(Physicist version of) Taylor expansions

[Jason Bennett]

Homework Statement

in the context of garnering Lie algebras from Lie groups

Relevant Equations

see below

3) Taylor expansion question in the context of Lie algebra elements:

Consider some n-dimensional Lie group whose elements depend on a set of parameters $$\alpha =(\alpha_1 ... \alpha_n)$$ such that $$g(0) = e$$ with e as the identity, and that had a d-dimensional representation $$D(\alpha)=D(g( \alpha),$$ such that $$D(0)=\mathbb{1}_{d \times d}$$. Then in some small neighborhood of $$\mathbb{1}$$, we can expand $$D(\alpha)$$ as,
$$D(d\alpha) = \mathbb{1} + i d \alpha_i X^i,$$ where $$X^a = -i \frac{\partial}{\partial \alpha_i} D(\alpha)|_{i=0}$$

I have always had trouble with this from quantum mechanics class and on ward. For instance, this process seems identical to the following, from Lancaster and Blundell's QFT for the gifted amateur:

Please see image [1] below.

Using this terminology on the Lie case:

$$\begin{eqnarray} D(0+d\alpha) &=& D(0) + \frac{ \partial D(\alpha)}{\partial \alpha_i}d\alpha \\ &=& \mathbb{1} + (i) (-i) \frac{ \partial D(\alpha)}{\partial \alpha_i}d\alpha \\ &=& \mathbb{1} + (i) X^i d\alpha \end{eqnarray}$$

is this correct? Also, why is the "taking the derivative at $$\alpha=0$$ important? And can you please point me towards a place to learn these types of Taylor expansions?

Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of $$(1+a)^x$$ as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?
[1]: https://i.stack.imgur.com/yAXum.png

 

[MichaelJ12]

Homework Statement: in the context of garnering Lie algebras from Lie groups
Homework Equations: see below

3) Taylor expansion question in the context of Lie algebra elements:

Consider some n-dimensional Lie group whose elements depend on a set of parameters $$\alpha =(\alpha_1 ... \alpha_n)$$ such that $$g(0) = e$$ with e as the identity, and that had a d-dimensional representation $$D(\alpha)=D(g( \alpha),$$ such that $$D(0)=\mathbb{1}_{d \times d}$$. Then in some small neighborhood of $$\mathbb{1}$$, we can expand $$D(\alpha)$$ as,
$$D(d\alpha) = \mathbb{1} + i d \alpha_i X^i,$$ where $$X^a = -i \frac{\partial}{\partial \alpha_i} D(\alpha)|_{i=0}$$

I have always had trouble with this from quantum mechanics class and on ward. For instance, this process seems identical to the following, from Lancaster and Blundell's QFT for the gifted amateur:

Please see image [1] below.

Using this terminology on the Lie case:

$$\begin{eqnarray} D(0+d\alpha) &=& D(0) + \frac{ \partial D(\alpha)}{\partial \alpha_i}d\alpha \\ &=& \mathbb{1} + (i) (-i) \frac{ \partial D(\alpha)}{\partial \alpha_i}d\alpha \\ &=& \mathbb{1} + (i) X^i d\alpha \end{eqnarray}$$

is this correct? Also, why is the "taking the derivative at $$\alpha=0$$ important? And can you please point me towards a place to learn these types of Taylor expansions?

Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of $$(1+a)^x$$ as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?
[1]: https://i.stack.imgur.com/yAXum.png

Your question is not specific to Lie algebras/groups. You simply don't know Taylor series.

A Taylor series is a polynomial expansion of a function around a certain point:

$f(x) =\displaystyle \sum _{n=0}^{\infty }{\frac {f^{(n)}(a)}{n!}}(x-a)^{n}$

$a$ is the point around which you are expanding.

Since you wrote $D(0+d\alpha) = D(0) + \frac{\partial D(\alpha)}{\partial \alpha_i}d\alpha$, the point you are expanding around is $\alpha = 0$, since the right hand side has the constant term $D(0)$. That's why the derivative is at zero.

In equations 1,2,3, $d\alpha$ should be changed to $d\alpha _i$ and there should be a sum over i.

As for the limit of $(1+a)^x$, you are forgetting that a depends on x. When $a = c/x$, you get $e^c$ as $x \to \infty$.

 

[vela]

Also having some trouble understanding the limit of N to infinity in eq. 9.13 of the included picture. In my mind the limit of $$(1+a)^x$$ as x goes to infinity, is infinity... Can someone help me grasp this limit in the case of going from infinitesimal variations with Taylor expansions, to finite variations?

Look at 9.13 a bit more carefully. It's not the limit of $(1+a)^N$ as $N \to \infty$; it's the limit of $(1+\delta a)^N$ where $\delta a = a/N$.