stevendaryl said:
I'm not sure that I understand the question.
Forgetting about matrices for a second, we can do a rotation of coordinate system (x,y) through an angle of \theta to get:
x' = x cos(\theta) - y sin(\theta)
y' = y cos(\theta) + x sin(\theta)
That can be written in matrix form as:
\left( \begin{array} \\ x' \\ y' \end{array} \right) = \left( \begin{array} \\ cos(\theta) & -sin(\theta) \\ +sin(\theta) & cos(\theta) \end{array} \right) \left( \begin{array} \\ x \\ y \end{array} \right)
The mathematical fact is that the square matrix can be written as an exponential:
\left( \begin{array} \\ cos(\theta) & -sin(\theta) \\ +sin(\theta) & cos(\theta) \end{array} \right) = e^{\theta L_z}
You can interpret the exponential in two different, equivalent ways:
- e^{\theta L_z} = lim_{n \rightarrow \infty} (1+ (\theta L_z)/n)^n
- e^{\theta L_z} = \sum_{n=0}^\infty(\theta L_z)^n/n!
I think it's easier to see the second one. Let's compute the first few terms:
e^{\theta L_z} = I + (\theta L_z) + (\theta L_z)^2/2! + (\theta L_z)^3/3! + ...
Note: (L_z)^2 = -I, (L_z)^3 = - L_z, (L_z)^4 = -I, etc. and so every other term is a multiple of the 2x2 identity matrix. So we can rearrange the terms to get:
e^{\theta L_z} = I[1 - (\theta)^2/2! + (\theta)^4/4! - ...] + L_z(\theta - (\theta)^3/3! + (\theta)^5/5! ...)
Those are just the taylor series for cosine and sine:
e^{\theta L_z} = I cos(\theta) + L_z sin(\theta)
I think it's a little harder with the first form of the exponential, but not too hard.
Thanks for the second one. However, I find the first one more intuitive in coming up with the exponential function that brings the infinitesimal transformation to a finite one.
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Lets say,
An infinitesimal rotation around the z-axis by δθ can be written as
$$\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} =
\begin{pmatrix}
x-δ\theta y \\
y+δ\theta x \\
\end{pmatrix} =
\begin{pmatrix}
1 & -δ\theta \\
δ\theta & 1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}=(I+δ\theta L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}$$ ...(1)
with
$$L_z=
\begin{pmatrix}
0 & -1 \\
1 & 0 \\
\end{pmatrix}
$$
A rotation by a finite angle θ is constructed as n consecutive rotations by θ/n each and taking the limit n→∞. Using (1), it can be written be
$$
\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} =\lim_{x \to \infty} (I + \frac{\theta}{n} L_z )^n
\begin{pmatrix}
x \\
y \\
\end{pmatrix} = e^{\theta L_z}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$...(2)
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The above makes sense to me as it shows me how to get to the exponential(which will subsequently bring me to the finite transformation) from the infinitesimal transformation.
However, how to get from (1) to (2) is a bit fuzzy to me.
How I rationalize this is that multiplying this (I + (θ/n) L
z ) to (x,y) to give it 1 unit of infinitesimal rotation. Multiply this (I + (θ/n) L
z ) twice gives (x,y) 2 units of infinitesimal rotation. Multiply this (I + (θ/n) L
z ) thrice gives (x,y) 3 units of infinitesimal rotation and so on.
In terms of matrix,
$$(I + \frac{\theta}{n} L_z )^2=(I + \frac{2\theta}{n} L_z+...)≈(I + \frac{2\theta}{n} L_z)$$
$$(I + \frac{\theta}{n} L_z )^3=(I + \frac{3\theta}{n} L_z+...)≈(I + \frac{3\theta}{n} L_z)$$ and so on.
I ignore the other power of (θ/n) cause they become negligible.
Hence, $$(I + \frac{\theta}{n} L_z )^n=(I + \frac{n\theta}{n} L_z+...)≈(I + \frac{n\theta}{n} L_z)$$ will means that there is n infinitesimal rotation which is equivalent to θ rotation.Here is the part where I am not very sure. Can I instead of doing step (2), I will just
$$
\begin{pmatrix}
x' \\
y' \\
\end{pmatrix} = (I + k*\frac{\theta}{n} L_z)
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
$$ where k is no. of transformation and n goes to infinity
So is something like this(n goes to infinity),
1 unit of infinitesimal rotation ⇒ (I + (θ/n) L
z ) , where k=1
2 units of infinitesimal rotation ⇒ (I + (2θ/n) L
z ) , where k=2
3 units of infinitesimal rotation ⇒ (I + (3θ/n) L
z ) , where k=3
and so on.
So is basically instead of multiplying (I + (θ/n) L
z ) to itself by n times, I will just sum (θ/n) by n times.
