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I am currently studying supersymmetry with the SUSY primer of Stephen P. Martin (https://arxiv.org/abs/hep-ph/9709356) and there seem to be not equally many bosonic and fermionic degrees of freedom (DOF) for a vector superfield as introduced in section 4.5.

The vector superfield $V$ is constructed in a general form by imposing the condition $V=V^*$ on a general superfield $S(x, \theta, \theta^{\dagger})$. From this construction, we get the following vector superfield:

$$ V(x, \theta, \theta^{\dagger}) = a + \theta\xi + \theta^{\dagger}\xi^{\dagger} + \theta\theta b + \theta^{\dagger}\theta^{\dagger} b^* + \theta^{\dagger}\bar{\sigma}^{\mu}\theta A_{\mu} + \theta^{\dagger}\theta^{\dagger}\theta\left(\lambda - \frac{i}{2}\sigma^{\mu}\partial_{\mu}\xi^{\dagger}\right) + \theta\theta\theta^{\dagger}\left(\lambda^{\dagger} - \frac{i}{2}\bar{\sigma}^{\mu}\partial_{\mu}\xi\right) + \theta\theta\theta^{\dagger}\theta^{\dagger}\left(\frac{1}{2}D+\frac{1}{4}\partial_{\mu}\partial^{\mu}a\right) $$

The fields appearing in $V(x, \theta,\theta^{\dagger})$ are:

  • A gauge boson $A^{\mu}$
  • A gaugino $\lambda$
  • An auxilliary real scalar field $D$
  • An auxilliary real scalar field $a$
  • An auxilliary complex scalar field $b$
  • An auxilliary two-component fermion field $\xi$

The fields $A^{\mu}$, $\lambda$ and $D$ are what we expect to obtain; these are also the fields we end up with if we construct a supersymmetric Lagrangian without the use of superfields. The fields $a$, $b$, and $\xi$ are new in that sense, they are not physical degrees of freedom and will not appear in the Lagrangian of a gauge supermultiplet because we can supergauge them away.

In supersymmetry, we want to have equally many fermionic and bosonic DOF, both on-shell and off-shell as I understand it. And if I am not mistaken, we get the following number of DOF for each type of field:

  • Gauge boson: on-shell: 2, off-shell: 3
  • Two-component spinor: on-shell: 2, off-shell: 4
  • Real scalar: on-shell: 1, off-shell: 1
  • Complex scalar: on-shell: 2, off-shell: 2

Additionally, we only get off-shell DOF for auxilliary fields.

If we now add up the DOF of all particles in the vector superfield $V$, we get a total of 4 on-shell bosonic and fermionic DOF and 4 off-shell bosonic and fermionic DOF from $A_{\mu}$, $\lambda$ and $D$. These are the fields that will appear in our Lagrangian and as expected, there is an equal number of fermionic and bosonic DOF. But we get an additional 3 off-shell bosonic DOF from $a$ and $b$ and 4 off-shell fermionic DOF from $\xi$. So in total, we have 7 off-shell bosonic DOF and 8 off-shell fermionic DOF which clearly aren't equally many.

How is this possible in a supersymmetric theory? I would expect that we get an equal number of fermionic and bosonic DOF both on-shell and off-shell, even for particles that can be supergauged away.

How can this discrepancy be resolved? Is there some reason why the fermionic DOF do not need to match the bosonic DOF for these auxilliary fields?

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  • $\begingroup$ I haven't looked at the details but physics only cares about on-shell degrees of freedom. In the Lagrangian formalism (which makes it meaningful to talk about on vs off shell), things are most convenient when off-shell degrees of freedom match but this is not always automatic. $\endgroup$ Nov 14, 2023 at 20:41
  • $\begingroup$ @Connor Behan In SUSY it actually matters that the off-shell DOF also match because otherwise, the Lagrangian would not be invariant under supersymmetry transformations. This is why the auxilliary field $D$ must be introduced. So it appears that for some fields it matters that on and off shell DOF match while for those that can be supergauged away it does not, which is confusing. $\endgroup$
    – FlavonBSM
    Nov 14, 2023 at 20:55
  • $\begingroup$ Theories whose symmetries cannot be made manifest in a Lagrangian are still completely fine. I agree with you that this isn't likely to happen here though. $\endgroup$ Nov 15, 2023 at 1:29

1 Answer 1

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I think the issue is that before gauge fixing, the vector $A^\mu$ has 4 offshell degrees of freedom, not 3. That is the extra bosonic degree of freedom you are looking for, to make 8=8. When you gauge-fix in Wess-Zumino gauge, as you would do if you work in components without superfields, then there will only be 3 vector degrees of freedom off-shell, but then only the gaugino $\lambda$ and the auxiliary field D are left, so again the degrees of freedom match, 4=4.

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  • $\begingroup$ I agree with you that if $A^{\mu}$ has 4 offshell degrees of freedom it will work out. But what happens to the one extra degree of freedom we seem to lose if we gauge fix in Wess-Zumino gauge? Somehow we need to get rid of one degree of freedom of $A^{\mu}$ and I don't see how this works... $\endgroup$
    – FlavonBSM
    Nov 30, 2023 at 14:12

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