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# Posts Tagged ‘rotation matrix’

## Infinitesimal rotations

Posted by peeterjoot on January 27, 2012

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Motivation.

In a classical mechanics lecture (which I audited) Prof. Poppitz made the claim that an infinitesimal rotation in direction $\hat{\mathbf{n}}$ of magnitude $\delta \phi$ has the form

\begin{aligned}\mathbf{x} \rightarrow \mathbf{x} + \delta \boldsymbol{\phi} \times \mathbf{x},\end{aligned} \hspace{\stretch{1}}(1.1)

where

\begin{aligned}\delta \boldsymbol{\phi} = \hat{\mathbf{n}} \delta \phi.\end{aligned} \hspace{\stretch{1}}(1.2)

I believe he expressed things in terms of the differential displacement

\begin{aligned}\delta \mathbf{x} = \delta \boldsymbol{\phi} \times \mathbf{x}\end{aligned} \hspace{\stretch{1}}(1.3)

This was verified for the special case $\hat{\mathbf{n}} = \hat{\mathbf{z}}$ and $\mathbf{x} = x \hat{\mathbf{x}}$. Let’s derive this in the general case too.

# With geometric algebra.

Let’s temporarily dispense with the normal notation and introduce two perpendicular unit vectors $\hat{\mathbf{u}}$, and $\hat{\mathbf{v}}$ in the plane of the rotation. Relate these to the unit normal with

\begin{aligned}\hat{\mathbf{n}} = \hat{\mathbf{u}} \times \hat{\mathbf{v}}.\end{aligned} \hspace{\stretch{1}}(2.4)

A rotation through an angle $\phi$ (infinitesimal or otherwise) is then

\begin{aligned}\mathbf{x} \rightarrow e^{-\hat{\mathbf{u}} \hat{\mathbf{v}} \phi/2} \mathbf{x} e^{\hat{\mathbf{u}} \hat{\mathbf{v}} \phi/2}.\end{aligned} \hspace{\stretch{1}}(2.5)

Suppose that we decompose $\mathbf{x}$ into components in the plane and in the direction of the normal $\hat{\mathbf{n}}$. We have

\begin{aligned}\mathbf{x} = x_u \hat{\mathbf{u}} + x_v \hat{\mathbf{v}} + x_n \hat{\mathbf{n}}.\end{aligned} \hspace{\stretch{1}}(2.6)

The exponentials commute with the $\hat{\mathbf{n}}$ vector, and anticommute otherwise, leaving us with

\begin{aligned}\mathbf{x} &\rightarrow x_n \hat{\mathbf{n}} + (x_u \hat{\mathbf{u}} + x_v \hat{\mathbf{v}}) e^{\hat{\mathbf{u}} \hat{\mathbf{v}} \phi} \\ &=x_n \hat{\mathbf{n}} + (x_u \hat{\mathbf{u}} + x_v \hat{\mathbf{v}}) (\cos\phi + \hat{\mathbf{u}} \hat{\mathbf{v}} \sin\phi) \\ &=x_n \hat{\mathbf{n}} + \hat{\mathbf{u}} (x_u \cos\phi - x_v \sin\phi) +\hat{\mathbf{v}} (x_v \cos\phi + x_u \sin\phi).\end{aligned}

In the last line we use $\hat{\mathbf{u}}^2 = 1$ and $\hat{\mathbf{u}} \hat{\mathbf{v}} = - \hat{\mathbf{v}} \hat{\mathbf{u}}$. Making the angle infinitesimal $\phi \rightarrow \delta \phi$ we have

\begin{aligned}\mathbf{x} &\rightarrow x_n \hat{\mathbf{n}} + \hat{\mathbf{u}} (x_u - x_v \delta\phi) +\hat{\mathbf{v}} (x_v + x_u \delta\phi) \\ &=\mathbf{x} + \delta\phi( x_u \hat{\mathbf{v}} - x_v \hat{\mathbf{u}})\end{aligned}

We have only to confirm that this matches the assumed cross product representation

\begin{aligned}\hat{\mathbf{n}} \times \mathbf{x}&=\begin{vmatrix}\hat{\mathbf{u}} & \hat{\mathbf{v}} & \hat{\mathbf{n}} \\ 0 & 0 & 1 \\ x_u & x_v & x_n\end{vmatrix} \\ &=-\hat{\mathbf{u}} x_v + \hat{\mathbf{v}} x_u\end{aligned}

Taking the two last computations we find

\begin{aligned}\delta \mathbf{x} = \delta \phi \hat{\mathbf{n}} \times \mathbf{x} = \delta \boldsymbol{\phi} \times \mathbf{x},\end{aligned} \hspace{\stretch{1}}(2.7)

as desired.

# Without geometric algebra.

We’ve also done the setup above to verify this result without GA. Here we wish to apply the rotation to the coordinate vector of $\mathbf{x}$ in the $\{\hat{\mathbf{u}}, \hat{\mathbf{v}}, \hat{\mathbf{n}}\}$ basis which gives us

\begin{aligned}\begin{bmatrix}x_u \\ x_v \\ x_n \end{bmatrix}&\rightarrow \begin{bmatrix}\cos\delta\phi & -\sin\delta\phi & 0 \\ \sin\delta\phi & \cos\delta\phi & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x_u \\ x_v \\ x_n \end{bmatrix} \\ &\approx\begin{bmatrix}1 & -\delta\phi & 0 \\ \delta\phi & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}\begin{bmatrix}x_u \\ x_v \\ x_n \end{bmatrix} \\ &=\begin{bmatrix}x_u \\ x_v \\ x_n \end{bmatrix} +\begin{bmatrix}0 & -\delta\phi & 0 \\ \delta\phi & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix}x_u \\ x_v \\ x_n \end{bmatrix} \\ &=\begin{bmatrix}x_u \\ x_v \\ x_n \end{bmatrix} +\delta\phi\begin{bmatrix}-x_v \\ x_u \\ 0\end{bmatrix} \end{aligned}

But as we’ve shown, this last coordinate vector is just $\hat{\mathbf{n}} \times \mathbf{x}$, and we get our desired result using plain old fashioned matrix algebra as well.

Really the only difference between this and what was done in class is that there’s no assumption here that $\mathbf{x} = x \hat{\mathbf{x}}$.

## Notes for Desai Chapter 26

Posted by peeterjoot on December 9, 2010

# Motivation.

Chapter 26 notes for [1].

# Guts

## Trig relations.

To verify equations 26.3-5 in the text it’s worth noting that

\begin{aligned}\cos(a + b) &= \Re( e^{ia} e^{ib} ) \\ &= \Re( (\cos a + i \sin a)( \cos b + i \sin b) ) \\ &= \cos a \cos b - \sin a \sin b\end{aligned}

and

\begin{aligned}\sin(a + b) &= \Im( e^{ia} e^{ib} ) \\ &= \Im( (\cos a + i \sin a)( \cos b + i \sin b) ) \\ &= \cos a \sin b + \sin a \cos b\end{aligned}

So, for

\begin{aligned}x &= \rho \cos\alpha \\ y &= \rho \sin\alpha \end{aligned} \hspace{\stretch{1}}(2.1)

the transformed coordinates are

\begin{aligned}x' &= \rho \cos(\alpha + \phi) \\ &= \rho (\cos \alpha \cos \phi - \sin \alpha \sin \phi) \\ &= x \cos \phi - y \sin \phi\end{aligned}

and

\begin{aligned}y' &= \rho \sin(\alpha + \phi) \\ &= \rho (\cos \alpha \sin \phi + \sin \alpha \cos \phi) \\ &= x \sin \phi + y \cos \phi \\ \end{aligned}

This allows us to read off the rotation matrix. Without all the messy trig, we can also derive this matrix with geometric algebra.

\begin{aligned}\mathbf{v}' &= e^{- \mathbf{e}_1 \mathbf{e}_2 \phi/2 } \mathbf{v} e^{ \mathbf{e}_1 \mathbf{e}_2 \phi/2 } \\ &= v_3 \mathbf{e}_3 + (v_1 \mathbf{e}_1 + v_2 \mathbf{e}_2) e^{ \mathbf{e}_1 \mathbf{e}_2 \phi } \\ &= v_3 \mathbf{e}_3 + (v_1 \mathbf{e}_1 + v_2 \mathbf{e}_2) (\cos \phi + \mathbf{e}_1 \mathbf{e}_2 \sin\phi) \\ &= v_3 \mathbf{e}_3 + \mathbf{e}_1 (v_1 \cos\phi - v_2 \sin\phi)+ \mathbf{e}_2 (v_2 \cos\phi + v_1 \sin\phi)\end{aligned}

Here we use the Pauli-matrix like identities

\begin{aligned}\mathbf{e}_k^2 &= 1 \\ \mathbf{e}_i \mathbf{e}_j &= -\mathbf{e}_j \mathbf{e}_i,\quad i\ne j\end{aligned} \hspace{\stretch{1}}(2.3)

and also note that $\mathbf{e}_3$ commutes with the bivector for the $x,y$ plane $\mathbf{e}_1 \mathbf{e}_2$. We can also read off the rotation matrix from this.

## Infinitesimal transformations.

Recall that in the problems of Chapter 5, one representation of spin one matrices were calculated [2]. Since the choice of the basis vectors was arbitrary in that exersize, we ended up with a different representation. For $S_x, S_y, S_z$ as found in (26.20) and (26.23) we can also verify easily that we have eigenvalues $0, \pm \hbar$. We can also show that our spin kets in this non-diagonal representation have the following column matrix representations:

\begin{aligned}{\lvert {1,\pm 1} \rangle}_x &=\frac{1}{{\sqrt{2}}} \begin{bmatrix}0 \\ 1 \\ \pm i\end{bmatrix} \\ {\lvert {1,0} \rangle}_x &=\begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix} \\ {\lvert {1,\pm 1} \rangle}_y &=\frac{1}{{\sqrt{2}}} \begin{bmatrix}\pm i \\ 0 \\ 1 \end{bmatrix} \\ {\lvert {1,0} \rangle}_y &=\begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix} \\ {\lvert {1,\pm 1} \rangle}_z &=\frac{1}{{\sqrt{2}}} \begin{bmatrix}1 \\ \pm i \\ 0\end{bmatrix} \\ {\lvert {1,0} \rangle}_z &=\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix} \end{aligned} \hspace{\stretch{1}}(2.5)

## Verifying the commutator relations.

Given the (summation convention) matrix representation for the spin one operators

\begin{aligned}(S_i)_{jk} = - i \hbar \epsilon_{ijk},\end{aligned} \hspace{\stretch{1}}(2.11)

let’s demonstrate the commutator relation of (26.25).

\begin{aligned}{\left[{S_i},{S_j}\right]}_{rs} &=(S_i S_j - S_j S_i)_{rs} \\ &=\sum_t (S_i)_{rt} (S_j)_{ts} - (S_j)_{rt} (S_i)_{ts} \\ &=(-i\hbar)^2 \sum_t \epsilon_{irt} \epsilon_{jts} - \epsilon_{jrt} \epsilon_{its} \\ &=-(-i\hbar)^2 \sum_t \epsilon_{tir} \epsilon_{tjs} - \epsilon_{tjr} \epsilon_{tis} \\ \end{aligned}

Now we can employ the summation rule for sums products of antisymmetic tensors over one free index (4.179)

\begin{aligned}\sum_i \epsilon_{ijk} \epsilon_{iab}= \delta_{ja}\delta_{kb}-\delta_{jb}\delta_{ka}.\end{aligned} \hspace{\stretch{1}}(2.12)

Continuing we get

\begin{aligned}{\left[{S_i},{S_j}\right]}_{rs} &=-(-i\hbar)^2 \left(\delta_{ij}\delta_{rs}-\delta_{is}\delta_{rj}-\delta_{ji}\delta_{rs}+\delta_{js}\delta_{ri} \right) \\ &=(-i\hbar)^2 \left( \delta_{is}\delta_{jr}-\delta_{ir} \delta_{js}\right)\\ &=(-i\hbar)^2 \sum_t \epsilon_{tij} \epsilon_{tsr}\\ &=i\hbar \sum_t \epsilon_{tij} (S_t)_{rs}\qquad\square\end{aligned}

## General infinitesimal rotation.

Equation (26.26) has for an infinitesimal rotation counterclockwise around the unit axis of rotation vector $\mathbf{n}$

\begin{aligned}\mathbf{V}' = \mathbf{V} + \epsilon \mathbf{n} \times \mathbf{V}.\end{aligned} \hspace{\stretch{1}}(2.13)

Let’s derive this using the geometric algebra rotation expression for the same

\begin{aligned}\mathbf{V}' &=e^{-I\mathbf{n} \alpha/2}\mathbf{V} e^{I\mathbf{n} \alpha/2} \\ &=e^{-I\mathbf{n} \alpha/2}\left((\mathbf{V} \cdot \mathbf{n})\mathbf{n}+(\mathbf{V} \wedge \mathbf{n})\mathbf{n}\right)e^{I\mathbf{n} \alpha/2} \\ &=(\mathbf{V} \cdot \mathbf{n})\mathbf{n}+(\mathbf{V} \wedge \mathbf{n})\Bne^{I\mathbf{n} \alpha}\end{aligned}

We note that $I\mathbf{n}$ and thus the exponential commutes with $\mathbf{n}$, and the projection component in the normal direction. Similarily $I\mathbf{n}$ anticommutes with $(\mathbf{V} \wedge \mathbf{n}) \mathbf{n}$. This leaves us with

\begin{aligned}\mathbf{V}' &=(\mathbf{V} \cdot \mathbf{n})\mathbf{n}\left(+(\mathbf{V} \wedge \mathbf{n})\mathbf{n}\right)( \cos \alpha + I \mathbf{n} \sin\alpha)\end{aligned}

For $\alpha = \epsilon \rightarrow 0$, this is

\begin{aligned}\mathbf{V}' &=(\mathbf{V} \cdot \mathbf{n})\mathbf{n}+(\mathbf{V} \wedge \mathbf{n})\mathbf{n}( 1 + I \mathbf{n} \epsilon) \\ &=(\mathbf{V} \cdot \mathbf{n})\mathbf{n} +(\mathbf{V} \wedge \mathbf{n})\mathbf{n}+\epsilon I^2(\mathbf{V} \times \mathbf{n})\mathbf{n}^2 \\ &=\mathbf{V}+ \epsilon (\mathbf{n} \times \mathbf{V}) \qquad\square\end{aligned}

## Position and angular momentum commutator.

Equation (26.71) is

\begin{aligned}\left[{x_i},{L_j}\right] = i \hbar \epsilon_{ijk} x_k.\end{aligned} \hspace{\stretch{1}}(2.14)

Let’s derive this. Recall that we have for the position-momentum commutator

\begin{aligned}\left[{x_i},{p_j}\right] = i \hbar \delta_{ij},\end{aligned} \hspace{\stretch{1}}(2.15)

and for each of the angular momentum operator components we have

\begin{aligned}L_m = \epsilon_{mab} x_a p_b.\end{aligned} \hspace{\stretch{1}}(2.16)

The commutator of interest is thus

\begin{aligned}\left[{x_i},{L_j}\right] &= x_i \epsilon_{jab} x_a p_b -\epsilon_{jab} x_a p_b x_i \\ &= \epsilon_{jab} x_a\left(x_i p_b -p_b x_i \right) \\ &=\epsilon_{jab} x_ai \hbar \delta_{ib} \\ &=i \hbar \epsilon_{jai} x_a \\ &=i \hbar \epsilon_{ija} x_a \qquad\square\end{aligned}

## A note on the angular momentum operator exponential sandwiches.

In (26.73-74) we have

\begin{aligned}e^{i \epsilon L_z/\hbar} x e^{-i \epsilon L_z/\hbar} = x + \frac{i \epsilon}{\hbar} \left[{L_z},{x}\right]\end{aligned} \hspace{\stretch{1}}(2.17)

Observe that

\begin{aligned}\left[{x},{\left[{L_z},{x}\right]}\right] = 0\end{aligned} \hspace{\stretch{1}}(2.18)

so from the first two terms of (10.99)

\begin{aligned}e^{A} B e^{-A}= B + \left[{A},{B}\right]+\frac{1}{{2}} \left[{A},{\left[{A},{B}\right]}\right] \cdots\end{aligned} \hspace{\stretch{1}}(2.19)

we get the desired result.

## Trace relation to the determinant.

Going from (26.90) to (26.91) we appear to have a mystery identity

\begin{aligned}\det \left( \mathbf{1} + \mu \mathbf{A} \right) = 1 + \mu \text{Tr} \mathbf{A}\end{aligned} \hspace{\stretch{1}}(2.20)

According to wikipedia, under derivative of a determinant, [3], this is good for small $\mu$, and related to something called the Jacobi identity. Someday I should really get around to studying determinants in depth, and will take this one for granted for now.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] Peeter Joot. Notes and problems for Desai Chapter V. [online]. http://sites.google.com/site/peeterjoot/math2010/desaiCh5.pdf.

[3] Wikipedia. Determinant — wikipedia, the free encyclopedia [online]. 2010. [Online; accessed 10-December-2010]. http://en.wikipedia.org/w/index.php?title=Determinant&oldid=400983667.