## My book (Geometric Algebra for Electrical Engineers) now available in paper.

January 29, 2019 Geometric Algebra for Electrical Engineers No comments

Edition 0.1.14 of my first book, Geometric Algebra for Electrical Engineers is now available, in a variety of pricing options:

Both paper versions are softcover, and have a 6×9″ format, whereas the PDF is formatted as letter size.  The leanpub version was made when I had the erroneous impression that it was a print on demand service like kindle-direct-publishing (aka createspace.) — it’s not, but the set your own price aspect of their service is kind of neat, so I’ve left it up.

If you download the free PDF or buy the black and white version, and feel undercharged, feel free to send some bitcoin my way.

## Book review: Based on a true story, by Norm Macdonald: 3/5 stars.

January 6, 2019 Incoherent ramblings No comments

It’s been a long time since I’ve had time to read anything fictional, so “Based on a true story” was a fun distraction, at least for a while.

This book has little bits of auto-biography mixed into a bizarre gambling win-big-or-die-trying story, as well as side visits with the Devil and God along the way.  I found that it held my attention until after Norm was released from his 40 year jail sentence for stalking Sarah Silverman and subsequently arranging a clumsy hit on her boyfriend.

There is a lot of funny content in this book, but the absurdity of it gets pretty tiresome about half way in.  The first half of the book is representative, and one need not read much further.

## Spinor solutions with alternate $$\gamma^0$$ representation.

January 2, 2019 phy2403 No comments , ,

This follows an interesting derivation of the $$u, v$$ spinors [2], adding some details.

In class (QFT I) and [3] we used a non-diagonal $$\gamma^0$$ representation
\label{eqn:spinorSolutions:20}
\gamma^0 =
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix},

whereas in [2] a diagonal representation is used
\label{eqn:spinorSolutions:40}
\gamma^0 =
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}.

This representation makes it particularly simple to determine the form of the $$u, v$$ spinors. We seek solutions of the Dirac equation
\label{eqn:spinorSolutions:60}
\begin{aligned}
0 &= \lr{ i \gamma^\mu \partial_\mu – m } u(p) e^{-i p \cdot x} \\
0 &= \lr{ i \gamma^\mu \partial_\mu – m } v(p) e^{i p \cdot x},
\end{aligned}

or
\label{eqn:spinorSolutions:80}
\begin{aligned}
0 &= \lr{ \gamma^\mu p_\mu – m } u(p) e^{-i p \cdot x} \\
0 &= -\lr{ \gamma^\mu p_\mu + m } v(p) e^{i p \cdot x}.
\end{aligned}

In the rest frame where $$\gamma^\mu p_\mu = E \gamma^0$$, where $$E = m = \omega_\Bp$$, these take the particularly simple form
\label{eqn:spinorSolutions:100}
\begin{aligned}
0 &= \lr{ \gamma^0 – 1 } u(E, \Bzero) \\
0 &= \lr{ \gamma^0 + 1 } v(E, \Bzero).
\end{aligned}

This is a nice relation, as we can determine a portion of the structure of the rest frame $$u, v$$ that is independent of the Dirac matrix representation
\label{eqn:spinorSolutions:120}
\begin{aligned}
u(E, \Bzero) &= (\gamma^0 + 1) \psi \\
v(E, \Bzero) &= (\gamma^0 – 1) \psi
\end{aligned}

Similarly, and more generally, we have
\label{eqn:spinorSolutions:140}
\begin{aligned}
u(p) &= (\gamma^\mu p_\mu + m) \psi \\
v(p) &= (\gamma^\mu p_\mu – m) \psi
\end{aligned}

also independent of the representation of $$\gamma^\mu$$. Looking forward to non-matrix representations of the Dirac equation ([1]) note that we have not yet imposed a spinorial structure on the solution
\label{eqn:spinorSolutions:260}
\psi
=
\begin{bmatrix}
\phi \\
\chi
\end{bmatrix},

where $$\phi, \chi$$ are two component matrices.

The particular choice of the diagonal representation \ref{eqn:spinorSolutions:40} for $$\gamma^0$$ makes it simple to determine additional structure for $$u, v$$. Consider the rest frame first, where
\label{eqn:spinorSolutions:160}
\begin{aligned}
\gamma^0 – 1 &=
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}

\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 \\
0 & 2
\end{bmatrix} \\
\gamma^0 + 1 &=
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
=
\begin{bmatrix}
2 & 0 \\
0 & 0
\end{bmatrix},
\end{aligned}

so we have
\label{eqn:spinorSolutions:280}
\begin{aligned}
u(E, \Bzero) &=
\begin{bmatrix}
2 & 0 \\
0 & 0
\end{bmatrix}
\begin{bmatrix}
\phi \\
\chi
\end{bmatrix} \\
v(E, \Bzero) &=
\begin{bmatrix}
0 & 0 \\
0 & 2
\end{bmatrix}
\begin{bmatrix}
\phi \\
\chi
\end{bmatrix}
\end{aligned}

Therefore a basis for the spinors $$u$$ (in the rest frame), is
\label{eqn:spinorSolutions:180}
u(E, \Bzero) \in \setlr{
\begin{bmatrix}
1 \\
0 \\
0 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
0 \\
0
\end{bmatrix}
},

and a basis for the rest frame spinors $$v$$ is
\label{eqn:spinorSolutions:200}
v(E, \Bzero) \in \setlr{
\begin{bmatrix}
0 \\
0 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
0 \\
0 \\
0 \\
1
\end{bmatrix}
}.

Using the two spinor bases $$\zeta^a, \eta^a$$ notation from class, we can write these
\label{eqn:spinorSolutions:220}
\begin{aligned}
u^a(E, \Bzero) &=
\begin{bmatrix}
\zeta^a \\
0
\end{bmatrix},
v^a(E, \Bzero) &=
\begin{bmatrix}
0 \\
\eta^a \\
\end{bmatrix}.
\end{aligned}

For the non-rest frame solutions, [2] opts not to boost, as in [3], but to use the geometry of $$\gamma^\mu p_\mu \pm m$$. With their diagonal representation of $$\gamma^0$$ those are
\label{eqn:spinorSolutions:240}
\begin{aligned}
\gamma^\mu p_\mu – m
&=
p_0
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
p_k
\begin{bmatrix}
0 & \sigma^k \\
– \sigma^k & 0
\end{bmatrix}

m
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
E – m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E – m
\end{bmatrix} \\
\gamma^\mu p_\mu + m
&=
p_0
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}
+
p_k
\begin{bmatrix}
0 & \sigma^k \\
– \sigma^k & 0
\end{bmatrix}
+
m
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
=
\begin{bmatrix}
E + m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E + m
\end{bmatrix} \\
\end{aligned}

Let’s assume that the arbitrary momentum solutions \ref{eqn:spinorSolutions:140} are each proportional to the rest frame solutions
\label{eqn:spinorSolutions:300}
\begin{aligned}
u^a(p) &= (\gamma^\mu p_\mu + m) u^a(E, \Bzero) \\
v^a(p) &= (\gamma^\mu p_\mu – m) u^a(E, \Bzero).
\end{aligned}

Plugging in \ref{eqn:spinorSolutions:240} gives
\label{eqn:spinorSolutions:320}
\begin{aligned}
u^a(p) &=
\begin{bmatrix}
(E + m) \zeta^a \\
(\Bsigma \cdot \Bp ) \zeta^a
\end{bmatrix} \\
v^a(p) &=
\begin{bmatrix}
(\Bsigma \cdot \Bp) \eta^a \\
(E + m) \eta^a
\end{bmatrix},
\end{aligned}

where an overall sign on $$v^a(p)$$ has been dropped. Let’s check the assumption that the rest frame and general solutions are so simply related
\label{eqn:spinorSolutions:340}
\begin{aligned}
\lr{ \gamma^\mu p_\mu – m } u^a(p)
&=
\begin{bmatrix}
E – m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E – m
\end{bmatrix}
\begin{bmatrix}
(E + m) \zeta^a \\
(\Bsigma \cdot \Bp ) \zeta^a
\end{bmatrix} \\
&=
\begin{bmatrix}
(E^2 – m^2 – \Bp^2) \zeta^a \\
0
\end{bmatrix} \\
&= 0,
\end{aligned}

and
\label{eqn:spinorSolutions:360}
\begin{aligned}
\lr{ \gamma^\mu p_\mu + m } v^a(p)
&=
\begin{bmatrix}
E + m & – \Bsigma \cdot \Bp \\
\Bsigma \cdot \Bp & -E + m
\end{bmatrix}
\begin{bmatrix}
(\Bsigma \cdot \Bp ) \eta^a \\
(E + m) \eta^a \\
\end{bmatrix} \\
&=
\begin{bmatrix}
0 \\
\Bp^2 + m^2 – E^2
\end{bmatrix} \\
&= 0.
\end{aligned}

Everything works out nicely. The form of the solution for this representation of $$\gamma^0$$ is much simpler than the Chiral solution that we found in class. We end up with an explicit split of energy and spatial momentum components in the spinor solutions, instead of factors involving $$p \cdot \sigma$$ and $$p \cdot \overline{\sigma}$$, which are arguably nicer from a Lorentz invariance point of view.

# References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[2] Claude Itzykson and Jean-Bernard Zuber. Quantum field theory. McGraw-Hill, 1980.

[3] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.