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In Particle Physics, I've seen Scalar potentials which look like this $$ V = a \Phi^2 + b \Phi^4$$

$\Phi$ is scalar (a number).

What about vector potentials, and spinor potentials? How are they constructed? I know that when constructing expressions, a number of symmetries should be respected but I don't understand the whole story.

Warrick
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user09876
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Potentials are always scalars, so there is no such thing as a vector or spinor potential. There are things called the "vector potential", but this is something else. You can learn the representation theory of the rotation group as follows:

The expression you give from particle physics is not constructed by symmetry, it is constructed from renormalizability. This is the reason that it looks the way it does, with quadratic and quartic pieces only.

The quick and useful way to construct things that are invariant is to understand Einstein summation convention and tensors. If you want to make something that is invariant, it should follow the Einstein convention for index contraction.

There are quantities $T_{abcd...}$ which are tensors, you contract the indices to make the appropriate indices left over. The rotation invariance is SU(2), so that the tensor indices range from 1 to 2 and take complex values. Lorentz invariance is two SU(2)s, so there are two kinds of indices both going from 1 to 2, and the values are complex. In particle physics, you need to understand bigger groups, at the very least SU(3) for the strong interaction, but the index contractions are always how you produce invariants.

The other thing you need are the invariant tensors, which are those tensors whose components are invariant under the group. For SU(N) you have $\delta_{ij}$ and $\epsilon_{i....z}$ where the number of indices is equal to N. You can use invariant tensors in appropriate contractions to make invariant objects.

Ron Maimon
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  • @BB1: I see--- you are considering the "scalar potential" to be the contribution to the particle energy, and then its true that the vector potential adds to the momentum in the same way. I was considering the example of a Lagrangian term. I'll take away my downvote, but it needs an edit on your question. – Ron Maimon Jun 20 '12 at 16:55
  • This is my understanding of how vector potentials arise in QFT from studying David Tong's Lectures (http://www.damtp.cam.ac.uk/user/tong/qft.html or), Susskind's Lectures on Youtube, and the first couple chapters of Zee's QFT in a Nutshell. I have no knowledge of vector potentials being involved in any other fashion. Therefore, I suspect the question is confused. But I admit that my knowledge is not as expansive as it could be. If my answer is wrong, then please explain why it is wrong because I'm curious. – MadScientist Jun 20 '12 at 17:52
  • @BB1: Your answer is ok, but it isn't usually the point of view people take regarding the vector potential. But it's a point of view I like very much! I only downvoted because I understood the question as being about group theory, not about vector potentials. The vector potential produces a potential momentum much like the scalar potential produces potential energy, and this is an important property, and one which comes across in your answer. I am sorry again for downvoting, I didn't understand what you were saying at first, and it looked completely unrelated to the question. – Ron Maimon Jun 20 '12 at 17:54
  • @ Ron Having reread your answer. I believe it is the same as mine, but more general. – MadScientist Jun 20 '12 at 18:09
  • @BB1: No it isn't. Your answer is still completely disconnected from the question, which is about group theory. It's just talking about the vector potential and how you introduce it in the point-particle action. This is a non-sequitor for this question, which is about how you make scalars out of vectors and tensors. – Ron Maimon Jun 20 '12 at 18:10
  • @BB1: I am talking about a completely classical thing, so no renormalization. You can see it in this answer: http://physics.stackexchange.com/questions/11768/does-a-static-electric-field-and-the-conservation-of-momentum-give-rise-to-a-rel/13837#13837 . It's the fact that p-eA is the kinetic momentum mv, so that the total momentum is the sum of the kinetic momentum mv and the "potential momentum" eA. You need to introduce potential momentum, because magnetic interactions aren't equal and opposite, and there is momentum carried in the field. – Ron Maimon Jun 20 '12 at 18:14
  • Isn't that answered, in the case of E&M, by this "Because E&M is relativistic, the Lagrangian should involve a 4-Vector potential and this should be contracted (this is the relativistic method of obtaining scalars from 4-Vectors)." If I changed it to "scalar under the Lorentz Group" would it be correct? – MadScientist Jun 20 '12 at 18:15
  • That is completely different. I agree. If that is what he meant, then I am wrong. But, from what he wrote, my interpretation seems reasonable. The $\phi^4$ and $\phi^2$ scalar potential theories are actually studied in QFT. If you feel the answer is wrong, then downvote it. I don't want to give someone faulty information and that is what the voting process is designed to eliminate. – MadScientist Jun 20 '12 at 18:19
  • @BB1: in a sense, yes, but gravity is also relativistic, and in this case you have a full tensor potential, and the generalization is different. The idea that the vector potential is the generalization of the concept of potential energy to momentum is valid, and is the way gauge invariance was formulated in the 19th century, before the proper quantum formulation was appreciated. As I said, your answer isn't wrong, it's just not an answer to the question OP asked. – Ron Maimon Jun 20 '12 at 19:22