$#*! I unlearned #2: causality and antiparticle

In the $#*! I unlearned #1, we saw that the propagator for a real Klein-Gordon scalar field is finite for two space-like separated points, which seems to violate causality. But not really. In the question of causality, we should not ask whether particles can propagate over space like intervals, but whether a measurement performed in one point can affect a measurement at another space-like separated point.

\[ \begin{aligned} [\phi(x), \phi(y)] &= \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2 p^0}} \int \frac{d^3q}{(2\pi)^3} \frac{1}{\sqrt{2 q^0}} \left[a_{\mathbf{p}} e^{-i p \cdot x} + a_{\mathbf{p}}^{\dagger} e^{i p \cdot x} ,\; a_{\mathbf{q}} e^{-i q \cdot y} + a_{\mathbf{q}}^{\dagger} e^{i q \cdot y} \right] \\ &= \int \frac{d^3p}{(2\pi)^3} \frac{1}{2 p^0} \left( e^{-ip(x-y)} - e^{ip(x-y)} \right) = D(x-y) - D(y-x) \end{aligned} \]

When \( (x - y)^2 < 0 \), we can perform a Lorentz transformation from \( (x-y) \) to \( - (x-y) \)--going around the light cone--so that the two propagators are actually equal and therefore exactly cancel; causality if preserved. When \( (x - y)^2 > 0 \), there is no such Lorentz transformation. In this case, the amplitude is nonzero, and roughly \( ( e^{-imt} - c.c ) \) for the special case \( \mathbf{x} - \mathbf{y} = 0 \), which we also calculated. Therefore: no measurement in the Klein-Gordon theory can affect another measurement outside the light-cone.

$#*! I unlearned #1: scalar field theory

The Hamiltonian:

\[ H = \frac12 \int d^3 x \left[ \pi^2 + (\nabla \phi)^2 + m^2\phi^2 \right] \]

where \( \pi(\mathbf{x}) \) is the conjugate momentum density of field \( \phi(\mathbf{x}) \) is quadratic. In particular by going to momentum space with:

\begin{align} \phi(\mathbf{x}) &= \int \frac{d^3p}{(2\pi)^3} \sqrt{\frac{1}{2\omega_{\mathbf{p}}}} \left(a_{\mathbf{p}} + a_{-\mathbf{p}}^{\dagger} \right) e^{i \mathbf{p} \cdot \mathbf{x}} \\ \pi(\mathbf{x}) &= \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{\omega_{\mathbf{p}}}{2}} \left(a_{\mathbf{p}} - a_{-\mathbf{p}}^{\dagger} \right) e^{i \mathbf{p} \cdot \mathbf{x}} \end{align}

\( H \) become just sum of uncouple oscillators:

\[ H = \int \frac{d^3p}{(2\pi)^3} \omega_{\mathbf{p}} \left( a_{\mathbf{p}}^{\dagger} a_{\mathbf{p}} + \frac12 [a_{\mathbf{p}}, a_{\mathbf{p}}^{\dagger} ] \right) \]

with well-known spectrum and stuff. More important are these commutation relations:

\[ [ H, a_{\mathbf{p}} ] = - \omega_{\mathbf{p}} a_{\mathbf{p}} \,,\quad [ H, a_{\mathbf{p}}^{\dagger} ] = \omega_{\mathbf{p}} a_{\mathbf{p}}^{\dagger} \]

TBBT Physics Bowl

In one of the episodes of The Big Bang Theory, the gang enters into the department physics bowl. Near the end, Sheldon was stunned by the following Feynman diagram problem:

Considering that this is one of the basic diagram for QED, it would be impossible for the character to not even recognize the problem. Anyway, the janitor then answered that the correct answer is \( -8 \pi \alpha \). It is incorrect. Well, at least not without additional conditions not specified by the problem.

4-force

Classes in relativity usually avoid discussing force even though every physics student started in non-relativistic physics with . It is not immediately obvious if this is still true under relativity and if not what is the correct formula. Of course, discussion of regular 3-vector in relativity is not very convenience. It is more natural to construct relativistic (i.e. covariant) relations using 4-force , 4-momentum , 4-acceleration , 4-velocity , and with proper time . There is a problem: should we define 4-force by acceleration or momentum? .

Momentum is more fundamental than acceleration, so:

We still want Newton's law to hold in some sense. Consider the collision of two particles. Newton's third law would read like this:

But this is strange because it means both particles have the same proper time even though they might be travelling with very different velocities.

Relativity and photons

It's a relatively free weekend, so I ended up thinking about bunch of random stuff. Another problem I pondered: is number of photon a Lorentz invariance? That is, do observers in different frame count the same number of photons? From intuition, it seems photon number should be an invariance.

Poles of GF (time-independent)

From college to graduate school, it took me years to get used to Green's function. I often wonder if it's the same case for others or am I just not too bright. GF is of course an extremely powerful method that takes many years of reading and practice until one is completely comfortable with it. Physics undergraduates probably learned about it---as I did---in a mathematics class, but it is usually unused by physics classes. For example, Griffiths' EM book---standard textbook on the subject---do not use and GF and his QM book only uses it in the scattering theory chapter (for which GF is very useful) at the very end.

My hand-waving understanding of GF is the following: it is the inverse of the Hamiltonian/Linear operator in the following sense:

with the parameter .

Two bands interaction: self-energy and effective low energy Hamiltonian.

Consider the following model of two-bands Hamiltonian with interaction between the bands:

$$ H = \left( \begin{array}{cc} H_{11} &H_{12} \\ H_{12}^{\dag} & H_{22} \end{array} \right)\,, $$

where $H_{11}$ and $H_{22}$ are some low and high energy bands that we know how to diagonalize and $H_{12}$ is some interaction between them. Since we know the spectrum of $H_{11}$, we can write down the GF. Here it is:

$$ G_{11}^{(0)} = \frac{1}{H_{11} - \epsilon^{0}} $$