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Eigenstates of hamiltonian

  1. Mar 22, 2015 #1
    1. The problem statement, all variables and given/known data
    I have a hamiltonian:
    [tex] \begin{pmatrix}
    a &0 \\
    0&d
    \end{pmatrix} + \begin{pmatrix}
    0 &ce^{i w t} \\
    ce^{-iwt}&0
    \end{pmatrix}=\begin{bmatrix}
    a & c e^{i w t} \\
    c e^{-i w t}&d \\
    \end{bmatrix}
    [/tex]
    Where the first hamiltonian can be labeled with states |1> and |2> corresponding to a and d.
    Now it says that it is in state |2> at time zero. Find the state at time t.
    2. Relevant equations


    3. The attempt at a solution
    I Found that looking at the first matrix ONLY, Its eigenstates are {{1},{0}} and {{0},{1}}
    but when you look for the eigenstates at t=0 from the total hamiltonian, you have to use quadratic formula and quickly gets messy.
    Im assuming I'm starting this all wrong
     
  2. jcsd
  3. Mar 22, 2015 #2

    Orodruin

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    It is a matter only of finding the eigenvectors of a 2x2 hermitian matrix, it should not be that messy. Why dont you show us what you get.
     
  4. Mar 22, 2015 #3
    [tex]H=\left(
    \begin{array}{cc}
    \text{a} & c e^{-i w t} \\
    c e^{i w t} & \text{d}
    \end{array}
    \right);
    [/tex]
    Then eigenvalues of:
    [tex]
    \text{$\lambda $1}=\frac{1}{2} \left(d+a-\sqrt{d^2-2a d+a^2+4 c^2}\right)\\
    \text{$\lambda $2}=\frac{1}{2} \left(d+a+\sqrt{d^2-2a d+a^2+4 c^2}\right)
    [/tex]
    Plugged those back in and solved and got an eigenvector elements of:
    [tex]
    \text{v1}\to -\frac{e^{-i t w} \text{v2} \left(d-a+\sqrt{d^2-2 a d+a^2+4 \epsilon ^2}\right)}{2 c}
    [/tex]
    for the first eigenvalue.
    So I set v2 = 1.
    [tex]
    V_1=\left(
    \begin{array}{c}
    -\frac{e^{-i t w} \left(d-a+\sqrt{d^2-2 a d+a^2+4 c^2}\right)}{2 c} \\
    1
    \end{array}
    \right)
    [/tex]
    The second eigenvector elements:
    [tex]
    \text{v3}\to \frac{e^{-i t w} \text{v4} \left(-\text{d}+\text{a}+\sqrt{\text{d}^2-2 \text{a} \text{d}+\text{a}^2+4c ^2}\right)}{2 c }
    [/tex]
    So i solved for this second eigenvector:
    [tex]
    V_2=\left(
    \begin{array}{c}
    \frac{e^{-i t w}\text{ }\left(-d+a+\sqrt{d^2-2 a d +a^2+4 c^2}\right)}{2 c} \\
    1
    \end{array}
    \right)

    [/tex]
    So, when i take the inner product with them, I get zero. Which is good since they are orthogonal.
    but how does this give me the state |2> at time zero??
     
  5. Mar 22, 2015 #4

    vela

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    It doesn't. {##\lvert 1 \rangle##, ##\lvert 2 \rangle##} are the eigenstates of the first Hamiltonian and form a basis. The vectors you found are representations of the eigenstates of the full Hamiltonian in that basis.

    Now why do you want to express a state in terms of the new eigenstates to find the time evolution of the state?
     
  6. Mar 22, 2015 #5
    Well, the problem says to find the state at time t>0.
    I know that it is initially in a state |2> at t=0.
     
  7. Mar 22, 2015 #6

    vela

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    What does that have to do with anything? I'm asking you what's the point of diagonalizing the full Hamiltonian. It's the right thing to do, but I don't get the feeling you know why you're doing it.
     
  8. Mar 22, 2015 #7
    The point is to get the correct eigenkets of the system given the hamiltonian.
    Then the eigenkets form a complete set and i can express any state as a linear combination of these eigenkets I found earlier
     
  9. Mar 22, 2015 #8

    vela

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    But you already had a perfectly good basis {|1>, |2>}. What's the reason for working in terms of the new basis instead of the old one?
     
  10. Mar 22, 2015 #9
    The old basis 1&2 where eigenstates of the first hamiltonian. Since we added in the new hamiltonian, It changed the eigenstates.
     
  11. Mar 22, 2015 #10

    vela

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    Sure. But you can still express any state in terms of the old basis |1>, |2> or in terms of the new basis |1'>, |2'>. What's the advantage of using the new basis?
     
  12. Mar 22, 2015 #11
    These states give us the energy eigenvalues possible from the full hamiltonian.
     
  13. Mar 22, 2015 #12

    Orodruin

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    Which is good for you because ...
     
  14. Mar 22, 2015 #13
    Not too sure?

    I know it gives us states that naturally describe the system.
    But i know that d & |2> ISNT an eigenvalue/state of the system. So how do these eigenstates allow me to say that it is initially in state two?
     
  15. Mar 23, 2015 #14

    Orodruin

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    How do states evolve with time?
     
  16. Mar 23, 2015 #15
    Well, with the Time evolution operator.
    When it acts on an eigenstate, it will simply replace the Hamiltonian in the exp to its corresponding eigenvalue.

    We use eigenstates so we can express the initial state as a linear combination of the two eigenstates.
    So, If I were to express the initial state as,
    [tex]
    \vert\alpha, t_{0} \rangle = a \vert\lambda_{1}\rangle +b \vert\lambda_[2]\rangle
    [/tex]
    We could solve for a and b since we know that it was initially in the state:
    [tex]
    \vert\alpha, t_{0} \rangle =\vert 2\rangle = \binom{0}{1}
    [/tex]
    Does that sound better?
     
  17. Mar 23, 2015 #16

    vela

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    Yup, you're on the right track.
     
  18. Mar 23, 2015 #17
    thank you guys.
    Im normally not this dense, I swear!
    Just been at it all week and I think my brain is fried..
     
  19. Mar 23, 2015 #18

    Orodruin

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    Just as a comment from someone who has been playing quite a lot with 2-state Hamiltonians: To make your states a bit easier to play with, split the Hamiltonian into a piece which is proportional to unity and a traceless part (the piece proportional to unity is just a shift of the eigenvalues). You can then introduce:
    $$
    a-d = E \cos(2\theta),\quad c = E \sin(2\theta)/2
    $$
    This will simplify your life considerably.

    Also, remember to normalise your states.
     
  20. Mar 27, 2015 #19
    Dear Orodruin

    We can't find eigenstates of a Hamiltonian depends on time. |t>=Htotal|2> in base of first Hamiltonian. <t|1> and <t|2> are probability amplitude of finding final state in |1> and |2> respectively.
     
  21. Mar 27, 2015 #20
    Oooooh
    Forgive me. I make a mistake in previous post. '|t>=Htotal|2>' is wrong we must replace 'i×h×(d/dt)|t>=H|t>' instead it and solve two coupled equations and with initial state find |t> in first Hamiltonian base.
     
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