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  Should the (On-shell) (2+1)d $N=2$ Chiral Multiplet Contain Two Scalars and Two Majorana Spinors?

+ 1 like - 0 dislike

In supermultiplets, the bosonic degrees of freedom and the fermionic degrees of freedom need to match in number. The number of degrees of freedom of a field corresponds to the number of independent components of that field.

Now, since the number of degrees of freedom of different fields depends on whether the theory is considered on-shell or off-shell the field content of a given supermultiplet depends on whether the theory is considered on-shell or off-shell. The following is my understanding of how the counting works:

Counting Degrees of Freedom

Off-shell Gauge Boson

In d-dimensions the gauge boson is a d-dimensional vector. Off-shell, a gauge condition reduces to the number of independent components to $d-1$ resulting in $d-1$ independent degrees of freedom.

On-shell Gauge Boson

On-shell, as well as the gauge condition reducing the number of independent components by one, the equation of motion reduces the number of independent components by a further one. This results in $d-2$ degrees of freedom.

Off-shell Dirac/Majorana Fermion

In d-dimensional spacetime, the number of components of a spinor is given by $2^{d/2}$ for even $d$, and by $2^{(d-1)/2}$ for odd $d$. For Majorana spinors these components are real, whilst for Dirac spinors they are complex. Off-shell, these components correspond to the number of degrees of freedom of the spinors.

On-shell Dirac/Majorana Fermion

On-shell, the Dirac equation relates half of these components to each other, and this means that the number of independent components is given by $2^{d/2}$ divided by two, or by $2^{(d-1)/2}$ divided by two, for even and odd dimensional spacetime respectively. This resulting number is the number of independent real and complex degrees of freedom of the Majorana and Dirac spinors respectively.

Now applying this counting to most off-shell or on-shell supermultiplets seems to work well, with the number of bosonic degrees of freedom matching the number of fermionic degrees of freedom. [hep-th/9908075v1, page 7] provides an exception. This source describes a 3d $N=2$ vector multiplet and a 3d $N=2$ chiral multiplet (both on-shell) with the following field content:

Vector Multiplet: Vector field, two Majorana fermions, real scalar field

Chiral Multiplet: Majorana fermion, complex scalar.

The on-shell counting works fine for the vector multiplet. The vector field provides 1 bosonic degree of freedom, the scalar provides 1 bosonic degree of freedom and the two Majorana fermions provide 1 fermionic degree of freedom each. The number of bosonic degrees of freedom matches the number of fermionic degrees of freedom.

However, for the chiral multiplet, the Majorana fermion provides 1 fermionic degree of freedom, whilst the complex scalar provides 2 bosonic degrees of freedom.

I believe this might be a typo, and that the authors of [hep-th/9908075v1] intended to write "two two-component Majorana fermions". In other words the chiral multiplet should contain one more Majorana spinor. Other evidence to support this comes from [hep-th/9703110v1]. On page 3 it states that the chiral and vector multiplets should each contain two bosonic and two fermionic degrees of freedom. On page 6 (of the same source) the authors describe how dualising a vector into a scalar (in 3d) turns the vector multiplet into the chiral multiplet. This would only be the case if the chiral multiplet had the same fermionic content as the vector multiplet (i.e. 2 Majorana fermions).

It could well be that my system for counting degrees of freedom is wrong, so feedback would be greatly appreciated.

This post imported from StackExchange Physics at 2015-07-04 21:44 (UTC), posted by SE-user Siraj R Khan
asked Jul 2, 2015 in Theoretical Physics by Siraj R Khan (105 points) [ no revision ]
Another way to come to the conclusion that the 3d $N=2$ chiral multiplet contains two Majorana fermions is to note that the on-shell $3d$ $N=4$ vector multiplet contains 1 vector, 4 Majorana fermions, and 3 real scalars. This decomposes into a 3d $N=2$ vector multiplet and a 3d $N=2$ chiral multiplet. The former contains a vector, 2 Majorana fermions and 1 real scalar. The two remaining scalars make up the complex scalar in the chiral multiplet, and the chiral multiplet must also contain the two remaining Majorana spinors.

This post imported from StackExchange Physics at 2015-07-04 21:44 (UTC), posted by SE-user Siraj R Khan

1 Answer

+ 1 like - 0 dislike

Yes, you are correct. $\mathcal{N}=2$ supersymmetry in 3d can be obtained by dimensionally reducing 4d $\mathcal{N}=1$. The 4d chiral superfield contains a complex scalar, a Weyl fermion, and a complex auxiliary field. Reducing this to 3d we get a compex scalar, a Dirac fermion, and the auxiliary.

You can also impose reality conditions to rewrite the 4d chiral in terms of two real scalars and a Majorana fermions (plus auxiliary fields). Reducing this to 3d you get two Majorana fermions.

Note that to get the off-shell counting of degrees of freedom right you need to take auxiliary fields into account. Without these the multiplet is only supersymmetric on shell.

This post imported from StackExchange Physics at 2015-07-04 21:44 (UTC), posted by SE-user Olof
answered Jul 4, 2015 by Olof (210 points) [ no revision ]
Thanks for the response Olof!

This post imported from StackExchange Physics at 2015-07-04 21:44 (UTC), posted by SE-user Siraj R Khan

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