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N-fermion space construction

  1. Aug 9, 2010 #1
    I'm considering a physical system of [tex]n[/tex]-species of fermions.
    The creation and annihilation operators satisfy
    [tex] \{b_i^-,b_j^-\} = 0 = \{ b_i^+ , b_j^+ \}, \quad \{ b_i^- , b_j^+ \} = \delta_{ij} [/tex]
    I want to construct the matrix representations of the state vectors and operators.
    I do it recursively, but, I can't find a way to systematically construct the representation, what I did is the following.

    Consider only one species of fermion, with one fermion state represented by [tex] \left(\begin{array}{c} 1\\ 0 \end{array}\right) [/tex], and empty state represented by [tex] \left(\begin{array}{c} 0\\ 1 \end{array}\right) [/tex]. We can construct the creation operator [tex]B^+[/tex] as [tex] B^+ = \left(\begin{array}{cc} 0 &1\\ 0 &0 \end{array}\right) [/tex] and annihilation operator [tex] B^- = \left( \begin{array}{cc} 0 &0 \\ 1 &0 \end{array}\right) [/tex].

    However, for two species fermion system, if the states should be the direct product of each single species states, i.e.
    [tex]|\frac{1}{2},\frac{1}{2}\rangle = \left(\begin{array}{c} 1\\0 \end{array}\right) \otimes \left(\begin{array}{c} 1\\0 \end{array}\right) [/tex] represents two-fermion states, likewise we have other states [tex] |\frac{1}{2},-\frac{1}{2}\rangle, |-\frac{1}{2},\frac{1}{2}\rangle, |-\frac{1}{2},-\frac{1}{2} \rangle [/tex] represents two possible 1-fermion states, and zero-fermion state.

    Now, what I was surprised is, the creation and annihilation operators cannot be constructed naively as the following direct products:
    [tex] b_1^+ = B^+ \otimes \mathbf{1}, b_1^- = B^- \otimes \mathbf{1} [/tex] and
    [tex] b_2^+ = \mathbf{1} \otimes B^+, b_2^- = \mathbf{1} \otimes B^- [/tex]
    The creation operators and annihilation operators constructed in this way would not satisfy the correct anti-commutation relation, for example, [tex] \{ b_1^+ , b_2^+ \} \neq 0 [/tex]

    I don't quite understand why this construction doesn't work?
  2. jcsd
  3. Aug 10, 2010 #2
    Don't Fermion operators for different species commute? I think you are trying to construct the matrix representations for the operator algebra for a single fermion species.
  4. Aug 10, 2010 #3
    Oh, that's right.
    Fermion operators for different species should anti-commute.
    The way I construct the creation and annihilation operators fails to have this anti-commute property.
    I found the way to solve this is to replace the identity matrix in one species by the following matrix
    [tex] \left( \begin{array}{cc} 1 &0\\ 0 &-1 \end{array}\right) [/tex]
  5. Aug 10, 2010 #4
    I'll echo what peteratcam said: shouldn't fermionic creation operators for different species commute? Isn't it fermion creation operators for different modes of the same species that do commute?

    I.e., shouldn't a proton creation operator and an electron creation operator commute? There's not (anti)symmetrization requirement for a two-particle state vector if the particles are distinguishable.
  6. Aug 10, 2010 #5
    To make sure.
    Do you mean "anti-commute?"
    Since we are working on the fermion creation and annihilation operators,
    we should use anti-commucation relation instead of commutation relations here.
  7. Aug 10, 2010 #6


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    Firstly do you mean "species" or "modes", i.e. an electron has two spin modes.

    Secondly w.r.t. whether different species creators/annihilators commute vs. anti-commute is a matter of taste (+/-1 phase factors) unless you want to consider superpositions between species, as with e.g. flavor mixing. In that case they need to anti-commute.

    Here is the trick to keep them anti-commuting:

    Define the two-by-two matrices:
    [tex]\sigma_+ = \left(\begin{array}{cc} 0 &1\\ 0 &0 \end{array}\right) [/tex]
    [tex] \sigma_- = \left( \begin{array}{cc} 0 &0 \\ 1 &0 \end{array}\right)[/tex]
    [tex] \eta =[\sigma_+,\sigma_-] =\left(\begin{array}{cc} 1 &0\\ 0 &-1 \end{array}\right) [/tex]
    and of course...
    [tex] \mathbf{1} = \left(\begin{array}{cc} 1 &0\\ 0 &1 \end{array}\right)[/tex]

    Now define your creators C and annihilators A in sequence:
    [tex] C_1 = \cdots\otimes \mathbf{1}\otimes\mathbf{1}\otimes \sigma_+[/tex]
    [tex] A_1 = \cdots\otimes \mathbf{1}\otimes\mathbf{1}\otimes \sigma_-[/tex]

    [tex] C_2 = \cdots\otimes \mathbf{1}\otimes \sigma_+\otimes\eta[/tex]
    [tex] A_2 = \cdots\otimes \mathbf{1}\otimes \sigma_-\otimes \eta[/tex]

    [tex] C_3 = \cdots\otimes \sigma_+\otimes\eta\otimes \eta[/tex]
    [tex] A_3 = \cdots\otimes \sigma_-\otimes \eta\otimes \eta[/tex]

    and so on.

    Then say the Fock mode corresponding to one each of species 1,3, and 4 is:
    [tex]|\cdots,0,1,1,0,1\rangle=\cdots \left(\begin{array}{c} 0\\ 1 \end{array}\right)\otimes\left(\begin{array}{c} 0\\ 1 \end{array}\right)\otimes \left(\begin{array}{c} 1\\ 0 \end{array}\right)\itimes \left(\begin{array}{c}1\\ 0 \end{array}\right)\otimes \left(\begin{array}{c} 0\\ 1 \end{array}\right)\otimes \left(\begin{array}{c} 1\\ 0 \end{array}\right)[/tex]

    That is up to a sign factor which depends on the order in which each fermion got created from the vacuum mode (and the sign of the vacuum mode you choose).

    For each mode (species) you have the number operator
    [tex]N_k=C_k A_k= ([C_k,A_k]+1)/2[/tex]
    Define [tex]\eta_k =[C_k,A_k] = 2N_k - 1[/tex] which in the representation is just an [tex]\eta[/tex] in the k-th position of the tensor product of factors with identities elsewhere. This operator is an occupation dependent sign factor and you can make two species k's creators and annihilators commute with species j's (j not equal to k) by e.g.
    [tex] A'_j = A_j \eta_k[/tex]
    [tex]C'_j = C_j \eta_k[/tex]

    As long as no mixing between species j and k occurs this will have no physical ramifications and should be viewed simply as a representation convention.
  8. Aug 10, 2010 #7
    Thank you very much for your reply.
    I mean species, but...could you explain also that how to construct the space of n-fermions each has two spin modes?

    I don't quite understand this.
    The creation and annihilation operators of fermions should be Grassman numbers.
    In this case, they should be anti-commutate, right?
    I don't understand why they need to anti-commute only if we need to consider flavor mixing.

    Yes. Replacing identity matrices in one side (the table is separated by the [tex]\sigma_+, \sigma_-[/tex]) by [tex]\eta[/tex] makes [tex]A_i, C_i [/tex] and [tex]A_k, C_k[/tex] anti-commute if [tex] i \neq [/tex].
    This table is what I want originally, thanks.

    In this way, [tex]A,C[/tex] between different species do commute:
    [tex] [ A_i \eta_k , A_k ] = [ A_i [C_k,A_k] , A_k ] = [A_iC_kA_k , A_k] - [ A_iA_kC_k , A_k ] = -A_iA_k + A_iA_k = 0 [/tex]
    If we want [tex] A_i [/tex] commute with [tex]A_j[/tex] too, we could just multiply [tex]\eta_j[/tex] on the [tex]A_i[/tex] again, i.e. [tex]A_i'' = A_i \eta_k \eta_j [/tex]
    But I still don't understand what's the relation between (anti-)commutators and flavor mixing?
  9. Aug 11, 2010 #8


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    The construction I gave, is for multiple modes (but also works with species by effectively treating them as modes). You can treat species as extensions of modes but then you must invoke an additional superselection rule (typically associated with a conserved quantity). For example we generally do not find coherent superpositions of a particle being an electron vs being a proton. Conservation of lepton number an baryon number, (oh yes and of electric charge) forbid such coherent modes. This then means we have a restricted set of observables, namely we have no observables associated with such forbidden superpositions.

    You can't build a particle detector to go "ping" if and only if a [tex] \alpha|electron\rangle+\beta|proton\rangle[/tex] mode particle hits it.

    Given no such observables are actualized then no physical distinction can be seen from invoking the occupation number dependent signs resulting from using say A vs A' annihilators I mentioned in the prior post.

    With modes, on the other hand you have the full unitary group mixing e.g. spin directions (rotation group), and you must use the anti-commuting case to get the proper representation of the respective creation or annihilation operators. The choice is physically meaningful because these group actions (rotating your spin measuring device) are physically actualizable. That is not the case e.g. with mixing of proton and electron modes (we can't build a device to detect a specific non-trivial superposition of being an electron vs being a proton.)

    More precisely they are generators of a Clifford algebra (on a neutral signature orthogonal space with complementary null metric.

    (Personally, I object to the idea of a "Grassman number", you have an element of a Grassman algebra which by virtue of non-commutivity cannot be treated as a classical number. It is at best a q-number outside the system under study. Thus say a 'photino' is simply a composite of a photon in the system and an auxiliary fermion in the theoretician's pocket. But I digress into personal biases.)

    A neutral metric (indefinite) orthogonal space will have two complementary null orthogonal spaces (zero sub-metric) and the extreme case of a clifford algebra over a null metric is a Grassman algebra. (I like to call the whole category Clifford-Grassman algebras allowing even mixing of +, - an null signature components.)

    Anyway the critical issue is what you do with the algebra. In particular if you have say two modes (a two dimensional Hilbert space) for a given fermionic particle then you have a continuum of possible bases any one of which you can use to define your creators and annihilators for the many particle system. Specifically for a given basis, define an arbitrary single particle mode:

    [tex] \psi = \psi_1 |1\rangle + \psi_2\rangle[/tex]

    Then to define the creator and annihilator for that mode you take the corresponding superposition:
    [tex] C_\psi = \psi_1 C_1 + \psi_2 C_2[/tex]
    [tex] A_\psi = \psi^*_1 A_1 + \psi^*_2 A_2[/tex]
    (One should double check that I've put the conjugates in the correct place. I may have it backwards.)

    For these operators to have the proper anticommutator relations, the two creators must anti-commute as do the two annihilators.

    It is very clear if you've played with general Clifford algebras but takes some worked out examples to see without such overview of the math.

    In a bigger (mathematical picture) consider the original single particle Hilbert space (dimension n) and the corresponding group U(n) of unitary transformations. U(n) is actually a real group (you don't take arbitrary complex exponentials of the generators, only real parameters are used.)

    Now U(n) has two conjugate representations one given by action on the Hilbert space, V the other by action on its dual space, V*.

    We can also embed U(n) in a larger orthogonal group SO(2n;C). The vector representations of the orthogonal groups are self-dual, but noting the doubling of dimension we see that this embedding U(n) in SO(2n;C) gives us both conjugate representations in a single orthogonal space. Basically this is V+V*.

    Now given an orthogonal group, you can define its spinor representation as the representation of the action of the grade two elements of its corresponding Clifford algebra.
    Given space X with metric g, associate a Clifford generator [tex]\gamma[/tex] for each basis element such that:
    [tex]\{\gamma_i,\gamma_j\} = g_{ij}[/tex]

    Now with real orthogonal groups we can have either definite or indefinite metrics (like with the Lorentz group SO(3,1)) but their complexifications all yield a single complex orthogonal group depending only on the dimension. Said another way, inside each complex orthogonal group SO(n;C) is all the various indefinite real orthogonal groups SO(p,q) p+q=n.
    This then means for the case above, SO(2n;C) we can pick a neutral metric (as for SO(n,n)) by suitable complex basis, and what's more we can pick that basis so the neutral metric has the peculiar form:

    [tex] g \sim \left( \begin{array}{cc} 0 & \mathbf{1}_n \\ \mathbf{1}_n & 0\end{array}\right)[/tex]

    This is the form where the unitary embedding maps V and V* into the two null half spaces. Visualize it as something like:
    [tex] X = \left( \begin{array}{c} V\\V^*\end{array}\right)[/tex]

    Now for the punch line. The spinors of this orthogonal group defines the Fock space for the many-fermion quantization. Many fermion modes are spinors of this orthogonal embedding of the single particle space. (This is a separate invocation of 'spinors' from the fact that fermions are also spin 1/2 with respect to the Lorentz group and its rotation subgroup.)

    So that's the Mathematical "big picture" though it doesn't give one much insight into the physics without a much deeper exposition.

    BTW I brought up flavor mixing because we might originally treat the distinct flavors (say of neutrinos) as different species and build separate algebras of creation and annihilation operators. As long as flavors don't mix we needn't worry about superpositions of say tau and electron neutrinos. But now that we observe neutrino oscillations we should properly use a unified set of (all mutually anti-commuting) creators and annihilators.

    No prob. I hope my explanation above was clear enough. If not I can clarify further.
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