downarrow \uparrow \rangle ##, which has no definite symmetry, is a superposition of the singlet state and one component of the triplet state.

    Not exactly. Assuming that spin-up is the lower energy state, then the ground state is ##|\uparrow \uparrow \rangle##, which is part of the triplet.

    The transition is dipole forbidden: to first order, there is no coupling between the electromagnetic field (vacuum in this case) and the spin, hence the total spin cannot change.

    No, because the state ##| \downarrow \uparrow \rangle ## (first neutron excited, second not) has some triplet character, so it can decay to ##|\uparrow \uparrow \rangle## through an allowed transition.

    This is all tied to way I wrote above. Write down ##| \downarrow \uparrow \rangle ## is terms of symmetrized spin states and calculate the probability of being in the singlet state, which can't decay.

    See above.

    I hope that part of this has already been clarified. The important point to note is that Dicke is making the distinction between the first neutron being excited and a neutron being excited, which is what you get when you consider the state mentioned, which corresponds to the middle state I gave above for the symmetric (triplet) state.

    I'll let you think about it in light of my answers above.
     
  4. Mar 13, 2015 #3
    Thank you for your reply. This is starting to make much more sense now.
    So for the first neutron in an excited state, with the second in the ground state, you have the state as you suggested:
    $$| \uparrow \downarrow \rangle = \frac{1}{\sqrt{2}} ( | \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle -| \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle) $$

    The probability of being in the singlet state is:

    $$ \frac{1}{\sqrt{2}} (\langle \uparrow \downarrow | - \langle \downarrow \uparrow | ) \frac{1}{2} [ ( | \uparrow \downarrow \rangle - | \downarrow \uparrow \rangle ) - (| \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle)] \\$$

    $$ \frac{1}{2\sqrt{2}} (
    \langle \uparrow \downarrow | \uparrow \downarrow \rangle
    - \langle \uparrow \downarrow | \downarrow \uparrow \rangle
    - \langle \uparrow \downarrow | \uparrow \downarrow \rangle
    - \langle \uparrow \downarrow | \downarrow \uparrow \rangle
    - \langle \downarrow \uparrow | \uparrow \downarrow \rangle
    +\langle \downarrow \uparrow | \downarrow \uparrow \rangle
    + \langle \downarrow \uparrow | \uparrow \downarrow \rangle
    +\langle \downarrow \uparrow | \downarrow \uparrow \rangle
    )
    \\ $$

    $$ \frac{1}{2\sqrt{2}} ( 1-0 -1-0-0+1+0+1) = \frac{1}{\sqrt{2}}\\ $$
    Which then gives you a probability of ## \frac{1}{2}\\ ##
    So good so far. However in the last paragraph that I posted, Dicke considers the initial state of a neutron being excited and in the triplet state.
    ##| \uparrow \downarrow \rangle + | \downarrow \uparrow \rangle \\ ## I agree that this state will then decay to the ##|\uparrow \uparrow \rangle ## state eventually, and that the probability of this decay would be twice that of the situation where you know specifically which neutron is excited, Just as you mention in your post. Because of the 50/50 probability that it is even in a triplet state from which it can decay.

    It seems to me that Dickes argument is saying that knowing which neutron is excited will half the transition rate. However Dicke then says that the transition probability would be double for a single lone neutron, which I do not understand at all.
     
    Last edited: Mar 13, 2015
  5. Mar 16, 2015 #4
    The knowledge of which neutron spin is in a particular state reduces the transition rate by 1/2 according to Dickes argument, but he claims that it increases the transition rate with respect to that of a single lone neutron. This can only be valid if the transition rate for a single neutron is that same as that of a pair of neutrons, where you know which neutron is in an excited state and which is not. This last part is then, seemingly, only valid when Dickes assumption that the neutrons are a distance far enough apart "such that the dipole-dipole interaction is negligible" is true.
     
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