# A Why are photons described by real field?

1. Nov 14, 2016

### xxfzero

Why are photons described by real field, while electrons by complex field?

2. Nov 15, 2016

### Simon Bridge

Define the fields, and do the math, and see.

(Aside: are particles described by fields - or are they understood as disturbances of the field?)

3. Nov 15, 2016

### xxfzero

Oh, sorry, field is a loose statement. I just refer to plane wave or other eigenfunctions by "field" (please see the attached figures, both from Scully's Quantum optics, one from pp5, the other pp27).

A complete statement should be that photons are always described by real function when treated as classical field, while are described by Hermitian field operator after quantization; in contrast, electrons are always described by complex wavefunction in quantum mechanics, while are described by non-Hermitian field operator after (second) quantization.

4. Nov 16, 2016

### samalkhaiat

Because they don't carry any kind of charges.
Because they are charged particles. In general, the quanta of real field have vanishing (Noether) charge, while those of complex field carry non-zero charge.

5. Nov 16, 2016

### xxfzero

Oh, thank you very much. So if I can roughly speak that when someone gives me a kind of particle, I should check if it has any kinds of Noether charge, then decide either write it as real field or complex field?

Another question is could you please recommend me some book/chapters where systematically discuss the correlation between real field/complex field and zero/nonzero Noether charges?

Thank you very much once again.

6. Nov 16, 2016

### Orodruin

Staff Emeritus
Gluons carry SU(3) charge, they are described by real gauge fields.

Although you may perfectly well have a complex field that is a gauge singlet.

In the end, it boils down to the adjoint representation being real.

7. Nov 16, 2016

### xxfzero

Oh, thank you very much, too

Could you please recommend me some book/chapters where systematically discuss the correlation between real field/complex field and real/complex adjoint representation?

8. Nov 16, 2016

### samalkhaiat

If you have to include non-Abelian gauge theories into the game, then you need to modify my statement by the phrase “gauge-invariant Noether charge”. The charge, $$Q_{a} = - C_{abc} \int d^{3}x \ G^{0j}_{b} (x) \ A_{cj}(x) ,$$ which is carried by (the self-interacting) non-Abelian gauge fields (i.e., your Gluons charge) is conserved but it is not gauge invariant. You can easily show that under arbitrary gauge transformation $U(x) \in SU(3)$, you get $$\mathbb{Q} \to \int dS_{j} \ U^{-1}(x) \mathbb{E}^{j}(x) U(x) ,$$ where $\mathbb{Q} = Q_{a}T^{a}$, $\mathbb{E}^{j} = T^{a} E^{j}_{a}$ and $E^{j}_{a}(x) = G^{j0}_{a}(x)$ is the chromo-electric field.

9. Nov 16, 2016

### samalkhaiat

No, we don't work this way. If we believe that $G$ is a symmetry of nature, then the irreducible representations of $G$ and experiment tell us about the allowed fields, Lagrangians and charges.

10. Nov 16, 2016

### xxfzero

I asked this question somewhere else and got a similar answer that real/complex field comes from real/complex representation (just experiment not mentioned), is this answer OK?

And, still, I want to know what the correlation between real field/complex field and real/complex representation is, could you please recommend some book/chapters?

11. Nov 17, 2016

### samalkhaiat

No, this is not what I said, and it is not correct. For example, $SU(2)$ has the property that all its representations are real, yet its fundamental representation $[2]$ can be realized in terms of doublets of complex spinor fields like $$q(x) = \begin{pmatrix} u(x) \\ d(x) \end{pmatrix} , l(x) = \begin{pmatrix} e(x) \\ \nu_{e}(x) \end{pmatrix} \ ,$$ and its adjoint representation $[3]$ can have multiplets consist of three real scalar fields (not realized in nature) some times called the phions $$\vec{\varphi} = \begin{pmatrix} \varphi_{1} \\ \varphi_{2} \\ \varphi_{3} \end{pmatrix} \ ,$$ and a triplet containing two complex and one real scalar fields, this is the usual pion triplet $(\pi^{\pm} , \pi^{0})$. Mathematically, the two triplets are equivalent to each other, but the experiment rules out the phions triplet in favour of the pions.
There is no such "correlation". Nothing prevent you from taking complex combinations if the dimension of the representation space is greater than one.
If you have not yet studied group theory, then you should start doing it now.

12. Nov 19, 2016

### xxfzero

OK, thank you. I'll firstly read group theory, esp. representation part, and then return.

13. Nov 21, 2016

### samalkhaiat

Very good, I am sure you will not regret it. Physicists have absolutely no excuse to be illiterates in group theory. However, the answer to your original question (which I gave in #4) requires no deep knowledge in group theory. Let me refine my answer in #4.
1) If $P$ is a particle carrying a conserved global-$U(1)$ charge $Q_{P}$, then its anti-particle $\bar{P}$ must have charge $Q_{\bar{P}} = - Q_{P}$. This means that you need a complex field to describe $P$.
2) If $P$ does not carry any $U(1)$ charge, then $P$ is also its own anti-particle, i.e., $P \equiv \bar{P}$. In this case, $P$ require a real field.
So, real fields cannot carry any $U(1)$ charge. We can actually prove this for mesons and fermions.
1) The mesons ($M$) case: This is easy, recall that the Noether current for meson field has the form $$J_{\mu}(M) = \left(\partial_{\mu}\varphi^{*}\right) \varphi - \varphi^{*} \partial_{\mu} \varphi .$$ Clearly $J_{\mu}(M) = 0$ when $\varphi^{*} = \varphi$.
2) The fermions $F$ case: This is a tricky one. Real spinor (Majorana) fields satisfy the following condition $$\Psi = C \bar{\Psi}^{T} , \ \ \ \ \ \ \ \ \ \ \ (1)$$ where $C = - C^{T} = - C^{-1}$ is the charge conjugation matrix, and the super-script $T$ stands for transpose. We also have the following relation $$\gamma_{\mu} = C^{T}\gamma_{\mu}^{T}C^{-1} . \ \ \ \ \ \ \ \ \ \ (2)$$ Now, the Noether current for spinor field is given by $$J_{\mu}(F) = \bar{\Psi} \gamma_{\mu} \Psi .$$ Substituting (1) and (2), we get $$J_{\mu}(F) = \bar{\Psi}C^{T} \gamma_{\mu}^{T} \bar{\Psi}^{T} .$$ Using the property $(AB)^{T} = B^{T}A^{T}$, we find $$J_{\mu}(F) = \left( C \bar{\Psi}^{T} \right)^{T} \left( \bar{\Psi} \gamma_{\mu}\right)^{T} .$$ Using Eq(1) again, we find $$J_{\mu}(F) = (\Psi)^{T} (\bar{\Psi} \gamma_{\mu})^{T} .$$ Now, one has to be careful when using the transpose property. Because the $\Psi$’s are anti-commuting numbers (real Grassmann numbers), they pick a minus sign when the order is switched. So, $$J_{\mu}(F) = - \left( \bar{\Psi} \gamma_{\mu} \Psi \right)^{T} = - J_{\mu}^{T}(F) .$$ But, $J_{\mu}$ is a number (not a matrix), so $J^{T}_{\mu} = J_{\mu}$. Thus $$J_{\mu}(F) = - J_{\mu}(F) , \ \Rightarrow \ J_{\mu}(F) = 0 .$$