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Particle with spin 1 in a non perturtuvative interaction

  1. May 31, 2013 #1
    Hi, i have been trying to figure how to solve this problem, i just don't know how to proceed

    1. The problem statement, all variables and given/known data
    A particle with spin s=1 has due to rotational invariance all three states [itex]|s_z\rangle=1[/itex], [itex]|s_z\rangle=0[/itex] and [itex]|s_z\rangle=-1[/itex] degenated with energy [itex]E_0>0[/itex]. Then its introduced an interaction non perturbative V

    [itex]V=-\Delta\left(\begin{array}{ccc}
    1 & 1 & 1\\
    1 & 1 & 1\\
    1 & 1 & 1
    \end{array}\right)
    [/itex]

    a) Obtain stationary states and energies of the system
    b) If in [itex]t=0[/itex], the particle has its third component with spin [itex]+\hbar[/itex] wich is the probability that after some time its [itex]-\hbar[/itex]


    2. Relevant equations



    3. The attempt at a solution
    a)
    So what i tried is this, since all three states have the same energy and those are eigenstates of the hamiltonian, the hamiltonian must have the form

    [itex]H=E_0\left(\begin{array}{ccc}
    1 & 0 & 0\\
    0 & 1 & 0\\
    0 & 0 & 1
    \end{array}\right)
    [/itex]

    So if we add H'=H+V and diagonalize we get

    [itex]\left|\begin{array}{ccc}
    E_0-\Delta-\lambda & -\Delta & -\Delta\\
    -\Delta & E_0-\Delta-\lambda & -\Delta\\
    -\Delta & -\Delta & E_0-\Delta-\lambda
    \end{array}\right|
    [/itex]
    So the characteristic equation is
    [itex](E_0-\lambda)(E_0-3\Delta-\lambda)=0[/itex]
    So we get the energies and the states
    [itex](E_0,{-1,0,1})[/itex]
    [itex](E_0,{-1,1,0})[/itex]
    [itex](E_0-3\Delta,{1,1,1})[/itex]

    I just want to know if this is the correct move or not

    b)
    i don't know how to aproach this, but i guess all i have to do is add the evolution term on each eigenstate [itex]e^{-i E_0 t/\hbar}[/itex] and do [itex]v^\dagger v[/itex]

    Thanks for your help :)
     
    Last edited: May 31, 2013
  2. jcsd
  3. May 31, 2013 #2

    TSny

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    Yes, that looks good!

    I think you might have the right idea. Can you "expand" the ##|s_z=+\hbar \rangle## state in terms of your eigenstates of the full Hamiltonian?
     
    Last edited: May 31, 2013
  4. May 31, 2013 #3
    I am lost on this, what i think may work is this
    I have to solve the equation

    [itex]i\hbar\frac{\partial}{\partial x} \chi = H \chi [/itex]

    where
    [itex]
    \chi=\left(\begin{array}{ccc}
    \alpha(t) \\
    \beta(t) \\
    \gamma(t)
    \end{array}\right)
    [/itex]

    The problem with this is that this gives a really complicated result with the system
    [itex]
    i \hbar \dot{\alpha}(t)=(E_0-\Delta)\alpha(t)-\Delta(\beta(t)+\gamma(t))
    [/itex]
    [itex]
    i \hbar \dot{\beta}(t)=(E_0-\Delta)\beta(t)-\Delta(\alpha(t)+\gamma(t))
    [/itex]
    [itex]
    i \hbar \dot{\gamma}(t)=(E_0-\Delta)\gamma(t)-\Delta(\beta(t)+\alpha(t))
    [/itex]

    And this seems wrong to me

    Thanks for help :)
     
  5. May 31, 2013 #4

    TSny

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    At time t = 0, you know the state is ##|\psi(0)\rangle = |s_z = +\hbar\rangle##. How would you represent this state as a column vector?

    Can you expand this column vector in terms of the eigenvetors
    [itex]|u1\rangle =
    \left(\begin{array}{ccc}
    -1 \\
    0 \\
    1
    \end{array}\right)
    [/itex] [itex],|u2\rangle =
    \left(\begin{array}{ccc}
    -1 \\
    1 \\
    0
    \end{array}\right)
    [/itex] [itex],|u3\rangle =
    \left(\begin{array}{ccc}
    1 \\
    1 \\
    1
    \end{array}\right)
    [/itex] ?
     
  6. May 31, 2013 #5
    So if if we normalize the vector we have

    [itex]
    |u_1\rangle = \frac{1}{\sqrt{2}} \left(\begin{array}{ccc} -1 \\ 0 \\ 1\end{array}\right)
    [/itex]

    [itex]
    |u_2\rangle = \frac{1}{\sqrt{2}} \left(\begin{array}{ccc} -1 \\ 1 \\ 0\end{array}\right)
    [/itex]

    [itex]
    |u_3\rangle = \frac{1}{\sqrt{3}} \left(\begin{array}{ccc} 1 \\ 1 \\ 1\end{array}\right)
    [/itex]
    so [itex]a|u_1\rangle+b|u_2\rangle+c|u_3\rangle=\left(\begin{array}{ccc} 1 \\ 0 \\ 0\end{array}\right)[/itex]

    gives us a system, when we solve this we have the vector

    [itex] |s_z{_h}\rangle=\left(\begin{array}{ccc} \frac{-\sqrt{2}}{3} \\ \frac{-\sqrt{2}}{3} \\ \frac{1}{3}\end{array}\right)
    [/itex]

    Now its turn for the time evolution
    [itex]|s_z=\hbar\rangle=\left(\begin{array}{ccc} \frac{-\sqrt{2}}{3}\exp{\frac{-i E_1 t}{\hbar}} \\ \frac{-\sqrt{2}}{3}\exp{\frac{-i E_2 t}{\hbar}} \\ \frac{1}{3}\exp{\frac{-i E_3 t}{\hbar}}\end{array}\right)
    [/itex]

    Since i am being asked for [itex]-\hbar[/itex], i think i need to do this with that vector

    [itex]|s_z=-\hbar\rangle=\left(\begin{array}{ccc} \frac{2\sqrt{2}}{3}\exp{\frac{-i E_1 t}{\hbar}} \\ \frac{-\sqrt{2}}{3}\exp{\frac{-i E_2 t}{\hbar}} \\ \frac{1}{3}\exp{\frac{-i E_3 t}{\hbar}}\end{array}\right)
    [/itex]

    Now ot obtain the evolution with time i have to get the expectation value i assume

    [itex] \langle s_z\rangle=\langle s_z=-\hbar|S_z|s_z=-\hbar\rangle [/itex]

    Is this good?

    Thanks again for your help :D
     
  7. May 31, 2013 #6

    TSny

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    Your expressions for a, b, and c look good.

    You didn't quite evolve in time correctly. It's only the eigenstates of H (i.e., |u1>|,u2>,|u3>) that have the simple time evolution of factors of exp(-iEt/[itex]\hbar[/itex]), for appropriate values of E.

    So, you found ψ(0) = a |u1> + b |u2> + c |u3>.

    To find ψ(t), let each of the |u> states evolve in time with it's own exponential factor. Then combine into one column vector.
     
  8. May 31, 2013 #7

    TSny

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    Your expressions for a, b, and c look good. But, you didn't quite evolve in time correctly. It's only the eigenstates of H (i.e., |u1>|,u2>,|u3>) that have the simple time evolution of factors of exp(-iEt/[itex]\hbar[/itex]), for appropriate values of E.

    So, you found ψ(0) = a |u1> + b |u2> + c |u3>.

    To find ψ(t), let each of the |u> states evolve in time with it's own exponential factor. Then combine into one column vector.

    [EDIT: For example, at time t = 0 we have a|u1> = a ##\left(\begin{array}{ccc}
    -1\\
    0\\
    1 \end{array}\right)##. To evolve this in time just multiply by a factor of ##e^{-iE_0 t/\hbar}##.]

    EDIT 2: Also, as I understand it, you are not looking for ##\langle s_z \rangle##. Rather your a looking for the probability that at time t the system will be in the state ##|s_z = -\hbar \rangle = \left(\begin{array}{ccc}
    0\\
    0\\
    1 \end{array}\right)##
     
    Last edited: May 31, 2013
  9. May 31, 2013 #8

    TSny

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    I have been assuming that we always write our column vectors with respect to the basis in which you wrote your matrices for ##H_o## and ##V##. This is the basis where

    ##\left(\begin{array}{ccc} 1\\ 0\\ 0 \end{array}\right) = |s_{z=\hbar}\rangle## = state where z-component of spin is ##+\hbar##

    ##\left(\begin{array}{ccc} 0\\ 1\\ 0 \end{array}\right) = |s_{z=0}\rangle## = state where z-component of spin is 0

    ##\left(\begin{array}{ccc} 0\\ 0\\ 1 \end{array}\right) = |s_{z=-\hbar}\rangle## = state where z-component of spin is ##-\hbar##

    It appears to me that you might be trying to switch to writing column vectors such that

    ##\left(\begin{array}{ccc} 1\\ 0\\ 0 \end{array}\right) = |u_1\rangle## , ##\left(\begin{array}{ccc} 0\\ 1\\ 0 \end{array}\right) = |u_2\rangle##, ##\left(\begin{array}{ccc} 0\\ 0\\ 1 \end{array}\right) = |u_3\rangle##

    You can't do that since ##|u1\rangle## and ##|u2\rangle## are not orthogonal.
     
  10. Jun 1, 2013 #9
    Ok, after some thinking i did this
    I get this vector after diagonalizing the full hamiltonian and now i make sure they are orthogonal to each other

    ##|u_1\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{ccc} 1\\ -1\\ 0 \end{array}\right) = |s_{z=\hbar}\rangle##

    ##|u_1\rangle=\frac{1}{\sqrt{6}}\left(\begin{array}{ccc} 1\\ 1\\ -2 \end{array}\right) = |s_{z=\hbar}\rangle##

    ##|u_1\rangle=\frac{1}{\sqrt{3}}\left(\begin{array}{ccc} 1\\ 1\\ 1 \end{array}\right) = |s_{z=\hbar}\rangle##

    Expanding the initial function and adding the time evolution terms

    ## |i\rangle = \sum_i \psi_i \langle \psi_i | i\rangle ##

    ##|i\rangle=\frac{1}{\sqrt{2}}|u_1\rangle e^{\frac{-i E_0 t}{\hbar}}+\frac{1}{\sqrt{6}}|u_2\rangle e^{\frac{-i E_0 t}{\hbar}}+\frac{1}{\sqrt{3}}|u_3\rangle e^{\frac{-i (E_0-3\Delta) t}{\hbar}} ##


    Now we have the initial state expanded using the full hamiltonian eigenstates, we must find the amplitude with the vector

    ## |s_z=-\hbar\rangle=\left(\begin{array}{ccc} 0\\ 0\\ 1 \end{array}\right) ##

    that is

    ## \langle s_z=-\hbar|i\rangle=\left(\frac{-1}{3}+\frac{e^{i 3 \Delta t/\hbar}}{3}\right)e^{-i E_0 t/\hbar} = c_i ##

    now its time to get the probability
    ##p(t)=|c_i|^2=\frac{1}{9}\left(2-2\cos(3\Delta t/\hbar\right) ##

    Wich i really hope its the right answer :/ , atleast ## p(0)=0 ## wich is a necessary condition.
     
  11. Jun 1, 2013 #10

    TSny

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    Yes. Very nicely done!
     
  12. Jun 1, 2013 #11
    i am really grateful, thanks for your patience with me :D
     
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