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Photon in cavity

  1. Jan 23, 2014 #1
    Hey,
    in my notes I have calculated the Eigenenergie of the Hamiltonian:
    [itex]

    H= \hbar \omega (n+\frac{1}{2}) \cdot

    \begin{pmatrix}
    1 & 0 \\
    0 & 1
    \end{pmatrix}

    +\hbar

    \begin{pmatrix}
    \frac{\Omega_0 -\omega_0}{2} & g \sqrt{n+1} \\
    g \sqrt{n+1}& -\frac{\Omega_0 -\omega_0}{2}
    \end{pmatrix}

    [/itex]

    We have just calculated:
    $$det\begin{pmatrix}
    \frac{\Omega_0 -\omega_0}{2} - \lambda & g \sqrt{n+1} \\
    g \sqrt{n+1}& -\frac{\Omega_0 -\omega_0}{2} - \lambda
    \end{pmatrix} => \lambda = \pm \sqrt{
    \frac{\Omega_0 -\omega_0}{4}+g^2(n+1)
    }$$

    And then we said the Eigenenergie is:

    [itex]E=\hbar \omega(n+\frac{1}{2}) \pm \hbar\sqrt{
    \frac{\Omega_0 -\omega_0}{4}+g^2(n+1)
    }[/itex]

    Why can I add [itex]\hbar \omega(n+\frac{1}{2})[/itex] to the result of the determinant?

    Or why is it possible to neglect the first term of the hamiltonian in the determinant?






    I also know that:

    [itex] [N,H]=[a^\dagger a + c_{1}^\dagger c_1 , H] =0[/itex]


    a is a photon annihilation operator and c a fermionic annihilation operator.

    The Hamiltonian in an other notation is [itex]H = \hbar \omega a^\dagger a + \frac{1}{2} \hbar \Omega(c^\dagger _1 c_1 -c^\dagger _0 c_0) +\hbar g(a c^\dagger _1 c _0 + a^\dagger c^\dagger _0 c_1)[/itex].

    I regard interaction of photons and fermions. So the states looks like [itex]|01>|n>[/itex] or [itex]|10>|n+1>[/itex] where n is the number of photons.

    Thank you very much!
     
  2. jcsd
  3. Jan 23, 2014 #2

    strangerep

    User Avatar
    Science Advisor

    That's not what's happening.

    What is ##\lambda##? Is it like E - <something>? :confused:

    The method for finding the energy eigenvalues ##E## is to find where the matrix ##H - E## is singular, i.e.,
    $$
    \det(H - E) ~=~ 0 ~.
    $$ So... work out this determinant explicitly and solve the resulting quadratic equation... :wink:
     
  4. Jan 24, 2014 #3

    Cthugha

    User Avatar
    Science Advisor

    That Hamiltonian is known as the Jaynes-Cummings Hamiltonian. The first term describes the field and the atom (or any other two-level system). The notation is somewhat non-standard though, but that does not matter. The second term describes the light matter coupling. The eigenstates of the first term are pretty easy to solve and of the type |n,e> or |n,g>. n is the number of photons, e and g mean that the atom is in the excited or ground state. Now the states of interest are those of the kind |n,e> and |n+1,g>. You either have n+1 photons and the atom is in the ground state or one photon gets absorbed and the atom is in the excited state. As these two states are degenerate for every n, it is sufficient to diagonalize the Jaynes-Cummings Hamiltonian (the second part about light matter interaction) in the subspaces given by these degenerate states, which gives you the result you get for every possible n.
     
  5. Jan 28, 2014 #4
    How to callculate the energy?

    But I don't understand why the energy can be callculated by:


    [itex]E=\hbar \omega(n+\frac{1}{2}) + \lambda
    [/itex]


    with


    [itex]

    det \lbrace



    \begin{pmatrix}
    \frac{\Omega_0 -\omega_0}{2}-\lambda & g \sqrt{n+1} \\
    g \sqrt{n+1}& -\frac{\Omega_0 -\omega_0}{2} -\lambda
    \end{pmatrix}
    \rbrace
    =0
    [/itex].



    Normally the Energy is callculated by:

    [itex]\det(H-\lambda \begin{pmatrix}
    1 & 0 \\
    0 & 1
    \end{pmatrix}) =0 [/itex]


    And normally: [itex]\det (A+B) \neq \det (A) +\det (B)[/itex]
     
  6. Jan 28, 2014 #5

    Cthugha

    User Avatar
    Science Advisor

    Remember that the operators/matrices are commuting. The spectral theorem helps here as commuting normal matrices can be diagonalized simultaneously. One can show that each eigenvalue of the sum of two commuting positive matrices is a sum of the eigenvalue of the two summands.

    If you want the "hard" math, you can find details in the standard book "Matrix Analysis" by Horn and Johnson. In the edition I have, you find the necessary details on page 51.
     
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