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).(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex] 0.955{|s>} + 0.2{|p_x>} + 0.2{|p_y>} + 0.1|p_z> [/tex]

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.

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# Is spin-2 a tensor?

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