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Simplified Heisenberg Hamiltonian; Linear combinations of basis states

  1. Aug 13, 2009 #1
    So, I'm doing some undergraduate research in quantum spin systems, looking at the ground states of the Heisenberg Hamiltonian, [tex]H=\sum{J_{ij}\textbf{S}_{i}\textbf{S}_{j}}[/tex]. But I think I have a critical misunderstanding of some fundamental quantum mechanics concepts. (I'm a math major, only had one introductory QM course...)

    Say you just have two interacting spin-1/2 particles. The Hamiltonian can be written as [tex]\textbf{S}_{1}\textbf{S}_{2}[/tex] (simplified by letting [tex]J_{1,2}=1[/tex]) which is equal to:

    [tex]\left( \begin{array}{cccc}

    1 & 0 & 0 & 0 \\

    0 & -1 & 2 & 0 \\

    0 & 2 & -1 & 0 \\

    0 & 0 & 0 & 1 \\\end{array} \right)[/tex]​

    which has one ground state, [tex]\frac{1}{\sqrt{2}}(\mid\downarrow\uparrow\rangle - \mid\uparrow\downarrow\rangle)[/tex], with energy eigenvalue -3.

    So let's say I have a system set up in this state. I make a measurement. I find the system is in state [tex]\mid\downarrow\uparrow\rangle[/tex] (which I would find half the time). Now, this is a basis state in my Hilbert space, but its not an eigenvector of my Hamiltonian.

    What does that mean, if the state after a measurement is no longer an eigenvector of the Hamiltonian? Is the energy of the system the same as before the measurement? I feel like I'm missing something important here.

    I guess I just don't understand what it means to have a superposition of states as an eigenvector of some observable. Like, I want to look at ground states, and I'm getting mostly linear combinations of basis states (especially for higher number of interacting particles) but physically I can't wrap my head around what this means.
  2. jcsd
  3. Aug 13, 2009 #2
    You're off by a factor of 1/4 in your Hamiltonian, (the matrix elements of S operators are proportionate to S, which is 1/2 in this case). Better to say you've set J = 4.

    Ok, so when you say "make a measurement" you have to say what exactly it is you are measuring. You have an operator which represents the quantity you are measuring and you will only measure eigenvectors of that operator. You say you find the system in the state [tex]|\downarrow\uparrow\rangle[/tex] but what operator is that an eigenstate of? As you have already found out, it is not an eigenstate of [tex]S_1 \cdot S_2[/tex], and from that you can easily show that it is not an eigenstate of [tex]S_1 + S_2[/tex]. I think it is quite challenging to find an operator with that state as an eigenvector without having the state [tex]|\uparrow\downarrow\rangle[/tex] as a degenerate eigenvector.

    Ok, so assuming you have such an operator and it's reasonable to measure it, and you find the system in state [tex]|\downarrow\uparrow\rangle[/tex]. This does mean that after the measurement the system is no longer in an eigenstate of the Hamiltonian, and the energy has changed due to the measurement.

    What you've done in this example is to construct basis states for a many particle system out of single particle states, which is easy to do because it is straightforward to enumerate the single particle states. The single particle states are constructed based on a particle which does not have anything to interact with, and now you use them to construct a Hamiltonian with an interaction, well the interaction is going to destroy the coherence of those single particle states. In other words, the interacting Hamiltonian does not commute with the spin at a single site S_i, so the m value for a single site is no longer a good quantum number. You will notice however that the Hamiltonian commutes with the total spin operator [tex]S = \sum_i S_i[/tex], so the total spin is still a good quantum number.
  4. Aug 17, 2009 #3
    Kanato, thanks for taking the time to reply, that helps a lot. I had to read it a couple times but I think I see now where I was going wrong. "Coherance" was something we covered right at the end of my QM course, and I didn't quite get it at the time, but your last paragraph here helps a lot.

    Thanks again.
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