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Notes for Desai Chapter 26

Posted by peeterjoot on December 9, 2010

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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.

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This entry was posted on December 9, 2010 at 10:59 pm and is filed under Math and Physics Learning.. Tagged: , , , , , . You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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