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Finding eigenfunctions for given Hamiltonian

  1. Jan 26, 2015 #1
    1. The problem statement, all variables and given/known data
    I am having too many troubles finding the eigenfunctions of a given Hamiltonian. I just never seem to know what exactly to do. My idea here is not for you to help me solve each problem below, but I would like to just set the equations. I know you guys don't like it when somebody posts 325235 different questions in the same topic, but please understand that I really don't want to open a new thread for each of the basic problems below:

    a) Particle with spin ##S=1## is in magnetic field ##\vec B=(0,0,B)##. At time ##t=0## the particle is in ground state. Find eigenfunctions and energies if ##H=-\lambda \vec S \vec B## where ##\lambda >0##.
    b) Two particles with spin 1/2 interact. This describes Hamiltonian ##H=\lambda(S_{1x}S_{2y}-S_{1y}S_{2x})##. Find eigenfunctions and energies for ##\lambda >0##.
    c) Two particles with spin 1/2 have Heisenberg interaction ##H=J\vec S_1 \vec S_2## where ##J>0##. Find eigenfunctions and energies.
    d) Or the same as c) only with ##S_1=1## and ##S_2=1/2##.

    2. Relevant equations


    3. The attempt at a solution
    For most of them I really don't know what to do. I know that the idea is to solve Schrodinger equation. But...
    a) Ok, because ##S=1## the basis is ##\left \{ |1,1>,|1,0>,|1,-1> \right \}##. Right?
    What is the ground state of a particle with spin 1? o_O
    So I would start like this: ##-\lambda B_zS_z(\alpha |1,1>+ \beta|1,0>+\gamma |1,-1>)=E(\alpha |1,1>+ \beta|1,0>+\gamma |1,-1>)##. but How do I determine the values of ##\alpha, \beta, \gamma ##?
    b) I don't know what the basis is? Is it ##\left \{ |\uparrow>,|\downarrow> \right \}## or is it ##\left \{ |\uparrow>|\uparrow>,|\uparrow>|\downarrow>,|\downarrow>|\downarrow>,|\downarrow>|\uparrow> \right \}##. And still, how would I determine the values of the coefficients before the basis vectors?

    ... As you can see, my questions are in all the cases more or less the same, therefore I published them in the same topic. :/ I hope you don't mind.
     
  2. jcsd
  3. Jan 26, 2015 #2

    Orodruin

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    The general approach is to find the eigenvalues of the Hamiltonian and then to find the corresponding eigenvectors. The ground state is the state with the lowest energy (ie, eigenvalue).

    In each case above, first identify the state space (if you have one particle spin, the state space is 2j+1 dimensional -where j is the spin, if you have more spins, the dimension is the product of the dimensions of the individual spins). Now, write down the Hamiltonian in a basis (with two spins, it may be good to use the basis of eigenvectors to the total spin operator). Once that is done, you can simply figure out the eigenvalues and eigenvectors as you would for any matrix.
     
  4. Jan 27, 2015 #3
    a) Particle with spin ##S=1## is in magnetic field ##\vec B=(0,0,B)##. At time ##t-> -\infty ## the particle is in ground state. Find eigenfunctions and energies if ##H=−λ\vec S\vec B## where ##λ>0##.
    (sorry, in my original post I made a typo. The particle is in ground state when ##t-> -\infty ## and not when ##t=0##.)

    Ok, I know what ground states are - like you said, states with the lowest energies. But what I don't know is which function is it? For example: If it was a harmonic oscillator then ##n=0##, if it was a hydrogen atom, than ##n=1## ,.... etc... And one can easily google the exact expression for a wavefunction in case of HO or hydrogen atom.
    So I imagined, that since the problem states that the particle is in ground state at given time (##t-> -\infty##), this should than reduce the number of possible wavefunctions at that time to ##1## and maybe some additional degenerated states. So, which one are they? o_O

    Now back to the problem:
    Ok, since the particle has spin ##S=1##, than my state space is (according to you) 3 dimensional. Basis vectors should be ##\left \{ |1,1>,|1,0>,|1,-1> \right \}##. Therefore my wavefunction will be a linear combination of basis vectors $$|\psi >=a|1,1>+b|1,0>+c|1,-1>$$ Is the idea now to solve stationary Schrodinger equation - I assume this should bring me to a ##3\times 3## matrix where eigenvectors would be my possible linear combinations of basis vectors and eigenvalues of the matrix their energies? Or...? Is that a complete nonsense?
     
  5. Jan 27, 2015 #4

    Orodruin

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    No, this is correct. Your choice of base states is also with respect to an axis (the spin projection on that axis and the total spin being the identifying quantum numbers). A good choice would be to use the same axis as that in which the magnetic field points. What does the resulting Hamiltonian do to each of the base states?
     
  6. Jan 27, 2015 #5
    huh, than let's continue with part a)

    So ##H|\psi >=E|\psi >## $$-\lambda B_zS_z(a|1,1>+b|1,0>+c|1,-1>)=E(a|1,1>+b|1,0>+c|1,-1>)$$ $$-\lambda B_z \hbar (a|1,1>-c|1,-1>)=E(a|1,1>+b|1,0>+c|1,-1>)$$ This than brings me to a ##3\times 3## matrix $$
    \begin{pmatrix}
    E+\lambda B_z\hbar & 0 & 0\\
    0& E & 0\\
    0 & 0& \lambda B_z\hbar-E
    \end{pmatrix}\begin{pmatrix}
    a\\
    b\\
    c
    \end{pmatrix}=\begin{pmatrix}
    0\\
    0\\
    0
    \end{pmatrix}$$ Now for eigenvalue ##E+\lambda B_z \hbar## the wavefunction is ##|\psi >=|1,1>##
    for eigenvalue ##E## the wavefunction is ##|\psi >=|1,0>## and
    for eigenvalue ##\lambda B_z \hbar-E## the wavefunction is ##|\psi >=|1,-1>##

    Now how does one determine the ground state of the system, if I don't know the value of ##E## ? Or do I?
     
  7. Jan 27, 2015 #6

    Orodruin

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    ##E## is the eigenvalue of ##H## by definition. The matrix equation you have is the eigenvector equation for the Hamiltonian given an eigenvalue ##E##. In order for a non-trivial solution to exist, the determinant of the matrix must be zero - which gives the characteristic polynomial with the eigenvalues as roots.
     
  8. Jan 27, 2015 #7
    Ahh, true. This should be correct that:

    Determinant of the matrix above, gives me three possible results for the energies of the states:
    ##E_1=0## which belongs to ##|1,0>##,
    ##E_2=-\lambda B_z\hbar## which belongs to ##|1,1>## and finally
    ##E_3=\lambda B_z\hbar## which belongs to ##|1,-1>##.

    Now for ##\lambda >0## this means that state ##|1,1>## is the ground state of the system. Right?
     
  9. Jan 27, 2015 #8

    Orodruin

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    Correct.
     
  10. Jan 27, 2015 #9
    Wow, that's great Orodruin, thank you!

    b)Two particles with spin ##1/2## interact. This describes Hamiltonian ##H=λ(S_{1x}S_{2y}−S_{1y}S_{2x})##. Find eigenfunctions and energies for ##λ>0##.

    Again, following your instructions, this means that each particle has a ##2## dimensional space, meaning both together have ##2\cdot 2=4## dimensional space. Basis vectors are ##\left \{|\uparrow>|\uparrow>,|\uparrow>|\downarrow>,|\downarrow>|\uparrow>,|\downarrow>|\downarrow> \right \}## or as you recommended written in the basis with total spin ##\left \{|1,1>,|1,0>,|1,-1>,|0,0> \right \}##.
    $$λ(S_{1x}S_{2y}−S_{1y}S_{2x})(a|1,1>+b|1,0>+c|1,-1>+d|0,0>)=E\psi $$ Where I couldn't find any easier or faster solution to this than using $$S_\pm |s,m>=\hbar \sqrt{s(s+1)-m(m\pm 1)}|s,m\pm 1>$$ and of course ##S_x=\frac 1 2 (S_++S_-)## and ##S_y=\frac {1}{2i}(S_+-S_-)##. If the idea is right, and if I did everything ok, than what I should get is $$\frac{\lambda \hbar ^2(c-a)}{2i}(|1,1>+|1,-1>)=E(a|1,1>+b|1,0>+c|1,-1>+d|0,0>)$$ The next steps are same as in a), so don't think we have to repeat them here.

    c) Two particles with spin ##1/2## have Heisenberg interaction ##H=J\vec S_1\cdot \vec S_2## where ##J>0##. Find eigenfunctions and energies.

    This one is a bit more confusing. I started by saying ##(\vec S_1+\vec S_2)^2=S^2=S_1^2+S_2^2+2\vec S_1\cdot \vec S_2##. This means $$\vec S_1\cdot \vec S_2=\frac 1 2(S^2-S_1^2-S_2^2).$$ Soo... since the basis is the same as in b) $$J\vec S_1\cdot \vec S_2(a|1,1>+b|1,0>+c|1,-1>+d|0,0>)=E(a|1,1>+b|1,0>+c|1,-1>+d|0,0>)$$ $$J\frac 1 2 (S^2-S_1^2-S_2^2)(a|1,1>+b|1,0>+c|1,-1>+d|0,0>)=E(a|1,1>+b|1,0>+c|1,-1>+d|0,0>)$$ Which should give me two possible solutions $$
    H|\psi >=\left\{\begin{matrix}
    \frac{\hbar ^2J}{4}|\psi > & ,S=1\\
    -\frac{3\hbar^2 J}{4}|\psi > &,S=0
    \end{matrix}\right.$$ Now I am confused how to continue. Since I already have form ##H|\psi >=E|\psi >## I already have the energy that depends on total spin ##S##. But what about wavefunctions?

    EDIT to c): Does this mean that a,b,c and d can be arbitrary? So why shouldn't they all be equal to ##1/\sqrt 4)## so the wavefunction is normalized??
     
    Last edited: Jan 27, 2015
  11. Jan 27, 2015 #10

    Orodruin

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    I did not check the arithmetics in detail, but in general it looks good. For (c), you have just found that all states with total spin 1 have the same energy, i.e., they form a degenerate subspace of the full state space (which makes sense, the Hamiltonian is invariant under rotations so it should not matter which axis you chose as your basis).
     
  12. Jan 27, 2015 #11
    No need to, as long as the general idea is ok! :) Thank you!
    Sorry to ask, but what exactly does this mean for my constants? I am supposed to explicitly write wavefunctions.
     
  13. Jan 27, 2015 #12

    Orodruin

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    Well, it means you can write down any orthonormal base for that subspace and it will span your state space (along with the spin 0 state) - the base in terms of the eigenstates of Sz is perfectly fine (but the eigenstates of Sx would also do).

    Also note that these are states, not really wave functions. In most cases, "wave function" specifically refers to the inner product of a state in position space with a state of definite position ##x##, i.e., ##\psi(x) = \langle x | \psi \rangle##. I guess in a more loose terminology, you could refer to the states as wave-functions, but the wave function really is more related to the components of the state in a given basis (from which the state can be reconstructed).
     
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