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leo.
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- TL;DR Summary
- In Quantum Mechanics a symmetry group acts upon the Hilbert space of a quantum system by unitary representations. Usually, though, one studies the representations of the Lie algebra to later get the unitary representations of the group out of the algebra. I want to understand the general procedure of getting the unitary representation of a Lie group out of the hermitian representation of its Lie algebra.
In Quantum Mechanics, by Wigner's theorem, a symmetry can be represented either by a unitary linear or antiunitary antilinear operator on the Hilbert space of states ##\cal H##. If ##G## is then a Lie group of symmetries, for each ##T\in G## we have some ##U(T)## acting on the Hilbert space and the association ##T\mapsto U(T)## is, in fact, a projective representation of the group ##G##. I am aware that such an issue can be dealt with by passing to the universal cover of ##G##, so for the moment let us focus on the case on which ##U(T)## is a de facto representation, obeying $$U(T)U(T′)=U(TT′).\tag{1}$$ Now take ##\{\theta^a\}## coordinates on a neighborhood of the identity ##e\in G## with ##\theta^a(e)=0##. Let further ##T(\theta)## be the group element with coordinates ##\theta^a##. Now expanding ##U(T(\theta))## to first order in ##\theta## we have $$U(T(\theta))=1+it_a\theta^a + O(\theta^2)\tag{2},$$
where we have defined these ##t_a## operators. One can show that the unitary condition on ##U(T(\theta))## corresponds to the ##t_a## being Hermitian operators. It so happens that the ##t_a## obey the commutation relations of the Lie algebra ##\mathfrak{g}\equiv T_e G## of ##G## defined as the tangent space at the identity. In that sense, if we define $$\pi (X) = X^a t_a,\quad \forall X = X^a\frac{\partial}{\partial \theta^a}\bigg|_e \in \mathfrak{g},\tag{3}$$ then ##\pi## is a Hermitian representation of the Lie algebra, derived from the unitary representation of the Lie group ##U##.
In that sense if we know the group unitary representation ##U## we find the Lie algebra hermitian representation ##\pi## essentially by expanding at first order around the identity of the group. I'm interested in the reverse of this: we know the Lie algebra hermitian representation ##\pi## and we wish to find ##U## from which it arises.
In the book "The Quantum Theory of Fields, Vol. 1", Chapter 2, Appendix B, Weinberg tries to reconstruct the ##U(T(\theta))## out of the ##t_a## by solving a differential equation which determines the ##U## along a path ##\Theta^a(s)## starting at the origin and discussing the dependence on the choice of path. The differential equation is Eq. (2.B.2) in his book: $$\dfrac{d}{ds}U_\theta(s)=it_aU_\theta(s) h^a_{\phantom{a}b}(\Theta_\theta(s))\dfrac{d\Theta^b_\theta(s)}{ds},\tag{4}$$ where ##U_\theta(s)## is just ##U(T(\Theta(s)))## and ##h^a_{\phantom{a}b}## is defined to be
$$[h^{-1}]^a_{\phantom{a}b}(\theta)=\left[\dfrac{\partial f^a(\bar{\theta},\theta)}{\partial \bar{\theta}^b}\right]_{\bar{\theta}=0}.\tag{5}$$
I don't understand, though, why solving this differential equation is the right way to get the ##U(T(\theta))## out of the ##t_a##.
So in summary: given a unitary representation of a Lie group we get a hermitian representation of the Lie algebra. If we know, instead, the representation of the Lie algebra, how do we find out the Lie group representation? In other words, how to get ##U(T(\theta))## out of the ##t_a##? Why Weinberg's differential equation (2.B.2) is what answers this question?
where we have defined these ##t_a## operators. One can show that the unitary condition on ##U(T(\theta))## corresponds to the ##t_a## being Hermitian operators. It so happens that the ##t_a## obey the commutation relations of the Lie algebra ##\mathfrak{g}\equiv T_e G## of ##G## defined as the tangent space at the identity. In that sense, if we define $$\pi (X) = X^a t_a,\quad \forall X = X^a\frac{\partial}{\partial \theta^a}\bigg|_e \in \mathfrak{g},\tag{3}$$ then ##\pi## is a Hermitian representation of the Lie algebra, derived from the unitary representation of the Lie group ##U##.
In that sense if we know the group unitary representation ##U## we find the Lie algebra hermitian representation ##\pi## essentially by expanding at first order around the identity of the group. I'm interested in the reverse of this: we know the Lie algebra hermitian representation ##\pi## and we wish to find ##U## from which it arises.
In the book "The Quantum Theory of Fields, Vol. 1", Chapter 2, Appendix B, Weinberg tries to reconstruct the ##U(T(\theta))## out of the ##t_a## by solving a differential equation which determines the ##U## along a path ##\Theta^a(s)## starting at the origin and discussing the dependence on the choice of path. The differential equation is Eq. (2.B.2) in his book: $$\dfrac{d}{ds}U_\theta(s)=it_aU_\theta(s) h^a_{\phantom{a}b}(\Theta_\theta(s))\dfrac{d\Theta^b_\theta(s)}{ds},\tag{4}$$ where ##U_\theta(s)## is just ##U(T(\Theta(s)))## and ##h^a_{\phantom{a}b}## is defined to be
$$[h^{-1}]^a_{\phantom{a}b}(\theta)=\left[\dfrac{\partial f^a(\bar{\theta},\theta)}{\partial \bar{\theta}^b}\right]_{\bar{\theta}=0}.\tag{5}$$
I don't understand, though, why solving this differential equation is the right way to get the ##U(T(\theta))## out of the ##t_a##.
So in summary: given a unitary representation of a Lie group we get a hermitian representation of the Lie algebra. If we know, instead, the representation of the Lie algebra, how do we find out the Lie group representation? In other words, how to get ##U(T(\theta))## out of the ##t_a##? Why Weinberg's differential equation (2.B.2) is what answers this question?