1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Matrix Representation for Combined Ladder Operators

  1. Jan 12, 2016 #1
    Due to the definition of spin-up (in my project ),
    \begin{eqnarray}
    \sigma_+ =
    \begin{bmatrix}
    0 & 2 \\
    0 & 0 \\
    \end{bmatrix}
    \end{eqnarray}
    as opposed to
    \begin{eqnarray}
    \sigma_+ =
    \begin{bmatrix}
    0 & 1 \\
    0 & 0 \\
    \end{bmatrix}
    \end{eqnarray}
    and the annihilation operator is
    \begin{eqnarray}
    \hat{a} =
    \begin{bmatrix}
    0 & \sqrt{1} & 0 & 0 & \dots\\
    0 & 0 & \sqrt{2} & 0 &\dots\\
    0 & 0 & 0 & \sqrt{3} & \dots\\
    0 & 0 & 0 & 0 &\dots\\
    \vdots & \vdots & \vdots & \vdots&\ddots\\
    \end{bmatrix}
    \end{eqnarray}

    The matrix elememts of \begin{eqnarray} \hat{a}\hat{\sigma_+} \end{eqnarray} were given to me and are:

    \begin{eqnarray}
    \hat{a}\hat{\sigma_+} =
    \begin{bmatrix}
    0 & 0 & 0 & 0 & 0\\
    0 & 0 & 2\sqrt{1} & 0 &0\\
    0 & 0 & 0 & 0 &0\\
    0 & 0& 0 & 0& 2\sqrt{2} \\
    0 & 0& 0 & 0 &0\\
    \end{bmatrix} \end{eqnarray}

    From this I need to find out what the matrix elements of
    \begin{eqnarray}
    \hat{a^{\dagger}}\hat{\sigma_-}
    \end{eqnarray}
    are?

    I suppose the issue is I don't know how to represent the atomic raising/lowering operator for dimenstions greater than 2.
     
  2. jcsd
  3. Jan 12, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    For any operator ##\hat{A}##, you find its matrix representation in the basis ##|n\rangle## by calculating the matrix elements using
    $$
    A_{mn} = \langle m | \hat{A} | n \rangle
    $$
    You simply need to apply this to the ladder operators.
     
  4. Jan 12, 2016 #3
    As ##\hat{a}## in my case is larger than a 2*2 matrix how does one find the elements for ## \hat{\sigma}_+## for a matrix of dimensions greater than 2*2?
     
  5. Jan 12, 2016 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    What basis are you working with in your project?
     
  6. Jan 12, 2016 #5
    The Fock states
     
  7. Jan 13, 2016 #6
    The ordering of the basis is very important, especially when you are working with tensor product states as in here. I would hazard a guess that you are working on atom-light interactions in some sort of cavity? (like Jaynes-Cummings?)
    In that case, your basis is a composite of the Fock states of the photon field and the spin of the atom: [itex]\left|n\right\rangle \otimes \left|\sigma\right\rangle [/itex]. Once you determine how the basis in Eq. (5) are ordered, you should be able to get the matrix elements of [itex]\hat{a}^{\dagger} \sigma_{-}[/itex]
    (There is actually a very fast method of doing so if you can see the relationship between [itex]\hat{a}^{\dagger} \sigma_{-}[/itex] and [itex]\hat{a} \sigma_{+}[/itex])
     
  8. Jan 13, 2016 #7

    ##(\sigma_- a^{\dagger})^{\dagger} = a \sigma_+##
    and
    ##(\sigma_+ a)^{\dagger} = a^{\dagger} \sigma_-##

    but how does that give ##a^{\dagger} \sigma_-## from ##a \sigma_+##
     
  9. Jan 13, 2016 #8
    Given a matrix [itex]M[/itex], how is [itex]M^{\dagger}[/itex] related to it?
     
  10. Jan 13, 2016 #9
    ## (M^{\dagger})^{\dagger}= M##
    so from:
    ##(\sigma_+ a)^{\dagger} = a^{\dagger} \sigma_-##
    ##((\sigma_-)^{\dagger} (a^{\dagger})^{\dagger})^{\dagger} = a^{\dagger} \sigma_-##
    but then you just go in circles, I am I missing something?
     
  11. Jan 13, 2016 #10
    If [tex]
    M =
    \begin{bmatrix}
    A & B \\
    C & D \\
    \end{bmatrix},
    [/tex]
    what is [itex]M^{\dagger}[/itex]?
     
  12. Jan 13, 2016 #11
    I think the question assumes that I know the matrix elements of ##\sigma_+ ##, for a matrix of dimensions greater than 2*2 , which I don't.
     
  13. Jan 13, 2016 #12

    DrClaude

    User Avatar

    Staff: Mentor

    The ##\sigma_+ ## matrix you gave in the OP is 2×2, so there is something I don't understand. Could you give more details about what you are working on?
     
  14. Jan 13, 2016 #13
    I am looking at the dynamics between a qubit and an electromagentic field in a cavity. I need to find the expectation value of ##a^{\dagger} \sigma_-## by contracting the density matrix and taking the trace. The dimensions of density matrix, in the project, can vary depending how many levels of the qubit and and how many levels of the cavity is chosen. All I need, is to set up ##a^{\dagger} \sigma_-## and the rest is done.
     
    Last edited: Jan 13, 2016
  15. Jan 13, 2016 #14
    You already have the matrix elements for [itex]a \sigma_{+}[/itex], which you have recognised to be the Hermitian conjugate of [itex]a^{\dagger} \sigma_{-}[/itex].
    The "big" matrix you see for [itex]a \sigma_{+}[/itex] is not the matrix element of "[itex]\sigma_{+}[/itex] for more than 2 x 2 dimensions", which you seem to be confused about. The matrix given in the problem is the tensor product of [itex]a[/itex] and [itex]\sigma_{+}[/itex].
     
  16. Jan 13, 2016 #15
    That makes sense. Thanks
     
  17. Jan 13, 2016 #16
    So:
    \begin{bmatrix}
    0 & \sqrt{1} & 0 & 0 & \\
    0 & 0 & \sqrt{2} & 0 \\
    0 & 0 & 0 & \sqrt{3} \\
    0 & 0 & 0 & 0 \\
    \end{bmatrix} ## \otimes## \begin{bmatrix} 0 & 2 & \\ 0 & 0 & \\ \end{bmatrix} =

    \begin{bmatrix}
    0 & 0 & 0 & 2 \sqrt{1} & 0& 0 \dots \\
    0 & 0 & 0 & 0 & 0 & 0 \dots \\
    0 & 0 & 0 & 0 & 0 & 2\sqrt{2} \dots \\
    0 & 0 & 0 & 0 & 0 & 0 \dots \\
    \vdots & \vdots & \vdots & \vdots & \vdots&\ddots\\
    \end{bmatrix}
    which is not the same as eqn(5)?
     
    Last edited: Jan 13, 2016
  18. Jan 13, 2016 #17
    That is because the basis in Eqn (5) was reordered (for some unknown reason)
    The direct tensor product you got in your previous post has the basis order [itex]|0,+\rangle, |0,-\rangle,|1,+\rangle,|1,-\rangle \cdots[/itex], while the basis for Eqn (5) has the order [itex]|0,-\rangle, |0,+\rangle,|1,-\rangle,|1,+\rangle \cdots[/itex].
    Both are equally valid - as long as you keep proper track of the basis order.
     
  19. Jan 14, 2016 #18
    So
    ##a^{\dagger} \sigma_-##=

    \begin{bmatrix}
    0 & 0 & 0 & 0 & 0& 0 &\dots \\
    0 & 0 & 0 & 0 & 0 & 0 &\dots \\
    0 & 2\sqrt{1}& 0 & 0 & 0 & 0& \dots \\
    0 & 0 & 0 & 0 & 0 & 0 & \dots \\
    0 & 0 & 0 & 2\sqrt{2} & 0& 0 &\dots \\
    \vdots & \vdots & \vdots & \vdots & \vdots & \vdots&\ddots\\
    \end{bmatrix}
     
  20. Jan 15, 2016 #19
    Yup, that seems correct to me.
    It'll probably be worthwhile to spend some time reviewing / learning about tensor product spaces, since you seemed to be confused at some points.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted