Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Motivation.

I was recently reviewing the section 3.10 in Sakurai's quantum mechanics in which he discusses tensor operators, and I was left desiring a more mathematically general/precise discussion. I then skimmed the Wikipedia page on tensor operators, and felt similarly dissatisfied. Here's why

In these discussions, one essentially defines an indexed set of operators $T_{i_1\cdots i_k}$ to be a "cartesian" tensor operator of rank $k$ provided $$ U(R)\, T_{i_1\cdots i_k}\, U^\dagger(R) = R_{i_1}^{\phantom{i_1}j_1}\cdots R_{i_1}^{\phantom{i_1}j_1}T_{j_1\cdots j_k} $$ for each rotation $R\in\mathrm{SO}(3)$ where $U$ is some unitary representation of $\mathrm{SO}(3)$ acting on a Hilbert space (usually that of some physical system whose behavior under rotations we with to study). Similarly one defines a "spherical" tensor operator of rank $n$ as an indexed set of operators $T^{(n)}_{q}$ with $-n<q,q'<n$ for which $$ U(R)\,T_q^{(n)}\,U^\dagger(R) = \sum_{q'=-n}^n D_{q'q}^{(n)}(R)T_{q'}^{(n)} $$ where $D^{(n)}$ is the irreducible representation of $\mathrm{SO}(3)$ of dimension $n$.

Based on these standard definitions, I would think that one could define something less "coordinate-dependent" and extended to representations of any group, not just $\mathrm{SO}(3)$, as follows.

Candidate Definition. Let a group $G$ be given. Let $U$ be a unitary representation of $G$ on a Hilbert space $\mathcal H$, and let $\rho$ be a representation of $G$ on a finite-dimensional, real or complex vector space $V$. A $k$-multilinear, linear operator-valued function $T:V^k\to \mathrm{Lin}(\mathcal H)$ is called a tensor operator relative to the pair of representations $U$ and $\rho$ provided \begin{align} U(g) T(v_1, \dots, v_k) U(g)^\dagger = T(\rho(g)v_1, \dots, \rho(g)v_k) \end{align} for all $g\in G$ and for all $v_1, \dots, v_k\in V$.

Notice that if a basis $u_1, \dots, u_N$ for $V$ is given, and if we define the components $T_{i_1,\dots i_k}$ of $T$ in this basis by \begin{align} T_{i_1 \dots i_k} = T(u_{i_1}, \dots, u_{i_k}) \end{align} and if $\rho(g)_i^{\phantom ij}$ denotes the matrix representation of $\rho(g)$ in this basis, then by using multilinearity the defining property of a tensor operator can be written as follows \begin{align} U(g) T_{i_1\cdots i_k} U^\dagger(g) = \rho(g)_{i_1}^{\phantom {i_1}j_1}\cdots \rho(g)_{i_k}^{\phantom {i_k}j_k} T_{j_1\cdots j_k} \end{align} So this definition immediately reproduces the cartesian tensor definition above if we take, $V =\mathbb R^3$, $G=\mathrm{SO}(3)$, and $\rho(R) = R$, and similarly for the spherical tensor definition if we take $V=\mathbb C^{2n+1}$, $G=\mathrm{SO}(3)$, $\rho = D^{(n)}$ and $k=1$.

Question.

Is the sort of object I just defined the "proper" formalization/generalization of the notion of tensor operators used in physics; it seems to contain the notion of tensor operator used in the physics literature? Is there any literature on the sort of object I define here? I would think that the answer would be yes since this sort of thing seems to me like a natural generalization a mathematically-minded physicist might like to study.

share|improve this question
    
For $SU(N)$, there is a direct link with fundamental representations and antisymmetric tensors. Idem for $SO(N)$, (not considering spinorial representations). (Note that they exist duality between these representations thanks to the Levi-Civita symbol) For $SP(N)$, there is a direct link with fundamental representations and symmetric tensors. Of corse, extending to all representations, you may gain correspondence with other tensors. For instance, the adjoint representation represents a mixed tensor $T^i_j$. For SU(N), representation $(20....)$ represents a symmetric traceless tensor. –  Trimok Aug 8 '13 at 7:25
    
@Trimok Thanks for the comment, but this is not intended to be a question on tensor representations of groups, but rather a question on the notion of a "tensor operator" on a Hilbert space, its formalization, and the existence of existing mathematical literature on such things. –  joshphysics Aug 8 '13 at 17:59
    
I just want to note that your definition is a bit too high-level, may be. In the sense that what you actually do is: you pick a representation $\rho$, then you 'tensor' it to the representation $\tau$ acting on tensors, and then define an object that is associated to $\tau$ rather than $\rho$. You could as well start with $\tau$. It seems to me that it is also usefull to think of 'linear object valued operators', the elements of $\mathrm{Hom}(\mathcal{H},L\otimes\mathcal{H})=L\otimes\mathrm{Hom}(\mathcal{H},‌​\mathcal{H})$, where $L$ is a vector space acted upon by some representation $\tau$. –  Peter Kravchuk Aug 8 '13 at 18:20
    
Second chance....But isn't $\rho(g)$ always (if it exists) the fundamental (vectorial) representation of the group $G$ ?. I do not understand your proposal with another representation, while I probably missed some point.... –  Trimok Aug 8 '13 at 18:59
    
@PeterKravchuk I'll think about what you say; thanks for the comment. –  joshphysics Aug 8 '13 at 19:09

6 Answers 6

up vote 4 down vote accepted

OP's candidate definition is a direct transcription of the tensor operator notion used in physics (and e.g. in Sakurai section 3.10) into a manifestly coordinate-independent mathematical construction. Tensor operators are e.g. used in the Wigner-Eckart theorem.

In this answer we suggest the following slight generalization of OP's candidate definition. Let the following five items be given:

  1. Let $G$ be a group.

  2. Let $H$ be a complex Hilbert space.

  3. Let $\rho: G \to GL(V,\mathbb{F})$ be a group representation.

  4. Let $R:G \to B(H)$ be a group representation.

  5. Let $T:V\to L(H;H)$ be a linear map.

Definition. Let us call $T$ for a $G$-equivariant map if $$ \forall g\in G, v\in V :\quad T(\rho(g)v)~=~ {\rm Ad}(R(g))T(v)~:=~R(g)\circ T(v)\circ R(g)^{-1}. \tag{*} $$

OP's candidate definition may be viewed as a special case of definition (*). For instance, if $\rho_0: G \to GL(V_0,\mathbb{F})$ is a group representation, then one may let $\rho: G \to GL(V,\mathbb{F})$ in point 3 be the tensor product representation $\rho=\rho_0^{\otimes m}$ with vector space

$$V~=~V_0^{\otimes m}~=~\underbrace{V_0\otimes \ldots \otimes V_0}_{m \text{ factors}}.$$

share|improve this answer
    
Thanks for the response/confirmation and +1 for the generalization. I'm going to wait on the checkmark in case anyone, including me, finds other, perhaps equivalent, formulations or generalizations that are interesting/relevant. –  joshphysics Oct 15 '13 at 1:26

The generalization of the "spherical tensor harmonics" of quantum mechanics to a general case of a compact Lie group is given as follows: Let $G$ be a compact Lie group and $H$ be a closed subgroup, then, the Hilbert space of square integrable functions on $G/H$ (which may be taken as the eigenfunctions of the Killing Laplacian) is a direct sum of $G$-representations called "spherical representations". These representations are characterized by having an $H$ singlet. Please see, for example, appendix B of HARMONIC ANALYSIS AND PROPAGATORS ON HOMOGENEOUS SPACES by Camporesi. This definition generalizes (and was named after) the spherical harmonics of quantum mechanics. In this case $G=SU(2)$, $H=U(1)$ and $G/H = SU(2)/U(1)$. The sphericality condition implies that the spherical harmonics can only be representations containing a $U(1)$ singlet, thus must be of integer spin.

share|improve this answer

If you want a more geometric viewpoint, this link is a good start. It's the first chapter of Applications of Classical Physics by Blandford and Thorne. The advantage of the formulation is that the transformation laws for tensors can be derived naturally from the transformation laws for vectors. Then, if you want your tensors to transform a certain way, just modify your transformation laws of your vectors.

Here's a very quick summary (for Cartesian tensors, ie, our vectors live in euclidean space): a rank-$k$ tensor is defined as a function from $k$ vectors to a real number, for example, a rank-two tensor might be written as

$$ T = T(\_,\_) $$

note that a vector can be regarded as a rank-1 tensor, $v(w) = v \cdot w$, and hence a rank-two tensor can also be regarded as a function from vectors to vectors, which is probably relevant for the specific application of tensors in Sakurai. Tensor product is defined as the product of the functions

$$ S(\_,\_) \otimes T(\_,\_,\_) = S(\_,\_)T(\_,\_,\_) $$

Once you choose a basis, you can write the tensor components

$$ T = T_{ijk} e_i e_j e_k $$

where $i,j,k$ are implicitly summed over. From this one can derive how the usual formula for how components $T_{ijk}$ transform under rotation. As an example, suppose $T$ is rank 2 and treat it as a function from vectors to vectors; then we can view the $T_ijk$ as components of 3 by 3 matrix. If $T$ carries a vector $v$ to $Tv$, then $T'$ must carry $Rv$ to $R(Tv)$, so $T' = RTR^{-1}$.

share|improve this answer
1  
This generalizes straightforwardly from rotation to the Lorentz group, and can also be generalized to arbitrary diffeomorphisms. –  Ben Crowell Sep 8 '13 at 20:35
    
You've simply described the standard formulation of tensors as multilinear maps that is used both in differential geometry and in algebra; I am well-aware of that stuff. Notice, in fact, that my candidate definition above is written precisely in terms of multilinear maps which is what you're describing here. The notion I want to define here, however, is a bit less straightforward than that I'm afraid. –  joshphysics Sep 10 '13 at 7:01
    
In your question I did notice that $U$ and $\rho$ are constrained to both be representations of the same underlying group, but in the standard formulation, the transformation law of the vectors (the $\rho$) completely determines the transformation law of the tensors (the $U$), so I was wondering if there were some inconsistency. Unfortunately I'm not familiar enough with group representations to be sure. –  xuanji Sep 10 '13 at 7:06
    
@zodiac Notice that $U$ is a representation of the group $G$ on a Hilbert space; it can potentially be a much different beast than $\rho$ which is a representation of $G$ on a finite-dimensional vector space. In particular, it need not be generated by the representation $\rho$ in the way you describe. –  joshphysics Sep 10 '13 at 8:30

In first chapter of Lie Groups for Pedestrians by Lipkin, a method of generalization of irreducible tensor operators (and other features of the quantum mechanical angular momentum algebra) is given.

The statement is that as long as one can find a finite number of operators $X_\rho$ satisfying analogous commutation relations to those of the angular momentum operators in quantum mechanics, i.e.

$$[X_\rho,\;X_\sigma]=C^\tau_{\rho\sigma}X_\tau,$$

it is always possible to find irreducible tensor operators. One can then, in analogy to $J_z$, choose one (or several) operators to be diagonal in the desired representation. Furthermore, one can extract the analogy of the ladder operators $J_x\pm iJ_y$.

For angular momentum ($SO(3)$), irreducible tensor operators are given in terms of the relation

$$[J_z,T_{kq}]=qT_{kq},$$

where $q$ is the number of components and $k$ is the rank of the tensor. There are $2k+1$ values for $q$, which ranges from $-k$ to $k$.

Analogous tensor operators can be constructed starting from any algebra of the above form. Note that the crucial object is the Lie algebra, not the Lie group, which can be formulated as the group of continuous transformations given by

$$\psi^\prime=(1+i\epsilon X_\rho)\psi.$$

This is not a rigorous answer since I haven't worked out the proof myself. I can only recommend you to read the book.

share|improve this answer
    
Frederic thanks for the response, but this is still not really what I'm looking for. I am aware that the standard treatment of tensor operators can be phrased in terms of the Lie algebra, but the procedure you've outlined doesn't address the fact that I'm hoping to find something less "coordinate-dependent" that characterizes the finite number of operators of which you speak as a single object in much the same way that tensor components in algebra can be thought of as components of multilinear maps. –  joshphysics Oct 10 '13 at 17:13

A tensor operator is a collection of operators that transform irreducibly under conjugation by group elements, i.e. that satisfy precisely the second condition of the OP. The individual elements of the set are the components of the tensors.

Irreducibility is not essential but an additional obvious requirement when dealing with groups for which representations are always fully reducible.

This can be generalized to any group - the group doesn't even need to be continuous. All you need is a group action on the set of components of the tensor operator.

share|improve this answer

I don't think that the candidate definition is sound because the notation $T(v_{1},\ldots ,v_{k})$ implies - by analogy with mathematicians' notation for tensors - that $T(v_{1},\ldots ,v_{k})$ is an operator which transforms trivially; but it clearly does not transform trivially as the RHS of the candidate definition shows. Nevertheless, there is a tensor which transforms trivially under the group G. For simplicity, consider the case $k=1$ and furthermore suppose that the tensor operators $T_{i}$ are Hermitian. So far, the indices of the Hermitian operators $T_{i}$ have been suppressed in the question and answer; putting in the indices, we are dealing with ordinary tensors, $$ T_{i\ \ B}^{\ \bar{A}} \ . $$ Suppose that these tensors transform trivially under the group G. In other words, $$ [D(g^{-T})]_{i}^{\ \ k}[D(g^{-\dagger})]^{\bar{A}}_{\ \ \bar{C}}[D(g^{-T})]_{B}^{\ \ D} T_{k\ \ D}^{\ \bar{C}}=T_{i\ \ B}^{\ \bar{A}} $$ Multiply both sides by $[D(g)]^{i}_{\ \ l}$. $$ [D(g^{-\dagger})]^{\bar{A}}_{\ \ \bar{C}}[D(g^{-T})]_{B}^{\ \ D} T_{l\ \ D}^{\ \bar{C}}=T_{i\ \ B}^{\ \bar{A}}[D(g)]^{i}_{\ \ l} $$ Suppressing the indices of the Hermitian operators recovers the standard definition of a set of tensor operators $T_{i}$, $$ D(g^{-\dagger})T_{l}D(g^{-1})=T_{i}[D(g)]^{i}_{\ \ l} $$ In other words, the ordinary tensor, $$ T_{i\ \ B}^{\ \bar{A}} \ . $$ transforms trivially under the group and a less coordinate-dependent definition needs to proceed from the fact that this tensor transforms trivially.

A further minor caveat is that the group matrices $[D(g)]^{A}_{\ \ B}$ are not necessarily unitary; if the indices $A,B$ are Lorentz spinor indices, the group matrices are the defining rep of SL(2,C); these are only unitary if the Lorentz transformation is a spatial rotation.

share|improve this answer
1  
Thanks for the response, but I think that you may have misunderstood the notation. The "indices of the tensors" are generated, in my notation, as is with the case in mathematics, by evaluating the multilinear map $T$ on a $k$-tuple of basis elements of the vector space $V$. Thus, for a tensor operator of rank $k$, this yields something like $T_{i_1\cdots i_k}$ as noted below the definition; there should be no extra indices. The intent is not for each $T_{i_1\cdots i_k}$ to be a tensor operator, it's the "collection" of these that is the tensor operator. –  joshphysics Aug 8 '13 at 21:39
    
Suppose we use $SU(n)$ and we want to get the tensor product of $2$ adjoints representations. Adjoint representation means, for objects upon which the transformation act : $$\phi^i_j \rightarrow U^i_k~(U^\dagger)_j^l~\phi^k_l$$ So, by tensoring this, we get : $$\phi^i_j \phi'^{i'}_{j'} \rightarrow U^i_k~(U^\dagger)_j^l~ U^{i'}_{k'}~(U^\dagger)_{j'}^{l'}~\phi^k_l ~\phi^{k'}_{l'}$$ But, in fact, it is obviously the same transformation as for the object $\Phi^{i i'}_{jj'}$ : $$\Phi^{i i'}_{jj'} \rightarrow U^i_k~(U^\dagger)_j^l~ U^{i'}_{k'}~(U^\dagger)_{j'}^{l'}~\Phi^{k k'}_{ll'}$$ –  Trimok Aug 9 '13 at 11:03
    
(continued.->...) And it is the same transformation as for the tensor representation : $$\phi^i \phi^{i'} \phi_j \phi_{j'} \rightarrow U^i_k~(U^\dagger)_j^l~ U^{i'}_{k'}~(U^\dagger)_{j'}^{l'}~\phi^k \phi^{k'} \phi_l \phi_{l'}$$ So, finally, while we begin with a tensorial product of 2 adjoint representations, it is the same law as tensorial product of 2 fundamental and 2 anti-fundamental (vectorial) representations. –  Trimok Aug 9 '13 at 11:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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