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So I want to show that a possible (infinite dimensional) representation of the generator of $\mathfrak{so}(2)$ is this: $$ T = i\bigg(y\frac{\partial}{\partial x} -x \frac{\partial}{\partial y} \bigg). $$

Attempt:

  1. $\mathfrak{so}(2)$ is one-dimensional (because $SO(2)$ is one dimensional, & the tangent space to a manifold has the same dimension as the manifold) subgroup so it has one generator. If $g(\phi) \in SO(2)$ is connected to the identity, where $\phi$ is the parameter: $$ g(\phi) = 1 + iT\phi \quad : \ T = -i\frac{\partial g(\phi)}{\partial \phi} $$

  2. Apparently the group action of $g$ on $\psi(\mathbf{x}) = \langle \mathbf{x}| \psi \rangle$ is $ g\psi = \psi(R^{-1}\mathbf{x})$

    • How do we prove this? Perhaps could argue, if $ |\psi'\rangle$ is the rotated state and $\mathbf{x'}$ the rotated position: $$ \langle \mathbf{r'}|\psi'\rangle = \langle \mathbf{r} | \psi \rangle \implies \psi'(\mathbf{r'}) = \psi(\mathbf{r})\implies \psi'(\mathbf{r}) = \psi(R^{-1}\mathbf{r}) $$ Where $R^{-1}$ is the inverse rotation matrix.Then define $g\psi = \psi'(\mathbf{r})$ - but what is the difference between $g$ and $R$? $R$ is the fundamental representation of the group element $g$?
    • This proof is completely useless for fields although Srednicki would seem to imply a generalization of this result??
  3. Assuming 2 fine: $$ g(\phi)\psi(\mathbf{x}) = (1+iT\phi)\psi(\mathbf{x}) = \psi(R^{-1}\mathbf{x})$$ $$ R^{-1} = \begin{pmatrix} cos\phi & -sin\phi \\ sin\phi & cos\phi \\ \end{pmatrix} \implies \psi(R^{-1}\mathbf{x}) = \psi\begin{pmatrix} x-\phi y \\ y+ \phi x \end{pmatrix} $$

-I now want to taylor expand this, but I'm not sure how?? If this were a simple multi variable functions $f: \mathbb{R} \rightarrow \mathbb{R} $ this would be obvious - but it's not.

-I'm also slightly concerned how I have ended up representing one generator as an infinite number of objects? - i.e this representation of $T$ is a member of $L^{2}(\mathbb{R}^{3})$, labelled by the tuple of parameters $(x,y)$

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  • $\begingroup$ 1. Why do you write subgroup? 2. Do you assume that $\psi$ is a scalar? If so, your transformation corresponds to the so-called passive transformation. 3. doesn't make too much sense unless $\phi$ is infinitesimal. If you leave $R^{-1}$ in its matrix form with the sin and cos entries, you can of course Taylor expand, and the expansion will yield powers of $T$. $\endgroup$ – user178876 Dec 24 '17 at 14:17
  • $\begingroup$ 1.Oops. This is a mistake (corrected) - I meant $SO(2)$ is one dimensional (meaning that the associated LA is also one dimensional) 2. Should be a scalar function - $|\psi \rangle$ should be an element of a hilbert space, and $\psi$ it's position representation 3. But how does that get me to the result I want to show? $\endgroup$ – thesundayscientist Dec 24 '17 at 15:05
  • $\begingroup$ 1.+2. Please updated your question accordingly. 3. You have all the ingredients, i.e. just insert your transformation into the equation for the generator. $\endgroup$ – user178876 Dec 24 '17 at 15:12
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    $\begingroup$ Can you be more specific what you want to achieve? Do you want a representation of the group or of the Lie Algebra. Do you want a unitary/self adjoint reprensentation, or what conditions? I mean shouldnt showing that T is self adjoint be enough? Then by the inverse of stones theorem your done, arent you? $\endgroup$ – lalala Dec 24 '17 at 15:48
  • $\begingroup$ @lalala I just want to show that the first expression is a completely valid representation of the lie algebra $\mathfrak{so}(2)$. A reference on how to handle infinite dimensional representations of lie algebras (for physicists!) would be fantastic. $\endgroup$ – thesundayscientist Dec 24 '17 at 20:08
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Alright, this is a new answer. First off, I'll fix some notation. It will be slightly different from the one in your question, but I believe it will help.

We will denote an element of $SO(2)$ by $R(\phi)$. To speak of "the" generator we need to choose a specific parametrization of $SO(2)$; it looks like the one we're using here is

$$R(\phi) = \begin{pmatrix} \cos \phi & \sin \phi \\ - \sin \phi & \cos \phi \end{pmatrix}.$$

This is the same as what you called $g(\phi)$, so let's not use two names for the same thing. The generator (i.e., basis element) of the algebra is

$$T = -i R'(0) = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix},$$

so that we have $R(\phi) = e^{i \phi T}$. The Lie algebra is a one-dimensional real vector space, isomorphic to $\mathbb{R}$. We would normally be interested in the commutators of elements, but since this is one dimensional this is pretty trivial: for any elements $A = \alpha T$ and $B = \beta T$ in the Lie algebra, we have $[A,B] = [\alpha T, \beta T] = 0$.

Next we have the vector space $V$ of smooth complex functions on $\mathbb{R}^2$, and we want to look at both group and algebra representations. A group representation is a function $\rho_G$ that takes an element of $SO(2)$ and returns a linear operator on $V$, such that $\rho_G(R_1) \rho_G(R_2) = \rho_G(R_1 R_2)$. There is a canonical representation on the space of functions, given by $(\rho_G(R)\psi)(\mathbf{x}) \equiv \psi(R^{-1} \mathbf{x})$. The inverse is necessary to make this obey the group composition law. Next, a Lie algebra representation is a function $\rho_A$ which takes an element $X$ of the algebra and returns a linear operator on $V$, such that $[\rho_A(X_1), \rho_A(X_2)] = \rho_A([X_1,X_2])$. We are given a representation defined by $\rho_A(T) = i \left(y \partial_x - x \partial_y \right)$, with $\rho_A(X)$ defined for all $X$ by linearity, since $\{T\}$ is a basis for the algebra. Showing that this is in fact a representation is easy because the commutators are always trivial. If our algebra wasn't one dimensional we would have some work to do.

The hard part here is showing that the group and algebra representations correspond to each other. In other words, we need to show that the relation $R(\phi) = e^{i \phi T}$ is also true for the representations; that is, $\rho_G(e^{i\phi T}) = e^{i \phi \rho_A(T)}$, or

$$\psi(e^{-i \phi T} \mathbf{x}) = (e^{i \phi \rho_A(T)} \psi) (\mathbf{x})$$

(remember that the left hand side is by definition the representation $\rho_G$). Now, the great thing about this is that because everything is an exponential function of $\phi$, we only need to check the equalities to first order in $\phi$, since the properties of the exponential guarantee that they will then be true for all $\phi$. We then have that, to first order,

$$\psi(e^{-i \phi T} \mathbf{x}) \approx \psi(\mathbf{x} - i \phi T \mathbf{x}) = \psi \begin{pmatrix} x - \phi y \\ y + \phi x \end{pmatrix}$$

and

$$(e^{i \phi \rho_A(T)} \psi) (\mathbf{x}) \approx (1 - \phi(y \partial_x + x \partial_y)) \psi(\mathbf{x}).$$

To Taylor expand the first expression, simply use the fact that $\psi(\mathbf{x} + \mathbf{d}) \approx \psi(\mathbf{x}) + \nabla \psi \cdot \mathbf{d}$. You will find that it is indeed equal to the second expression.

I want to highlight something, though. If you just want to show that the given differential operator gives a Lie algebra representation, all of this is unnecessary. Since the Lie algebra is one dimensional, you can represent the matrix $T$ by anything and the commutator relations will be automatically satisfied, since they're trivial.

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  • $\begingroup$ Yes this is what I want to do, though I didn't think this clearly! To summarise what you said - LA representation is a LA homomorphism. So you're saying that bc $\mathfrak{so}(2)$ is 1-d for any $x \in \mathfrak{so}(2) \; span\{x\} = \mathfrak{so}(2)$ So I can represent any element I like and call it a generator. The $(g\psi)(x)$ isn't a result I need to prove - it's just a choice of a representation for $SO(2)$ on $L^{2}(\mathbb{R}^{3})$ [easily checked is group homomorphism]. I then want the corresponding LA homomorphism (rep'n) to this - which I have to show is as boxed? $\endgroup$ – thesundayscientist Dec 29 '17 at 22:43
  • $\begingroup$ and yes I still don't know how to do this! Sorry. $\endgroup$ – thesundayscientist Dec 29 '17 at 22:43
  • $\begingroup$ @thesundayscientist That's correct (though we need $x \neq 0$). The tricky part is showing that the algebra representation corresponds to the group representation; here the choice of $x$ actually matters. But the box doesn't say to prove this! Do you actually need to show it? $\endgroup$ – Javier Dec 29 '17 at 23:00
  • $\begingroup$ No, I suspect what I'm expected to do is just to Taylor expand the end of (3) - but I imagine you'll tell me there is a much better way than this (and the Taylor expansion doesn't look very pretty) $\endgroup$ – thesundayscientist Dec 29 '17 at 23:16
  • $\begingroup$ You didn't answer my question - is the boxed statement all you need to show, or is there more? $\endgroup$ – Javier Dec 29 '17 at 23:23
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I can't claim I understood the flow of your question to be helpful, but here are two basic facts manifest whenever you are looking at abelian rotations:

  • In polar coordinates, r,θ, you just have $T=-i\frac{\partial}{\partial \theta}$, a mere plane rotation obviously commuting with itself. It is a one-dimensional representation. (The Lie group itself is an infinite-dimensional group, but that only counts the infinite $\phi$ angles.)

$$ g(\phi) = 1 + i\phi T\quad +O(\phi^2)= 1+\phi\frac{\partial}{\partial\theta}+O(\phi^2) \\ T = -i\frac{\partial g(\phi)}{\partial \phi} | _{\phi=0} ~~.$$ The rotation merely translates the angle θ of your representation by the parameter $\phi$.

  • So, explicitly, $$ g(\phi)\psi(\mathbf{x}) \sim (1+iT\phi)~\psi( e^{i\theta}r)=\left(1+\phi\frac{\partial}{\partial\theta}\right )\psi( e^{i\theta}r)=\psi( e^{i\theta}r)+i\phi~ r e^{i\theta} \psi'( e^{i\theta}r).$$

This is but the infinitesimal version of your final expression, $\psi( e^{i(\theta+\phi)}r)=\psi(x-\phi y, y+\phi x)\sim (1+\phi(x\partial_y-y\partial_x))\psi(x,y)$.

Two refs: WP and Woit 2017 (Quantum Theory, Groups and Representations--An Introduction).

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  • $\begingroup$ Sorry for the delay in replying! The form of the generator you state is exactly what I'm trying to prove - I don't know how beforehand it should have this form? Shouldn't the Lie group $SO(2)$ should have dimension 1? I can represent it in an infinite dimensional vector-space, but there is only ever one parameter involved in the group itself? $\endgroup$ – thesundayscientist Dec 29 '17 at 22:21
  • $\begingroup$ Let's call it U(1), isomorphic to SO(2), basically. It is a Lie group, so infinite dimensional, acting on a 1d vector space, or 2d if you wished to use matrices. The one parameter θ can take infinite values. $\endgroup$ – Cosmas Zachos Dec 29 '17 at 22:33
  • $\begingroup$ But I thought the dimension of the lie group should be defined by the number of continuous parameters required to describe the group? The LG is just a manifold with a group operation (and group properties) - and the LA a tangent space. If the LG were always infinite dimensional, then you're LA would also have to always be an infinite dimensional vector space.. which isn't true, given that e.g $\mathfrak{so}(N)$ has finite number of generators? $\endgroup$ – thesundayscientist Dec 29 '17 at 22:52
  • $\begingroup$ The dimension of the LG is the number of elements, so always infinite. The dimension of the algebra is the number of independent directions of the tangent space, so generators of the LA. In this case, we are dealing with a circle: it has infinite points but only one dimension--it is a line. Are we on the same page? Maybe Woit's chapter linked could help? $\endgroup$ – Cosmas Zachos Dec 29 '17 at 23:36
  • $\begingroup$ Right I'll have a read through that... though I'm very confused as to how this fits in with respect to e.g this question $\endgroup$ – thesundayscientist Dec 29 '17 at 23:48

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