2
$\begingroup$

Consider the following two understandings of tensors:

  • Given $k$ vector spaces $V_1,\dots,V_k$ one can define the tensor product $V_1\otimes\cdots \otimes V_k$ by means of the universal property: it allows any multilinear map $g : V_1\times\cdots\times V_k\to W$ to be written as

    $$g(v_1,\dots,v_k)=f(v_1\otimes\cdots\otimes v_k)$$

    for a linear $f$. Furthermore, it can be shown that $V^{\otimes k} $ can be identified with the space of multilinear functions $f : V^\ast\times\cdots\times V^\ast\to \mathbb{K}$ while ${V^\ast}^{\otimes k}$ can be identified with the space of all multilinear functions $f : V\times\cdots\times V\to \mathbb{K}$.

    This allows tensors to be presented usually just as follows: a tensor of type $(r,s)$ in the vector space $V$ over $\mathbb{K}$ is a multilinear map $T : V\times\cdots\times V\times V^\ast\times\cdots\times V^\ast \to \mathbb{K}$ where there are $r$ copies of $V$ and $s$ copies of $V^\ast$.

  • In Quantum Mechanics, on the other hand, tensors seems to be an entirely different thing.

    They are somehow related to the representations of the rotation group $SO(3)$. Furthermore, sometimes one talks about "irreducible tensors". More than that, I've seem for example in Sakurai, the author decomposing

    $$U_iV_j=\dfrac{\mathbf{U}\cdot\mathbf{V}}{3}\delta_{ij}+\dfrac{(U_iV_j-U_jV_i)}{2}+\left(\dfrac{U_iV_j+U_jV_i}{2}-\dfrac{\mathbf{U}\cdot\mathbf{V}}{3}\delta_{ij}\right)$$

    where the author says that $U_iV_j$ are the components of a reducible tensor which is reduced according to the above formula.

This doesn't seem to match the usual definitions of representation theory, where a representation of a group $G$ is a pair $(\rho,V)$ with $\rho : G\to GL(V)$ a homomorphism and where such representation is irreducible when there is no proper invariant subspace. In other words, representations can be reducible or irreducible, not tensors.

To make things worse there are the so-called spherical tensors which are objects $T_k^q$ with two indices and that are somehow related to the irreducible representations of $SO(3)$. It seems all tensors in QM are of this type.

But again, in the standard lore, tensors can have lots of indices (see for example the Riemann curvature tensor in Differential Geometry, with four indices).

So there seems to be a bunch of things mixed together and I'm unable to understand what is actually going on here.

My doubts are:

  • How are tensors from QM and tensors from linear algebra widely used in geometry related?

  • What do tensors have to do with representations of the rotation group anyway?

  • What is the meaning of reducibility of a tensor opposed to the reducibility of a representation of a group?

  • Why these spherical tensors seem to be all tensors in QM and how do they relate to the usual tensors?

$\endgroup$
5
$\begingroup$

The first of these is easy:

How are tensors from QM and tensors from linear algebra widely used in geometry related?

They're the same thing ─ though sometimes QM will choose to look only at some specific subset of tensors, e.g. sets which carry group-theoretic representations of the rotation group.

The apparent conflict arises because there's a wide spectrum of ways to talk about tensors, and you're pulling examples from two complete extremes of that spectrum.

Let's bridge the gap by taking the universal-property tensor product, as in the first understanding,

  • Theorem: Given two vector spaces $U$ and $V$ over $F$, there exists a vector space $U\otimes V$ and a bilinear product $\otimes:U\times V\to U\otimes V$ such that for every bilinear $f:U\times V\to F$ there exists a $g:U\otimes V$ such that $f(u,v) = g(u\otimes v)$. Moreover, this vector space and mapping are unique up to a canonical isomorphism.

and provide an instantiation of that abstract tensor-product state:

  • Claim. Let $U,V$ be vectors spaces over $F$, and choose coordinate representations $i_U:U\to F^n$ and $i_V:V\to F^m$ for them. Then $F^{n\times m}$ and $\otimes: U\times V\to F^{n\times m}$, defined via $$\otimes:(u,v)\mapsto \left(i_U(u)_ji_V(v)_k\right)_{j,k=1}^{n,m},$$ are an instantiation of the abstract tensor-product as defined above.

Put another way, this just says that you can do tensor products in a coordinate-wise way, in a manner that is reasonably straight-forward to work out.


Quantum mechanics uses tensors in the second sense, in that a $\boldsymbol U\bf\otimes \boldsymbol V$-tensor-valued operator is defined as an $(n\times m)$-tuple of operators $\hat w_{jk}:\mathcal H \to\mathcal H$, with the understanding that if we have $U$-vector and $V$-vector operator tuples $\hat u_j$ and $\hat v_k$ we can form their tensor product (where order now matters) as $\hat u_j \hat v_k$.

Of course, now that we've done this, we need to walk back some distance, since we're just trying to talk about $U$-vector and $U\otimes V$-tensor operators, and those spaces don't come equipped with canonical coordinate maps $i_U$ and so on. Thus, we also need to demand that if we change our choices of coordinate maps, then the operator tuples will change to the same linear combinations that they would if they were plain coordinates instead of operators. This is what the usual requirement that 'tensor operators need to transform as tensors' means.

(I should also mention that the traditional treatment is a bit more obtuse than it strictly needs to be. This great answer shows that you can define $U$-vector operators to be simply linear operators $\hat u:\mathcal H \to \mathcal H \otimes U$, and it is easy to extend that formalism to $U\otimes V$-tensor operators defined as linear operators $\hat w:\mathcal H \to \mathcal H \otimes U\otimes V$.)


As you can see, then, quantum mechanics is perfectly happy to talk about tensor operators occupying any abstract tensor product you wish to pull from classical mechanics. However, when one is actually out and about doing quantum mechanics, one usually doesn't care about arbitrary tensor products - we specifically care about tensor products of $\mathbb R^3$ with itself, and we care about how those tensor products interact with the additional structure carried by our vector spaces, including in particular its inner-product structure and with it the symmetry group of that structure, the rotation group.

This is where the representation theory comes in, and it does so in the most natural way. Let me phrase this in a general setting:

  • Let $U$ and $V$ be vector spaces over $F$, $G$ be a group, and $R:G\to \mathrm{GL}(U)$ and $S:G\to \mathrm{GL}(V)$ be representations of $G$. There is then a natural representation $T:G\to \mathrm{GL}(U\otimes V)$ in the tensor product space, which is uniquely specified by its action on tensor-product vectors, $$T(g)(u\otimes v) = R(g)(u)\otimes S(g)(v).$$

Normally, of course, we have $U=V=\mathbb R^3$ and $G=\mathrm{SO}(3)$. Typically, even if the single factor representations $R$ and $S$ are irreducible, the tensor-product representation will not be irreducible. When we speak of tensors being reducible or irreducible, we're talking about the word in the representation-theoretic way: a reducible tensor lives in a tensor-product space that carries a reducible representation of $\mathrm{SO}(3)$, while an irreducible tensor lives in a restricted subspace such which carries an irreducible representation.

This is also where spherical tensors come in: they are simply a convenient basis for the restricted subspaces that carry irreducible representations. When they are notated as $T_q^{(k)}$, it normally means that you have a tensor-valued operator (i.e. living in some bigger tensor product space, whose size and number of factors is not that relevant) that's been restricted to a subspace $\mathrm{span} \mathopen{}\left(\{T_{-k}^{(k)}, T_{-k+1}^{(k)},\ldots, T_{k-1}^{(k)}, T_{k}^{(k)}\}\right)\mathclose{}$ that carries the $k$ representation of $\mathrm{SO}(3)$.

In this sense, your final question (why are all tensors in QM spherical tensors?) can be rephrased as follows: why are all tensors in QM separated into irreducible representations of the rotation group? The answer, of course, is that they aren't, and there's nothing intrinsic about QM that requires tensors to live in irreducible representations ─ it's just that they're more useful so we use them more often.

$\endgroup$
2
$\begingroup$

You're confusing tensors and irreducible tensors. The $U_iV_j$ quantities you give above have two indices but subsets of them transforms irreducibly under $SO(3)$. If you ignore the irreducibility aspect, then a rotation $T$ will transform $$ TU_i=\sum_k U_k T_{ki} $$ and $U_iV_j$ will transform to $$ U_iV_j \to \sum_{kp}U_kV_p T_{ki}T_{pj} $$ like the linear algebra tensor, except the transformations are $SO(3)$. Since the product $T_{ki}T_{pj}$ is reducible it is often more convenient to work with linear combo of operators which transform irreducibly.

$\endgroup$
  • $\begingroup$ But still the tensors from linear algebra, as I've outlined, are built in any vector space $V$ which need not carry a representation of $SO(3)$. So implicitly it is assumed in QM that (i) the tensors are defined in a vector space which does carry a representation of $SO(3)$ and (ii) these tensors are elements of the tensor product, which is considered to actually carry the representation induced on the tensor product of spaces by the one on the original vector space? $\endgroup$ – user1620696 Jun 27 '17 at 19:10
  • 1
    $\begingroup$ $SO(3)$ tensors, or $SU(5)$ tensors or whatever other type of is indeed assumed to transform under the appropriate group. Since for the compact case any rep can be expressed as a direct sum of irreps, there is little harm in assuming the $U_i$ lives in an irrep. $\endgroup$ – ZeroTheHero Jun 27 '17 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.