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Homework Help: 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.
     
  22. Mar 27, 2015 #21

    Orodruin

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    This is wrong. You can find instantaneous eigenstates of any Hamiltonian. This does not mean that an initial eigenstate will continue being in that eigenstate as the system evolves. There is an additional term related to the change in the eigenstates with time, but the OP has not gotten there yet.
     
  23. Mar 27, 2015 #22
    We must find a eigenstate which changes with time, it is true, but your procedure isn't true. Matrix form of H|a>=E|a> is deduced from time independent Schrödinger equation. if Hamiltonian be time independent we can find eigenstates from above equation. Here we have time dependent Hamiltonian and must use time dependent Schrödinger equation which i writed in previous post.
     
  24. Mar 27, 2015 #23

    Orodruin

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    No, this is still correct and is not dependent on the time independent Schrödinger equation to hold. It is the very definition of an instantaneous eigenstate. What is not correct is assuming that a state originally in the eigenstate ##|a(0)\rangle## at a later time will be in the eigenstate ##|a(t)\rangle## along with an additional phase. This is only true if the adiabatic approximation holds.
     
  25. Mar 27, 2015 #24
    Therefore there are two answer from two procedure and they must be equal. If you calculate them,you find this two answer don't match. Also in your procedure two eigenvalues are obtained, what is the meaning of this eigenvalues?
    Is this measurable?
     
  26. Mar 27, 2015 #25

    Orodruin

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    I believe you have a fundamental misunderstanding regarding what you call "my procedure". It so far has nothing to do with the time evolution, but only with findin the instantaneous eigenstates of the Hamiltonian. Do you understand that these are different things? Do you understand that what is typically called the "time independent Schrödinger equation" is simply an eigenvalue equation and a priori has nothing to do with the time evolution of states?
    You can perfectly well use the basis of instantaneous eigenstates to write down the time evolution of states (and it is often convenient to do so). If the Hamiltonian is time dependent it will not be diagonal, but that is a different issue that can be dealt with.
    And yes, the instantaneous Hamiltonian is a hermitian operator and you can create an observable from it. Naturally, this observable will depend on time, but that does not stop it from being observable. It corresponds to the instantaneous energy of the system. Not having constant expectation value of this observable is perfectly fine if the Hamiltonian depends on time.
     
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