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Consider a complex scalar field $\phi$ with the Lagrangian:
$L = \partial_\mu\phi^\dagger\partial^\mu\phi - m^2 \phi^\dagger\phi$

Consider also two real scalar fields $\phi_1$ and $\phi_2$ with the Lagrangian:
$L = \frac12\partial_\mu\phi_1\partial^\mu\phi_1 - \frac12m^2 \phi_1^2 +\frac12\partial_\mu\phi_2\partial^\mu\phi_2 - \frac12m^2 \phi_2^2$

Are these two systems essentially the same? If not -- what is the difference?

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$\phi = \phi_1 + i \phi_2$ maps the two to each other, these are identical. – user566 Jan 21 '11 at 16:17
What about $C$? – Kostya Jan 21 '11 at 16:25
Actually, $\phi \equiv \left( \phi_1 + i \phi_2 \right)/\sqrt{2}$. – QGR Jan 21 '11 at 16:30
Kostya: the charge conjugation exchanges $\phi$ with $\phi^\dagger$. Because $\phi=(\phi_1+i\phi_2)/\sqrt{2}$ and $\phi^\dagger=(\phi_1-i\phi_2)/\sqrt{2}$, it follows that the exchanging of $\phi$ and $\phi^\dagger$ in this case is simply $\phi_2\to-\phi_2$ while $\phi_1$ is kept fixed. We say that $\phi_1$ is C-even while $\phi_2$ is C-odd. However, if $\phi$ is charged under any continuous symmetry, such as $U(1)$, it would be silly to decompose it into two parts. However, the message that the C-conjugation may look ad hoc is completely valid. C is not a God-given symmetry. – Luboš Motl Jan 21 '11 at 19:23
Well, it doesn't matter when one discusses the equivalence of theories. But it does matter if you want to find a particular field redefinition that makes two expressions equal. – Luboš Motl Jan 22 '11 at 8:10
up vote 18 down vote accepted

There are some kind of silly answers here, except for QGR who correctly says they are identical. The two Lagrangians are isomorphic, the fields have just been relabeled. So anything you can do with one you can do with the other. The first has manifest $U(1)$ global symmetry, the second manifest $SO(2)$ but these two Lie algebras are isomorphic. If you want to gauge either global symmetry you can do it in the obvious way. You can use a complex scalar to represent a single charged field, but you could also use it to represent two real neutral fields. If you don't couple to some other fields in a way that allows you to measure the charge there is no difference.

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And normalization conditions? Are they the same? – Vladimir Kalitvianski Jan 21 '11 at 19:28
Normalization conditions for one particle field is different from that for two independent other ones. Besides, if the neutral scalars are different (in some other quantum number than mass), you cannot make their superposition (no SU(2)). U(1) assumes such a superposition. – Vladimir Kalitvianski Jan 21 '11 at 19:55

They're identical. Typically, we use complex fields if we have a $U(1)$ symmetry, or some more complicated gauge group with complex representations.

Incidentally, the same comment applies to whether we use Majorana spinors or Weyl spinors.

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I thought the mass terms were different for Majorana and Weyl spinors--in particular, if the neutrino masses were Majorana masses, we could identify left-handed neutrinos with antineutrinos, while if they were Weyl masses, it is necessary that there be sterile neutrinos. – Jerry Schirmer Jan 21 '11 at 19:22
@Jerry; You are confusing Dirac masses with Weyl masses. A Weyl mass is the same as a Majorana mass in 4d. – Ron Maimon Aug 13 '11 at 2:40

A complex scalar field represents a single charged particle whereas two real scalar fields may represent two independent neutral particles. The difference is easy to note while imposing physical initial, boundary and/or normalization conditions which essentially depend on what you are describing - one charged or two different neutral particles. Two independent neutral scalars do not obey a superposition principle, one cannot mix them in one field.

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This is a great observation. +1 – user346 Jan 21 '11 at 18:04
I've got -2 for my explanation. Would you be so kind to point out where I am wrong, please? – Vladimir Kalitvianski Jan 21 '11 at 21:34
Interesting. You're saying essentially the something quite similar to what Dr. Motl said in the comments. – Carl Brannen Jan 21 '11 at 23:36
It's not true that the two ways to describe the complex field differ in the ability to interfere. The two expressions are fully equivalent. Charged particles have to be excitations of complex fields but that's an entirely different question than interference. Much like $\phi_1$ and $\phi_2$ don't interfere with each other, $\phi$ and $\phi^\dagger$ don't interfere with one another. It's the same thing, just a different basis. – Luboš Motl Jan 22 '11 at 8:12
I agree that a charged particle described with a complex field $\phi$ which is decomposed into two real components like $\phi_1 + i\phi_2$ . I disagree that two neutral independent particles describe a charged one. – Vladimir Kalitvianski Jan 22 '11 at 10:41

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