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According to my skript:
Quantum mechanic states $ψ ∈ \mathcal H$ changes under a rotation $R ∈ \text{SO(3)}, \vec{x} \rightarrow R\vec{x}$ according to $ψ \rightarrow U(R)ψ$, whereas $U(R)$ is a unitary representation of $\text{SO(3)}$ is, that means: $$U: \text{SO(3)} \rightarrow \mathcal L(\mathcal H) = \{\text{linear transformation } \mathcal H \rightarrow \mathcal H\} = \text{GL}(\mathcal H,\mathcal H)$$ $$R \longmapsto U(R) $$
is a homomorphism, i.e. $U(R_1)U(R_2) = U(R_1R_2), U(1) = \mathbb I$ which is unitary $U(R)^{-1} = U(R)^*$.

Infinitesimal rotations are elements $Ω$ of the tangent space $T_{\mathbb I}SO(3) = \{\dot{γ}(0)|γ:[-ε,ε] \rightarrow \text{SO(3)}, γ(0) = \mathbb I\}$, where $γ(ε) = e^{εΩ} ∈ \text{SO(3)},γ(0) = e^{0Ω} = \mathbb I$, on $\text{SO(3)}$ at the point $\mathbb I$: $$Ω = \frac{d}{dt}R(t)\bigg|_{t=0},$$
whereas $t \longmapsto R(t)$ is a differentiable curve in $\text{SO(3)}$ through $R(0) = \mathbb I$.

Every Lie group representation of $\text{SO(3)}$ on $\mathcal H$ corresponds to a Lie algebra representation of $\text{so(3)}$ (but not vice versa):
$$U(Ω):= \frac{d}{dt}U(R(t))\bigg|_{t=0}.$$

The transformation $Ω \longmapsto U(Ω)$ is a homomorphism of $\text{so(3)}$ $(\alpha_1,\alpha_2 ∈ ℝ)$: $$U(\alpha_1Ω_1 + \alpha_2Ω_2) = \alpha_1U(Ω_1) + \alpha_2U(Ω_2)$$ $$U([Ω_1,Ω_2]) = [U(Ω_1),U(Ω_2)],$$ whereas the last follows from $U(RΩR^{-1}) = U(R)U(Ω)U(R)^{-1} \quad (R ∈ \text{SO(3)}).$


I want to check the last statement, i.e. that $Ω \longmapsto U(Ω)$ is a homomorphism.
Calculation Questions:
1) If I just consider $\alpha_1U(Ω_1)$, is it then correct to state:
$$\alpha_1U(Ω_1) = \alpha_1 U(\frac{d}{dt}R_1(t)\bigg|_{t=0})$$ $$\alpha_1U(Ω_1) = \alpha_1\frac{d}{dt}U(R_1(t))\bigg|_{t=0},$$
by simply plugging in the definitions?
2) If this holds, can I then conclude: $$U(\alpha_1Ω_1 + \alpha_2Ω_2) = U(\frac{d}{dt}(R_1(\alpha_1t)R_2(\alpha_2t))\bigg|_{t=0})\\ \quad= \frac{d}{dt}U(R_1(\alpha_1t)R_2(\alpha_2t))\bigg|_{t=0} = \frac{d}{dt}(U(R_1(\alpha_1t))U(R_2(\alpha_2t)))\bigg|_{t=0} =\alpha_1Ω_1 + \alpha_2Ω_2.$$
3) I ask this, because someone else wrote down that:
$$U(Ω_1+\alphaΩ_2) := \frac{d}{dt}U(R_1(t)R_2(\alpha t)\bigg|_{t=0}.$$
4) So in general I am a bit confused about the notation, maybe one could clarify this a bit.


General Questions & Remarks:
I still don't get the main concept behind representations.
Let me place a few words into this room: Quantum Mechanics; Born's rule; Symmetries; Projective Representations; Wigner's Theorem; Irreducible Representations; Eigenstates; Composite Systems & Clebsch-Gordan-Coefficients; Wigner-Eckart-Theorem.
It would be great if one could show me his Idea in a few lines using the words given above. I myself will later try (so far I have only a long version in another language which I can poste later on). Thank you in advance! :)

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    $\begingroup$ A representation is simply a set of linear transformations over a vector space. Usually they are expressed as matrices with ordinary matrix multiplication being the group operation. This isn't related to QM, at all. You can use them in classical mechanics just as well as in computer graphics. This is usually very well spelled out in books on group theory, including the results about irreducible representation (matrices which can't be factorized into smaller sub-matrices). $\endgroup$ – CuriousOne Dec 29 '14 at 21:35
  • $\begingroup$ The key idea for groups here is homomorphism, so that transformations wrought on e.g. spacetime compose in the same way as the corresponding transformations that happen on e.g. quantum state space. Indeed Wigner's theorem says that projective homomorphisms are allowed. See whether you have any joy with my answer here. I may get time to address your detailed questions later. $\endgroup$ – WetSavannaAnimal Dec 30 '14 at 0:50
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The answer is yes to all your questions. In general one assumes the representation to be strongly continuous (i.e. continuous in the strong operator topology) so expressions make sense when evaluated against a vector from the Hilbert space. The existence of the "derivative" of a 1-parameter subgroup like $U(R_1(t))$ is given by Stone theorem which says that there exists a self-adjoint operator $H_1$ such that $U(R_1(t)) = e^{iH_1t}$ (through functional calculus) and $H_1$ is precisely the "derivative" of $U(R_1(t))$ at $t=0$.

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  • $\begingroup$ Cheers! Could you further explain me this fact: Every unitary representation of $\text{SO(3)}$ on $\mathcal H$ corresponds to a unitary representation of $\text{so(3)}$ (but not vice verca). $\endgroup$ – Red Pencil Dec 29 '14 at 23:04
  • $\begingroup$ I think this is related to the fact that not all the representations of a Lie algebra can be integrated to a representation of the Lie group $\endgroup$ – Phoenix87 Dec 29 '14 at 23:08
  • $\begingroup$ Maybe it goes into this direction: If one considers $U(R) = e^{U(Ω)t}$ than it could be that $e^0 = \mathbb I = e^{2πiΩ}$, therefore different $U(Ω)$ could lead to the same $U(R)$. However the definition $U(Ω) = \frac{d}{dt}U(R(t))\bigg|_{t=0}$ leads always to a unique $U(Ω)$ for different $U(R)$. $\endgroup$ – Red Pencil Dec 29 '14 at 23:24
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    $\begingroup$ @RedPencil It doesn't work the other way around because $SO(3)$ is not the only Lie group with $\mathfrak{so}(3)$ as its Lie algebra: $SU(2)$ is the only other connected Lie group with this algebra, but you also have $O(3)$ as another Lie group - this one with two connected components. $\endgroup$ – WetSavannaAnimal Dec 30 '14 at 0:47

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