In page 168 in Ref. [1], the authors search for a suitable order parameter for the nematic phase in liquid crystal. If $\vec v^\alpha$ is the direction of a single molecule, than due to the inversion symmetry, both $\vec v^\alpha$ and $-\vec v^\alpha$ contribute to order so the order parameter must be even in $\vec v^\alpha$. In addition, they want the order parameter to be $0$ in the disordered state. Since, I quote,

a symmetric traceless tensor will yield zero when averaged over all directions,

they define the following tensor

$$Q_{ij} = \frac{V}{N} \sum_{\alpha} (v_i^\alpha v_j^\alpha - \frac{1}{3} \delta_{ij}) \delta(\vec x -\vec x_{\alpha}) .$$

Now, I understant that this tensor is symmetric and traceless. I can also insert $v_i^\alpha = [\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta]$ and see that the expression above is zero when averaged over all directions (but it does seem like magic).

So my question is, how general is the statement in the quote above?

For instance, in general, what does it mean to average a tensor over all directions? Does the quote require that the tensor has been defined in terms of quantity that has a direction in space (which is then the subject of averaging)? If this is the correct interpretation, is there an easy answer to why symmetric traceless tensors are zero when averaged over all directions?

[1] Chaikin, P. M., & Lubensky, T. C. (1995). Principles of condensed matter physics. Cambridge University Press.

  • $\begingroup$ ...Because the said averaged value is just the trace up to a factor. $\endgroup$ Commented Nov 10, 2020 at 10:50
  • $\begingroup$ How do you show that the average value of a tensor is proportional to its trace? Also, what does it even mean, in the general case, to take the average of a tensor? $\endgroup$ Commented Nov 10, 2020 at 11:08
  • $\begingroup$ Look at my answer. $\endgroup$ Commented Nov 10, 2020 at 11:16

4 Answers 4


The order parameter is a meso- or macroscopic quantity which is formed by adding together the tensor contributions from a bunch of different molecules: if each molecule $\alpha$ has a tensor $T^{(\alpha)}$, then the total order parameter is the sum of the $T^{(\alpha)}$ from all the molecules in the relevant sample of material, i.e., $$ T = \sum_\alpha T^{(\alpha)}, $$ or, when seen as a bilinear function with explicit arguments, $$ T(u,v) = \sum_\alpha T^{(\alpha)}(u,v), $$ for arbitrary $u,v\in\mathbb R^3$.

The book's claim is that they want $T$ to be zero in the disordered state, in which the $T^{(\alpha)}$ themselves are pointing in every direction: that is,

  • you fix some arbitrary reference tensor $T^{(0)}$,
  • you assert that each of the $T^{(\alpha)}$s is a rotated version $R[T^{(0)}]$ of the reference tensor,
  • and you assert that the ensemble combination $\sum_\alpha$ samples over all possible orientation rotations $R$, i.e., $$ T(u,v) = \int_\mathrm{SO(3)} R[T^{(0)}](u,v) \:\mathrm dR. $$

Here the integral over $\mathrm{SO}(3)$ must be taken over the Haar measure of the group, which is the only definition of integration which is invariant w.r.t. the group action.

You might also be wondering what the rotated tensor "$R[T^{(0)}](u,v)$" means. In the abstract, it means that you rotate the tensor, but in practice, you can flip that "active transformation" view on its head into a "passive" one, by simply rotating the whole system ($R[T^{(0)}]$, $u$ and $v$) back to the original orientation, which gives you $$R[T^{(0)}](u,v) = T^{(0)}(R^{-1}u,R^{-1}v)$$ as a concrete recipe for evaluating $R[T^{(0)}]$.

The central claim here is the following:

Claim: this orientation averaging reduces the tensor to a multiple of the isotropic tensor $\delta(u,v) = u\cdot v$, with components proportional to the Kronecker delta $\delta(e_i,e_j) = \delta_{ij}$. Moreover, the coefficient is proportional to the trace of the reference tensor:

\begin{align} T(u,v) & = \int_\mathrm{SO(3)} R[T^{(0)}](u,v) \:\mathrm dR \\ & = \frac13 \mathrm{tr}(T^{(0)}) \delta(u,v) . \end{align}

(As pointed out by Valter Moretti, this is basically a whittled-down version of Schur's lemma, a central result in group representation theory which is well worth learning about if you feel like digging deeper.)

There are two main ways to understand this result: there is a fundamental core intuition that makes it work, but you can also prove it rigorously.


The core of the result is that the orientation averaging means that $T(u,v)$ must be an isotropic tensor, i.e., it must be such that $$ T(Ru,Rv) = T(u,v) $$ for every $R\in\mathrm{SO}(3)$. This is an extremely strong constraint: up to a constant multiple, there is only one isotropic tensor of rank two $-$ the Kronecker delta. This implies that $T(u,v) = \lambda \delta(u,v)$, and you just need to determine $\lambda$. This is extremely simple to do, by just taking the trace: $$ \mathrm{tr}(T) = \lambda \: \mathrm{tr}(\delta) = \lambda \sum_{i=1}^3 \delta(e_i,e_i) = 3\lambda, $$ so $\lambda = \frac 13 \mathrm{tr}(T)$.


To complete the rigorous proof, thus, we only need to show that the Kronecker delta is the only isotropic tensor. As in V. Moretti's answer, the simplest way to do this is to consider the action of $T$ on a basis $\{ e_i\}$.

  • For the off-diagonal elements, $T(e_i,e_j)$ with $i\neq j$, there exists a rotation $R$ (specifically, a rotation by $180°$ about $e_i$) for which $Re_i=e_i$ and $Re_j=-e_j$, which means that $$ T(e_i,e_j) = T(Re_i,Re_j) = T(e_i,-e_j) = -T(e_i,e_j), $$ and therefore all the off-diagonal elements must vanish.
  • With this in hand, we only need to show that all the diagonal elements are equal. Choosing $e_i$ and $e_j$ with $i\neq j$, a rotation by $\pm 90°$ about $e_k$ does $Re_i=e_j$, so $$ T(e_i,e_i) = T(Re_i,Re_i) = T(e_j,e_j). $$

This completes the proof: $T(u,v)$ and $\lambda\:\delta(u,v) = T(e_1,e_1)\delta(u,v)$ coincide on a basis, so they must coincide everywhere.

  • $\begingroup$ +1 Emilio, however I produced another proof, using Shur's lemma... $\endgroup$ Commented Nov 10, 2020 at 14:21
  • $\begingroup$ can this proof extend if we define isotropic tensors to be tensors,whose components are the same in every basis, not only those bases, which are related by orthogonal transformations? $\endgroup$
    – ProphetX
    Commented Aug 25, 2023 at 18:51
  • $\begingroup$ @ProphetX There is no such tensor. There is no way to have a tensor have the same components in the basis $\{e_1,e_2,e_3\}$ and in the basis $\{2e_1,2e_2,2e_3\}$. $\endgroup$ Commented Aug 25, 2023 at 20:12
  • $\begingroup$ Isn't your statement true only for (0,2) and (2,0) tensors? We should be able to find invariant tensor wrt all bases of valence (1,1) $\endgroup$
    – ProphetX
    Commented Aug 25, 2023 at 23:40

OK, another proof according to the Remark by Emilio that made clear the interpretation of the question. We have to average the tensor (as a bilinear map) in the space of the rotations. The only way to do it, as Emilio wrote, is to take an integral with respect to the normalized (left-invariant) Haar measure of $SO(3)$: $$\langle T\rangle_{SO(3)}(u,v) := \int_{SO(3)} T(R^tu,R^tv) dR$$ Since $dB^tR = dR$ for every $B\in SO(3)$, $$\langle T\rangle_{SO(3)}(Bu,Bv) = \int_{SO(3)} T(R^tBu,R^tBv) dR = \int_{SO(3)} T((B^tR)^tu,(B^tR)^tv) dR $$ $$ = \int_{SO(3)} T((B^tR)^tu,(B^tR)^tv) dB^{t}R = \int_{SO(3)} T(R'^tu,R'^tv) dR' = \langle T\rangle_{SO(3)}(u,v)\:.$$ In matricial form, writing $R$ for $B$, $$u^tR \langle T\rangle_{SO(3)} R^tv = u^t\langle T\rangle_{SO(3)}v\:. $$ Since $u,v$ are arbitrary, $$R \langle T\rangle_{SO(3)} R^t = \langle T\rangle_{SO(3)}\:. $$ That is $$R \langle T\rangle_{SO(3)}= \langle T\rangle_{SO(3)}R \quad \forall R \in SO(3)\:.$$ Since the fundamental representation of $SO(3)$ is irreducible, Schur's lemma implies that, for some constant, $t\in \mathbb{R}$ $$ \langle T\rangle_{SO(3)} = tI\:.$$ In particular, since every unit vector can be transformed to any other unit vector with a rotation, an orthonormal a basis is transformed into an orthonormal basis, and the trace does not depend on the basis, $$3t = \sum_{i=1}^3\langle T\rangle_{SO(3)}(e_i,e_i)$$ $$ = tr\langle T\rangle_{SO(3)} = \int_{SO(3)} \sum_{i=1}^3T(R^te_i,R^te_i) dR = \int_{SO(3)} tr(T) dR = tr(T)\:.$$ In summary $$ (\langle T\rangle_{SO(3)})_{ab} = \frac{1}{3}tr(T)\delta_{ab}\:.$$

  • $\begingroup$ I agree -- the central claim at stake here is little more than a baby version of Schur's lemma. $\endgroup$ Commented Nov 10, 2020 at 15:19

Symmetry requirement is not necessary. Let us take an orthonormal basis $e_1,e_2,e_3$ and consider all the unit vectors $n \in S^2$. $$\int_{S^2} T(n,n) d n = \sum_{i,j=1}^3T(e_i,e_j) \int_{S^2} n^i n^j dn\:.$$ $dn$ is the standard rotation-invariant measure on $S^2$ with total value $4\pi$, i.e., referring to standard spherical coordinates $$dn = \sin \theta d\theta d\phi\:.$$ Now, for every choice of $i=1,2,3$ and $j=1,2,3$ with $i\neq j$ there is a rotation $R$ such that $(Rn)^i=-n^i$ but $(Rn)^j = n^j$. Since $dn = dRn$ we immediately have that $$\int_{S^2} n^i n^j dn= -\int_{S^2} n^i n^j dn= 0 $$ if $i\neq j$. Therefore \begin{align} \sum_{i,j=1}^3T(e_i,e_j) \int_{S^2} n^i n^i dn & = \sum_{i=1}^3T(e_i,e_i) \int_{S^2} n^i n^i dn \\ & = T(e_1,e_1)\int_{S^2} (n^1)^2 dn + T(e_2,e_2)\int_{S^2} (n^2)^2 dn \\ & \qquad \quad + T(e_3,e_3)\int_{S^2} (n^3)^2 dn\:. \end{align} To conclude, observe that rotational invariance (as above) $$\int_{S^2} n^i n^i dn = d \quad \mbox{independently of $i=1,2,3$}$$ and thus $$4\pi = \int_{S^2} \sum_{i=1}^3 (n^i)^2 dn = 3d$$ which implies $$d = \frac{4\pi}{3}\:.$$ Inserting in the formula above \begin{align} \int_{S^2} T(n,n) d n & = \sum_{i,j=1}^3T(e_i,e_j) \int_{S^2} n^i n^j dn \\ \frac{4\pi}{3} \sum_{i=1}^3 T(e_i,e_i) & = \frac{4\pi}{3} \mathrm{tr}(T)\:. \end{align} In other words, averaging over all directions we obtain the trace up to the factor $1/3$: $$\langle T\rangle_{S^2} := \frac{1}{4\pi}\int_{S^2} T(n,n) d n = \frac{1}{3} \mathrm{tr}(T)\:.$$ Your hypothesis also requires $\mathrm{tr}(T)=0$ concluding the proof.

  • $\begingroup$ I don't think this is quite the right calculation -- the OP's book is orientation-averaging the tensor itself, not its inputs. The result should be a (Kronecker delta) tensor, not a scalar. $\endgroup$ Commented Nov 10, 2020 at 13:26
  • $\begingroup$ I see, one should take the averaged value in with respect to the rotations using the Haar measure. It is a similar computation... $\endgroup$ Commented Nov 10, 2020 at 13:50

The author here maybe concentrate on how to construct a tensor by a vector, so the average over direction means the average over direction of vector,namely $$\langle T_{ij}\rangle=\frac{1}{4\pi}\int T_{ij}d\Omega $$ where $d\Omega$ is the solid angle

The argument

a symmetric traceless tensor will yield zero when averaged over all directions

has some background on group theory -- the representation of SO(3) group. We could consider two vector $U,V$ and their tensor product $U_iV_j$ the problem is $U_iV_j$ could be divided into several part obeying different rules under rotation (that is to say, this is a reducible representation)

$$U_iV_j=\frac{1}{3}U\cdot V\delta_{ij}+\frac{1}{2}(U_iV_j-U_jV_i)+(\frac{U_iV_j-U_jV_i}{2}-\frac{U\cdot V}{3}\delta_{ij})$$

where the number of independent variable is $3\times3=1+3+5$

the first part is a scalar-- it's invariant under rotation. The second part is $(U\times V)_k\epsilon_{ijk}$ which transform like vector. In fact, the tensor is divided into terms called spherical tensor. The last part ,not that simple, tranform as l=2 sphere harmonic function. From the orthogonality of nonequivalent irreducible representation you could find that the integral (interpreted as the inner product of the l=2 with l=0 component) is zero.

Spherical tensor will be useful if you want to discuss items concerning angular momentum ,though they seems do little thing with what you ask. You may refer to books like Modern Quantum Mechanics J.J Sakurai if you're interested in this topic.


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