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.

The integral stated in the problem is:

\[ (2\pi)^4 \int \left[ \bar{u}^{s_3}(p_3) \left( i\sqrt{4\pi\alpha} \, \gamma^{\mu} \right) u^{s_1}(p_1) \right]
\frac{ig_{\mu\nu}}{q^2} \left[ \bar{u}^{s_4}(p_4) \left( i\sqrt{4\pi\alpha} \, \gamma^{\nu} \right) u^{s_2}(p_2) \right]
\delta^{4}(p_1 - p_3 - p) \delta^{4} (p_2 - p_4 - p) d^4 p \]

Integrating over delta function is easy by definition. We get the usual definition of scattering matrix element:

\[ i\mathcal{M} = \frac{-i4\pi\alpha}{(p_1 - p_3)^2} g_{\mu\nu}
\left[ \bar{u}^{s_3}(p_3) \gamma^{\mu} u^{s_1} (p_1) \right] \cdot
\left[ \bar{u}^{s_4}(p_4) \gamma^{\mu} u^{s_2} (p_2) \right] \]

after we dropped the delta function term

\[ (2\pi)^4 \delta^{4}(p_1 + p_2 - p_3 -p_4) \]

It gets messy from this point on, since the Dirac wave functions are not simple and frame dependent. Let's go to the simplest frame possible: the center of mass frame. Here all the three momenta has the same magnitude, \( p \). In this frame, due to conservation of momentum, we also have \( (p_1 - p_3)^2 = 2 p^2 ( 1- \cos\theta) \). \( \theta \) is the scattered angle of the particle.

In the CM frame, one of the square bracket is:

\[ \bar{v}(p^{\prime}) \gamma^{\mu} v(p) = \left( m_{v} - E_{v} - \frac{p^2 \cos\theta}{m_{v} + E_{v}}, p \sin\theta, i p \sin\theta, p ( 1 - \cos\theta) \right) \]

where \( m_v \) can be e.g. the electron mass, \( E_v \) the CM frame energy. \( p \) and \( p^{\prime} \) are shorthand for incoming and outgoing three-momentum. \( \theta \) is therefore the angle between them.

Assuming both particles are ulta-relativistic, \( E_{e} \approx E_{\mu} \gg m_{\mu} \), this expression simplify to:

\[ \bar{v}(p^{\prime}) \gamma^{\mu} v(p) \to p \left( 1 - \cos\theta, \sin\theta, i\sin\theta, 1 - \cos\theta \right) \]

If both electron and muon are in the same (positive) helicity state, then both square bracket gives the same answer in the ultra-relativistic limit. So, we have the four-scalar:

\[ \left[ \bar{u}(-p^{\prime}) \gamma^{\mu} u(-p) \right] g_{\mu\nu} \left[ \bar{v}(p^{\prime}) \gamma^{\nu} v(p) \right] \simeq 2 p^2 (1 - \cos\theta)^2 \]

Substituting back into the scattering matrix element:

\[ \mathcal{M} = - 4\pi \alpha (1 - \cos\theta) \]

For complete back scatter, \( \theta = \pi \),

\[ \mathcal{M} = - 8 \pi \alpha \,. \]

Hence the janitor's answer.

To summarize, we had to

• Drop a delta function factor to arrive at usual definition of scattering matrix element
• Work in CM frame
• Chose same helicity state for electron and muon.
• Work in the ultra-relativistic limit
• Assume full back scatter angle.

before arriving at this answer.