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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?

Qmechanic
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Kostya
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    Actually, $\phi \equiv \left( \phi_1 + i \phi_2 \right)/\sqrt{2}$. – QGR Jan 21 '11 at 16:30
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    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

5 Answers5

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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.

David Z
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pho
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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
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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.

QGR
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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. – Zo the Relativist Jan 21 '11 at 19:22
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    @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 02:40
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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.

Vladimir Kalitvianski
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    This is a great observation. +1 –  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
  • I do not know. He posted his comments two hours later. – Vladimir Kalitvianski Jan 21 '11 at 23:45
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    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 08:12
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    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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    Dear downvoters, tell me where I am wrong, please. I would like to learn. – Vladimir Kalitvianski Jan 31 '11 at 16:22
  • You are wrong that there is a difference because the Lagrangians are exactly the same on the most fundamental mathematical level by setting $\phi=\phi_1+i\phi_2$, which is the completely general expression for this complex scalar field in terms of real scalar fields – Michael Sep 16 '18 at 15:12
  • @Michael: read my answer carefully and think it over. – Vladimir Kalitvianski Sep 16 '18 at 19:05
  • I disagree because there is no mathematical difference in any way and therefore they describe exactly the same thing. Take the following two problems 1.ax+b=c solve for x and 2.ay+b=c solve for y Mathematically they are the same describing a straight line. The solutions are the same.

    Arguing that the two real scalar fields are different than one complex scalar field for the reasons you give above is the same as arguing that the two example equations I wrote are different because one has y in them and the other x.

    Does this clarify the issue?

     – Michael Oct 01 '18 at 03:53
  • @Michael: No, it does not. If I take two different scalar fields describing different particles, I am not allowed to make any superposition. Superselection rules forbid that. You do not make a superposition of electron and proton, do you? – Vladimir Kalitvianski Oct 01 '18 at 14:46
  • This is a misunderstanding you follow and I think I figured out your misunderstanding and it is in no way related to interference. Sure two complex fields act differently than two real ones because they correspond to 4 real ones. But here they are talking about two real valued fields that mathematically and therefore in phenomenology are exactly the same as one complex. Any complex field can be decomposed in two real ones. This is the reason why for the complex klein Gordon field when quantizing it you end up with creation operators for two different species. Does this clear it up? – Michael Oct 02 '18 at 16:26
  • @Michael: No. Your "mathematically" is different from my "mathematically" because of quite different "boundary" conditions implied by me. For example, in a problem when one neutral particle is absent and another one is scattered or interacts with something else. – Vladimir Kalitvianski Oct 03 '18 at 06:55
  • Expanding out a complex field $\psi$ by writing it in terms of two real fields $\psi=\psi_1+i\psi_2$ neither changes boundary conditions nor anything else and at the most fundamental level is a simple rewriting. There is just no other way to spin it. Please note that boundary conditions for your fields have to be introduced only after you write down your field equations and attempt to solve them. These boundary conditions can then be freely chosen to fit your needs. For example if you have fields confined to a container you may want the fields to vanish on the inner container wall etc. – Michael Oct 11 '18 at 04:59
  • @Michael: Yes, there is another way "to spin it", if you like, for example: $\psi=a\psi_1+b\psi_2$, where $a$ and $b$ are independent complex numbers. But I do not see any meaning in such a $\psi$. – Vladimir Kalitvianski Oct 12 '18 at 18:11
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    I just read this conversation; I note that Michael keeps returning to the complex field. But it is not disputed that a complex field can be treated as pair of real ones. What is disputed is whether or not the Lagrangian on its own dictates the physics, and thus whether or not it you have two real fields whose Lagrangian maps to that of a single complex field, it amounts to the same physics. Vladimir is saying that the Lagrangian alone does not dictate all the physics. Whether or not he is right, that is the question one has to address. – Andrew Steane Dec 05 '19 at 13:21
  • As I stated in my answer, many things depend on the boundary and initial conditions. – Vladimir Kalitvianski Dec 05 '19 at 16:37
  • Yes, and I agree with you. – Andrew Steane Dec 05 '19 at 22:58
  • Very late to the party, but I think the issue here is that $\psi=\psi_1+i \psi_2$ only holds true if all the quantum numbers of psi equal those of $\psi_1 and \psi_2$. Michael assumes this implicitly while Vladimir says this is not a given. More generally $\psi(q_1,...,q_n) = \psi_1(r_1,...,r_n) + i \psi_2(s_1,...,s_n)$ only if $q_i = r_i = s_i$. While it is always possible to deconstruct $\psi(q_1,...q_n) = \psi_1(q_1,...,q_n) + i \psi_2(q_1,...,q_n)$ this clearly falls apart for the other direction if the second lagrangian in the OP would make a difference between the masses. – infinitezero Oct 26 '22 at 10:05
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they are equivalent from a physics point of view and can be mapped into each other.

Tania Robens
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    Hi and welcome to the Physics SE! Please note that you should try to write fuller answers, at least when other answers, such as pho's, have already given the same information as yours in a more comprehensive manner. – stafusa Dec 05 '19 at 12:25
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I think the free Lagrangian alone does not give the physical content. We can also alternatively represent $\phi = \phi_0 \exp(i \theta)$. Then we have $$ L = { 1 \over 2} \partial^\mu \phi_0 \partial_\mu \phi_0 + m^2 \phi_0^2 + {1 \over 2} \partial^\mu \theta \partial_\mu \theta $$ Here we can also ask whether we have one charged massive field or one massive neutral field and one massless one. In order to decide the field content, one must couple the scalar field with the vector field or spinor field. The complex scalar representation can have a coupling with the vector gauge field, while the two real scalar representation does not have one. Now I have another question: If we want to couple a complex scalar field with a Dirac spinor, how can we choose from the two following $$ L_1 = \bar \psi (\phi^\dagger + \phi) \psi $$ or alternatively $$ L_2 = i \bar \psi (\phi^\dagger - \phi) \psi $$ And what is the physical meaning of the above two interactions?

user43442
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  • Your Lagrangian in polar coordinates is not right, you need a $\rho^2$ in front of the $\theta$ kinetic term. It must be different from what you wrote, since the spectrum doesn't depend on the choice of fields. – Javier Sep 16 '18 at 14:24