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Low dimensional anticommutation example

  1. Jan 25, 2010 #1
    Can somebody give an example of two matrices [itex]a_1[/itex] and [itex]a_2[/itex] which would satisfy the relation

    [tex]
    a_i a_j^{\dagger} + a_j^{\dagger} a_i = \delta_{ij}
    [/tex]

    I know that

    [tex]
    \left(\begin{array}{cc}
    0 & 1 \\ 0 & 0 \\
    \end{array}\right)
    \left(\begin{array}{cc}
    0 & 0 \\ 1 & 0 \\
    \end{array}\right)
    + \left(\begin{array}{cc}
    0 & 0 \\ 1 & 0 \\
    \end{array}\right)
    \left(\begin{array}{cc}
    0 & 1 \\ 0 & 0 \\
    \end{array}\right)
    = 1
    [/tex]

    but I found myself unable to modify this for the index [itex]i\in\{1,2\}[/itex].
     
    Last edited: Jan 25, 2010
  2. jcsd
  3. Jan 25, 2010 #2
    One obvious example is creation and annihilation operators for electrons. These operators act in the Fock space, which is a direct sum of N-electron Hilbert spaces, where N runs from 0 to infinity. So, the corresponding matrices are infinite-dimensional.

    Eugene.
     
  4. Jan 25, 2010 #3
    Raising and lowering operators for fermions were my original motivation for this problem. I didn't feel comfortable by the way books only give the properties of the operators, but I also want to see how these operators could be defined. I decided to make the problem simpler by allowing "only two Fourier modes" for the particles.

    My first idea was that I should use four dimensional vector space, denote the basis vectors as [itex]|00\rangle[/itex], [itex]|01\rangle[/itex], [itex]|10\rangle[/itex], [itex]|11\rangle[/itex], and then define [itex]a_1[/itex] so that it maps [itex]|01\rangle\mapsto |00\rangle[/itex] and [itex]|11\rangle\mapsto |10\rangle[/itex], and [itex]a_2[/itex] so that it maps [itex]|10\rangle\mapsto |00\rangle[/itex] and [itex]|11\rangle\mapsto |01\rangle[/itex]. With the following notation choice:

    [tex]
    |00\rangle = \left(\begin{array}{c}
    1 \\ 0 \\ 0 \\ 0 \\
    \end{array}\right),\quad
    |01\rangle = \left(\begin{array}{c}
    0 \\ 1 \\ 0 \\ 0 \\
    \end{array}\right),\quad
    |10\rangle = \left(\begin{array}{c}
    0 \\ 0 \\ 1 \\ 0 \\
    \end{array}\right),\quad
    |11\rangle = \left(\begin{array}{c}
    0 \\ 0 \\ 0 \\ 1 \\
    \end{array}\right)
    [/tex]

    the operators would be

    [tex]
    a_1 = \left(\begin{array}{cccc}
    0 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 \\
    0 & 0 & 0 & 0 \\
    \end{array}\right),\quad\quad
    a_2 = \left(\begin{array}{cccc}
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    \end{array}\right)
    [/tex]

    But this does not work as wanted. These work like this:

    [tex]
    a_1 a_2^{\dagger} + a_2^{\dagger} a_1 = \left(\begin{array}{cccc}
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 2 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    \end{array}\right)
    [/tex]

    Because [tex]a_1 a_2^{\dagger}[/tex] maps [itex]|01\rangle\mapsto |11\rangle\mapsto |10\rangle[/itex], and [tex]a_2^{\dagger} a_1[/tex] maps [itex]|01\rangle\mapsto |00\rangle\mapsto |10\rangle[/itex].
     
  5. Jan 25, 2010 #4
    I was informed elsewhere, that matrices

    [tex]
    a_1 = \frac{1}{\sqrt{2}}\left(\begin{array}{cc}
    1 & 0 \\ 0 & -1 \\
    \end{array}\right),\quad\quad
    a_2 = \frac{1}{\sqrt{2}}\left(\begin{array}{cc}
    0 & 1 \\ 1 & 0 \\
    \end{array}\right)
    [/tex]

    satisfy the property I was asking for. This did not solve my problem though, because it only made me realize that I should have also demanded that [itex]a_1 a_1 = 0[/itex] and [itex]a_2 a_2 = 0[/itex] in the beginning.
     
  6. Jan 25, 2010 #5

    SpectraCat

    User Avatar
    Science Advisor

    Well, using your 4-D vector space from post #3, I think the following matrices fit the bill:

    [tex]
    \left(\begin{array}{cccc}
    0 & 1 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    \end{array}\right),\quad\quad

    and

    \left(\begin{array}{cccc}
    0 & 0 & 0 & 0 \\
    1 & 0 & 0 & 0 \\
    0 & 0 & 0 & 1 \\
    0 & 0 & 0 & 0 \\
    \end{array}\right)
    [/tex]

    but they are adjoints of each other, so perhaps that is not what you are looking for. I guess it's just a block diagonal generalization of your original 2x2 case, now that I look at it.
     
  7. Jan 25, 2010 #6
    That's right. I'm surprised to notice only now, that the matrices

    [tex]
    a_1 = \left(\begin{array}{cc}
    0 & 1 \\ 0 & 0 \\
    \end{array}\right),\quad\quad
    a_2 = \left(\begin{array}{cc}
    0 & 0 \\ 1 & 0 \\
    \end{array}\right)
    [/tex]

    answer my original question :surprised

    But there still seems to be something that doesn't make sense in this. Perhaps I'll need to come up with a better formulation for the problem...
     
  8. Jan 29, 2010 #7
    this is one example of dirac algebra, and can be generalized to clifford algebra.
     
  9. Feb 10, 2010 #8
    I have now understood that not every non-zero number has to be one.

    So I want

    [tex]a_1 |00\rangle = 0[/tex]
    [tex]a_1 |01\rangle \propto |00\rangle[/tex]
    [tex]a_1 |10\rangle = 0[/tex]
    [tex]a_1 |11\rangle \propto |10\rangle[/tex]

    [tex]a_2 |00\rangle = 0[/tex]
    [tex]a_2 |01\rangle = 0[/tex]
    [tex]a_2 |10\rangle \propto |00\rangle[/tex]
    [tex]a_2 |11\rangle \propto |01\rangle[/tex]

    and

    [tex]a_ia_j^{\dagger} + a_j^{\dagger}a_i = \delta_{ij}[/tex]

    If I choose any four complex numbers [itex]\alpha,\beta,\gamma,\delta[/itex] such that [itex]|\alpha|=|\beta|=|\gamma|=|\delta|=1[/itex] and [itex] \beta\delta^* + \gamma^*\alpha = 0[/itex], then the desired operators can be defined as

    [tex]
    a_1 = \left(\begin{array}{cccc}
    0 & \alpha & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & \beta \\
    0 & 0 & 0 & 0 \\
    \end{array}\right),\quad\quad\quad
    a_2 = \left(\begin{array}{cccc}
    0 & 0 & \gamma & 0 \\
    0 & 0 & 0 & \delta \\
    0 & 0 & 0 & 0 \\
    0 & 0 & 0 & 0 \\
    \end{array}\right)
    [/tex]

    For example [itex](\alpha,\beta,\gamma,\delta)=(1,1,1,-1)[/itex] is one possibility.

    But there is something inelegant about this. It seems laborous to try to generalise this for three Fourier modes next. Would I need to solve again even more equations then?
     
    Last edited: Feb 11, 2010
  10. Feb 11, 2010 #9
    Suppose we have a fermion system with N Fourier modes. Then for [itex]i\neq j[/itex] the raising operator [tex]a_j^{\dagger}[/tex] and lowering operator [tex]a_i[/tex] will commute modulo some coefficient, so that

    [tex]
    a_i a_j^{\dagger} |\delta_1\delta_2 \cdots \underset{(i)}{1} \cdots \underset{(j)}{0} \cdots \delta_N\rangle
    \propto |\delta_1\delta_2 \cdots \underset{(i)}{0} \cdots \underset{(j)}{1} \cdots \delta_N\rangle
    [/tex]

    [tex]
    a_j^{\dagger} a_i|\delta_1\delta_2 \cdots \underset{(i)}{1} \cdots \underset{(j)}{0} \cdots \delta_N\rangle
    \propto |\delta_1\delta_2 \cdots \underset{(i)}{0} \cdots \underset{(j)}{1} \cdots \delta_N\rangle
    [/tex]

    It seems, that we can always choose these coefficients so that [tex]a_i a_j^{\dagger} + a_j^{\dagger} a_i = 0[/tex] (when [itex]i\neq j[/itex]).

    Why do physicist insist on getting the anticommutativity property holding? If the crucial property of fermion system is that each Fourier mode always has only one excitation state, the anticommutativity of raising and lowering operators does not seem to be inherent to the system. We might as well have fermion systems without anticommuting raising and lowering operators too, right?
     
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