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What is the expression of N=2 SUSY transformation

  1. Sep 30, 2013 #1
    I have checked many textbooks and papers on SUSY and it seems that none of them mentions anything about the infinitesimal susy transformation on component fields in the case [itex]N\neq 1[/itex]. So I am wondering what does it looks like, say for N=2 vector multiplet?
    Another related question is, do we need auxiliary fields? I would say no since there are equal number of boson and fermion degrees of freedom.
     
    Last edited: Sep 30, 2013
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  3. Oct 1, 2013 #2

    fzero

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    You can find the states constructed in section 2.3.2 of Quevedo's lectures, for example. The N=2 vector multiplet has a scalar, 2 fermions and a vector field. There are two supercharges ##Q^1_\alpha## and ##Q^2_\alpha##. From the scalar, acting with ##(Q^1_\alpha)^\dagger## gives one fermion, while the other comes from acting with ##(Q^2_\alpha)^\dagger##. The vector comes from acting with ##(Q^1_\alpha)^\dagger (Q^2_\alpha)^\dagger## on the scalar, or alternatively by acting with the other supercharge on either of the fermions. It shouldn't be especially hard to work out the explicit SUSY transformations.

    Adding auxiliary fields to supermultiplets is not done to equalize the boson and fermion degrees of freedom. If the representation is supersymmetric, the number of physical degrees of freedom will always match. The reason for introducing auxiliary fields is to have the SUSY algebra close off-shell. In terms of the physical degrees of freedom, the SUSY algebra always closes on-shell, but this can be inconvenient for things like loop amplitudes where we will have off-shell degrees of freedom entering into intermediate stages of the calculation. So with extended SUSY, we will want to introduce auxiliary fields for the same reasons as for N=1 SUSY.
     
  4. Oct 1, 2013 #3
    I have another question. We can derive the expressions of susy transformation in superspace formalism, and what we get is in terms of partial derivative rather than covariant derivative. But in the presence of gauge field, all partial derivatives are converted into covariant derivatives. How to explain it with the language of superspace? since it is not evident that there is still supersymmetry with the infinitesimal susy transformations defined with covariant derivative. Maybe we can make the susy transformation operator gauge covariant. Do you happen to know this kind of discussion?
    Another thing is the extra term in [itex]\delta F[/itex]. Terning's textbook said this is to cancel terms in gaugino interaction, which I am not sure. Is it possible to be explained in superspace?
     
    Last edited: Oct 1, 2013
  5. Oct 1, 2013 #4

    fzero

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    There is a discussion of a completely covariant formulation of SUSY gauge theories in sect 4.2 of Superspace. That reference is classic and will almost always contain whatever details are left out of other references.

    I'm a bit confused whether your ##F## is the auxiliary field for a chiral multiplet or the gauge field strength. In either case, I would say that yes, the transformation rules of all components are explained in superspace, because all transformation properties of the components of a linear superfield ##S## are defined by ##\delta S = i (\epsilon Q + \bar{\epsilon} \bar{Q})S##. The chiral and vector superfields are defined by placing constraints on a linear superfield.

    If you want to clarify what you are specifically asking about, perhaps I can be a bit clearer.
     
  6. Oct 1, 2013 #5
    I was referring to the auxiliary field for a chiral multiplet. The extra term takes the form:
    [tex]\sqrt{2}g(T^a\phi)_j\epsilon^\dagger\lambda^{\dagger a}[/tex]
    following Terning's Modern SUSY. It's like [itex]\mathcal{F}[/itex] can be susy-transformed into a gaugino, which sounds weird to me.
     
  7. Oct 1, 2013 #6

    fzero

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    I'm not familiar with Terning's text and can't recall having done this particular calculation myself. In any case, it's clear that this is due to the ##\bar{\psi}\lambda \phi## terms, which are SUSY related to the ##\bar{\psi}A_\mu \psi## terms that are present in any gauge theory. Perhaps it is just convenient to use ##F## to handle the off-shell SUSY of these terms. Maybe things are less wierd if we don't choose Wess-Zumino gauge before working out the transformation laws. Presumably the manifestly covariant formulation in the Superspace book makes this look a bit cleaner, though they probably don't point out this specific issue.
     
  8. Oct 1, 2013 #7
    It's clear that there are 4 physical states for N=2 vector multiplet, which can be constructed from one N=1 chiral multiplet ([itex]\phi[/itex], [itex]\psi[/itex]) and one N=1 vector multiplet ([itex]A_\mu[/itex], [itex]\lambda[/itex]). Some paper mentioned that the second set of susy transformation can be obtained from the simple rotation of ([itex]\psi[/itex], [itex]\lambda[/itex]) by [itex]\pi[/itex]. But in this way I got some very strange result about the closure relation.
    I calculated the commutation of the two sets of susy transformation on different states:
    [tex][\delta_{\epsilon_2},\delta_{\epsilon_1}][/tex]
    where the two [itex]\epsilon[/itex]s correspond to the two generators. From the SUSY algebra, we should find zero (ignoring central charge). But according to the transformation law mentioned above, it seems to me there is no way we can get exact cancellation. Am I missing something?

    You mentioned we can work out the explicit expressions of susy transformation. So how do we do that?
    Thanks so much for your replies!
     
  9. Oct 1, 2013 #8

    fzero

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    It is true that an ##N=2## SUSY theory has a ##U(2)## R-symmetry, which is not always manifest for a given choice of fields. In this case, there is a natural ##U(2)## action on the pair ##(\psi,\lambda)##, which is why that paper made the statement. However, it seems a bit easier to just follow the discussion in Terning, specifically eqs (1.58) and (1.59). If we put the SUSY labels back on the supercharges and reorganize, we find that

    $$ |\Omega_{-1}\rangle,~~ Q^\dagger_1 |\Omega_{-1}\rangle, ~~Q^\dagger_2 |\Omega_0\rangle (= |\Omega_{1/2}\rangle) ,~~Q^\dagger_1 Q^\dagger_2|\Omega_0\rangle, $$

    is the ##N=1## vector multiplet, while

    $$ |\Omega_0\rangle,~~ Q^\dagger_1 |\Omega_0\rangle, ~~Q^\dagger_2 |\Omega_{-1}\rangle (= |\Omega_{-1/2}\rangle) ,~~Q^\dagger_1 Q^\dagger_2|\Omega_{-1}\rangle, $$

    is the ##N=1## chiral multiplet. So we have identified ##Q_1## as the ordinary supercharge that takes bosons to fermions and vice-versa, while ##Q_2## is a supercharge that furnishes the CPT conjugate states for one multiplet by acting on the Clifford vacuum of the other multiplet.

    We could probably decompose this into something more symmetric by identifying ##Q_1## with the identity operator in ##U(2)## and ##Q_2## with the traceless diagonal generator of the ##SU(2)\subset U(2)## subgroup.

    If we use the states constructed above, we have, for example,

    $$ \{ Q^\dagger_1 , Q^\dagger_2 \} | \Omega_0\rangle = (Q^\dagger_1 Q^\dagger_2 + Q^\dagger_2 Q^\dagger_1) | \Omega_0\rangle = (Q^\dagger_1 Q^\dagger_2 - Q^\dagger_1 Q^\dagger_2) | \Omega_0\rangle =0,$$

    so you should find the right algebra.

    I think you can put the action of an individual ##Q_a## together with the representation above to work this out. I can probably try to help if you get stuck.
     
  10. Oct 4, 2013 #9

    So before deriving the transformation laws myself, I choose to trust hep-th/9701069, which mentioned the second set of susy transformation can be obtained by the simple replacement of [itex]\lambda \rightarrow \psi[/itex] and [itex]\psi \rightarrow -\lambda[/itex]. Since for N=1 we have
    [tex]\delta A^a_\mu =-\frac{1}{\sqrt{2}}(\epsilon^\dagger \bar{\sigma}_\mu \lambda^a+\lambda^{\dagger a} \bar{\sigma}_\mu \epsilon)[/tex]
    [tex]\delta\phi_j=\epsilon \psi_j[/tex]
    does it mean that for N=2, the infinitesimal transformations corresponding to the second generator look like:
    [tex]\delta A^a_\mu =-\frac{1}{\sqrt{2}}(\epsilon^\dagger \bar{\sigma}_\mu \psi^a+\psi^{\dagger a} \bar{\sigma}_\mu \epsilon)[/tex]
    [tex]\delta\phi_j=-\epsilon \lambda_j[/tex]
     
  11. Oct 4, 2013 #10

    fzero

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    I believe that's correct. If you compare with my post above, you should be able to work out that the first transformation corresponds to ##Q_1##, while the 2nd is related to ##Q_2##.
     
  12. Oct 28, 2013 #11
    Sorry I didn't make my question clear. Here I use two-component notation. So the N=2 SUSY algebra looks like:
    [tex]\{Q^a_\alpha, Q^\dagger_{\dot{\alpha}b}\}=2 \sigma^\mu_{\alpha\dot{\alpha}}P_\mu \delta^a_b, \{Q^a_\alpha,Q^b_\beta\}=2\sqrt{2}\epsilon_{\alpha\beta}\epsilon^{ab}Z[/tex]
    Then we can find that [tex][\delta_{\epsilon_2}, \delta_{\epsilon_1}]=-\frac{1}{2}[\epsilon_2 Q_2+\epsilon^\dagger_2 Q^\dagger_2, \epsilon_1 Q_1+\epsilon^\dagger_1 Q^\dagger_1]=\sqrt{2}(\epsilon_2\epsilon_1 Z^{21}-\epsilon^\dagger_2\epsilon^\dagger_1 Z^*_{21})[/tex] which equals to zero in the absence of central charge.

    This is the result I am trying to verify in terms of the infinitesimial susy transformations in the N=2 case. In other words, I expect to obtain, for example, [itex](\delta_{\epsilon_2}\delta_{\epsilon_1}-\delta_{\epsilon_1}\delta_{\epsilon_2}) \phi = 0,[/itex]. But I didn't get what I expected. Instead, I obtained [itex](\delta_{\epsilon_2}\delta_{\epsilon_1}-\delta_{\epsilon_1}\delta_{\epsilon_2}) \phi=\sqrt{2}\epsilon_1 \epsilon_2 D[/itex]. I also tried [itex]\lambda, A^\mu[/itex] and the results looked even worse. So I don't know what's missing in this logic.
     
  13. Oct 28, 2013 #12

    fzero

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    I'll use (80) and (81) of the Alvarez-Gaume review, namely

    $$\begin{split}
    & \delta_{\epsilon_1} A^a = \sqrt{2} \epsilon_1 \psi^a, \\
    & \delta_{\epsilon_1} \psi^a = \sqrt{2} \epsilon_1 F^a + i \sqrt{2} \sigma^\mu \bar{\epsilon}_1 D_\mu A^a, \\
    & \delta_{\epsilon_1} \lambda^a = \frac{1}{2} \sigma^{\mu\nu}\epsilon_1 {F^a}_{\mu\nu} + i \epsilon_1 D^a.
    \end{split}$$

    The other transformation is obtained from ##\psi\rightarrow \lambda, \lambda\rightarrow \psi##, so

    $$\begin{split}
    & \delta_{\epsilon_2} A^a = \sqrt{2} \epsilon_2 \lambda^a, \\
    & \delta_{\epsilon_2} \lambda^a = \sqrt{2} \epsilon_2 F^a + i \sqrt{2} \sigma^\mu \bar{\epsilon}_2 D_\mu A^a, \\
    & \delta_{\epsilon_2} \psi^a = - \frac{1}{2} \sigma^{\mu\nu}\epsilon_2 {F^a}_{\mu\nu} - i \epsilon_2 D^a.
    \end{split}$$

    Then

    $$ \delta_{\epsilon_2} \delta_{\epsilon_1} A^a = - \sqrt{2}\left( \frac{1}{2} \sigma^{\mu\nu}\epsilon_1\epsilon_2 {F^a}_{\mu\nu} + i \epsilon_1\epsilon_2 D^a\right) ,$$

    while

    $$ \delta_{\epsilon_1} \delta_{\epsilon_2} A^a = \sqrt{2}\left(\frac{1}{2} \sigma^{\mu\nu}\epsilon_2\epsilon_1 {F^a}_{\mu\nu} + i \epsilon_2\epsilon_1 D^a\right).
    $$

    The difference vanishes using the anticommutativity of ##\epsilon_{1,2}##.
     
  14. Oct 28, 2013 #13
    I can get the same equations. The problem is: [itex]\epsilon_1 \epsilon_2 =\varepsilon^{\alpha\beta}\epsilon_{1 \beta}\epsilon_{2 \alpha}=\epsilon_2 \epsilon_1[/itex] (Page 22 in Alvarez-Gaume). So the terms containing [itex]D^a[/itex] don't cancel.
     
  15. Oct 29, 2013 #14

    fzero

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    I see my mistake and, after cleaning up the slopiness in my calculation, I get the same answer up to an overall minus sign. I also verified that


    $$ \{ Q^{(1)}_\alpha, Q^{(2)}_\beta \} A = - i \sqrt{2} \epsilon_{\alpha \beta} D.$$

    I had an idea and double-checked with the Physics Reports by Sohnius (see eq (12.7) if you can find a copy) that this result for the commutator is an artifact of the Wess-Zumino gauge. We can write the RHS as an infinitesimal gauge transformation with ##\Lambda = \sqrt{2} D ##. So the RHS is gauge-equivalent to zero, which is all we can expect in WZ gauge.
     
    Last edited: Oct 29, 2013
  16. Oct 29, 2013 #15
    I am not sure whether gauge transformation works in this case, since [itex]D[/itex] should be gauge-invariant. But if the [itex]D[/itex] on the right-hand side of the commutator is just the coefficient of [itex]\theta^2\bar{\theta}^2[/itex] in vector superfield rather than the one defined as in Page 28 of Alvarez ( one of the two terms in the coefficient of [itex]\theta^2\bar{\theta}^2[/itex]) , gauge transformation would be perfect in this case. Well, I am inclined to believe this [itex]D[/itex] should be understood in the latter way, since it's from susy transformation of the spin one-half fields.
     
  17. Oct 29, 2013 #16

    fzero

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    ##D^a## is adjoint valued, the same as ##A^a_\mu##. The idea here is that we can make a gauge transformation where we set the parameters equal to the ##D^a## (up to sign) to remove the offending term from the SUSY transformation.

    There are two terms in the [itex]\theta^2\bar{\theta}^2[/itex] component of the vector superfield before taking WZ gauge. The ##D## showing up on the RHS of the commutator is the same ##D##, since the WZ gauge left it alone, setting ##C## and the other auxilary fields to zero.
     
  18. Oct 29, 2013 #17
    I mean ##D## doesn't change under a gauge transformation. According to Martin's SUSY Primer:
    for the gauge transformation ##V \rightarrow V+i(\Omega^*-\Omega)## where ##\Omega=\phi+\sqrt{2}\theta\psi+\theta^2 F+……##,
    [tex]
    A_\mu → A_\mu+\partial_\mu(\phi+\phi^*),\\
    \lambda_\alpha \rightarrow \lambda_\alpha,\\
    D \rightarrow D
    [/tex]
    so ##D## couldn't be set to zero after this gauge transformation.
     
  19. Oct 29, 2013 #18

    fzero

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    That is for an abelian gauge theory. In general, ##\delta D^a = i f^{abc} \Lambda^c D^b##. Though it might not be possible to set ##D=0##. In any case, do not set ##D## to zero, we gauge transform the chiral field by ##\delta_g A \sim D## so that

    $$ \left( [\delta_1,\delta_2] + \delta_g \right) A =0.$$

    So the idea is not to make ##D=0##, but to make this commutator zero by mixing SUSY and gauge transformations. I should have written the above equation down before to be clear.
     
  20. Oct 30, 2013 #19
    Oh yes, you are right. Thanks soooooo much, you are of great help.
     
  21. Mar 8, 2014 #20
    A related question. In Alvarez-Gaume, the discrete R transformation is ##\psi\rightarrow -\lambda, \lambda\rightarrow \psi##. There is an extra minus sign. So do you think this is just convention or what? Actually from the lagrangian eq(72) of Alvarez-Gaume, there should be a relative minus sign in order to keep the Lagrangian invariant. But when it comes to abelian case, what we need is the replacement without the minus sign. So I am confused.
    This R symmetry is SU(2) which has three generators. I expect all three kinds of transformations should be the symmetry of the Lagrangian, but it seems this is not the case.
     
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