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QM: Master equations and derivatives

  1. Jul 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Hi

    I have a technical question regarding the following paper: http://arxiv.org/pdf/quant-ph/0602170v1.pdf

    In it they derive an equation for [itex]\left\langle a \right\rangle[/itex] (equation #5) from the master equation (equation #2). My question is how they do this. Here is what I know
    [tex]
    \left\langle {\dot a} \right\rangle = \text{Tr}[a\dot\rho]
    [/tex]
    My problem comes when I have to evaluate this expression. The trace is to be taken over the system states, but what are these?

    I would be very happy to get a push in the right direction.

    Best,
    Niles.
     
  2. jcsd
  3. Jul 25, 2012 #2

    fzero

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    The system states here are the ground state ##|g\rangle## and the excited state ##|e\rangle##, but the expectation values are taken in the state defined by the density matrix:

    $$ \langle O \rangle_\rho = \mathrm{Tr}[\rho O].$$

    To obtain eq (5), we don't need to know the specific density matrix, though. We only need to manipulate the individual terms appearing in ##\mathrm{Tr}[ O \dot{\rho}]## to write them in terms of expectation values. Eq (5) is obtained by applying basic commutation relations to these expressions.
     
  4. Jul 26, 2012 #3
    Hi

    Thanks for taking the time to reply. I'm not still quite sure how to do this, but I'll try it out on the first two terms. Will you comment on my procedure? So the terms I am looking at are [itex]-i[H,\rho][/itex] and [itex]\kappa 2a\rho a^\dagger [/itex]. Starting with the equation for [itex]a[/itex] I get

    [tex]
    \text{Tr}(-ai[H, \rho]) + \text{Tr}(2\kappa a a\rho a^\dagger) \\
    \text{Tr}(-aiH\rho + ai\rho H) + \text{Tr}(2\kappa a a\rho a^\dagger)
    [/tex]
    Now I guess I need to insert the expression for H. But lets say I do that:

    The first term in H does not have a nonzero expectation value
    Nor does the second term
    The third term might as it contains [itex]aa^\dagger = 1+a^\dagger a[/itex]. So now this term says
    [tex]
    g_0 ((1+a^\dagger a)\sigma_- - aa\sigma_+)
    [/tex]
    I'm not sure how to evaluate the trace of these two terms. I would really appreciate a hint.

    Best,
    Niles.
     
  5. Jul 26, 2012 #4

    fzero

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    I'm afraid I was too quick before and misled you. The Jaynes-Cummings model describes both the two-level atom and the radiation bath that it's in. The states ##|g\rangle## and ##|e\rangle## describe the atom, but we should also specify the state describing the radiation field, which is just a harmonic oscillator state ##|n\rangle##. The total state is a tensor product of these, which we could write in terms of ##|n,g\rangle## and ##|n,e\rangle##.

    Now the density matrix has not been specified, so it can have off-diagonal terms ##\rho_{ng,me} |n,g\rangle \langle m,e|##. Expectation values like ##\langle a \rangle## (that would otherwise vanish in diagonal states) do not necessarily vanish in the presence of off-diagonal terms. So we should generally keep all possible terms.

    Getting back to ##\langle \dot{a}\rangle ##, the first term in

    $$\text{Tr}(-aiH\rho + ai\rho H) = i \left\langle (Ha - aH) \right\rangle $$

    is

    $$ i \Delta_c \left\langle (a^\dagger a a - a a^\dagger a) \right\rangle = - i \Delta_c \langle a \rangle,$$

    where we used the commutation relation ##[a,a^\dagger]=1##. This term is the one proportional to ##\Theta## in (5).

    The second term, proportional to ##\Delta_a## vanishes because ##[a,\sigma_\pm]=0##.

    It helps to write all similar terms down at once using the commutator. There's a term involving ##\langle [a^\dagger,a] \sigma_-\rangle## and a term with ##\langle [a,a] \sigma_+\rangle##. We don't need any details of the state to reduce these to expressions involving just ##\langle\sigma_\pm\rangle## .
     
  6. Jul 26, 2012 #5
    Thanks for all of that! Very helpful. I should be able to take it from here, however there is one thing I can't sort out: Generally [itex]\left\langle O \right\rangle = \text{Tr}[\rho O][/itex], but how do you know that
    $$\text{Tr}(-aiH\rho + ai\rho H) = i \left\langle (Ha - aH) \right\rangle $$
    is true?

    Best,
    Niles.
     
  7. Jul 26, 2012 #6

    fzero

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    We can cyclicly permute operators around inside the trace, so, for example,

    $$\text{Tr}(ai\rho H ) = i \text{Tr}(\rho H a) = i \left\langle H a\right\rangle .$$
     
  8. Jul 26, 2012 #7
    Ah, I see. I just looked it up in my linear algebra book, and it also says that (in a footnote.. ?!).

    Thanks.

    Best,
    Niles.
     
  9. Jul 26, 2012 #8
    I believe I have run into a problem. It is regarding the equation for ##\dot \sigma_-##, the second term of the commutator. So I have $$i\left\langle {H\sigma _ - - \sigma _ - H} \right\rangle $$ and the second term of this yields
    [tex]
    \Delta_a(\sigma_+\sigma_-\sigma_- - \sigma_-\sigma_+\sigma_-) = \Delta_a\sigma_z\sigma_-
    [/tex]
    but the paper says that it gives
    [tex]
    \Delta\sigma_-
    [/tex]
    Do you agree with me?

    Best,
    Niles.
     
  10. Jul 26, 2012 #9

    fzero

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    You're just missing one step. If we work in a basis where

    $$ \sigma_+ = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, ~~~\sigma_- = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},$$

    then

    $$ \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$

    and ##\sigma_z \sigma_\pm = \pm \sigma_\pm##.
     
  11. Jul 27, 2012 #10
    Ah, I see. Thanks for taking the time to help me.

    Best,
    Niles.
     
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