# (1/2,1/2) representation and photon

## Main Question or Discussion Point

I'm considering that, the $$(1/2,1/2)$$ representation of Lorentz group is the four-vector representation, and this four-vector representation has spin 0,1.
As we know, this four-vector representation can serve for the gauge field $$A_\mu$$, i.e. photon. However, the photon has spin 1. So we face a problem---how to get rid of the spin 0 part of (1/2,1/2) representation.

I found in a QFT book by Maggiore, after quantizing the EM field, she computed the angular momentum operator from Noether theorem, then extracting out the intrinsic spin part (throwing away the orbital angular momentum part); finally, computing the eigenvalues of intrinsic spin operator on one particle state, she showed that the circular polarized one particle states are eigenstates of the intrinsic spin operator and the eigenvalues are 1 and -1. This shows that the spin of photon is 1.

My question is, is there a concrete way to realize that the spin-0 part of (1/2,1/2) field representation is ruled out by some mechanism? (gauge symmetry or...) Because by explicit quantizing the theory and compute the spin eigenvalues of one particle state is a tough work. Moreover, in other theory, we also take the four-vector field $$A_\mu$$, i.e. (1/2,1/2) representation as a spin-1 particle, so probably there is a way to understand that spin-0 part is ruled out?

Any ideas or instructions will be appreciated.
Sincerely

## Answers and Replies

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Avodyne
Science Advisor
Take a look at volume I of Weinberg. This has a good discussion of this issue. I don't have it in front of me, but if IIRC the spin-zero part is removed by writing the kinetic term as $F^{\mu\nu}F_{\mu\nu}$ rather than as a more general linear combination of $\partial^\mu A^\nu\partial_\mu A_\nu$ and $\partial^\mu A^\nu\partial_\nu A_\mu$; note that $F^{\mu\nu}$ transforms as $(3,1)\oplus(1,3)$, so it only includes spin one. Gauge symmetry then removes the $m{=}0$ component of the spin-one part, so that only helicities $m{=}{\pm}1$ remain.

Take a look at volume I of Weinberg. This has a good discussion of this issue. I don't have it in front of me, but if IIRC the spin-zero part is removed by writing the kinetic term as $F^{\mu\nu}F_{\mu\nu}$ rather than as a more general linear combination of $\partial^\mu A^\nu\partial_\mu A_\nu$ and $\partial^\mu A^\nu\partial_\nu A_\mu$; note that $F^{\mu\nu}$ transforms as $(3,1)\oplus(1,3)$, so it only includes spin one. Gauge symmetry then removes the $m{=}0$ component of the spin-one part, so that only helicities $m{=}{\pm}1$ remain.
I glanced over volumeI of Weinberg's, the subject seems to be at the end of chap 5. I will study it soon. Thank you!

However, I have another related question.
Before I carefully study the field representation of Lorentz group, I always obtain the spin of a certain field by counting the number of Lorentz indices of that field, for example, I thought $$\partial_\mu A_{\nu\rho}$$ is a spin-3 field.
But things seem to be more subtle than that.
Is the counting of spin by the number of Lorentz indices correct?

Avodyne
Science Advisor
Is the counting of spin by the number of Lorentz indices correct?
No. You need to know the rep in the (m,n) format, where m and n are integers or half-integers. Then the allowed spins are |m-n|, ..., m+n. So the vector, which is (1/2,1/2), has spins 0 and 1. The symmetric two-index tensor, which is in (0,0)+(1,1), has spins 0,1,2.

Oh, and when I wrote (3,1)+(1,3) before, I was using the notation (2n+1,2m+1), so I should have written (1,0)+(0,1) in (m,n) notation.

samalkhaiat
Science Advisor
Any ideas or instructions will be appreciated.
(1/2,1/2) is an IRREDUCIBLE representation of SO(1,3) = SU(2) X SU(2). So, by definition, there is no (0,0) representation in (1/2,1/2). See my posts (only my posts) in

https://www.physicsforums.com/showthread.php?t=192572

When you have irreducible rep. (K,J), the spin of this rep. is S = K + J, its dimension is d = [(2J + 1)(2K + 1)] and the eigenvalues of $S_{z}$ are
(J + K), (J + K) - 1, ...., -(J + K) for massive representations and (J + K), -(J + K) for massless representations.
Remember, when we mention the mass, the group is no longer SO(1,3), it is Poincare. However, we can still do the business using SO(1,3) + field equations, i.e., without using the Casimir operators of the Poincare algebra.

Since the dimension of (1/2,1/2) is d = 4, a 4-component field $A_{\mu}$ can be used to describe the spin = 1 rep. of SO(1,3) provided we forbid the existence of any scalar function (i.e., spin-0 object) built out of $A_{\mu}$. Hence the condition

$$\partial^{\mu}A_{\mu} = 0$$

To find the field equation for an arbitrary vector field $A_{\mu}$, we assume;
1) there exists no other vector field independent from $A_{\mu}$
2) free fields satisfy linear PDE's of order not greater than 2.
3) covariance, i.e., tensorial equations

Thus the most general field equation for an arbitrary vector field has the form

$$\partial^{2}A_{\mu} + a \partial_{\mu}(\partial . A) + b A_{\mu} = 0$$

where (a,b) are real constants. Therefore, for (1/2,1/2), it is

$$\partial^{2}A_{\mu} + m^{2}A_{\mu} = 0 \ \ \ \ (1)$$

with

$$\partial^{\mu}A_{\mu} = 0 \ \ \ \ \ \ (2)$$

When one considers Poincare' group, one can show that m is the mass of (1/2,1/2);

If we put $A_{\mu} = c_{\mu}e^{ip.x}$ in eq(1) and (2), we find

$$p^{2} = m^{2}$$

and

$$p^{\mu}A_{\mu} = p^{\mu}c_{\mu} = 0 \ \ \ \ (3)$$

Now, in the rest frame $p_{\mu} = ( m,\vec{0})$, eq(3) gives $A_{0} = c_{0}= 0$. Therefore, in the rest frame of a massive vector field, there are only three independent components corresponding to the three eigenvalues of the massive spin-1 rep. of Poincare group.
Notice that eq(1) and eq(2) are equivalent to

$$\partial^{\nu}F_{\nu \mu} + m^{2}A_{\mu} = 0 \ \ \ \ (4)$$

where $F_{\mu \nu}$ is the antisymmetric tensor rep. $(0,1) \oplus (1,0)$ of SO(1,3). Eq(4) is the Proca equation which describes massive vector meson.

The case of massless spin-1 field can be obtained from eq(4) by putting m = 0, the result is Maxwell equation $\partial^{\nu}F_{\nu \mu} = 0$. This means that the EM field, when its quantum nature is fully exploited, will correspond to massless, spin one particles. The only difference from the massive case is the invariance of Maxwell's equation under the gauge transformation

$$\bar{A}_{\mu} = A_{\mu} + \partial_{\mu}f \ \ \ \ (5)$$

The choice of the scalar gauge function f imposes extra condition on $A_{\mu}$, and this reduces the physically significant components to only two. To see this, note first that Maxwell's equations allow the non-trivial solutions (i.e., cannot be completely gauged away)

$$A_{\mu} = c_{\mu} e^{ip.x}$$

provided $p^{2} = 0$ (massless photons), and

$$p^{\mu}c_{\mu} = 0 \ \ \ \ (6)$$

This means that some components of the polarization vector $c_{\mu}$ do not carry physical information. To see this, we choose the gauge function

$$f(x) = i a e^{ip.x}$$

This leads to

$$\bar{A}_{\mu} = (c_{\mu} - a p_{\mu}) e^{ip.x}$$

Thus, under gauge transformation, the polarization vector changes according to

$$c_{\mu} \rightarrow \bar{c}_{\mu} = c_{\mu} - a p_{\mu} \ \ \ (7)$$

Now take a photon traveling along the z-axis, i.e.,

$$p_{\mu} = \left( -|k|, 0 , 0 , k \right)$$

From eq(6), we find

$$c_{0} = c_{3}$$

Then, from eq(7) we see that

$$\bar{c}_{\mu} = \left( c_{0} + a|k|, c_{1}, c_{2}, c_{0} + a|k|\right)$$

Thus, if we choose

$$a = - \frac{c_{0}}{|k|}$$

we have gauged away the scalar photon $c_{0}$ and the longitudinal photon $c_{3}$ completely. This means that all physical information is carried by the transverse photons $( c_{1},c_{2})$.

By preforming a Lorentz transformation, we can go one step further and show that the field for a scalar or a longitudinal photon carries spin projection $S_{z} = 0$, while the fields for transverse photons carry spin projection $S_{z} = \pm 1$. We therefore say that the (observed = physical) photon has helicity $\pm 1$.

regards

sam

Apologize that I don't have enough time now, so I can only reply your nice discussion shortly.
(1/2,1/2) is an IRREDUCIBLE representation of SO(1,3) = SU(2) X SU(2). So, by definition, there is no (0,0) representation in (1/2,1/2). See my posts (only my posts) in

https://www.physicsforums.com/showthread.php?t=192572
I thought (1/2,1/2) is a reducible representation of SO(1,3), since it is the direct product of two SU(2)'s. Therefore, (1/2,1/2) representation can be divided into total spin = 1 and 0, right?
Where did I got lost?
When you have irreducible rep. (K,J), the spin of this rep. is S = K + J, its dimension is d = [(2J + 1)(2K + 1)] and the eigenvalues of $S_{z}$ are
(J + K), (J + K) - 1, ...., -(J + K) for massive representations and (J + K), -(J + K) for massless representations.
Actually I finished the study of Weinberg's section 5.9 just now, and he said that for massless particles, the representation (K,J) can only describe massless particles with helicity (eigenvalue of $$S_z$$) J - K. That's the reason why (1/2,1/2) representation, i.e. four vector representation can not describe photon directly, sine photon is a massless particle with helicity $$\pm 1$$. The way he solved this is, constructing the antisymmetric fields, $$\partial_\mu A_\nu - \partial_\nu A_\mu$$.

But he mentioned that, higher spin representation would produce short distance interaction. So, in order to satisfy the inverse squared law, we still choose the four-vector representation to describe the photon.

Actually, I didn't get it much.
By the way, many thanks for your nice article.

samalkhaiat
Science Advisor
I thought (1/2,1/2) is a reducible representation of SO(1,3), since it is the direct product of two SU(2)'s. Therefore, (1/2,1/2) representation can be divided into total spin = 1 and 0, right?
No, (1/2,1/2) is a direct product of two inequivalent fundamental representations

$$(1/2,0)\otimes (0,1/2) = (1/2,1/2) \ \ \ (1)$$

So, it is irreducible because it contains no invariant subspaces, i.e., we cannot write it as a direct sums of irreducible representations.

Where did I got lost?
You are confusing (1/2,1/2) with the direct product of two identical fundamental representations which is reducible. we can make two of them;

$$(1/2,0) \otimes (1/2,0) = (0,0) \oplus (1,0) \ \ \ (2)$$
$$(0,1/2) \otimes (0,1/2) = (0,0) \oplus (0,1) \ \ \ (2')$$

Indeed, we do not have a bracket convention for reducible representation. So, by (j,k) we mean irreducible representation of spin

$$S(j,k) = j + k$$

and dimension

$$D(j,k) = (2j + 1)(2k + 1)$$

On Minkowski space, geometrical object with (2j + 1)(2k + 1) independent components is called spin = (j + k) field. Indeed, the geometrical object

$$\Psi_{(r_{1}...r_{2j})(\dot{r}_{1}...\dot{r}_{2k})}(x)$$

symmetrical under the permutation of the undotted indices among themselves and of the dotted indices among themselves separately, span an irreducible representation space,(j,k), of the Lorentz group SU(2)XSU(2).

So,

S(0,0) = 0, D(0,0) = 1, and $\Psi_{0\dot{0}}$ is a single component (scalar) field.
S(1/2,0) = 1/2, D(1/2,0) = 2, and $\Psi_{r}$ is a 2-component undotted (left-handed) spinor.
S(0,1/2) = 1/2, D(0,1/2) = 2, and $\Psi_{\dot{r}}$ is the dotted (right-handed) spinor field.
S(1/2,1/2) = 1, D(1/2,1/2) = 4, and $\Psi_{r\dot{s}}$ is a 4-component multi-spinor equivalent to the spacetime vector

$$V_{\mu} = \frac{1}{2}(\sigma_{\mu})^{s\dot{s}}\Psi_{s\dot{s}}$$

S(1,0) = 1, D(1,0) = 3, and $\Psi_{rs} = \Psi_{sr}$ is 3-component spinor-tensor. It corresponds to a self-dual antisymmetric tensor;

$$F^{+}_{\mu\nu} \propto (\sigma_{\mu\nu})^{rs}\Psi_{rs}$$

S(0,1) = 1, D(0,1) = 3, and $\Psi_{\dot{r}\dot{s}} = \Psi_{\dot{s}\dot{r}}$ corresponds to antiself-dual antisymmetric tensor field

$$F^{-}_{\mu\nu} \propto (\sigma_{\mu\nu})^{\dot{r}\dot{s}}\Psi_{\dot{r}\dot{s}}$$

So, any rank-2 antisymmetric tensor

$$F_{\mu\nu} = F^{+}_{\mu\nu} + F^{-}_{\mu\nu}$$

transforms like

$$(1,0) \oplus (0,1)$$

This way, you can make a dictionary full of irreducible representations.
In tensor language, Eq(1) says that the tensor product of 2 inequivalent spinors transforms as a 4-vector under SO(1,3). And Eq(2) says that the tensor product of 2 left-handed spinors decomposes into a scalar and a self-dual antisymmetric tensor.

Actually I finished the study of Weinberg's section 5.9 just now, and he said that for massless particles, the representation (K,J) can only describe massless particles with helicity (eigenvalue of $$S_z$$) J - K. That's the reason why (1/2,1/2) representation, i.e. four vector representation can not describe photon directly, sine photon is a massless particle with helicity $$\pm 1$$. The way he solved this is, constructing the antisymmetric fields, $$\partial_\mu A_\nu - \partial_\nu A_\mu$$.
For $P^{2} = m^{2} \neq 0$, the above representations of SO(1,3) are also representations of Poicare' group.
For example; the massive (1/2,1/2) can still be represented by a genuine 4-vector $A_{\mu}$ subject to the condition $\partial^{\mu}A_{\mu}= 0$ [we did this in post#5].

In the following, I will summarize the results derived in section (5.9) og Weinberg's;

When m = 0, it is impossible to construct a genuine 4-vector with helicity $\pm 1$. Indeed, the only admissible vector field of mass zero is a gradient of a massless scalar

$$V_{\mu}^{(1/2,1/2)}(x) \equiv \partial_{\mu}\phi^{(0,0)}(x)$$

This means that massless, spin = 1 particle (photon) cannot be described by a genuine 4-vector field! You could say that photons belong not to (1/2,1/2) but to the antisymmetric rep. $(0,1)\oplus (1,0)$.

But, who cares about genuine vector field? In post #5, we saw that a 4-component potential (which is not a genuine 4-vector) does a pretty good job in describing the physical photon, thanks to the gauge principle.
Lorentz and gauge invariant electrodynamics can be based on the potential $A_{\mu}$ because

1) the gauge invariant field strength $F_{\mu\nu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu}$ transforms as a tensor, even though $A_{\mu}$ does not transform as a genuine vector under SO(1,3).

2) the potential couples to the conserved current of the gauge symmetry (Noether).

But he mentioned that, higher spin representation would produce short distance interaction. So, in order to satisfy the inverse squared law, we still choose the four-vector representation to describe the photon.
Be careful, the gauge potential is not a vector! If it was, we wouldn't have had any problem in constructing (1/2,1/2) with helicity $\pm 1$.

Actually, I didn't get it much.
I am afraid the whole subject is not pretty!

regards

sam