## Updated statistical mechanics notes.

February 13, 2019 math and physics play No comments I’ve posted a minor update to my old stat mech notes, plus instructions on how to clone the github repos and the latex, should somebody wish to attempt to fork these notes for their own purposes.  Enjoy!

## Mathematica notebooks updated, and a bivector addition visualization.

February 10, 2019 math and physics play No comments , , ,

This blog now has a copy of all my Mathematica notebooks (as of Feb 10, 2019), complete with a chronological index.  I hadn’t updated that index since 2014, and it was quite stale.

I’ve also added an additional level of per-directory indexing.  For example, you can now look at just the notebooks for my book, Geometric Algebra for Electrical Engineers.  That was possible before, but you would have had to clone the entire git repository to be able to do so easily.

This update includes a new notebook written today, which has a Manipulate visualization of 3D bivector addition that is kind of fun. Bivector addition, at least in 3D, can be done graphically almost like vector addition.  Instead of trying to add the planes (which can be done, as in the neat illustration in Geometric Algebra for Computer Science), you can do the task more simply by connecting the normals head to tail, where each of the normals are scaled by the area of the bivector (i.e. it’s absolute magnitude).  The resulting bivector has an area equal to the length of that sum of normals, and a “direction” perpendicular to that resulting normal.  This fun little Manipulate lets you interactively visualize this process, by changing the radius of a set of summed bivectors, each oriented in a different direction, and observing the effects of doing so.

Of course, you can interpret this visualization as nothing more than a representation of addition of cross products, if you were to interpret the vector representing a cross product as an oriented area with a normal equal to that cross product (where the normal’s magnitude equals the area, as in this bivector addition visualization.)  This works out nicely because of the duality relationship between the cross and wedge product, and the duality relationship between 3D bivectors and their normals.

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

January 2, 2019 phy2403 No comments , ,

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

In class (QFT I) and  we used a non-diagonal $$\gamma^0$$ representation
\begin{equation}\label{eqn:spinorSolutions:20}
\gamma^0 =
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix},
\end{equation}
whereas in  a diagonal representation is used
\begin{equation}\label{eqn:spinorSolutions:40}
\gamma^0 =
\begin{bmatrix}
1 & 0 \\
0 & -1
\end{bmatrix}.
\end{equation}
This representation makes it particularly simple to determine the form of the $$u, v$$ spinors. We seek solutions of the Dirac equation
\begin{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}
\end{equation}
or
\begin{equation}\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}
\end{equation}
In the rest frame where $$\gamma^\mu p_\mu = E \gamma^0$$, where $$E = m = \omega_\Bp$$, these take the particularly simple form
\begin{equation}\label{eqn:spinorSolutions:100}
\begin{aligned}
0 &= \lr{ \gamma^0 – 1 } u(E, \Bzero) \\
0 &= \lr{ \gamma^0 + 1 } v(E, \Bzero).
\end{aligned}
\end{equation}
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
\begin{equation}\label{eqn:spinorSolutions:120}
\begin{aligned}
u(E, \Bzero) &= (\gamma^0 + 1) \psi \\
v(E, \Bzero) &= (\gamma^0 – 1) \psi
\end{aligned}
\end{equation}
Similarly, and more generally, we have
\begin{equation}\label{eqn:spinorSolutions:140}
\begin{aligned}
u(p) &= (\gamma^\mu p_\mu + m) \psi \\
v(p) &= (\gamma^\mu p_\mu – m) \psi
\end{aligned}
\end{equation}
also independent of the representation of $$\gamma^\mu$$. Looking forward to non-matrix representations of the Dirac equation () note that we have not yet imposed a spinorial structure on the solution
\begin{equation}\label{eqn:spinorSolutions:260}
\psi
=
\begin{bmatrix}
\phi \\
\chi
\end{bmatrix},
\end{equation}
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
\begin{equation}\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}
\end{equation}
so we have
\begin{equation}\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}
\end{equation}
Therefore a basis for the spinors $$u$$ (in the rest frame), is
\begin{equation}\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}
},
\end{equation}
and a basis for the rest frame spinors $$v$$ is
\begin{equation}\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}
}.
\end{equation}
Using the two spinor bases $$\zeta^a, \eta^a$$ notation from class, we can write these
\begin{equation}\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}
\end{equation}

For the non-rest frame solutions,  opts not to boost, as in , but to use the geometry of $$\gamma^\mu p_\mu \pm m$$. With their diagonal representation of $$\gamma^0$$ those are
\begin{equation}\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}
\end{equation}

Let’s assume that the arbitrary momentum solutions \ref{eqn:spinorSolutions:140} are each proportional to the rest frame solutions
\begin{equation}\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}
\end{equation}
Plugging in \ref{eqn:spinorSolutions:240} gives
\begin{equation}\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}
\end{equation}
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
\begin{equation}\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}
\end{equation}
and
\begin{equation}\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}
\end{equation}
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

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

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

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

## Final first draft of complete notes for UofT PHY2403, QFT I .

December 27, 2018 phy2403 No comments , ,

I’ve now uploaded a new version of my class notes for PHY2403, the UofT Quantum Field Theory I course, taught this year by Prof. Erich Poppitz.

This update adds notes for all remaining lectures (up to and including lecture 23.)  I’ve made a pass with a spellchecker to correct some of the aggregious spelling erorss, and also redrawn three figures, replacing photos, which cuts the size in half!

I’ve posted the redacted version (316 pages).  The full version, with my problem set solutions (including errors) is 409 pages.

Feel free to contact me for the complete version (i.e. including my problem set solutions, with errors) of any of these notes, provided you are not asking because you are taking or planning to take this course.

Contents:

• Preface
• Contents
• List of Figures
• 1 Fields, units, and scales.
• 1.1 What is a field?
• 1.2 Scales.
• 1.2.2 Compton wavelength.
• 1.2.3 Relations.
• 1.3 Natural units.
• 1.4 Gravity.
• 1.5 Cross section.
• 1.6 Problems.
• 2 Lorentz transformations.
• 2.1 Lorentz transformations.
• 2.2 Determinant of Lorentz transformations.
• 2.3 Problems.
• 3 Classical field theory.
• 3.1 Field theory.
• 3.2 Actions.
• 3.3 Principles determining the form of the action.
• 3.4 Principles (cont.)
• 3.4.1 d = 2.
• 3.4.2 d = 3.
• 3.4.3 d = 4.
• 3.4.4 d = 5.
• 3.5 Least action principle.
• 3.6 Problems.
• 4 Canonical quantization, Klein-Gordon equation, SHOs, momentum space representation, raising and lowering operators.
• 4.1 Canonical quantization.
• 4.2 Canonical quantization (cont.)
• 4.3 Momentum space representation.
• 4.4 Quantization of Field Theory.
• 4.5 Free Hamiltonian.
• 4.6 QM SHO review.
• 4.7 Discussion.
• 4.8 Problems.
• 5 Symmetries.
• 5.1 Switching gears: Symmetries.
• 5.2 Symmetries.
• 5.3 Spacetime translation.
• 5.4 1st Noether theorem.
• 5.5 Unitary operators.
• 5.6 Continuous symmetries.
• 5.7 Classical scalar theory.
• 5.8 Last time.
• 5.9 Examples of symmetries.
• 5.10 Scale invariance.
• 5.11 Lorentz invariance.
• 5.12 Problems.
• 6 Lorentz boosts, generators, Lorentz invariance, microcausality.
• 6.1 Lorentz transform symmetries.
• 6.2 Transformation of momentum states.
• 6.3 Relativistic normalization.
• 6.4 Spacelike surfaces.
• 6.5 Condition on microcausality.
• 7 External sources.
• 7.1 Harmonic oscillator.
• 7.2 Field theory (where we are going).
• 7.3 Green’s functions for the forced Klein-Gordon equation.
• 7.4 Pole shifting.
• 7.5 Matrix element representation of the Wightman function.
• 7.6 Retarded Green’s function.
• 7.7 Review: “particle creation problem”.
• 7.8 Digression: coherent states.
• 7.9 Problems.
• 8 Perturbation theory.
• 8.1 Feynman’s Green’s function.
• 8.2 Interacting field theory: perturbation theory in QFT.
• 8.3 Perturbation theory, interaction representation and Dyson formula.
• 8.4 Next time.
• 8.5 Review.
• 8.6 Perturbation.
• 8.7 Review.
• 8.8 Unpacking it.
• 8.9 Calculating perturbation.
• 8.10 Wick contractions.
• 8.11 Simplest Feynman diagrams.
• 8.12 Phi fourth interaction.
• 8.13 Tree level diagrams.
• 8.14 Problems.
• 9 Scattering and decay.
• 9.2 Definitions and motivation.
• 9.3 Calculating interactions.
• 9.4 Example diagrams.
• 9.5 The recipe.
• 9.6 Back to our scalar theory.
• 9.7 Review: S-matrix.
• 9.8 Scattering in a scalar theory.
• 9.9 Decay rates.
• 9.10 Cross section.
• 9.11 More on cross section.
• 9.12 d(LIPS)_2.
• 9.13 Problems.
• 10 Fermions, and spinors.
• 10.1 Fermions: R3 rotations.
• 10.2 Lorentz group.
• 10.3 Weyl spinors.
• 10.4 Lorentz symmetry.
• 10.5 Dirac matrices.
• 10.6 Dirac Lagrangian.
• 10.7 Review.
• 10.8 Dirac equation.
• 10.9 Helicity.
• 10.10 Next time.
• 10.11 Review.
• 10.12 Normalization.
• 10.13 Other solution.
• 10.14 Lagrangian.
• 10.15 General solution and Hamiltonian.
• 10.16 Review.
• 10.17 Hamiltonian action on single particle states.
• 10.18 Spacetime translation symmetries.
• 10.19 Rotation symmetries: angular momentum operator.
• 10.20 U(1)_V symmetry: charge!
• 10.21 U(1)_A symmetry: what was the charge for this one called?
• 10.22 CPT symmetries.
• 10.23 Review.
• 10.24 Photon.
• 10.25 Propagator.
• 10.26 Feynman rules.
• 10.27 Example: muon pair production
• 10.28 Measurement of intermediate quark scattering processes.
• 10.29 Problems.
• A Useful formulas and review.
• A.1 Review of old material.
• A.2 Useful results from new material.
• B Momentum of scalar field.
• B.1 Expansion of the field momentum.
• B.2 Conservation of the field momentum.
• C Reflection using Pauli matrices.
• D Explicit expansion of the Dirac u,v spinors.
• D.1 Compact representation of
• E Mathematica notebooks
• Bibliography

## PHY2403H Quantum Field Theory. Lecture 23: QED and QCD interaction Lagrangian, Feynman propagator and rules for Fermions, hadron pair production, scattering cross section, quark pair production. Taught by Prof. Erich Poppitz

Here is a link to [a PDF with my notes for the final QFT I lecture.] That lecture followed  section 5.1 fairly closely (filling in some details, leaving out some others.)

This lecture

• Introduced an interaction Lagrangian with QED and QCD interaction terms
\begin{equation*}
\LL_{\text{QED}}
=
– \inv{4} F_{\mu\nu} F^{\mu\nu}
+
\overline{\Psi}_e \lr{ i \gamma^\mu \partial_\mu – m } \Psi_e

e \overline{\Psi}_e \gamma_\mu \Psi_e A^\mu
+
\overline{\Psi}_\mu \lr{ i \gamma^\mu \partial_\mu – m } \Psi_\mu

e \overline{\Psi}_\mu \gamma_\mu \Psi_\mu A^\mu,
\end{equation*}
as well as the quark interaction Lagrangian
\begin{equation*}
\LL_{\text{quarks}} = \sum_q \overline{\Psi}_q \lr{ i \gamma^\mu – m_q } \Psi_q + e Q_q \overline{\Psi}_q \gamma^\nu \Psi_q A_\nu.
\end{equation*}
• The Feynman propagator for Fermions was calculated
\begin{equation*}
\expectation{ T( \Psi_\alpha(x) \Psi_\beta(x) }_0
=
\lr{ \gamma^\mu_{\alpha\beta} \partial_\mu^{(x)} + m } D_F(x – y)
=
\int \frac{d^4 p}{(2 \pi)^4 } \frac{ i ( \gamma^\mu_{\alpha\beta} p_\mu + m ) }{p^2 – m^2 + i \epsilon} e^{-i p \cdot (x – y)}.
\end{equation*}
• We determined the Feynman rules for Fermion diagram nodes and edges.
The Feynman propagator for Fermions is
\begin{equation*}
\frac{ i \lr{ \gamma^\mu p_\mu + m } }{p^2 – m^2 + i \epsilon},
\end{equation*}
whereas the photon propagator is
\begin{equation*}
\expectation{ A_\mu A_\nu } = -i \frac{g_{\mu\nu}}{q^2 + i \epsilon}.
\end{equation*}
• We then studied muon pair production in detail, and determined the form of the scattering matrix element
\begin{equation*}
i M
=
i \frac{e^2}{q^2}
\overline{v}^{s’}(p’) \gamma^\rho u^s(p)
\overline{u}^r(k) \gamma_\rho v^{r’}(k’),
\end{equation*}
where the $$(2 \pi)^4 \delta^4(…)$$ term hasn’t been made explicit, and detemined that the average of its square over all input and output polarization (spin) states was
\begin{equation*}
\inv{4} \sum_{ss’, rr’} \Abs{M}^2
=
\frac{e^4}{4 q^4}
\textrm{tr}{ \lr{
\lr{ \gamma^\alpha {k’}_\alpha – m_\mu }
\gamma_\nu
\lr{ \gamma^\beta {k}_\beta + m_\mu }
\gamma_\mu
}}
\times
\textrm{tr}{ \lr{
\lr{ \gamma^\kappa {p}_\kappa + m_e }
\gamma^\nu
\lr{ \gamma^\rho {p’}_\rho – m_e }
\gamma^\mu
}}.
\end{equation*}.
In the CM frame (neglecting the electron mass, which is small relative to the muon mass), this reduced to
\begin{equation*}
\inv{4} \sum_{\text{spins}} \Abs{M}^2
=
\frac{8 e^4}{q^4}
\lr{
p \cdot k’ p’ \cdot k
+ p \cdot k p’ \cdot k’
+ p \cdot p’ m_\mu^2
}.
\end{equation*}

• We computed the differential cross section
\begin{equation*}
{\frac{d\sigma}{d\Omega}}_{\text{CM}}
=
\frac{\alpha^2}{4 E_{\text{CM}}^2 }
\sqrt{ 1 – \frac{m_\mu^2}{E^2} }
\lr{
1 + \frac{m_\mu^2}{E^2}
+ \lr{ 1 – \frac{m_\mu^2}{E^2} } \cos^2\theta
},
\end{equation*}
and the total cross section
\begin{equation*}
\sigma_{\text{total}}
=
\frac{4 \pi \alpha^2}{3 E_{\text{CM}}^2 }
\sqrt{ 1 – \frac{m_\mu^2}{E^2} }
\lr{
1 + \inv{2} \frac{m_\mu^2}{E^2}
},
\end{equation*}
and compared that to the cross section that we was determined with the dimensional analysis handwaving at the start of the course.
• We finished off with a quick discussion of quark pair production, and how some of the calculations we performed for muon pair production can be used to measure and validate the intermediate quark states that were theorized as carriers of the strong force.

# References

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

## Dirac spinor relations after rest frame boost

In , Prof Osmond explicitly boosts a $$u^s(p_0)$$ Dirac spinor from the rest frame with rest frame energy $$p_0$$.
After doing so he claims the identification
\begin{equation}\label{eqn:squarerootpsigma:20}
\begin{aligned}
\sqrt{m} e^{-\inv{2} \eta \sigma^3} &= \sqrt{ p \cdot \sigma } \\
\sqrt{m} e^{\inv{2} \eta \sigma^3} &= \sqrt{ p \cdot \overline{\sigma} },
\end{aligned}
\end{equation}
for the components of $$u^s(\Lambda p_0)$$.

Let’s verify this by squaring. First
\begin{equation}\label{eqn:squarerootpsigma:40}
e^{\pm \inv{2} \eta \sigma^3 }
=
\cosh\lr{ \inv{2} \eta \sigma^3 }
\pm
\sinh\lr{ \inv{2} \eta \sigma^3 } \sigma^3,
\end{equation}
which squares to (FIXME: link to uvspinor.nb)
\begin{equation}\label{eqn:squarerootpsigma:60}
\lr{ e^{\pm \inv{2} \eta \sigma^3 } }^2
=
\begin{bmatrix}
e^{\pm \eta} & 0 \\
0 & e^{\mp \eta}
\end{bmatrix}.
\end{equation}

Explicitly boosting the rest energy $$p_0$$ gives
\begin{equation}\label{eqn:squarerootpsigma:80}
\begin{bmatrix}
p_0 \\
0
\end{bmatrix}
\rightarrow
\begin{bmatrix}
\cosh\eta & \sinh\eta \\
\sinh\eta & \cosh\eta \\
\end{bmatrix}
\begin{bmatrix}
p_0 \\
0
\end{bmatrix}
=
p_0
\begin{bmatrix}
\cosh\eta \\
\sinh\eta
\end{bmatrix},
\end{equation}
so after the boost
\begin{equation}\label{eqn:squarerootpsigma:100}
\begin{aligned}
p \cdot \sigma
&\rightarrow
p_0 \lr{ \cosh \eta – \sinh \eta \sigma^3 } \\
&= p_0
\begin{bmatrix}
\cosh\eta – \sinh\eta & 0 \\
0 & \cosh\eta + \sinh\eta
\end{bmatrix} \\
&=
p_0
\begin{bmatrix}
e^{-\eta} & 0 \\
0 & e^{\eta}
\end{bmatrix},
\end{aligned}
\end{equation}
where $$p_0 = m$$ is still the rest frame energy. However, according to
\ref{eqn:squarerootpsigma:60} this is exactly
\begin{equation}\label{eqn:squarerootpsigma:120}
\lr{\sqrt{m} e^{-\inv{2} \eta \sigma^3 }}^2
\end{equation}

Since $$p \cdot \overline{\sigma}$$ flips the signs of the spatial momentum, we have shown that
\begin{equation}\label{eqn:squarerootpsigma:140}
\begin{aligned}
\lr{\sqrt{m} e^{-\inv{2} \eta \sigma^3 }}^2 &= p \cdot \sigma \\
\lr{\sqrt{m} e^{\inv{2} \eta \sigma^3 }}^2 &= p \cdot \overline{\sigma},
\end{aligned}
\end{equation}
which isn’t a full proof of the claimed result (i.e. the most general orientation isn’t considered), but at least validates the claim.

# References

 Dr. Tobias Osborne. Qft lecture 15, dirac equation, boost from stationary frame. Youtube. URL https://youtu.be/J2lV8uNx0LU?list=PLDfPUNusx1EpRs-wku83aqYSKfR5fFmfS&t=4328. [Online; accessed 18-December-2018].

## PHY2403H Quantum Field Theory. Lecture 22: Dirac sea, charges, angular momentum, spin, U(1) symmetries, electrons and positrons. Taught by Prof. Erich Poppitz

This post is a synopsis of the material from the second last lecture of QFT I. I missed that class, but worked from notes kindly provided by Emily Tyhurst, and Stefan Divic, filling in enough details that it made sense to me.

[Click here for an unabrided PDF of my full notes on this day’s lecture material.]

Topics covered include

• The Hamiltonian action on single particle states showed that the Hamiltonian was an energy eigenoperator
\begin{equation}\label{eqn:qftLecture22:140}
H \ket{\Bp, r}
=
\omega_\Bp \ket{\Bp, r}.
\end{equation}
• The conserved Noether current and charge for spatial translations, the momentum operator, was found to be
\begin{equation}\label{eqn:momentumDirac:260}
\BP =
\int d^3 x
\end{equation}
which could be written in creation and anhillation operator form as
\begin{equation}\label{eqn:momentumDirac:261}
\BP = \sum_{s = 1}^2
\int \frac{d^3 q}{(2\pi)^3} \Bp \lr{
a_\Bp^{s\dagger}
a_\Bp^{s}
+
b_\Bp^{s\dagger}
b_\Bp^{s}
}.
\end{equation}
Single particle states were found to be the eigenvectors of this operator, with momentum eigenvalues
\begin{equation}\label{eqn:momentumDirac:262}
\BP a_\Bq^{s\dagger} \ket{0} = \Bq (a_\Bq^{s\dagger} \ket{0}).
\end{equation}
• The conserved Noether current and charge for a rotation was found. That charge is
\begin{equation}\label{eqn:qftLecture22:920}
\BJ = \int d^3 x \Psi^\dagger(x) \lr{ \underbrace{\Bx \cross (-i \spacegrad)}_{\text{orbital angular momentum}} + \inv{2} \underbrace{\mathbf{1} \otimes \Bsigma}_{\text{spin angular momentum}} } \Psi,
\end{equation}
where
\begin{equation}\label{eqn:qftLecture22:260}
\mathbf{1} \otimes \Bsigma =
\begin{bmatrix}
\Bsigma & 0 \\
0 & \Bsigma
\end{bmatrix},
\end{equation}
which has distinct orbital and spin angular momentum components. Unlike NRQM, we see both types of angular momentum as components of a single operator. It is argued in  that for a particle at rest the single particle state is an eigenvector of this operator, with eigenvalues $$\pm 1/2$$ — the Fermion spin eigenvalues!
• We examined two $$U(1)$$ global symmetries. The Noether charge for the “vector” $$U(1)$$ symmetry is
\begin{equation}\label{eqn:qftLecture22:380}
Q
=
\int \frac{d^3 q}{(2\pi)^3} \sum_{s = 1}^2
\lr{
a_\Bp^{s \dagger} a_\Bp^s

b_\Bp^{s \dagger}
b_\Bp^s
},
\end{equation}
This charge operator characterizes the $$a, b$$ operators. $$a$$ particles have charge $$+1$$, and $$b$$ particles have charge $$-1$$, or vice-versa depending on convention. We call $$a$$ the operator for the electron, and $$b$$ the operator for the positron.
• CPT (Charge-Parity-TimeReversal) symmetries were also mentioned, but not covered in class. We were pointed to , ,  to start studying that topic.

# References

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

 Dr. Michael Luke. Quantum Field Theory., 2011. URL https://www.physics.utoronto.ca/ luke/PHY2403F/References_files/lecturenotes.pdf. [Online; accessed 05-Dec-2018].

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

 Dr. David Tong. Quantum Field Theory. URL http://www.damtp.cam.ac.uk/user/tong/qft.html.

## Explicit form of the square root of p . sigma.

December 10, 2018 phy2403 No comments , , , ,

With the help of Mathematica, a fairly compact form was found for the root of $$p \cdot \sigma$$
\begin{equation}\label{eqn:DiracUVmatricesExplicit:121}
\sqrt{ p \cdot \sigma }
=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
\omega_\Bp- p^3 + \sqrt{ \omega_\Bp^2 – \Bp^2 } & – p^1 + i p^2 \\
– p^1 – i p^2 & \omega_\Bp+ p^3 + \sqrt{ \omega_\Bp^2 – \Bp^2 }
\end{bmatrix}.
\end{equation}
A bit of examination shows that we can do much better. The leading scalar term can be simplified by squaring it
\begin{equation}\label{eqn:squarerootpsigma:140}
\begin{aligned}
\lr{ \sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} } }^2
&=
\omega_\Bp- \Norm{\Bp} + \omega_\Bp+ \Norm{\Bp} + 2 \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
&=
2 \omega_\Bp + 2 m,
\end{aligned}
\end{equation}
where the on-shell value of the energy $$\omega_\Bp^2 = m^2 + \Bp^2$$ has been inserted. Using that again in the matrix, we have
\begin{equation}\label{eqn:squarerootpsigma:160}
\begin{aligned}
\sqrt{ p \cdot \sigma }
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\begin{bmatrix}
\omega_\Bp- p^3 + m & – p^1 + i p^2 \\
– p^1 – i p^2 & \omega_\Bp+ p^3 + m
\end{bmatrix} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0
-p^1 \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
-p^2 \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
-p^3 \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}
} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0
-p^1 \sigma^1
-p^2 \sigma^2
-p^3 \sigma^3
} \\
&=
\inv{\sqrt{ 2 \omega_\Bp + 2 m }}
\lr{
(\omega_\Bp + m) \sigma^0 – \Bsigma \cdot \Bp
}.
\end{aligned}
\end{equation}

We’ve now found a nice algebraic form for these matrix roots
\begin{equation}\label{eqn:squarerootpsigma:180}
\boxed{
\begin{aligned}
\sqrt{p \cdot \sigma} &= \inv{\sqrt{ 2 \omega_\Bp + 2 m }} \lr{ m + p \cdot \sigma } \\
\sqrt{p \cdot \overline{\sigma}} &= \inv{\sqrt{ 2 \omega_\Bp + 2 m }} \lr{ m + p \cdot \overline{\sigma}}.
\end{aligned}}
\end{equation}

As a check, let’s square one of these explicitly
\begin{equation}\label{eqn:squarerootpsigma:101}
\begin{aligned}
\lr{ \sqrt{p \cdot \sigma} }^2
&= \inv{2 \omega_\Bp + 2 m }
\lr{ m^2 + (p \cdot \sigma)^2 + 2 m (p \cdot \sigma) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ m^2 + (\omega_\Bp^2 – 2 \omega_\Bp \Bsigma \cdot \Bp + \Bp^2) + 2 m (p \cdot \sigma) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ 2 \omega_\Bp^2 – 2 \omega_\Bp \Bsigma \cdot \Bp + 2 m (\omega_\Bp – \Bsigma \cdot \Bp) } \\
&= \inv{2 \omega_\Bp + 2 m }
\lr{ 2 \omega_\Bp \lr{ \omega_\Bp + m } – (2 \omega_\Bp + 2 m) \Bsigma \cdot \Bp } \\
&=
\omega_\Bp – \Bsigma \cdot \Bp \\
&=
p \cdot \sigma,
\end{aligned}
\end{equation}
which validates the result.

## Explicit expansion of the Dirac u,v matrices

We found that the solution of the $$u(p), v(p)$$ matrices were
\begin{equation}\label{eqn:DiracUVmatricesExplicit:20}
\begin{aligned}
u(p) &=
\begin{bmatrix}
\sqrt{p \cdot \sigma} \zeta \\
\sqrt{p \cdot \overline{\sigma}} \zeta \\
\end{bmatrix} \\
v(p) &=
\begin{bmatrix}
\sqrt{p \cdot \sigma} \eta \\
-\sqrt{p \cdot \overline{\sigma}} \eta \\
\end{bmatrix},
\end{aligned}
\end{equation}
where
\begin{equation}\label{eqn:DiracUVmatricesExplicit:40}
\begin{aligned}
p \cdot \sigma &= p_0 \sigma_0 – \Bsigma \cdot \Bp \\
p \cdot \overline{\sigma} &= p_0 \sigma_0 + \Bsigma \cdot \Bp.
\end{aligned}
\end{equation}
It was pointed out that these square roots can be conceptualized as (in the right basis) as the diagonal matrices of the eigenvalue square roots.

It was also pointed out that we don’t tend to need the explicit form of these square roots.We saw that to be the case in all our calculations, where these always showed up in the end in quadratic combinations like $$\sqrt{ (p \cdot \sigma)^2 }, \sqrt{ (p \cdot \sigma)(p \cdot \overline{\sigma})}, \cdots$$, which nicely reduced each time without requiring the matrix roots.

I encountered a case where it would have been nice to have the explicit representation. In particular, I wanted to use Mathematica to symbolically expand $$\overline{\Psi} i \gamma^\mu \partial_\mu \Psi$$ in terms of $$a^s_\Bp, b^r_\Bp, \cdots$$ representation, to verify that the massless Dirac Lagrangian are in fact the energy and momentum operators (and to compare to the explicit form of the momentum operator found in eq. 3.105 ). For that mechanical task, I needed explicit representations of all the $$u^s(p), v^r(p)$$ matrices to plug in.

It happens that $$2 \times 2$$ matrices can be square-rooted symbolically (FIXME: link to squarerootOfFourSigmaDotP.nb notebook). In particular, the matrices $$p \cdot \sigma, p \cdot \overline{\sigma}$$ have nice simple eigenvalues $$\pm \Norm{\Bp} + \omega_\Bp$$. The corresponding unnormalized eigenvectors for $$p \cdot \sigma$$ are
\begin{equation}\label{eqn:DiracUVmatricesExplicit:60}
\begin{aligned}
e_1 &=
\begin{bmatrix}
– p_x + i p_y \\
p_z + \Norm{\Bp}
\end{bmatrix} \\
e_1 &=
\begin{bmatrix}
– p_x + i p_y \\
p_z – \Norm{\Bp}
\end{bmatrix}.
\end{aligned}
\end{equation}
This means that we can diagonalize $$p \cdot \sigma$$ as
\begin{equation}\label{eqn:DiracUVmatricesExplicit:80}
p \cdot \sigma
= U
\begin{bmatrix}
\omega_\Bp+ \Norm{\Bp} & 0 \\
0 & \omega_\Bp- \Norm{\Bp}
\end{bmatrix}
U^\dagger,
\end{equation}
where $$U$$ is the matrix of the normalized eigenvectors
\begin{equation}\label{eqn:DiracUVmatricesExplicit:100}
U =
\begin{bmatrix}
e_1′ & e_2′
\end{bmatrix}
=
\inv{ \sqrt{ 2 \Bp^2 + 2 p_z \Norm{\Bp} } }
\begin{bmatrix}
-p_x + i p_y & -p_x + i p_y \\
p_z + \Norm{\Bp} & p_z – \Norm{\Bp}
\end{bmatrix}.
\end{equation}

Letting Mathematica churn through the matrix products \ref{eqn:DiracUVmatricesExplicit:80} verifies the diagonalization, and for the roots, we find
\begin{equation}\label{eqn:DiracUVmatricesExplicit:120}
\sqrt{ p \cdot \sigma }
=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
\omega_\Bp- p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } & – p_x + i p_y \\
– p_x – i p_y & \omega_\Bp+ p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 }
\end{bmatrix}.
\end{equation}
Now we can plug in $$\zeta^{1\T} = (1,0), \zeta^{2\T} = (0,1), \eta^{1\T} = (1,0), \eta^{2\T} = (0,1)$$ to find the explicit form of our $$u$$’s and $$v$$’s
\begin{equation}\label{eqn:DiracUVmatricesExplicit:140}
\begin{aligned}
u^1(p) &=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
\omega_\Bp- p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
– p_x – i p_y \\
\omega_\Bp+ p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
p_x + i p_y \\
\end{bmatrix} \\
u^2(p) &=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
– p_x + i p_y \\
\omega_\Bp+ p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
p_x – i p_y \\
\omega_\Bp- p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
\end{bmatrix} \\
v^1(p) &=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
\omega_\Bp- p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
– p_x – i p_y \\
-\omega_\Bp- p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
-p_x – i p_y \\
\end{bmatrix} \\
v^2(p) &=
\inv{
\sqrt{ \omega_\Bp- \Norm{\Bp} } + \sqrt{ \omega_\Bp+ \Norm{\Bp} }
}
\begin{bmatrix}
– p_x + i p_y \\
\omega_\Bp+ p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
-p_x + i p_y \\
-\omega_\Bp+ p_z + \sqrt{ \omega_\Bp^2 – \Bp^2 } \\
\end{bmatrix}.
\end{aligned}
\end{equation}
This is now a convenient form to try the next symbolic manipulation task. If nothing else this takes some of the mystery out of the original compact notation, since we see that the $$u,v$$’s are just $$4$$ element column vectors, and we know their explicit should we want them.

Also note that in class we made a note that we should take the positive roots of the eigenvalue diagonal matrix. It doesn’t look like that is really required. We need not even use the same sign for each root. Squaring the resulting matrix root in the end will recover the original $$p \cdot \sigma$$ matrix.

# References

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

## Momentum operator for the Dirac field?

In the borrowed notes I have for last Monday’s lecture (which I missed) I see the momentum operator defined by
\begin{equation}\label{eqn:momentumDirac:20}
\BP = \sum_{s = 1}^2
\int \frac{d^3 q}{(2\pi)^3} \Bp \lr{
a_\Bp^{s\dagger}
a_\Bp^{s}
+
b_\Bp^{s\dagger}
b_\Bp^{s}
}.
\end{equation}

There’s a “use Noether’s theorem” comment associated with this. For the scalar field, using Noether’s theorem, we identified the conserved charge of a spacetime translation as the momentum operator
\begin{equation}\label{eqn:momentumDirac:40}
P^i = \int d^3 x T^{0i} = – \int d^3 x \pi(x) \spacegrad \phi(x),
\end{equation}
and if we plugged in the creation and anhillation operator representation of $$\pi, \phi$$, out comes
\begin{equation}\label{eqn:momentumDirac:60}
\BP =
\inv{2} \int \frac{d^3 q}{(2\pi)^3} \Bp \lr{ a_\Bp^\dagger a_\Bp + a_\Bp a_\Bp^\dagger},
\end{equation}
(plus $$e^{\pm 2 i \omega_\Bp t}$$ terms that we can argue away.)

It wasn’t clear to me how this worked with the Dirac field, but it turns out that this does follow systematically as expected. For a spacetime translation
\begin{equation}\label{eqn:momentumDirac:80}
x^\mu \rightarrow x^\mu + a^\mu,
\end{equation}
we find
\begin{equation}\label{eqn:momentumDirac:100}
\delta \Psi = -a^\mu \partial_\mu \Psi,
\end{equation}
so for the Dirac Lagrangian, we have
\begin{equation}\label{eqn:momentumDirac:120}
\begin{aligned}
\delta \LL
&= \delta \lr{ \overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } \Psi } \\
&=
(\delta \overline{\Psi}) \lr{ i \gamma^\mu \partial_\mu – m } \Psi
+
\overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } \delta \Psi \\
&=
(-a^\sigma \partial_\sigma \overline{\Psi}) \lr{ i \gamma^\mu \partial_\mu – m } \Psi
+
\overline{\Psi} \lr{ i \gamma^\mu \partial_\mu – m } (-a^\sigma \partial_\sigma \Psi ) \\
&=
-a^\sigma \partial_\sigma \LL \\
&=
\partial_\sigma (-a^\sigma \LL),
\end{aligned}
\end{equation}
i.e. $$J^\mu = -a^\mu \LL$$.
To plugging this into the Noether current calculating machine, we have
\begin{equation}\label{eqn:momentumDirac:160}
\begin{aligned}
\PD{(\partial_\mu \Psi)}{\LL}
&=
\PD{(\partial_\mu \Psi)}{} \lr{ \overline{\Psi} i \gamma^\sigma \partial_\sigma \Psi – m \overline{\Psi} \Psi } \\
&=
\overline{\Psi} i \gamma^\mu,
\end{aligned}
\end{equation}
and
\begin{equation}\label{eqn:momentumDirac:180}
\PD{(\partial_\mu \overline{\Psi})}{\LL} = 0,
\end{equation}
so
\begin{equation}\label{eqn:momentumDirac:140}
\begin{aligned}
j^\mu
&=
(\delta \overline{\Psi}) \PD{(\partial_\mu \overline{\Psi})}{\LL}
+
\PD{(\partial_\mu \Psi)}{\LL} (\delta \Psi)
– a^\mu \LL \\
&=
\overline{\Psi} i \gamma^\mu (-a^\sigma \partial_\sigma \Psi)
– a^\sigma {\delta^{\mu}}_{\sigma} \LL \\
&=
– a^\sigma
\lr{
\overline{\Psi} i \gamma^\mu \partial_\sigma \Psi
+ {\delta^{\mu}}_{\sigma} \LL
} \\
&=
-a_\nu
\lr{
\overline{\Psi} i \gamma^\mu \partial^\nu \Psi
+ g^{\mu\nu} \LL
}.
\end{aligned}
\end{equation}

We can now define an energy-momentum tensor
\begin{equation}\label{eqn:momentumDirac:200}
T^{\mu\nu}
=
\overline{\Psi} i \gamma^\mu \partial^\nu \Psi
+ g^{\mu\nu} \LL.
\end{equation}
A couple things are of notable in this tensor. One is that it is not symmetric, and there’s doesn’t appear to be any hope
of making it so. For example, the space+time components are way different
\begin{equation}\label{eqn:momentumDirac:220}
\begin{aligned}
T^{0k} &= \overline{\Psi} i \gamma^0 \partial^k \Psi \\
T^{k0} &= \overline{\Psi} i \gamma^k \partial^0 \Psi,
\end{aligned}
\end{equation}
so if we want a momentum like creature, we have to use $$T^{0k}$$, not $$T^{k0}$$. The charge associated with that current is
\begin{equation}\label{eqn:momentumDirac:240}
\begin{aligned}
Q^k
&=
\int d^3 x
\overline{\Psi} i \gamma^0 \partial^k \Psi \\
&=
\int d^3 x
\Psi^\dagger (-i \partial_k) \Psi,
\end{aligned}
\end{equation}
or translating from component to vector form
\begin{equation}\label{eqn:momentumDirac:260}
\BP =
\int d^3 x
\end{equation}
which is the how the momentum operator is first stated in . Here the vector notation doesn’t have any specific representation, but it is interesting to observe how this is directly related to the massless Dirac Lagrangian

\begin{equation}\label{eqn:momentumDirac:280}
\begin{aligned}
\LL(m = 0)
&=
\overline{\Psi} i \gamma^\mu \partial_\mu \Psi \\
&=
\Psi^\dagger i \gamma^\mu \partial_\mu \Psi \\
&=
\Psi^\dagger i (\partial_0 + \gamma_0 \gamma^k \partial_k) \Psi \\
&=
\Psi^\dagger i (\partial_0 – \gamma_0 \gamma_k \partial_k ) \Psi,
\end{aligned}
\end{equation}
but since $$\gamma_0 \gamma_k$$ is a $$4 \times 4$$ representation of the Pauli matrix $$\sigma_k$$ Lagrangian itself breaks down into
\begin{equation}\label{eqn:momentumDirac:300}
\LL(m = 0)
=
\Psi^\dagger i \partial_0 \Psi
+
\Bsigma \cdot \lr{ \Psi^\dagger (-i\spacegrad) \Psi },
\end{equation}
components, and lo and behold, out pops the momentum operator density! There is ambiguity as to what order of products $$\gamma_0 \gamma_k$$, or $$\gamma_k \gamma_0$$ to pick to represent the Pauli basis ( uses $$\gamma_k \gamma_0$$), but we also have sign ambiguity in assembling a Noether charge from the conserved current, so I don’t think that matters. Some part of this should be expected this since the Dirac equation in momentum space is just $$\gamma \cdot p – m = 0$$, so there is an intimate connection with the operator portion and momentum.

The last detail to fill in is going from \ref{eqn:momentumDirac:260} to \ref{eqn:momentumDirac:20} using the $$a, b$$ representation of the field. That’s an algebraically messy looking job that I don’t feel like trying at the moment.

# References

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

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