Series expansion of the Lorentz Transformation

[JD_PM]

Homework Statement

Proper Lorentz Transformations' expansion is given to be

$$\Lambda^{\mu}_{ \ \ \nu}(\epsilon)=(\exp(\epsilon))^{\mu}_{ \ \ \nu} = \delta^{\mu}_{\nu} + \epsilon^{\mu}_{ \ \ \nu} +\frac{1}{2!} \epsilon^{\mu}_{ \ \ \rho}\epsilon^{\rho}_{ \ \ \nu}+ \ ..., \tag1$$

Where

$$\epsilon_{\mu \nu}=-\epsilon_{\nu \mu} \tag2$$

a) Show that the Lorentz Transformation condition $\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho} (\epsilon)\Lambda^{\nu}{}_{\sigma}(\epsilon) = \eta_{\rho \sigma}$ holds for $(1)$

b) Show (by working out the matrix exponential explicitly) that only when $\epsilon_{12}$ does not vanish we get a rotation around the 3-axis over an angle $\epsilon_{12}$

c) Show that only when $\epsilon_{03}$ does not vanish we get a boost along the 3-axis with rapidity $\epsilon_{03}$

Relevant Equations

$$\Lambda^{\mu}_{ \ \ \nu}(\epsilon)=(\exp(\epsilon))^{\mu}_{ \ \ \nu} = \delta^{\mu}_{\nu} + \epsilon^{\mu}_{ \ \ \nu} +\frac{1}{2!} \epsilon^{\mu}_{ \ \ \rho}\epsilon^{\rho}_{ \ \ \nu}+ \ ..., \tag1$$

a) I think I got this one (I have to thank samalkhaiat and PeroK for helping me with the training in LTs :) )

$$\eta_{\mu\nu}\Big(\delta^{\mu}_{\rho} + \epsilon^{\mu}_{ \ \ \rho} +\frac{1}{2!} \epsilon^{\mu}_{ \ \ \lambda}\epsilon^{\lambda}_{ \ \ \rho}+ \ ...\Big)\Big(\delta^{\nu}_{\sigma} + \epsilon^{\nu}_{ \ \ \sigma} +\frac{1}{2!} \epsilon^{\nu}_{ \ \ \alpha}\epsilon^{\alpha}_{ \ \ \sigma}+ \ ...\Big)=$$

$$=\eta_{\mu \nu} \delta^{\mu}_{\rho}\delta^{\nu}_{\sigma}+\eta_{\mu\nu} \epsilon^{\mu}_{ \ \ \rho}\epsilon^{\nu}_{ \ \ \sigma}+\frac 1 4 \eta_{\mu \nu}\epsilon^{\mu}_{ \ \ \lambda}\epsilon^{\lambda}_{ \ \ \rho} \epsilon^{\nu}_{ \ \ \alpha}\epsilon^{\alpha}_{ \ \ \sigma} +\eta_{\mu \nu}\delta^{\mu}_{\rho}\epsilon^{\nu}_{ \ \ \sigma}+\frac{1}{2!}\eta_{\mu\nu}\delta^{\mu}_{\rho}\epsilon^{\nu}_{ \ \ \alpha}\epsilon^{\alpha}_{ \ \ \sigma}+\eta_{\mu\nu}\epsilon^{\mu}_{ \ \ \rho}\delta^{\nu}_{\sigma}$$

$$+\frac{1}{2!}\eta_{\mu\nu}\epsilon^{\mu}_{ \ \ \rho}\epsilon^{\nu}_{ \ \ \alpha}\epsilon^{\alpha}_{ \ \ \sigma} + \frac{1}{2!} \eta_{\mu\nu} \epsilon^{\mu}_{ \ \ \lambda}\epsilon^{\lambda}_{ \ \ \rho}\delta^{\nu}_{\sigma}+\frac{1}{2!} \eta_{\mu\nu} \epsilon^{\mu}_{ \ \ \lambda}\epsilon^{\lambda}_{ \ \ \rho}\epsilon^{\nu}_{ \ \ \sigma} + \ ... =$$

$$ = \eta_{\mu \nu} \delta^{\mu}_{\rho}\delta^{\nu}_{\sigma}+\eta_{\mu\nu} \epsilon^{\mu}_{ \ \ \rho}\epsilon^{\nu}_{ \ \ \sigma}+\eta_{\mu \nu}\delta^{\mu}_{\rho}\epsilon^{\nu}_{ \ \ \sigma}+\frac{1}{2!}\eta_{\mu\nu}\delta^{\mu}_{\rho}\epsilon^{\nu}_{ \ \ \alpha}\epsilon^{\alpha}_{ \ \ \sigma}+\eta_{\mu\nu}\epsilon^{\mu}_{ \ \ \rho}\delta^{\nu}_{\sigma}+\frac{1}{2!} \eta_{\mu\nu} \epsilon^{\mu}_{ \ \ \lambda}\epsilon^{\lambda}_{ \ \ \rho}\delta^{\nu}_{\sigma}+\mathcal O(\epsilon^3)=$$ $$=\eta_{\rho \sigma} + \epsilon_{\nu \rho}\epsilon^{\nu}_{ \ \sigma}+ \epsilon_{\rho \sigma}+\frac{1}{2!}\epsilon_{\rho \alpha}\epsilon^{\alpha}_{ \ \sigma}+\epsilon_{\sigma \rho}+\frac{1}{2!}\epsilon_{\sigma \lambda}\epsilon^{\lambda}_{ \ \rho} + \mathcal O(\epsilon^3)$$

We now massage one of the terms to get

$$\epsilon_{\sigma \lambda}\epsilon^{\lambda}_{ \ \rho}=\epsilon^{\lambda}_{ \ \rho}\epsilon_{\sigma \lambda}=-\epsilon^{\lambda}_{ \ \rho}\epsilon_{\lambda \sigma}=\epsilon_{\rho \lambda}\epsilon^{\lambda}_{ \ \sigma}$$

Neglecting order three and higher we get

$$\eta_{\mu\nu}\Lambda^{\mu}{}_{\rho} (\epsilon)\Lambda^{\nu}{}_{\sigma}(\epsilon)=\eta_{\rho \sigma} - \epsilon_{\rho \beta}\epsilon^{\beta}_{ \ \sigma}+ \epsilon_{\rho \sigma}+\frac{1}{2!}\epsilon_{\rho \beta}\epsilon^{\beta}_{ \ \sigma}-\epsilon_{\rho \sigma}+\frac{1}{2!}\epsilon_{\rho \beta}\epsilon^{\beta}_{ \ \sigma}=\eta_{\rho \sigma}$$

QED.

For b) and c) I am afraid I need hints. What do they mean by saying 'a rotation around the 3-axis over an angle $\epsilon_{12}$' and 'a boost along the 3-axis with rapidity $\epsilon_{03}$'?

Thanks

 

[strangerep]

For b) and c) I am afraid I need hints. What do they mean by saying 'a rotation around the 3-axis over an angle $\epsilon_{12}$' and 'a boost along the 3-axis with rapidity $\epsilon_{03}$'?

Think about "rotation around the z-axis by an angle $\theta$". I.e., a rotation in the x-y plane by an angle $\theta$. What is the general (matrix) expression for such a 2D rotation of the x,y coordinates? Try to express that matrix in terms of an exponential of a generator matrix. (Hint: think about the Taylor series for sin and cos.)

A similar approach applies to part (c), except that there you work with a boost transformation involving the $z,t$ coordinates instead of a rotation of the $x,y$ coordinates.

 

[JD_PM]

Think about "rotation around the z-axis by an angle $\theta$". I.e., a rotation in the x-y plane by an angle $\theta$. What is the general (matrix) expression for such a 2D rotation of the x,y coordinates? Try to express that matrix in terms of an exponential of a generator matrix. (Hint: think about the Taylor series for sin and cos.)

b) Alright so the Rotation matrix in 2D

$$R(\theta)=
\begin{pmatrix}
\cos(\theta) & -\sin(\theta) \\
\sin(\theta) & \cos(\theta) \\
\end{pmatrix}
$$

Can also be written as follows (I checked out the rotation case here)

$$R(\theta)=e^{G \theta}= 1_{2\times2} + G\sin(\theta) + G^2(1-\cos\theta)$$

Where

$$G=ba^T -ab^T, \ \ \ \ a^T b=0, \ \ \ \ G^2=-aa^T-bb^T$$

In our problem we have

$$R(\epsilon^{\mu}_{\ \ \nu})=e^{G \epsilon^{\mu}_{\ \ \nu}}= 1_{2\times2} + G\sin(\epsilon^{\mu}_{\ \ \nu}) + G^2(1-\cos\epsilon^{\mu}_{\ \ \nu})$$

Applying Taylor series for $\cos$ and $\sin$ up to second order we get

$$R(\epsilon^{\mu}_{\ \ \nu}) = 1_{2\times2} + G \epsilon^{\mu}_{\ \ \nu}+G^2\frac{\epsilon^{\mu}_{\ \ \lambda}\epsilon^{\lambda}_{\ \ \nu}}{2!} \tag{*}$$

We now multiply both sides of $(*)$ by $\eta_{\sigma \mu}$ to get

$$\eta_{\sigma \mu}R(\epsilon^{\mu}_{\ \ \nu}) = \eta_{\sigma \mu}1_{2\times2} + G \epsilon_{\sigma\nu}+G^2\frac{\epsilon_{\sigma\lambda}\epsilon^{\lambda}_{\ \ \nu}}{2!} \tag{**}$$

OK so far.

However, I do not see why Eq $(**)$ only holds when $\epsilon_{12} \neq 0$.

 

[strangerep]

Actually, I was hoping you would see the correct matrix $G$ almost by inspection.

Hint: expand $e^{i \sigma_2\, \theta}$, where $\sigma_2$ is the 2nd Pauli matrix.

 

[JD_PM]

Hint: expand $e^{i \sigma_2\, \theta}$, where $\sigma_2$ is the 2nd Pauli matrix.

$$e^{i \sigma_2\, \theta}=1+i\sigma_2\theta-\frac{(\sigma_2\theta)^2}{2!}+ \ ...$$

Actually, I was hoping you would see the correct matrix $G$ almost by inspection.

I see that if we pick $a= \begin{pmatrix} 1 \\ 0 \\ \end{pmatrix}$ and $b= \begin{pmatrix} 0 \\ 1 \\ \end{pmatrix}$ then we get $G= \begin{pmatrix} 0 & -1 \\ 1 & 0 \\ \end{pmatrix}$ and $G^2= \begin{pmatrix} -1 & 0 \\ 0 & -1 \\ \end{pmatrix}$

Thus we see that

$$\sigma_2=iG$$

Thus $(**)$ can be rewritten as follows (note that in my previous message I forgot to explicitly state that $(1_{2\times2})^{\mu}_{ \ \ \ \nu}$)

$$\eta_{\sigma \mu}R(\epsilon^{\mu}_{\ \ \nu}) = \eta_{\sigma \mu}(1_{2\times2})^{\mu}_{ \ \ \ \nu} -i \sigma_2 \epsilon_{\sigma\nu}-\sigma_2^2 \frac{\epsilon_{\sigma\lambda}\epsilon^{\lambda}_{\ \ \nu}}{2!} \tag{***}$$

 

[JD_PM]

I was a bit stuck once I got $(***)$ but then I recalled I came across rotation block matrices $S^{ij}$ containing Pauli matrices while studying Tong's notes (page 85). They are given to be

$$S^{ij} = \frac 1 2 \begin{pmatrix}
0 & \sigma^i \\
-\sigma^i & 0 \\
\end{pmatrix} \begin{pmatrix}
0 & \sigma^j \\
-\sigma^j & 0 \\
\end{pmatrix}=-\frac{i}{2} \mathscr{E}^{ijk}\begin{pmatrix}
\sigma^k & 0 \\
0 & \sigma^k \\
\end{pmatrix} \tag{4.25}$$

Where $\mathscr{E}^{ijk}$ is the Levi-Civita symbol .

We note that $(4.25)$ only holds when $i \neq j$. Thus we could define a rotation parameter $\Omega_{ij} = -\mathscr{E}_{ijk} \varphi^k$. Then Tong asserts that the rotation matrix $S^{ij}$ becomes

$$S[\Lambda]=\exp(\frac 1 2 \Omega_{\rho \sigma}S^{\rho \sigma})=\begin{pmatrix}
e^{\frac 1 2 i \vec \varphi \cdot \vec \sigma} & 0 \\
0 & e^{\frac 1 2 i \vec \varphi \cdot \vec \sigma} \\
\end{pmatrix} \tag{4.26}$$

OK.

Coming back to the original problem b)

I was hoping to be able to use the above to solve the b) but I noticed that in $(4.25)$ we need to be dealing with $i \neq j$ if we want to get $S^{ij} \neq 0$ and in $(***)$ we're dealing all the time with $\sigma_2$; so either $(***)$ is wrong or the rotation matrices provided by Tong are not useful to solve this particular problem.

PS: I may be overcomplicating things though.

 

[vanhees71]

Usually it's tough to do literally the infinite matrix-exponential calculation. It's easy for spin-1/2 with Pauli matrices (which also can be used for the Lorentz transformations for the Weyl-spinor representations of the Lorentz group, or rather the covering group of the original Lorentz group, $\mathrm{SL}(2,\mathbb{C})$).

Often it's easier to define the vector
$$\vec{x}(t)=\exp(t \hat{\epsilon})$$
and then solve for
$$\dot{\vec{x}}(t)=\hat{\epsilon} \vec{x}.$$
Here $\vec{x}$ is of course a vector appropriate for the problem, i.e., $\mathbb{R}^n$ or $\mathbb{C}^n$ when you have a real or complex $(n \times n)$ matrix to exponentiate.

 

[strangerep]

I was hoping to be able to use the above to solve the b) but I noticed that in $(4.25)$ we need to be dealing with $i \neq j$ if we want to get $S^{ij} \neq 0$ and in $(***)$ we're dealing all the time with $\sigma_2$; so either $(***)$ is wrong or the rotation matrices provided by Tong are not useful to solve this particular problem.

Take another look at eq(4.9) in Tong's notes, and the following paragraph.
A finite Lorentz transformation takes the form of Tong's eq(4.10). Which elements in Tong's $\omega^\mu_{~\nu}$ must be nonzero to get a rotation in the $(x^1,x^2)$-plane?

So yes, you were indeed overcomplicating it again. The answer is (almost) staring you in the face in Tong's eqns (4.9) and (4.10). Take the square of the matrix $M^{12}$. Which terms remain zero? Do the same with the cube of $M^{12}$. You'll find simply that the Taylor series I hinted at involving just the 2D matrix $\sigma_2$ remains nestled inside the 4D matrix.

 

[JD_PM]

OK I see we can construct a basis for the antisymmetric matrix $\omega^\mu_{~\nu}$. Such a basis $(\mathcal{M}^{\rho \sigma})^{\mu}_{ \ \ \nu}$ contains 6 independent elements (3 boosts and 3 rotations).

Rotations basis elements are $(\mathcal{M}^{1 3})^{\mu}_{ \ \ \nu}$, $(\mathcal{M}^{2 3})^{\mu}_{ \ \ \nu}$ and $(\mathcal{M}^{1 2})^{\mu}_{ \ \ \nu}$

I see that $(\mathcal{M}^{1 2})^{\mu}_{ \ \ \nu}$ generates rotations around the $x^3$ axis, which are the ones they are asking at b). Then by Tong's $(4.10)$

$$\omega^{\mu}_{ \ \ \nu}= \frac 1 2 \Omega_{12} (\mathcal{M}^{1 2})^{\mu}_{ \ \ \nu}$$

And the rotation parameter is indeed $\Omega_{12}$

I'd rather use the series expansion method to solve the problem though.

So yes, you were indeed overcomplicating it again. The answer is (almost) staring you in the face in Tong's eqns (4.9) and (4.10). Take the square of the matrix $M^{12}$. Which terms remain zero? Do the same with the cube of $M^{12}$. You'll find simply that the Taylor series I hinted at involving just the 2D matrix $\sigma_2$ remains nestled inside the 4D matrix.

Mmm thanks to your comment I've noticed that I gave $G$ (which is $(\mathcal{M}^{12})^{\mu}_{ \ \ \nu}$ indeed) and $G^2$ incorrectly; they are $4\times4$ matrices instead. Let's use $(\mathcal{M}^{12})^{\mu}_{ \ \ \nu}$ notation from now on.$$(\mathcal{M}^{12})^{\mu}_{ \ \ \nu}=
\begin{pmatrix}
0 & 0 & 0 & 0\\
0 & 0 & -1 & 0 \\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix}, \ \ \ \ \ \
\Big((\mathcal{M}^{12})^{\mu}_{ \ \ \nu}\Big)^2=
\begin{pmatrix}
0 & 0 & 0 & 0\\
0 & -1& 0& 0 \\
0 & 0 & -1 & 0\\
0 & 0 & 0 & 0\\
\end{pmatrix}
$$

Now I am trying to set up the correct series expansion (including the right generator); I am thinking of something of the form

$$R(\epsilon^{\mu}_{\ \ \nu}) = (1_{4\times4})^{\mu}_{ \ \ \nu} + (\mathcal{M}^{12})\epsilon^{\mu}_{\ \ \nu}+\Big(\mathcal{M}^{12}\Big)^2\frac{\epsilon^{\mu}_{\ \ \lambda}\epsilon^{\lambda}_{\ \ \nu}}{2!} \tag{****}$$

But I'm a bit stuck...

 

[strangerep]

I'd rather use the series expansion method to solve the problem though.
[...]
Mmm thanks to your comment I've noticed that I gave $G$ (which is $(\mathcal{M}^{12})^{\mu}_{ \ \ \nu}$ indeed) and $G^2$ incorrectly; they are $4\times4$ matrices instead. Let's use $(\mathcal{M}^{12})^{\mu}_{ \ \ \nu}$ notation from now on.

Do you still not get the point that you can concentrate just on $\sigma_2$ (which is nestled inside the 4x4 $M^{12}$ matrix)? $(M^{12})^n$ (for any power $n$) always has zeroes around the outside.

Go back and do what I suggested in my post #4 (you have not yet computed the matrices properly). Do the expansion up to 4th order and then compare with the Taylor series for $\sin\theta$ and $\cos\theta$. Relate this back to the expression for a finite rotation in the x-y place by angle $\theta$. (You still haven't actually done this properly.)

 

[vanhees71]

I'd solve the 1st-order system of ODE's. It's much simpler. You've done the calculation when you solved for the relativistic motion of particles in homogeneous electric fields ("boost") or magnetic fields ("rotation").

 

[JD_PM]

I'd solve the 1st-order system of ODE's. It's much simpler.

Hi vanhees71.

That’s the approach you suggested at #7, right? I have to acknowledge I don’t really understand it. Could you please give an explicit worked-out example? (Maybe I’m asking too much).

Solving the problem using both ways sound great to me.

 

[vanhees71]

Well, let's take the example of a rotation around a fixed axis $\vec{n}$ in $\mathbb{R}^3$. Writing
$$\vec{x}(\phi)=\hat{D}_{\text{n}}(\phi) \vec{x}_0,$$
you know that
$$\vec{x}'(\phi)=\vec{n} \times \vec{x}(\phi).$$
It's most easy to choose a Cartesian reference frame with $\vec{n}=\vec{e}_3$. Then you have
$$\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix} = \begin{pmatrix} -y \\ x \\0 \end{pmatrix}.$$
This set of ODE's you can solve with the initial condition $\vec{x}(0)=\vec{x}_0$.

Taking the derivative of the first component of the equation leads to
$$x''=-y'=-x \; \Rightarrow \; x=A \cos \varphi + B \sin \varphi.$$
From this
$$y=-x'=-A \sin \varphi + B \cos \varphi.$$
The 3rd equation is trivial:
$$z'=0 \; \Rightarrow \; z=z_0=\text{const}.$$
The initial condition for $x$ and $y$ leads to
$$x(0)=A=x_0, \quad y(0)=B=y_0.$$
From that you have
$$\vec{x}(\varphi)=\begin{pmatrix} \cos \varphi & \sin \varphi & 0 \\
-\sin \varphi & \cos \varphi & 0\\
0 & 0 & 1 \end{pmatrix} \vec{x}_0.$$
The matrix is $\hat{D}_{\vec{e}_3}(\varphi)$. Now to express this with invariant vectors just take
$$\vec{n} \times \vec{x}_0=\vec{e}_3 \times \vec{x}_0=\begin{pmatrix}-y_0 \\ x_0 \\ 0 \end{pmatrix}$$
and
$$\vec{n} \times (\vec{n} \times \vec{x}_0) = - \begin{pmatrix} x_0 \\ y_0 \\ 0 \end{pmatrix}.$$
This leads to
$$\hat{D}_{\varphi} \vec{x}_0=\vec{n} (\vec{n} \cdot \vec{x}_0) - \sin \varphi \vec{n} \times \vec{x}_0 - \cos \varphi \vec{n} \times (\vec{n} \times \vec{x}_0).$$

 

[JD_PM]

Do you still not get the point that you can concentrate just on $\sigma_2$ (which is nestled inside the 4x4 $M^{12}$ matrix)? $(M^{12})^n$ (for any power $n$) always has zeroes around the outside.

Yes, I see it.

Go back and do what I suggested in my post #4 (you have not yet computed the matrices properly). Do the expansion up to 4th order and then compare with the Taylor series for $\sin\theta$ and $\cos\theta$. Relate this back to the expression for a finite rotation in the x-y place by angle $\theta$. (You still haven't actually done this properly.)

Let's go step by step.

Go back and do what I suggested in my post #4 (you have not yet computed the matrices properly).

You suggested to write the Taylor expansion of $e^{i \sigma_2\, \theta}$. Let's now do it up to fourth order

$$e^{i \sigma_2\, \theta}=1+i\sigma_2\theta+i^2 \frac{(\sigma_2\theta)^2}{2!}+i^3 \frac{(\sigma_2\theta)^3}{3!}+i^4 \frac{(\sigma_2\theta)^4}{4!} \ +...$$

Where

$$\sigma_2 =

\begin{pmatrix}

0 & -i \\

i & 0 \\

\end{pmatrix}$$

$$\sigma_2^2 =

\begin{pmatrix}

1 & 0 \\

0 & 1 \\

\end{pmatrix}=1_{2\times2}$$

$$\sigma_2^3 =

\begin{pmatrix}

0 & -i \\

i & 0 \\

\end{pmatrix}=\sigma_2$$

$$\sigma_2^4 =

\begin{pmatrix}

1 & 0 \\

0 & 1 \\

\end{pmatrix}=1_{2\times2}$$

then compare with the Taylor series for $\sin\theta$ and $\cos\theta$. Relate this back to the expression for a finite rotation in the x-y place by angle $\theta$. (You still haven't actually done this properly.)

I see how to relate the expansion of the exponential and $\sin\theta$ and $\cos\theta$ through Euler's formula $e^{\pm i \sigma_2\, \theta}=\cos(\sigma_2\, \theta) \pm i\sin(\sigma_2\, \theta)$. However I do not see how to relate this to the expression for a finite rotation in the x-y plane by angle $\theta$.

I know once I see it I'll say oh, it was really easy but unfortunately I did not 'click' yet. I was thinking that an example could help me out to see your point.

Thanks for your patience.

 

[JD_PM]

Well, let's take the example of a rotation around a fixed axis $\vec{n}$ in $\mathbb{R}^3$. Writing
$$\vec{x}(\phi)=\hat{D}_{\text{n}}(\phi) \vec{x}_0,$$
you know that
$$\vec{x}'(\phi)=\vec{n} \times \vec{x}(\phi).$$
It's most easy to choose a Cartesian reference frame with $\vec{n}=\vec{e}_3$. Then you have
$$\begin{pmatrix}x' \\ y' \\ z' \end{pmatrix} = \begin{pmatrix} -y \\ x \\0 \end{pmatrix}.$$
This set of ODE's you can solve with the initial condition $\vec{x}(0)=\vec{x}_0$.

Taking the derivative of the first component of the equation leads to
$$x''=-y'=-x \; \Rightarrow \; x=A \cos \varphi + B \sin \varphi.$$
From this
$$y=-x'=-A \sin \varphi + B \cos \varphi.$$
The 3rd equation is trivial:
$$z'=0 \; \Rightarrow \; z=z_0=\text{const}.$$
The initial condition for $x$ and $y$ leads to
$$x(0)=A=x_0, \quad y(0)=B=y_0.$$
From that you have
$$\vec{x}(\varphi)=\begin{pmatrix} \cos \varphi & \sin \varphi & 0 \\
-\sin \varphi & \cos \varphi & 0\\
0 & 0 & 1 \end{pmatrix} \vec{x}_0.$$
The matrix is $\hat{D}_{\vec{e}_3}(\varphi)$. Now to express this with invariant vectors just take
$$\vec{n} \times \vec{x}_0=\vec{e}_3 \times \vec{x}_0=\begin{pmatrix}-y_0 \\ x_0 \\ 0 \end{pmatrix}$$
and
$$\vec{n} \times (\vec{n} \times \vec{x}_0) = - \begin{pmatrix} x_0 \\ y_0 \\ 0 \end{pmatrix}.$$
This leads to
$$\hat{D}_{\varphi} \vec{x}_0=\vec{n} (\vec{n} \cdot \vec{x}_0) - \sin \varphi \vec{n} \times \vec{x}_0 - \cos \varphi \vec{n} \times (\vec{n} \times \vec{x}_0).$$

Thanks vanhees71.

 

[strangerep]

$$e^{i \sigma_2\, \theta}=1+i\sigma_2\theta+i^2 \frac{(\sigma_2\theta)^2}{2!}+i^3 \frac{(\sigma_2\theta)^3}{3!}+i^4 \frac{(\sigma_2\theta)^4}{4!} \ +...$$

Continue this a bit more. E.g., replace $i\sigma_2$ by the corresponding matrix with real entries only (I'll call it ``"$\alpha$"). Rewrite your 4th order expansion in terms of $\alpha$ and $1_{2\times 2}$. I.e., get rid of all powers on your matrices by substituting what they evaluate to, e.g., $\alpha^3 = -\alpha$.

Then rearrange the expansion so that even and odd powers of $\theta$ are grouped together separately (say, by putting those respective terms on 2 separate lines).

Then WRITE OUT the Taylor series for $\sin\theta$ and $\cos\theta$ (each on a separate line) and compare with the 2-line expansion). Notice anything?

I know once I see it I'll say oh, it was really easy.

Yep.

Late Edit: I've decided to make it easier for you. Prove that $$e^{-i \sigma_2\, \theta} ~=~ \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} ~.$$

 

[JD_PM]

Yes, love the 'I got it' feeling! (thanks for your patience )

Continue this a bit more. E.g., replace $i\sigma_2$ by the corresponding matrix with real entries only (I'll call it ``"$\alpha$"). Rewrite your 4th order expansion in terms of $\alpha$ and $1_{2\times 2}$. I.e., get rid of all powers on your matrices by substituting what they evaluate to, e.g., $\alpha^3 = -\alpha$.

$$e^{i \sigma_2\, \theta}=\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}\theta^0 +\begin{pmatrix}
0 & 1 \\
-1 & 0 \\
\end{pmatrix}\theta
+
\frac{1}{2!} \begin{pmatrix}
-1 & 0 \\
0 & -1 \\
\end{pmatrix}\theta^2
+
\frac{1}{3!} \begin{pmatrix}
0 & -1 \\
1 & 0 \\
\end{pmatrix}\theta^3
+
\frac{1}{4!} \begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}\theta^4
\ +...$$

Then rearrange the expansion so that even and odd powers of $\theta$ are grouped together separately (say, by putting those respective terms on 2 separate lines).

$$e^{i \sigma_2\, \theta}=\begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}\theta^0 +
\frac{1}{2!} \begin{pmatrix}
-1 & 0 \\
0 & -1 \\
\end{pmatrix}\theta^2 +
\frac{1}{4!} \begin{pmatrix}
1 & 0 \\
0 & 1 \\
\end{pmatrix}\theta^4$$ $$+
\begin{pmatrix}
0 & 1 \\
-1 & 0 \\
\end{pmatrix}\theta
+
\frac{1}{3!} \begin{pmatrix}
0 & -1 \\
1 & 0 \\
\end{pmatrix}\theta^3 \ + ...$$

Then WRITE OUT the Taylor series for $\sin\theta$ and $\cos\theta$ (each on a separate line) and compare with the 2-line expansion).

$$\cos \theta = 1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} \ - ...$$

$$\sin \theta = \theta - \frac{\theta^3}{3!} \ + ...$$

Notice anything?

Ahhh so the key was to focus on matrix entries! For instance if we focus on $x_{11}$ position we indeed get $\cos \theta$

$$\sum_{i=0}^n x_{i 11} = \theta^0 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} \ - ... = \cos \theta$$

For the other three we get

$$\sum_{i=0}^n x_{i 12} = \theta - \frac{\theta^3}{3!} \ + ... = \sin \theta$$

$$\sum_{i=0}^n x_{i 21} = -\theta + \frac{\theta^3}{3!} \ - ... = -\sin \theta$$

$$\sum_{i=0}^n x_{i 22} = \theta^0 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} \ - ... = \cos \theta$$

Where $i=0$ refers to the matrix $\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}\theta^0$ and so on

Thus we have just proven that

$$e^{i \sigma_2\, \theta} ~=~ \begin{pmatrix} \cos\theta & \sin\theta \\ -\sin\theta & \cos\theta \end{pmatrix} ~.$$

By analogous work we can prove

$$e^{-i \sigma_2\, \theta} ~=~ \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} ~.$$

 

[JD_PM]

OK. I see how to relate the exponential to the 2D rotation matrix.

But once we've worked out the matrix exponential, how can we see that only when $\epsilon_{12}$ does not vanish we get a rotation around the 3-axis over an angle $\epsilon_{12}$?

 

[strangerep]

By analogous work we can prove
$$e^{-i \sigma_2\, \theta} ~=~ \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix} ~.$$

It doesn't to be "analogous work". Just substitute $\theta \to -\theta$.

 

[strangerep]

OK. I see how to relate the exponential to the 2D rotation matrix.

But once we've worked out the matrix exponential, how can we see that only when $\epsilon_{12}$ does not vanish we get a rotation around the 3-axis over an angle $\epsilon_{12}$?

You've got the general 2D rotation transformation $x \to x'(\theta) = x \cos\theta - y\sin\theta; ~~~ y \to y'(\theta) = x \sin\theta + y\cos\theta$.

So now write down the general 3D rotation transformation (in terms of coordinates $x,y,z$) that rotate $x,y$ around the $z$ axis, i.e., which leave $z$ unchanged. This will be 3 equations $x' = ?;~$ $y' = ?;~$ $z' = ?\;$. Then combine these 3 equations into a single 3D matrix equation.

What are the corresponding 3D matrix equations for $x - z$ rotations (which preserve $y$) and for $y - z$ rotations (which preserve $x$) ?

 

[JD_PM]

So now write down the general 3D rotation transformation (in terms of coordinates $x,y,z$) that rotate $x,y$ around the $z$ axis, i.e., which leave $z$ unchanged. This will be 3 equations $x' = ?;~$ $y' = ?;~$ $z' = ?\;$. Then combine these 3 equations into a single 3D matrix equation.

The rotation transformations around the $z$ axis

$$x \to x'(\theta) = x \cos\theta + y\sin\theta; ~~~ y \to y'(\theta) = -x \sin\theta + y\cos\theta; ~~~ z \to z' = z$$

Thus the z-rotation matrix is

$$R_z(\theta)=
\begin{pmatrix}
\cos\theta & \sin\theta & 0 \\
-\sin\theta & \cos\theta & 0 \\
0 & 0 & 1 \\
\end{pmatrix}
$$

What are the corresponding 3D matrix equations for $x - z$ rotations (which preserve $y$) [...] ?

The rotation transformations around the $y$ axis

$$x \to x'(\theta) = x \cos\theta - z\sin\theta; ~~~ y \to y' = y; ~~~ z \to z'(\theta) = x \sin\theta + z\cos\theta$$

Thus the y-rotation matrix is

$$R_y(\theta)=
\begin{pmatrix}
\cos\theta & 0 & -\sin\theta \\
0 & 1 & 0 \\
\sin\theta & 0 & \cos\theta \\
\end{pmatrix}
$$

[...] for $y - z$ rotations (which preserve $x$) ?

The rotation transformations around the $x$ axis

$$x \to x' = x; ~~~ y \to y'(\theta) = y \cos\theta + z\sin\theta; ~~~ z \to z'(\theta) = -y \sin\theta + z\cos\theta$$

Thus the x-rotation matrix is

$$R_x(\theta)=
\begin{pmatrix}
1 & 0 & 0 \\
0 & \cos\theta & \sin\theta \\
0 & -\sin\theta & \cos\theta \\
\end{pmatrix}
$$

 

[JD_PM]

OK I now understand the statement 'only when $\epsilon_{12}$ does not vanish we get a rotation around the 3-axis over an angle $\epsilon_{12}$'; they mean that only when the z-coordinate remains unchanged we get a rotation around the z-axis.

However, they want us to show it explicitly in terms of $\epsilon$ and matrix notation. Does that mean we have to introduce the Levi-Civita symbol and proceed like below:

?

 

[strangerep]

However, they want us to show it explicitly in terms of $\epsilon$ and matrix notation. Does that mean we have to introduce the Levi-Civita symbol and proceed like below:

View attachment 266675
?

Similar, but not exactly like that. The text you've reproduced looks like how one would represent 3D rotations that act on a Dirac 4-spinor. But in previous posts you've been working rotations acting on ordinary 3-space vectors. To understand this distinction better, I suggest you take your $R_i(\theta)$ matrices, (where $i=x,y,z$) and compute their corresponding generators.

If you don't know what a "generator" is, it's the thing you use in the exponential version of a rotation matrix. So to recover a generator from the corresponding finite transformation, just differentiate wrt $\theta$ and then set $\theta=0$.

So for each of your $R_i(\theta)$ matrices, you'll get a "$J_i$" generator matrix.

Further exercises:
1) Show that the set of the 3 $J_i$ matrices form a Lie algebra. (If you don't know what a Lie algebra is, then consider it part of this exercise to look that up.)

2) Show that the set of the 3 $J_i$ matrices form a vector space, in the sense that any linear combination of them is in a fact a generator for some 3D rotation.

3) Show that ANY finite 3D rotation can be obtained by exponentiating a linear combination of the $J_i$'s, (meaning that exponentiating any such linear combination yields a finite 3D rotation matrix).

[Btw, if you haven't yet acquired a copy of Greiner's "Quantum Mechanics -- Symmetries", I can assure you it would be money well spent. I learned a vast amount from that book when I was getting started into serious study of QM. Greiner's textbooks are good for a beginner student because they give lots of worked examples, and he doesn't skip steps in the detailed computations.]