# Is spin-2 a tensor?

1. Apr 4, 2010

### conway

This thread picks up on a discussion that started in another thread ("What exactly is spin?") and drifted off topic. The gist of the question is: how do you line up the 5-component representation of spin-2 along the z axis with the 9 components of a 3x3 tensor? The corresponding question was asked first by haael with regard to spin-1 (3 components along the z-axis) versus ordinary vectors (3 components).

I answered the first question by using the p orbitals of a hydrogen atom to track the algebra of spin-1 systems. SpectraCat rightly pointed out that in the usual terminology "spin" refers to the intrinsic spin of a particle, while the quantities I was working with should be called "orbital angular momentum" (l=1,2,3 etc.). The point is well taken; however, since for integer spin particles the algebra of spin is exactly the same as the algebra of the spherical harmonics, I won't get any inconsistent results by using the hydrogen orbitals for my visualisation. In the end, I'm not going to have any useful insights on hypothetical spin-2 particles, but I think I will have at least identified some mathematically correct correspondences.

I noted first that by using what I called the "chemist's orbitals" for l=1, you can see that small values of p_x, p_y, and p_z, when added to the s-orbital, give you vector displacements of the s orbital. So a superposition of the type

$$0.955{|s>} + 0.2{|p_x>} + 0.2{|p_y>} + 0.1|p_z>$$

basically pushes the cloud, to a good approximation, in the direction (2,2,1). The p orbitals can in fact do more that this because I have only used real values for the coefficients, while in general complex values may be used. But the p-orbitals can do at least this much.

Now I want to generalize this line of thinking to the case of l=2. And the generalisation is: if the p orbitals give you vector diplacements, then the d orbitals give you distortions.

(Which d orbitals am I talking about? There is a helpful picture on the Wikipedia page http://en.wikipedia.org/wiki/Atomic_orbital
I call these the "chemists orbitals" because they are purely real-valued as opposed to the physicist's l=2, m=(2,1,0, -1, -2) z-axis states. So for example the "chemist's" d_xy and d_xz are simply the sums and differences of the "physicists" m=+/-1 states.)

The most general way you can distort a sphere, to first order, is to deform it into an ellipsoid. In a piece of steel subjected to stress, a tiny spherical volume will in general be distorted into an ellipsoid. (If that's not a tensor I don't know what is!) How many numbers does it take to specify the ellipsoid? Six, actually: three for the length of its principal axes, and three to describe the its orientation in space. You might think two numbers is enough to describe the direction, e.g. a unit vector where the third number is redundant since it lies on the surface of a sphere, but that isn't enough to describe the orientation of the ellipsoid! It's not a vector, it's a little more than than, because once you line up the long axis along an arbitrary vector, you can hold it fixed and still rotate the other two axes within their plane.

What I am want to demonstrate is that the five d orbitals, when added in small quantities to the s orbital, are exactly what is needed to deform the spherically symmetric cloud into an arbitrary ellipsoid.

There is already an apparent problem with my argument: I showed that it takes six numbers to specify an arbitrary ellipsoid, but the representation of orbital angular momentum requires only five independent numbers. What about the coefficient of the s-orbital being your sixth number? No, because the sum-of-the-squares of the coefficients has to equal 1, so there are still only five independent numbers. The answer to this little conundrum: in fact, with the five d-orbitals, we can only create ellipsoids of deformation which leave the original volume of the sphere unchanged. In engineering terms, we can say that the d orbitals allow us to put the s orbital into an arbitrary state of pure shear (no compression). And that is specified by five numbers.

As with the vector case, I am again limiting this to real numbers. Wth complex coefficients, you can actually create more general "deformations" than what a lump of steel is capable of twisting itself into.

I suppose I could do an example of showing how an arbitrary ellipsoid might be created using a superposition of the five d orbitals, but that might be an awful lot of work. So I think I'll stop here for now.

Last edited: Apr 4, 2010
2. Apr 4, 2010

### haael

So, my current picture of angular momentum 2: It's a 9-component tensor with 4 constraints, so that 5 degrees of freedom remain. Am I correct?

Also, would someone show me the actual equations that restrict the tensor? From the wikipedia article on orbitals, my guessing is that this tensor has to be self-symmetric and SpectraCat said it cannot be symmetric under space rotations. Is that right? This would give as many equations as is needed.

3. Apr 4, 2010

### Frame Dragger

Although it degenerated into a squabble, everyone does agree that spin-2 = 9-component with 5 degrees of freedom. No question about it.

Here is the math: http://xxx.lanl.gov/PS_cache/astro-ph/pdf/0006/0006423v1.pdf (check section 2.2)

4. Apr 4, 2010

### haael

Wow, but they said in this thread, that spin-2 is actually 16-component 4x4 tensor with 5 degrees of freedom. Or I'm missing something.

5. Apr 4, 2010

### Frame Dragger

Sorry, it seems you're right, 16-component, although for some reason that's nagging at me. I was citing the link more for the mathematics of the Spin-2, and the thread to confirm 5 degrees of freedom in the tensor.

6. Apr 4, 2010

### SpectraCat

In non-relativistic representations, the generalized spin-2 tensor (which is identical in form to the quadrupole tensor in cartesian coords) has 9 components, but in most cases there are symmetry restrictions so that 3 of the off-diagonal elements are equivalent to the other 3 (i.e. $$\sigma_{xz}=\sigma_{zx}$$, etc.). In addition, the trace of the tensor (i.e. the sum of the diagonal elements) is also typically conserved (i.e. it is zero), so that is the fourth constraint. (I think this is equivalent to conway's constant volume argument). Anyway, you can look at the wikipedia page for "quadrupole" to get some more mathematical details if you are interested.

I think the reason the other thread was talking about 16 components for the spin-2 tensor was that they were using relativistic notation, where a 4th dimension proportional to time is added to the position vector. I am not so familiar with that treatment, but I think we can safely ignore it in the context of this thread.

7. Apr 4, 2010

### Frame Dragger

...And THAT is what was nagging at me. Damn. I really hate missing these things.

8. Apr 4, 2010

### conway

I started off by saying I wasn't going to have any insights into spin-two particles, but after following the discussion I'm moved to speculate that the five degrees of freedom correspond to the gravitational tensor in free space (no compression, pure shear) and that in the presence of matter, you get the compression (the sixth degree). The compression in the presence of matter would force the neighboring space to go into shear to geometrically accomodate the compressed region.

I'm going to further speculate (with the stipulation that I really don't know if this even makes sense) that just like in electrostatics you have the potential (1/r for a central charge) which you differentiate to get the field (1/r^2, vector), in gravity the 1/r^2 vector appears to play the role of the potential, because when you differentiate a vector field you get a tensor field. You even get the correct tensor field at least for the special case of the sphere of uniform density, because the deriviative of the force vector in that geometry gives you the uniform compression, zero-shear tensor

And that would be why the "graviton" is spin two...because it describes a tensor field, just like the "photon" (spin one) describes a vector field.

9. Apr 17, 2010

### haael

OK, I read Weinberg and some Wikipedia and now I understand the following:
For each type of geometric object there are some instances of eigenstates of that type (for certain momentum).

There is only one scalar that is an eigenstate of spin.
There are three 4d vectors that are eigenstates of spin.
There are five 4x4 tensors that are eigenstates of spin.
And so on. I believe, this holds also for spinors.

In 4d space, angular momentum operator has 6 components (it's antisymmetric 4x4 matrix of operators). These operators split up into 2 sets: 3 space-space rotations and 3 space-time rotations.

For one space-space rotation, say $$J_{z}$$ we can write its eigenstate equation:
$$J_{z} \phi_{*} = j \phi_{*}$$
For scalars, we just get:
$$J_{z} \phi = 0$$
so there's only one scalar eigenstate and its eigenvalue is always 0.

Now, for vectors the spin operator replaces x and y components, changing sign one of them.
$$J_{z} [\phi_{x}, \phi_{y}, \phi_{z}, \phi_{t}] = [-\phi_{y}, \phi_{x}, \phi_{z}, \phi_{t}]$$
Let's find vector eigenstate:
$$j [\phi_{x}, \phi_{y}, \phi_{z}, \phi_{t}] = [-\phi_{y}, \phi_{x}, \phi_{z}, \phi_{t}]$$
$$\left\{\begin{array}{lr} j \phi_{x} = -\phi_{y}\\ j \phi_{y} = \phi_{x} \end{array}\right.$$
So:
$$j^2 = -1$$
So, we can have 3 vectors that are eigenstates of spin. One, where $$j = i$$ and x and y components are related as written above. Second, where $$j = -i$$. The third possibility to solve these equations is to put x and y components to zero. There is however the z component, so we can have another non-zero eigenstate.

The best part is for time-space rotations. These operators replace time and space, but due to metric they do not change sign of one component.
$$J_{tx} [\phi_{x}, \phi_{y}, \phi_{z}, \phi_{t}] = [\phi_{t}, \phi_{y}, \phi_{z}, \phi_{x}]$$
We can write similar equations as above, but now we get:
$$\left\{\begin{array}{lr} j \phi_{x} = \phi_{t}\\ j \phi_{t} = \phi_{x} \end{array}\right.$$
This equation has only one answer: $$j = 1$$. So, angular momentum operators related to time-space rotations do not give us more eigenstates. It's only space-space rotations that matter.

Plus, space-space angular momentum operators do not commute, so when we compute eigenstates for one of them, the other ones can not produce more eigenstates.

So, in 4d spacetime, we have 3 vector eigenstates, 5 tensor eigenstates and so on.

It was harder than I expected, but now I feel satisfied :).

10. Apr 17, 2010

### jeblack3

This can be made a lot easier by employing spinor indices. We write a spin-1/2 object as $$X_a$$, where the index a can be 1 or 2. A spin-n object is a fully symmetric object with 2n indices:

$$A_{abc} = A_{acb} = A_{bac} = A_{bca} = A_{cab} = A_{cba}$$

$$A_{111}$$ is the $$s_z = +3/2$$ spin component, $$A_{211}=A_{121}=A_{112}$$ is the $$s_z = +1/2$$ component, etc. You have to put in some normalization factors when going back and forth between this form and the column-vector form.

We use the Pauli spin matrices $${\sigma_{ia}}^b$$ to relate spinors to vectors, and the antisymmetric objects $$\epsilon^{ab} = - \epsilon^{ba}$$ and $$\epsilon_{ab} = - \epsilon_{ba}$$ to raise and lower indices: $$X^a = \epsilon^{ab} X_b = \epsilon^{ab} \epsilon_{bc} X^c.$$ A possible convention is $$\epsilon^{12} = - \epsilon^{21} = - \epsilon_{12} = \epsilon_{21} = +1,$$ with all other components zero.

So we can convert a spin-2 object into a tensor by contracting it with the Pauli matrices:

$$A_{ij} = {\sigma_i}^{ab} {\sigma_j}^{cd} A_{abcd}$$

It's immediately obvious that this object is symmetric ($$A_{ij} = A_{ji}$$). For the traceless part ($${A_i}^i = 0$$) we first need to work out an identity involving the Pauli matrices. If I've worked it out correctly:

$${\sigma_{ia}}^b {{\sigma^i}_c}^d = \delta_a^b \delta_c^d + 2 \epsilon_{ac} \epsilon^{bd}$$

Raising the indices gives us

$${\sigma_i}^{ab} \sigma^{icd} = \epsilon^{ab} \epsilon^{cd} - 2 \epsilon^{ac} \epsilon^{bd},$$

and even if I've made an odd number of sign errors, the antisymmetry property of the epsilons means that it will vanish when contracted with the fully symmetric $$A_{abcd}$$.

All of the above was in three dimensions, of course. You can find a brief description of the four-dimensional version in Section 2 of Stephen Martin's http://arxiv.org/abs/hep-ph/9709356" [Broken], Appendix A of Wess and Bagger's Supersymmetry and Supergravity, Appendix E of Zee's Quantum Field Theory in a Nutshell, etc. Don't worry about the supersymmetry part; this notation is a prerequisite for learning SUSY, rather than the other way around.

In 4D you have two kinds of spinor indices, dotted and undotted, so in order to categorize the 4D spinors, you have to do a separate count of their dotted and undotted indices. But if you pick a reference frame, you can transform the dotted indices into undotted ones by taking $$X^a = \bar{\sigma}^{0 \.a a} X_{\.a}$$ and interpret the result as a 3D spinor, thus transforming any 4D spinor with m undotted and n dotted indices into a 3D spinor with m+n indices. If the original 4D spinor was fully symmetric in each set of indices, the resulting 3D spinor can be decomposed into components with spins ranging from |m-n|/2 to (m+n)/2 by breaking it down into symmetric and antisymmetric parts.

Last edited by a moderator: May 4, 2017