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This question is about the motivation for Weinberg's approach in "The Quantum Theory of Fields" to obtain unitary representations of Lie groups out of its generators.

One is dealing with a Lie group $G$. We have coordinates $\{\theta^a\}$ on a neighborhood of the identity and $T(\theta)$ is the group element with coordinates $\theta$. Group multiplication is encoded in a function $f$ as $$T(\bar{\theta})T(\theta)=T(f(\bar{\theta},\theta)).$$

If $U(T(\theta))$ is a unitary representation on a Hilbert space, the generators of the representation are defined by the expansion $$U(T(\theta))=1+it_a\theta^a+O(\theta^2).$$

The problem then is:

If we know the $t_a$ and how they act on the Hilbert space, how can we find $U(T(\theta))$?

This is dealt with in Weinberg's Appendix 2B:

To prove this theorem, let us recall the method by which we construct the operators corresponding to symmetry transformations. As described in Section 2.2, we introduce a set of real variables $\theta^a$ to parameterize these transformations, in such a way that the transformation satisfy the composition rule (2.2.15): $$T(\bar{\theta})T(\theta)=T(f(\bar{\theta},\theta)).$$ We want to construct operators $U(T(\theta))\equiv U[\theta]$ that satisfy the corresponding condition $$U[\bar{\theta}]U[\theta]=U\left[f(\bar{\theta},\theta)\right].\tag{2.B.1}$$ To do this, we lay down arbitrary 'standard' paths $\Theta_\theta^a(s)$ in group parameter space, running from the origin to each point $\theta$, with $\Theta^a_\theta(0)=0$ and $\Theta_\theta^a(1)=\theta^a$, and define $U_\theta(s)$ along each such path by the differential equation $$\dfrac{d}{ds}U_\theta(s)=it_aU_\theta(s) h^a_{\phantom{a}b}(\Theta_\theta(s))\dfrac{d\Theta^b_\theta(s)}{ds}\tag{2.B.2}$$ with the initial condition $$U_\theta(0)=1,$$ where $$[h^{-1}]^a_{\phantom{a}b}(\theta)=\left[\dfrac{\partial f^a(\bar{\theta},\theta)}{\partial \bar{\theta}^b}\right]_{\bar{\theta}=0}.$$

The basic claim is that if we know the generators $t_a$ of the representation we can find the unitary representation $U[\theta]$ by defining $U_\theta(s)$ through Eq. (2.B.2) and identifying $U[\theta]=U_\theta(1)$.

What is the motivation for Weinberg's approach? What is the motivation to define $U_\theta(s)$ by (2.B.2)? How one would even think about defining this $U_\theta(s)$ through (2.B.2) in order to obtain $U[\theta]$ out of the generators?

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    $\begingroup$ That should be an edit to your question, not a comment, but (2.B.2) is really just the (typically for Weinberg shrouded in incomprehensible notation) infinitesimal version of that. $\endgroup$ – ACuriousMind Dec 17 '19 at 23:45
  • $\begingroup$ @ACuriousMind I have transformed the comments into an edit to the question. My initial idea has been really that (2.B.2) should somehow be the infinitesimal version of $$E(\exp(t X)) = \exp (t D(X))$$ and I'm trying to explicitly see this (perhaps notation is an issue as you point out). $\endgroup$ – user1620696 Dec 18 '19 at 0:18
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    $\begingroup$ 2.B.2 is just a rephrasing with modern notation of Lie's equations. Do reading from a math book about Lie's three theorems. $\endgroup$ – DanielC Dec 18 '19 at 0:22
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    $\begingroup$ You might, or might not, profit from WP applied to the eminently acceptable definition you read in Ticciati. @ACuriousMind's warning about top-heavy Weinbergese is the very first most students get: Learn the stuff first, and only then recast it in Weinbergese. My sense is much would become clearer if you used su(2) and Pauli matrices. $\endgroup$ – Cosmas Zachos Dec 18 '19 at 0:41
  • $\begingroup$ Well, first I notice that if $R_g$ is the right translation by $g$ we can identify that $$it_ah^a_{\phantom{a}b}(\Theta_\theta(s))\dfrac{d\Theta^b_\theta(s)}{ds}=D\bigg([R_{T(\Theta_\theta(s))}]_{\ast e}^{-1} (T\circ\Theta_\theta)'(s)\bigg).$$ Now taking the path on group manifold $\exp(tX)$, by definition of the exponential, that term collapses to $D(X)$ and resulting eq. is $dU/dt=D(X)U(t)$. So differentiating Ticciati's equation seems to give Weinberg's equation for the specific choice of path $\exp(tX)$. But Weinberg is taking one arbitrary path, that is what seems hard to reconcile. $\endgroup$ – user1620696 Dec 18 '19 at 13:09
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The idea behind this equation is the following:

  • associate the identity operator $1$ to the identity element of the Lie Group.

$$ U(0)=1 $$

  • represent an element infinitesimally close to the identity by

$$ U(\delta\theta)=1+i\delta\theta^aT_a $$

where $\delta\theta^a$ is infinitesimal and $T_a$ are the generators in some representation of the Lie Algebra. Now by repeated applications of $U(\delta\theta)$ we can "walk" on the Lie group from the identity to any element that is connected to the identity. If we hit $U(\delta\theta)$ in some $U(\theta)$ this will "move" $U(\theta)$ to

$$ U(\delta\theta)U(\theta)=U(f(\delta\theta,\theta))= U(\theta^{a}+(h^{-1})^{a}\,_{b}(\theta)\delta \theta^{b}) $$

so if we have a representation $U(\theta)$ of the Lie group element associated with the coordinates $\theta^{a}$, hitting $U(\delta\theta)$ will lead to a representation $U(\theta^{a}+(h^{-1})^{a}\,_{b}(\theta)\delta \theta^{b})$ of the Lie group element associated with the coordinates $\theta^{a}+(h^{-1})^{a}\,_{b}(\theta)\delta \theta^{b}$.

Now, we trace a path in the Lie Group that starts ($s=0$) at the identity and reach the Lie Group element associated to the coordinates $\theta^{a}$ at the end ($s=1$). This path will be in coordinates denoted by $\Theta_{\theta}^a(s)$ and should satisfy

$$ \Theta_{\theta}^{a}(0)=0,\qquad \Theta_{\theta}^{a}(1)=\theta^{a} $$

where $0$ is the identity element of the Lie group in coordinates. We are going to attach at each point in the curve $\Theta_{\theta}(s)^{a}$ a representation given by $U_{\theta}(s)\equiv U(\Theta_{\theta}(s))$ by following the idea presented above, such that at the end of the path we obtain $U[\theta]\equiv U_{\theta}(1)$.

The equation (2.B.2) can be obtained from the composition rule of (2.B.1)

$$ (1+i\delta\theta^{a}T_a)U(\Theta_{\theta}(s))=U(\Theta^a_{\theta}(s) + (h^{-1}) ^a\,_b(\Theta_{\theta}(s))\delta\theta^b) $$

by requiring that $\Theta^a_{\theta}(s) + (h^{-1}) ^a\,_b(\Theta_{\theta}(s))\delta\theta^b$ lies in the path, i.e.

$$ \delta \Theta^a_{\theta}(s)\equiv \Theta^{a}_{\theta}(s+\delta s)-\Theta^{a}_{\theta}(s)= (h^{-1})^a\,_b(\Theta_{\theta}(s))\delta\theta^b $$

which is the same as $\delta\theta^{a}=h^{a}\,_{b}(\Theta_{\theta}(s))\delta\Theta^{a}_{\theta}(s)$. The result is

$$ U(\Theta_{\theta}^a(s+\delta s)) - U(\Theta^a_{\theta}(s))= iT_{a}h^a\,_b(\Theta_{\theta}(s)) U(\Theta_{\theta}(s))\left( \Theta^b_{\theta}(s+\delta s)-\Theta^b_{\theta}(s)\right) $$

If we divide by $\delta s$ we get the equation (2.B.2).

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I believe I have found a coordinate-free answer to this question in which only in the end coordinates are invoked. I'm posting it here too, in case anyone also likes this approach. Corrections are highly appreciated.

There will be a slight shift in notation here. The unitary Lie Group representation will be denoted $\pi : G\to {\rm U}(\cal H)$. Its Lie algebra derived representation will be denoted $d\pi : \mathfrak{g}\to {\operatorname{End}}(\cal H)$.

Straight answer to the question: the motivation to use (2.B.2) to look for $\pi$ given $d\pi$ is that it is the "in-representation" version of the equation saying that along any curve in $G$ the Lie algebra element $X(s)=[R_{\gamma(s)}]_{\ast e}^{-1}\gamma'(s)$ generates a transformation that infinitesimally moves $\gamma(s)$ towards $\gamma(s+\delta s)$.

Given any curve $\gamma :[0,1]\to G$ starting off of the identity its tangent vector can always be seem as the image of something in the Lie algebra by right-translation: $$\gamma'(s)=[R_{\gamma(s)}]_{\ast e}X(s),\quad X(s)=[R_{\gamma(s)}]_{\ast e}^{-1}\gamma'(s).\tag{1}$$

On the contrary to specify any such $\gamma$ we can instead give such $X : [0,1]\to \mathfrak{g}$ and solve (1) with initial condition $\gamma(0)=e$.

The idea is to translate (1) into the representation and find a differential equation for $\pi(\gamma(s))$. We must recall how to define the derivative of $\pi(\gamma(s))$, which is a curve on ${\rm U}(\cal H)$ which has no obvious smooth structure turning it into a Lie group. For that we go with the idea of smooth vectors of a representation. We take ${\cal H}^\infty_\pi$ the space of all $\Psi\in \cal H$ for which $\Pi_\Psi(g)=\pi(g)\Psi$ is smooth. We then define the derivative of $\pi(\gamma(s))$ pointwise on ${\cal H}^\infty_\pi$, i.e.

$$\left[\dfrac{d}{ds}\pi(\gamma(s))\right]\Psi\equiv\dfrac{d}{ds}\left[\pi(\gamma(s))\Psi\right]=\dfrac{d}{ds}\Pi_\Psi(\gamma(s))\tag{2}$$

Since ${\cal H}^\infty_\pi$ may be shown to be dense in the Hilbert space, this defines the derivative of $\pi(\gamma(s))$ everywhere.

Now since $\Pi_\Psi : G\to {\cal H}^\infty_\pi$ we can safely do $$\dfrac{d}{ds}\Pi_\Psi(\gamma(s))=[\Pi_\Psi]_{\ast \gamma(s)}(\gamma'(s))=[\Pi_\Psi]_{\ast \gamma(s)}([R_{\gamma(s)}]_{\ast e} X(s))=(\Pi_\Psi\circ R_{\gamma(s)})_{\ast e}X(s)\tag{3}$$

Now notice that everything happens at fixed $s$, so we are left with the problem of evaluating $(\Pi_\Psi\circ R_g)_{\ast e}Z$ for $Z\in \mathfrak{g}$. To do so we take a short curve $\sigma : (-\epsilon,\epsilon)\to G$ with $\sigma(0)=e$ and $\sigma'(0)=Z$. The obvious such curve is $\sigma(\lambda)=\exp \lambda Z$. We then have $$(\Pi_\Psi\circ R_g)_{\ast e}Z =\dfrac{d}{d\lambda}\bigg|_{\lambda =0}\Pi_\Psi(R_g(\exp \lambda Z))=\dfrac{d}{d\lambda}\bigg|_{\lambda =0}\pi(\exp \lambda Z)\pi(g)\Psi=d\pi(Z)\pi(g)\Psi\tag{4}$$

where the last equality is the definition of the derived representation which may be invoked because if $\Psi$ is a smooth vector so is $\pi(g)\Psi$.

Going back to (3) and (2) this means that $$\left[\dfrac{d}{ds}\pi(\gamma(s))\right]\Psi=d\pi(X(s))\pi(\gamma(s))\Psi,\tag{5}$$

equality for all such smooth vectors then imply equality of the operators and we find $$\dfrac{d}{ds}\pi(\gamma(s))=d\pi(X(s))\pi(\gamma(s)).\tag{6}$$

Eq. (6) is just the "in representation" version of equation (1).

If one introduces coordinates $\theta^a$ centered at the identity it turns out that it is not hard to see that $d\pi(X(s))$ becomes $$d\pi(X(s))=it_ah^a_{\phantom{a}b}(\Theta(s))\dfrac{d\Theta^b(s)}{ds}\tag{7}.$$

Combination of (6) and (7) produces Weinberg's equation (2.B.2) for $U(s)=\pi(\gamma(s))$:

$$\dfrac{d}{ds}U(s)=it_aU(s)h^a_{\phantom{a}b}(\Theta(s))\dfrac{d\Theta^b(s)}{ds}\tag{8}$$

Now Weinberg has the $t_a$, hence he picks standard paths to define each $g\in G$ and uses (8) as a starting point to try defining $\pi(g) = U_g(1)$ where $U_g$ is defined from (8) using the standard path defining $g$.

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