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Dimensions of fields in QFT

  1. May 1, 2009 #1
    I have a question about the dimensions of quantum fields. In natural units, the dimensions of bosonic fields (both scalar and vector) is 1. The dimension of spin=1/2 fermion fields is 3/2. This is all very good, but I have never read any explanation anywhere why we cannot have other types of fields where the dimension of the field is different from those mentioned above. For example, is there a reason why a field (of any spin) cannot have dimension= 1/2? For such a scalar field f, for example, there will be a term f^8 in the Lagrangian density with a dimensionless coupling constant in 4 space-time dimensions. For such a term, naive dimensional counting appears not to rule out renormalizibility outright. (In contrast, ordinary scalar fields of dimension=1 cannot have a renormalizable f^8 interaction.)

    How about something even more exotic: why can't we have spin=1/2 fermion fields of dimension=1, instead of 3/2? Is there some simple argument or symmetry in QFT that forbids such exotics?
    Last edited: May 1, 2009
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  3. May 1, 2009 #2


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    The action must be dimensionless, which means the Lagrangian density must have dimensions of mass^d in d dimensional spacetime. The standard kinetic terms for scalars and vectors are of the form [itex] (\partial phi)^2[/itex], and since a partial derivative has dimensions of mass, this fixes the dimension of the fields at (d-2)/2. A similar argument shows the dimension of a spinor field with the usual kinetic term is (d-1)/2.

    If you don't have a kinetic term, your fields don't propagate, that is, the equations of motion are algebraic, and the fields basically don't enter into the theory. If you have a weird kinetic term, such as one with three derivatives of more than two powers of the field, things get strange. For example, the field equations might become differential equations of higher than second order, and so initial values for the fields and their first time derivatives don't give unique evolution. Other than this, I don't know much about non-standard kinetic terms, other than to avoid them.
  4. May 1, 2009 #3
    Not having a kinetic term is not an option. Having more than two derivatives may not be so bad. What is wrong if the initial conditions and the first derivatives do not give unique classical solutions? Still, I am not really interested in these really weird cases. There are many interesting cases that do not fall into these two categories you mention. The most interesting options involve one or two powers of derivatives multiplied by other powers of fields. For example, for dimension=1/2 scalar field, a term in the form [tex]\phi[/tex]2 [itex] (\partial \mu[/tex][tex]\phi[/tex] [itex]) (\partial \mu[/tex][tex]\phi[/tex]) appears perfectly OK.

    I can think of a very rich set of non-linear terms, especially if these are multiplets transforming under some gauge group.
  5. May 2, 2009 #4


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    It's true, there are things called non-linear sigma models, where the kinetic term looks like:

    [tex] \int d^d x g^{\mu \nu}(\phi) \partial_\mu \phi \partial_\nu \phi [/tex]

    where [itex]\phi[/itex] are coordinates on a target manifold, and g is the pull-back of the metric. In general, g depends non-trivially on [itex]\phi[/itex], and you get terms like the one you mentioned. But the leading term is still the usual kinetic term, with the others typically treated as a perturbation. I've never seen a model without such a leading term. It seems to me like this would come about from expanding around a singular point on the manifold, which doesn't seem like a good idea, but I'm sure it's come up somewhere.

    In fact, it seems like all such bad kinetic terms come because of expanding around a bad point in field space. In all these cases, couldn't you just redefine the fields by shifting them by a constant and get the usual kinetic term back? Of course, this constant would have the same dimensions as your field, so would basically mean the kinetic term is weighted by a dimensionful coupling constant, another thing I haven't seen, but which seems somewhat more reasonable, and would get around the argument I gave in my earlier post.
    Last edited: May 2, 2009
  6. May 2, 2009 #5
    Is the metric a function or functional of [tex] \phi [/tex]? And when you say g is the pullback of the metric, do you mean a pull-back to a flat space [tex]d^dx[/tex] (or what's the manifold you're pulling back to?). Don't all the your phi's and partials in your Lagrangian have to be pull-backs and not just the metric?
  7. May 2, 2009 #6


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    The "fields" are maps from d dimensional spacetime into the target manifold M of, say, n dimensions. In other words, in doing the path integral, we integrate over all maps f:Rd->M. The field I called [itex]\phi[/itex] is really a collection of scalar fields [itex]\phi^i[/itex] which represent coordinates on M, so that a map f is represented by n scalar fields.

    Then the kinetic term is obtained as follows (what I said above was a little sloppy). First look at the differential of f, which is a map between the tangent spaces. Now, given two tangent vectors in spacetime, this map gives two vectors tangent to M, whose inner product we can take using the metric on M. But this procedure can be thought of as a bilinear map on tangent vectors in spacetime, and we can take the trace of this map using the Minkowski metric. In coordinates we get:

    [tex] g_{ij}(\phi) \eta^{\mu \nu} \partial_\mu \phi^i \partial_\nu \phi^j [/tex]

    For example, if we go to Riemann normal coordinates, the metric expands in such a way that the above becomes something like (I probably have coefficients wrong):

    [tex] \left( \delta_{ij} + R_{ikjl} \phi^k \phi^l \right) \eta^{\mu \nu} \partial_\mu \phi^i \partial_\nu \phi^j [/tex]

    So we see we get the normal kinetic term plus terms like [itex]\phi^2 (\partial \phi)^2[/itex].
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