When we do a transformation (norm preserving one) for a given quantity, from what I have understood it seems like there is a representation of the group element for each quantity depending how they transform (eg : scalar, vector, rank 2 tensor spinor.. ).

Considering a normal rotation in 2-D plane (associated to $\mathrm{SO(2)}$ group), if a scalar field $\psi(x,y) \rightarrow \bar\psi(\bar x, \bar y)$ transforms, the representation for the group (which I derived from my arguements of infinitesimal rotations) : $$ \psi(x,y) = (1+ L_z\delta\theta)\bar\psi(\bar x, \bar y) $$

Now I am just wondering what exactly remains unchanged here (since it is a scalar field) or am I just understanding it in the wrong way ?

Also is there a general way of finding a representation given a quantity (a field, vector, tensor ...) in group theory. Since I derived the above scalar representation unlike how find the representation in Lie groups.

PS : Is it probably the form of the field remains invariant, (like how Lorentz covariance gives us spinor transformations)

$$ x^2 +y^2 \rightarrow \bar x^2 +\bar y^2 $$

EDIT 1 : If I wanted to derive this infinite dimensional representation from its Lie group (like I do for the vector case), how can I do it ?

EDIT 2 : I am not able to see the Representation for transformation of Quantum mechanical operators from group directly.


1 Answer 1


General Remarks.

In general, you cannot "derive" a representation of a given group $G$ on the objects you're considering, but there are some really standard definitions of certain group representations which are given special names like "scalar," "vector," and so on.

However, given the representation of a Lie group $G$, this induces a representation of its Lie algebra $\mathfrak g$, and determining an explicit formula for this Lie algebra representation is precisely what we do when we find the so-called "infinitesimal generators" of the corresponding group representation.

An example. $\mathrm{SO}(2)$

Let $C^\infty(\mathbb R^2)$ denote the vector space of smooth functions on the plane $\mathbb R^2$. The scalar representation $\rho$ of $\mathrm{SO}(2)$ acting on $C^\infty(\mathbb R^2)$ is defined as \begin{align} (\rho_0(R)\phi)(\mathbf x) = \phi(R^{-1}\mathbf x). \end{align} for each $\phi\in C^\infty(\mathbb R^2)$ and for each $R\in\mathrm{SO}(2)$. What the heck is going on here? Well, notice that this can also be written as follows: \begin{align} (\rho_0(R)\phi)(R\mathbf x) = \phi(\mathbf x) \end{align} So this definition encapsulates the intuitive idea that the transformed field $\rho(R)\phi$ evaluated at the transformed point $R\mathbf x$ agrees with the untransformed field $\phi$ evaluated at the untransformed point $\mathbf x$. In physics, it is common to see "primed" notations for the transformed field and transformed point; \begin{align} \rho_0(R)\phi = \phi', \qquad R\mathbf x = \mathbf x' \end{align} in which case the definition of the scalar representation can be written as \begin{align} \phi'(\mathbf x') = \phi(\mathbf x) \end{align} This probably looks familiar. So basically the "invariance" that's happening is that the value of the field doesn't change provided the transformed field is evaluated at the transformed point.

Infinitesimal generators.

To find the infinitesimal generators of a given representation, we are really just trying to find a certain representation of the Lie algebra of the group. This Lie group representation $\rho$ naturally induces a Lie algebra representation $\bar \rho$ as follows: \begin{align} \bar \rho(X) = \frac{d}{dt}\rho(e^{tX})\Big|_{t=0} \end{align} So, for the $\mathrm{SO}(2)$ example, we know that the Lie algebra $\mathfrak{so}(2)$ is generated by the single element \begin{align} J = \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}, \end{align} and we can determine how this element is represented in representation induced by the scalar representation defined above as follows: \begin{align} (\bar\rho_0(J) \phi)(\mathbf x) &= \frac{d}{dt}\phi(e^{-tJ}\mathbf x)\Big|_{t=0} \\ &= \frac{d}{dt}\phi(x-ty, y+tx)\Big|_{t=0} \\ &= -y\partial_x\phi(x,y) + x\partial_y\phi(x,y) \\ &= (-y\partial_x + x\partial_y)\phi(\mathbf x) \end{align} In other words, in the scalar representation, the generator of rotations on the plane is represented by a differential operator; \begin{align} \bar\phi_0(J) = -y\partial_x + x\partial_y. \end{align} This same procedure can be extended to find infinitesimal generators of other representations as well, like the vector representation $\rho_1$ of $\mathrm{SO}(2)$ which is defined to act on vector fields $\mathbf v$ on the plane as follows: \begin{align} (\rho_1(R)\mathbf v)(\mathbf x) = R\mathbf v(R^{-1}\mathbf x) \end{align}

By the way, you might find the following links interesting and/or helpful as well:

Tensor Operators

Representations of Lie algebras in physics

Differential Realizations of certain algebras

Generators of Poincare Groups

Idea of Covering Group

Unitary spacetime translation operator

Rigorous underpinnings of infinitesimals in physics

  • $\begingroup$ Couple of things, firstly R in the first equation itself seems like vector representation, is it right ? Secondly, all quantities that transform, the representation is motivated by how the quantity changes or remains constant. $\endgroup$
    – user35952
    Commented Feb 14, 2014 at 4:24
  • $\begingroup$ @user35952 Basically yes; the representation in which an element of $\mathrm{SO}(2)$ is simply given by itself, namely a $2\times 2$ matrix, is often called the "defining representation" or the "fundamental representation" by physicists, and you can think of it as a "vector" representation. As for the second point, I'm not entirely clear on what you're saying/asking, but from how I would interpret it, the answer is yes. For example, the representation of $\mathrm{SO}(2)$ on a vector field rotates both the point where the field is applied, and the indices on the vector field. $\endgroup$ Commented Feb 14, 2014 at 6:47
  • 1
    $\begingroup$ @user35952 Here is also an exposition for you from one of the better technical writers I have seen in the last fifteen years: Chapters 1 and 2 of Rossmann,W. ,"Lie Groups: An introduction Through Linear Groups" (unfortunately its not on the web for free, but I'd highly recommend reading up to and including section 2.6 (the most relevant section). Joshphysics is talking about the Lie Map induced on the algebra induced by a group homomorphism and it's always calculated by the exponentiate-differentiate-at-identity method explained above $\endgroup$ Commented Feb 15, 2014 at 3:58
  • $\begingroup$ @WetSavannaAnimalakaRodVance : Thanks ! I am still taking time to convince myself about these. $\endgroup$
    – user35952
    Commented Feb 15, 2014 at 4:15
  • $\begingroup$ @WetSavannaAnimalakaRodVance Thansk for that comment. In particular, I wasn't aware of the terminology "Lie map". $\endgroup$ Commented Feb 17, 2014 at 3:29

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