Explicit form of the square root of p . sigma.

December 10, 2018 phy2403 No comments , , , ,

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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

December 9, 2018 phy2403 No comments , , , ,

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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 [1]). 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

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

Momentum operator for the Dirac field?

December 8, 2018 phy2403 No comments , , , , ,

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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
\Psi^\dagger (-i \spacegrad) \Psi,
\end{equation}
which is the how the momentum operator is first stated in [2]. 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 ([1] 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

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

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

QFT Lecture 23: Part I. Raw notes.

December 5, 2018 phy2403 No comments , , , ,

I’ve now uploaded raw notes from the first portion of today’s class, edited enough to compile and no more. I won’t try to tidy these up, or type up my paper notes for the second portion of the lecture, until after the exam is done.

Topics included:

  • Gauge symmetries
  • QED Lagrangian
  • quark Lagrangian terms
  • Fermi interaction
  • Muon fields

New aggregate collection of notes for PHY2403, QFT I : up to lecture 21.

December 2, 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 version includes the following content.  Fermion related content is new since the last upload, as well as some work to make the scattering content (lecture 17,18) somewhat sensible (it was in it’s raw post-class form before, which made little sense)

  • Chapter 1: Fields, units, and scales.
  • 1.1: What is a field?
  • 1.2: Scales.
  • 1.3: Natural units.
  • 1.4: Gravity.
  • 1.5: Cross section.
  • 1.6: Problems.
  • Chapter 2: Lorentz transformations.
  • 2.1: Lorentz transformations.
  • 2.2: Determinant of Lorentz transformations.
  • 2.3: Problems.
  • Chapter 3: Classical field theory.
  • 3.1: Field theory.
  • 3.2: Actions.
  • 3.3: Principles determining the form of the action.
  • 3.4: Principles (cont.)
  • 3.5: Least action principle.
  • 3.6: Problems.
  • Chapter 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.
  • Chapter 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.
  • Chapter 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.
  • Chapter 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.
  • Chapter 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.
  • Chapter 9: Scattering and decay.
  • 9.1: Definitions and motivation.
  • 9.2: Definitions and motivation (cont.)
  • 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.
  • Chapter 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: Problems.
  • Appendix A: A Useful formulas and review.
  • A.1: Review of old material.
  • A.2: Useful results from new material.
  • Appendix B: Momentum of scalar field.
  • B.1: Expansion of the field momentum.
  • B.2: Conservation of the field momentum.
  • Appendix C: Reflection using Pauli matrices.

Problem set 1-3 solutions are redacted.  If interested (and not a future student of PHY2403), feel free to contact me for an un-redacted copy.

PHY2403H Quantum Field Theory. Lecture 21, Part II: Dirac Hamiltonian, Hamiltonian eigenvalues, general solution, creation and anhillation operators, Dirac Sea, antielectrons. Taught by Prof. Erich Poppitz

December 1, 2018 phy2403 No comments , , , , ,

This post contains a summary of my lecture notes for the second half of last Wednesday’s QFT-I lecture.
[Click here for a PDF with the full notes for this portion of the lecture.]

DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz, fall 2018.

Summary:

  • We found that the Dirac Hamiltonian is
    \begin{equation*}
    H
    =
    \int d^3 x
    \Psi^\dagger
    \lr{
    – i \gamma^0 \gamma^j \partial_j \Psi + m \gamma^0
    }
    \Psi.
    \end{equation*}
  • We found that our plane wave solutions
    \(\Psi_u = u(p) e^{-i p \cdot x}\), and \( \Psi_v = v(p) e^{i p \cdot x} \), were eigenvectors of the operator portion of the Hamiltonian
    \begin{equation*}
    \begin{aligned}
    -\gamma^0 \lr{ i \gamma^j \partial_j – m } \Psi_u &= p_0 \Psi_u \\
    -\gamma^0 \lr{ i \gamma^j \partial_j – m } \Psi_v &= -p_0 \Psi_v.
    \end{aligned}
    \end{equation*}
  • We formed a linear superposition of our plane wave solutions
    \begin{equation}\label{eqn:qftLecture21b:800}
    \Psi(\Bx, t)
    =
    \sum_{s = 1}^2
    \int \frac{d^3 p}{(2 \pi)^3 \sqrt{ 2 \omega_\Bp } }
    \lr{
    e^{-i p \cdot x} u^s_\Bp a_\Bp^s
    +
    e^{i p \cdot x} v^s_\Bp b_\Bp^s
    }.
    \end{equation}
  • and expressed the Dirac Hamiltonian in terms of creation and anhillation operators
    \begin{equation*}
    H_{\text{Dirac}}
    =
    \sum_{r = 1}^2
    \int \frac{d^3 p }{(2\pi)^3 }
    \omega_\Bp
    \lr{
    a^{r \dagger}_\Bp
    a^r_\Bp

    b^{r \dagger}_{-\Bp}
    b^r_{-\Bp}
    }.
    \end{equation*}
  • Finally, we interpreted this using the Dirac Sea argument

    Dirac Sea

  • It was claimed that the \( a, b\)’s satisfied anticommutator relationships
    \begin{equation}\label{eqn:qftLecture21b:940}
    \begin{aligned}
    \symmetric{a^s_\Bp}{a^{r \dagger}_\Bq} &= \delta^{sr} \delta^{(3)}e(\Bp – \Bq) \\
    \symmetric{b^s_\Bp}{b^{r \dagger}_\Bq} &= \delta^{sr} \delta^{(3)}(\Bp – \Bq),
    \end{aligned}
    \end{equation}
    where all other anticommutators are zero
    \begin{equation}\label{eqn:qftLecture21b:960}
    \symmetric{a^r}{b^s} =
    \symmetric{a^r}{b^{s\dagger}} =
    \symmetric{a^{r\dagger}}{b^s} =
    \symmetric{a^{r\dagger}}{b^{s\dagger}} = 0.
    \end{equation}
    and used these to algebraically remove the negative energy states of the Hamiltonian.

PHY2403H Quantum Field Theory. Lecture 21, Part I: Dirac equation solutions, orthogonality conditions, direct products. Taught by Prof. Erich Poppitz

November 29, 2018 phy2403 No comments , , , , ,

[Click here for a PDF of this notes with full details.]

DISCLAIMER: Rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz, fall 2018.

Overview.

See the PDF above for full notes for the first part of this particular lecture. We covered

  • Normalization:
    \begin{equation*}
    u^{r \dagger} u^{s}
    = 2 p_0 \delta^{r s}.
    \end{equation*}
  • Products of \( p \cdot \sigma, p \cdot \overline{\sigma} \)
    \begin{equation*}
    (p \cdot \sigma) (p \cdot \overline{\sigma})
    =
    (p \cdot \overline{\sigma}) (p \cdot \sigma)
    = m^2.
    \end{equation*}
  • Adjoint orthogonality conditions for \( u \)
    \begin{equation*}
    \overline{u}^r(\Bp) u^{s}(\Bp) = 2 m \delta^{r s}.
    \end{equation*}
  • Solutions in the \( e^{i p \cdot x} \) “direction”
    \begin{equation}\label{eqn:qftLecture21:99}
    v^s(p)
    =
    \begin{bmatrix}
    \sqrt{p \cdot \sigma} \eta^s \\
    -\sqrt{p \cdot \overline{\sigma}} \eta^s \\
    \end{bmatrix},
    \end{equation}
    where \( \eta^1 = (1,0)^\T, \eta^2 = (0,1)^\T \).
  • \(v\) normalization
    \begin{equation*}
    \begin{aligned}
    \overline{v}^r(p) v^s(p) &= – 2 m \delta^{rs} \\
    v^{r \dagger}(p) v^s(p) &= 2 p^0 \delta^{rs}.
    \end{aligned}
    \end{equation*}
  • Dirac adjoint orthogonality conditions.
    \begin{equation*}
    \begin{aligned}
    \overline{u}^r(p) v^s(p) &= 0 \\
    \overline{v}^r(p) u^s(p) &= 0.
    \end{aligned}
    \end{equation*}
  • Dagger orthogonality conditions.
    \begin{equation*}
    \begin{aligned}
    v^{r \dagger}(-\Bp) u^s(\Bp) &= 0 \\
    u^{r\dagger}(\Bp) v^s(-\Bp) &= 0.
    \end{aligned}
    \end{equation*}
  • Tensor product.

    Given a pair of vectors
    \begin{equation*}
    x =
    \begin{bmatrix}
    x_1 \\
    \vdots \\
    x_n \\
    \end{bmatrix},
    \qquad
    y =
    \begin{bmatrix}
    y_1 \\
    \vdots \\
    y_n \\
    \end{bmatrix},
    \end{equation*}
    the tensor product is the matrix of all elements \( x_i y_j \)

    \begin{equation*}
    x \otimes y^\T =
    \begin{bmatrix}
    x_1 \\
    \vdots \\
    x_n \\
    \end{bmatrix}
    \otimes
    \begin{bmatrix}
    y_1 \cdots y_n
    \end{bmatrix}
    =
    \begin{bmatrix}
    x_1 y_1 & x_1 y_2 & \cdots & x_1 y_n \\
    x_2 y_1 & x_2 y_2 & \cdots & x_2 y_n \\
    x_3 y_1 & \ddots & & \\
    \vdots & & & \\
    x_n y_1 & \cdots & & x_n y_n
    \end{bmatrix}.
    \end{equation*}

  • Direct product relations.
    \begin{equation*}
    \begin{aligned}
    \sum_{s = 1}^2 u^s(p) \otimes \overline{u}^s(p) &= \gamma \cdot p + m \\
    \sum_{s = 1}^2 v^s(p) \otimes \overline{v}^s(p) &= \gamma \cdot p – m \\
    \end{aligned}
    \end{equation*}

PHY2403H Quantum Field Theory. Lecture 20: Dirac Lagrangian, spinor solutions to the KG equation, Dirac matrices, plane wave solution, helicity. Taught by Prof. Erich Poppitz

November 27, 2018 phy2403 No comments , , , , , , , , ,

[Here is another PDF only post, containing my notes for Lecture 20 of the UofT QFT I (quantum field theory) course.]

In this lecture Professor Poppitz derived a rest frame solution of the Dirac equation, then demonstrated that the generalization to non-zero momentum satisfied the equation. We also saw that Dirac spinor solutions of the Dirac equation are KG equation solutions, and touched on the relation of some solutions to the helicity operator.

This post doesn’t have a web version, since my latex -> wordpress-mathjax script doesn’t have support for the theorem/lemma environments that I used for Monday’s notes, and I don’t have time to implement that right now.

PHY2403H Quantum Field Theory. Lecture 19: Pauli matrices, Weyl spinors, SL(2,c), Weyl action, Weyl equation, Dirac matrix, Dirac action, Dirac Lagrangian. Taught by Prof. Erich Poppitz

November 24, 2018 phy2403 No comments , , , , , , ,

[Here are my notes for lecture 19 of the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz, fall 2018.] For this lecture my notes are pdf only, due to length. While the after-class length was 8 pages, it ended up expanded to 17 pages by the time I finished making sense of the material.

These also include a portion of the notes from Lecture 18 (not yet posted), as it made sense to group all the Pauli matrix related content.  This particular set of notes diverges from the format presented in class, as it made sense to me to group things in this particular lecture in a more structured definition, theorem, proof style.  I’ve added a number of additional details that I found helpful, as well as a couple of extra problems (some set as formal problems at the end, and others set as theorem or lemmas in with the rest.)

Reflection using Pauli matrices.

November 22, 2018 phy2403 No comments , , , , ,

[Click here for a PDF of this post with nicer formatting]

In class yesterday (lecture 19, notes not yet posted) we used \( \Bsigma^\T = -\sigma_2 \Bsigma \sigma_2 \), which implicitly shows that \( (\Bsigma \cdot \Bx)^\T \) is a reflection about the y-axis.
This form of reflection will be familiar to a student of geometric algebra (see [1] — a great book, one copy of which is in the physics library). I can’t recall any mention of the geometrical reflection identity from when I took QM. It’s a fun exercise to demonstrate the reflection identity when constrained to the Pauli matrix notation.

Theorem: Reflection about a normal.

Given a unit vector \( \ncap \in \mathbb{R}^3 \) and a vector \( \Bx \in \mathbb{R}^3 \) the reflection of \( \Bx \) about a plane with normal \( \ncap \) can be represented in Pauli notation as
\begin{equation*}
-\Bsigma \cdot \ncap \Bsigma \cdot \Bx \Bsigma \cdot \ncap.
\end{equation*}

To prove this, first note that in standard vector notation, we can decompose a vector into its projective and rejective components
\begin{equation}\label{eqn:reflection:20}
\Bx = (\Bx \cdot \ncap) \ncap + \lr{ \Bx – (\Bx \cdot \ncap) \ncap }.
\end{equation}
A reflection about the plane normal to \( \ncap \) just flips the component in the direction of \( \ncap \), leaving the rest unchanged. That is
\begin{equation}\label{eqn:reflection:40}
-(\Bx \cdot \ncap) \ncap + \lr{ \Bx – (\Bx \cdot \ncap) \ncap }
=
\Bx – 2 (\Bx \cdot \ncap) \ncap.
\end{equation}
We may write this in \( \Bsigma \) notation as
\begin{equation}\label{eqn:reflection:60}
\Bsigma \cdot \Bx – 2 \Bx \cdot \ncap \Bsigma \cdot \ncap.
\end{equation}
We also know that
\begin{equation}\label{eqn:reflection:80}
\begin{aligned}
\Bsigma \cdot \Ba \Bsigma \cdot \Bb &= a \cdot b + i \Bsigma \cdot (\Ba \cross \Bb) \\
\Bsigma \cdot \Bb \Bsigma \cdot \Ba &= a \cdot b – i \Bsigma \cdot (\Ba \cross \Bb),
\end{aligned}
\end{equation}
or
\begin{equation}\label{eqn:reflection:100}
a \cdot b = \inv{2} \symmetric{\Bsigma \cdot \Ba}{\Bsigma \cdot \Bb},
\end{equation}
where \( \symmetric{\Ba}{\Bb} \) is the anticommutator of \( \Ba, \Bb \).
Inserting \ref{eqn:reflection:100} into \ref{eqn:reflection:60} we find that the reflection is
\begin{equation}\label{eqn:reflection:120}
\begin{aligned}
\Bsigma \cdot \Bx –
\symmetric{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap
&=
\Bsigma \cdot \Bx –
{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap

{\Bsigma \cdot \Bx}{\Bsigma \cdot \ncap}
\Bsigma \cdot \ncap \\
&=
\Bsigma \cdot \Bx –
{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap

{\Bsigma \cdot \Bx} \\
&=

{\Bsigma \cdot \ncap}{\Bsigma \cdot \Bx}
\Bsigma \cdot \ncap,
\end{aligned}
\end{equation}
which completes the proof.

When we expand \( (\Bsigma \cdot \Bx)^\T \) and find
\begin{equation}\label{eqn:reflection:n}
(\Bsigma \cdot \Bx)^\T
=
\sigma^1 x^1 – \sigma^2 x^2 + \sigma^3 x^3,
\end{equation}
it is clear that this coordinate expansion is a reflection about the y-axis. Knowing the reflection formula above provides a rationale for why we might want to write this in the compact form \( -\sigma^2 (\Bsigma \cdot \Bx) \sigma^2 \), which might not be obvious otherwise.

References

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