Timeline for What is the difference between a complex scalar field and two real scalar fields?
Current License: CC BY-SA 2.5
24 events
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| Oct 26, 2022 at 10:05 | comment | added | infinitezero | 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. | |
| Dec 5, 2019 at 22:58 | comment | added | Andrew Steane | Yes, and I agree with you. | |
| Dec 5, 2019 at 16:37 | comment | added | Vladimir Kalitvianski | As I stated in my answer, many things depend on the boundary and initial conditions. | |
| Dec 5, 2019 at 13:21 | comment | added | Andrew Steane | 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. | |
| Oct 12, 2018 at 18:11 | comment | added | Vladimir Kalitvianski | @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 $\psi$. | |
| Oct 11, 2018 at 4:59 | comment | added | Michael | 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. | |
| Oct 3, 2018 at 6:55 | comment | added | Vladimir Kalitvianski | @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. | |
| Oct 2, 2018 at 16:26 | comment | added | Michael | 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? | |
| Oct 1, 2018 at 14:46 | comment | added | Vladimir Kalitvianski | @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? | |
| Oct 1, 2018 at 3:53 | comment | added | Michael | 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? | |
| Sep 16, 2018 at 19:05 | comment | added | Vladimir Kalitvianski | @Michael: read my answer carefully and think it over. | |
| Sep 16, 2018 at 15:12 | comment | added | Michael | 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 | |
| Jan 31, 2011 at 16:22 | comment | added | Vladimir Kalitvianski | Dear downvoters, tell me where I am wrong, please. I would like to learn. | |
| Jan 22, 2011 at 20:07 | history | edited | Vladimir Kalitvianski | CC BY-SA 2.5 | added 1 characters in body |
| Jan 22, 2011 at 10:41 | comment | added | Vladimir Kalitvianski | 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. | |
| Jan 22, 2011 at 8:12 | comment | added | Luboš Motl | 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. | |
| Jan 21, 2011 at 23:45 | comment | added | Vladimir Kalitvianski | I do not know. He posted his comments two hours later. | |
| Jan 21, 2011 at 23:36 | comment | added | Carl Brannen | Interesting. You're saying essentially the something quite similar to what Dr. Motl said in the comments. | |
| Jan 21, 2011 at 21:34 | comment | added | Vladimir Kalitvianski | I've got -2 for my explanation. Would you be so kind to point out where I am wrong, please? | |
| Jan 21, 2011 at 21:23 | history | edited | Vladimir Kalitvianski | CC BY-SA 2.5 | added 105 characters in body |
| Jan 21, 2011 at 21:03 | history | edited | Vladimir Kalitvianski | CC BY-SA 2.5 | added 10 characters in body |
| Jan 21, 2011 at 18:04 | comment | added | user346 | This is a great observation. +1 | |
| Jan 21, 2011 at 17:33 | history | edited | Vladimir Kalitvianski | CC BY-SA 2.5 | added 19 characters in body |
| Jan 21, 2011 at 17:23 | history | answered | Vladimir Kalitvianski | CC BY-SA 2.5 |