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Jaynes-Cummings Hamiltonian

  1. Aug 18, 2012 #1
    Hi

    I have a question regarding the pump-term in the Hamiltonian on page 9 (equation 2.9b) of this thesis: http://mediatum2.ub.tum.de/download/652711/652711.pdf. This term is not very intuitive to me. [itex]a[/itex] is followed by a CCW phase, whereas [itex]a^\dagger[/itex] is followed by a CW phase. How does this represent pumping? I have seen many thesis presenting this term, but no one ever explains it.

    Best,
    Niles.
     
  2. jcsd
  3. Aug 18, 2012 #2

    DrDu

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    If you neglect the sigma containing terms, the Hamiltonian with the source term can easily be diagonalized. The new ground state of the photon field will now contain a finite number of photons.
    That's what you want: perturbations about a state containing a finite number of photons.
    You may also regard it as kind of a chemical potential term for the photons to fix the ground state expectation value.
     
  4. Aug 18, 2012 #3
    Hi

    Thanks for replying. I think there is a misunderstanding. My question is solely regarding equation (2.9b) on page 9 -- what is the physical argument for writing the pump term like that?

    Best,
    Niles.
     
  5. Aug 18, 2012 #4

    DrDu

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    I think I have answered that question.
     
  6. Aug 18, 2012 #5

    Physics Monkey

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    Hi Niles,

    If I may try to clarify, what DrDu is saying is that you can understand the added term, in the absence of coupling to the atom, as forcing the ground state of the EM field to have non-zero number of photons. For example, if you ignore the time dependence for a moment, and convert back to x- and p-like variables you will see that the Hamiltonian is [itex] H \sim p^2 + x^2 + \eta x [/itex] and hence has a new ground state at [itex] x \sim - \eta [/itex]. Since the electric field is [itex] E \sim a + a^\dagger \sim x [/itex] one sees that the ground state of the pumped Hamiltonian has [itex] E \sim \eta [/itex] i.e. a non-zero field. You should be able to fill in the details from here.

    Here is another perspective on the origin of the pump term. Think about the source of the pump field. The source may be a collection of other atoms being driven by some source of energy or whatever. However, the important thing is that the source couples to field mode described by [itex] a [/itex] (since it can produce photons of that frequency). Thus think about the simplest coupling of a source atom to the field, something like [itex] U \sim e x \cdot E [/itex] (U is the potential energy, x is the electron coordinate). If we now imagine that this atom, along with many others, is being driven coherently by the internal mechanisms of the laser, then it would be reasonable to replace the dipole moment operator [itex] e x [/itex] by a classical expectation value oscillating at the laser frequency. Now if you look at the energy U in some kind of rotating wave approximation (again, I've not been careful with all the details), and if you replace the atom variables with classical oscillating expressions, you will see that the remaining dependence on the EM field variables is precisely the pump term you are interested in.

    Hope this helps.
     
  7. Aug 19, 2012 #6
    I think I understand it now, sorry about my previous post then. OK, so I write the Hamiltonian as
    [tex]
    H = \hbar \omega_c a^\dagger a + \hbar \eta (a^\dagger + a)
    [/tex]
    and this I want to diagonalize. If I want to write it in matrix form, then I need to find the eigenstates. I guess these are just the typical number states [itex]\left| n \right\rangle[/itex], but there are infinitely many of these. But by acting with [itex]\left| n \right\rangle[/itex] from right and [itex]\left\langle n \right|[/itex] from left I see that the diagonal elements are proportional to [itex]\hbar \omega_c + \hbar \eta[/itex]. All the off-diagonal elements must be zero, so I guess this constitutes the diagonalization. Is this correct reasoning?


    Hi, thanks for that. I tried doing something similar above, but your procedure is probably a little more professional than mine. I don't know how you can see that [itex] H \sim p^2 + x^2 + \eta x [/itex] has the ground state [itex] x \sim - \eta [/itex], but if that really is the case, isn't it sufficient to stop here? I mean, you've already shown that the pump term has shifted the ground state, and I guess that is what characterizes a pump.

    Thanks for the other explanation, that makes it intuitively clear.

    Best,
    Niles.
     
  8. Aug 19, 2012 #7

    DrDu

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    I set hbar=1 and get
    [tex]
    H = \omega a^\dagger a + \eta (a^\dagger + a)= \omega( (a^\dagger+\eta/\omega)(a+\eta/\omega))-\eta^2/\omega=\omega b^\dagger b -\eta^2/\omega
    [/tex]
    The constant term is not important. Hence the new ground state is an eigenstate [itex] b|0>=0[/itex] or [itex] a|0>=-\eta/\omega |0>[/itex]. That is the definition of a coherent state.
    Hence particle number is not fixed in this new ground state but only it's average [itex]<n>=-\eta/\omega[/itex].
     
  9. Aug 19, 2012 #8
    Thanks for that, but something is unclear to me. So we have to look at [itex]b|0>[/itex], and we get
    [tex]
    b|0> = (a+\eta/\omega)|0> = \eta/\omega|0>
    [/tex]
    So indeed the new groundstate has energy [itex]\eta/\omega[/itex]. But I don't see why the constant term in the Hamiltonian is unimportant - is it just because it shifts the total energy?

    Best,
    Niles.
     
  10. Aug 19, 2012 #9

    DrDu

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    Yes, it only shifts the total energy. But a|0> is not zero. The ground state |0> of the b operators is different from the ground state of the a operators. You should look up the concept "coherent state".
     
  11. Aug 19, 2012 #10
    I've heard of coherent states before, defined as eigenstates of the creation/annihilation operators. But we defined b in terms of a,
    [tex]
    b \equiv a + \eta/\omega
    [/tex]
    so
    [tex]
    b|0> = (a+\eta/\omega)|0> = \eta/\omega |0>
    [/tex]
    How can this be wrong?
     
    Last edited: Aug 19, 2012
  12. Aug 19, 2012 #11

    Cthugha

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    If I am not off completely, it is [itex]<n>=(-\eta/\omega)^2[/itex].
    Although that may be obvious, it may help the thread opener to explicitly state that |0> is not the zero photon Fock state in this case, but the system ground state - typically the vacuum.

    That is correct if you choose to define the ground state in terms of a as zero. Typically it is rather the vacuum expectation value. If you have a look at the Wigner functions of the light field, you will find that the two fields according to a and b will look similar, but will be displaced by an amount of [itex]\eta/\omega [/itex]. This displacement is just the amplitude of the coherent field which you now have. See for example:
    http://iqis.org/quantech/wiggalery.php. Wigner functions give quite an intuitive explanation why vacuum states are coherent states, too and also give an intuitive picture for a coherent pumping process.
     
  13. Aug 20, 2012 #12

    DrDu

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    Oh, yes, if you say so it is certainly correct.
    As regards the ground states, if [itex] a|zero>=0 [/itex] and [itex]b|0>=0[/itex], then [itex] |0>=\exp(-|\eta/\omega| a^\dagger) |zero> [/itex], up to normalization.
     
  14. Aug 20, 2012 #13
    This part is a little abstract to me. The Hamiltonian describes the same system, diagonalized or not. Diagonalizing is just writing it in a different basis. So it doesn't make sense to me that the two ground states are different.

    Another thing, how do I know that [itex]b|0> = 0|0>[/itex] is the groundstate and not [itex]b|20> = 20|20>[/itex]? If we can just see from the diagonalized Hamiltonian that zero-amplitude coherent states are the ground state, then wouldn't it be sufficient to just diagonalize it and see that the pump adds [itex]-\eta^2/\omega[/itex] to it? In other words, wouldn't the "proof" just stop there?

    Best,
    Niles.
     
  15. Aug 20, 2012 #14

    DrDu

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    Come on, b and b^\dagger fulfill the same commutation relations as do a and a^\dagger. If you accept that a|zero>=0 defines the ground state of a^\dagger a , then b|0> will define the ground state of b^\dagger b.
     
  16. Aug 23, 2012 #15
    You are right, thanks for that and thanks to all for the help.

    Best,
    Niles.
     
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