Negative square amplitude for a decay process

• simonisr
In summary: Homework Equations(1) ## \displaystyle\sum_s u^s(p) \bar{u}^s(p) = \gamma_{\mu} p^{\mu} + m ##(2) ## \displaystyle\sum_{polarizations} \epsilon_{\mu}^*(k) \epsilon_{\nu}(k) \rightarrow - g_{\mu \nu} ##, in the case that the vector boson is massless.(3) ## \displaystyle\sum_{polarizations} \epsilon_{\mu}^*(k)\epsilon_{\nu}(k) \rightarrow - g_{\mu \nu} + \frac{k_{\mu
simonisr

Homework Statement

I'm trying to compute the square amplitude ## |\mathcal{M}|^2 ## for a decay process in which a Majorana fermion, call it ## \chi_2 ##, decays into another Majorana fermion, ## \chi_1 ##, and a vector boson denoted by ## A^{\mu} ##. The model is such that the mass of the two fermions differ slightly by a value ## \delta ##, i.e., ## m_2 - m_1 = \delta \ll m_1, m_2 ##, and thus for the decay to be possible we also have that ## m_A < \delta ##. The decay is taken from an interaction term in a Lagrangian density of the form ## \mathcal{L}_{int} = \frac{ig}{2}( \bar{\chi}_2 \gamma^{\mu} \chi_1 - \bar{\chi}_1 \gamma^{\mu} \chi_2 )A_{\mu} ##. I want to compute the square amplitude to tree level, i.e., corresponding to the Feynman diagram simply consisting of one vertex with two external fermions and one external vector boson. I'm using the conventions found in "An Introduction to Quantum Field Theory" by Peskin and Schroeder.

Homework Equations

Equations used throughout my attempted solution, taken from Peskin and Schroeder.

(1) ## \displaystyle\sum_s u^s(p) \bar{u}^s(p) = \gamma_{\mu} p^{\mu} + m ##
(2) ## \displaystyle\sum_{polarizations} \epsilon_{\mu}^*(k) \epsilon_{\nu}(k) \rightarrow - g_{\mu \nu} ##, in the case that the vector boson is massless.
(3) ## \displaystyle\sum_{polarizations} \epsilon_{\mu}^*(k)\epsilon_{\nu}(k) \rightarrow - g_{\mu \nu} + \frac{k_{\mu} k_{\nu}}{m^2} ##, in the case of a vector boson with mass m.
(4) ## Tr(\gamma^{\mu} \gamma^{\nu}) = 4 g^{\mu \nu} ##.
(5) ## Tr(\gamma^{\mu} \gamma^{\nu} \gamma^{\rho} \gamma^{\sigma}) = 4(g^{\mu \nu}g^{\rho \sigma} - g^{\mu \rho}g^{\nu \sigma} + g^{\mu \sigma}g^{\nu \rho}) ##.
(6) Trace of an odd number of gamma matrices yield zero.
(7) ## (\gamma^{\mu})^{\dagger} = \gamma^0 \gamma^{\mu} \gamma^0 ##

The Attempt at a Solution

First I attempt to rewrite the interaction term in a more clear way:
## \mathcal{L}_{int} = \frac{ig}{2}( \bar{\chi}_2 \gamma^{\mu} \chi_1 - \bar{\chi}_1 \gamma^{\mu} \chi_2 )A_{\mu} = \frac{ig}{2}( \chi_2^{\dagger} \gamma^0 \gamma^{\mu} \chi_1 - \bar{\chi}_1 \gamma^{\mu} \chi_2 )A_{\mu} = \frac{ig}{2}( (\chi_1^{\dagger} (\gamma^{\mu})^{\dagger} \gamma^{0} \chi_2 )^{\dagger} - \bar{\chi}_1 \gamma^{\mu} \chi_2 )A_{\mu} \\
= \frac{ig}{2}( (\chi_1^{\dagger} \gamma^{0} \gamma^{\mu} \chi_2 )^{\dagger} - \bar{\chi}_1 \gamma^{\mu} \chi_2 )A_{\mu} = \frac{ig}{2}( (\chi_1^{\dagger} \gamma^{0} \gamma^{\mu} \chi_2 )^* - \bar{\chi}_1 \gamma^{\mu} \chi_2 )A_{\mu} = \frac{ig}{2}( (\bar{\chi}_1 \gamma^{\mu} \chi_2 )^* - \bar{\chi}_1 \gamma^{\mu} \chi_2 )A_{\mu} \\
= \frac{ig}{2} 2i \text{Im}(\bar{\chi}_1 \gamma^{\mu} \chi_2) A_{\mu} = ig \bar{\chi}_1 \gamma^{\mu} \chi_2 A_{\mu} ##
To get the last equality I argue that in the Majorana basis the fields are completely real and the gamma matrices are completely imaginary, and thus the product I'm taking the imaginary part of is a completely imaginary number. This should not depend on the representation of the gamma matrices I'm using so it should be safe to say that the quantity is imaginary and make the replacement.

From this I conclude that the vertex factor for the interaction vertex should be (up to a sign): ## g \gamma^{\mu} ##

Now to the matrix element. I'm a bit lazy about writing out spin indices, but u's carrying the same momenta carry the same spin index.

Let the incoming ## \chi_2 ## fermion have four-momenum ##p##, the outgoing ## \chi_1 ## have four momentum ##q##, and the outgoing vector boson have four-momentum ##k##. We should then get
## i \mathcal{M} = g \bar{u}(q) \gamma^{\mu} u(p) \epsilon_{\mu}^*(k) ## or
## \mathcal{M} = -i g \bar{u}(q) \gamma^{\mu} u(p) \epsilon_{\mu}^*(k) ##.

I compute the complex conjugate of this matrix element, and when doing this we need the following relation.

## (\bar{u}(q)\gamma^{\mu}u(p))^* = \bar{u}(p)\gamma^{\mu}u(q) ##

## \mathcal{M}^* = ig \bar{u}(p) \gamma^{\mu} u(q) \epsilon_{\mu}(k) ##

With this in place I get

## | \mathcal{M} |^2 = g^2 \bar{u}(p) \gamma^{\mu} u(q) \bar{u}(q) \gamma^{\nu} u(p) \epsilon_{\mu}(k) \epsilon_{\nu}^*(k) ##

I then perform the usual spin and polarization sums.

## |\bar{\mathcal{M}}|^2 = \frac{1}{2} \displaystyle\sum_{spins} \displaystyle\sum_{polarizations} |\mathcal{M}|^2 = \frac{g^2}{2} \displaystyle\sum_{spins} \bar{u}(p) \gamma^{\mu} u(q) \bar{u}(q) \gamma^{\nu} u(p) \displaystyle\sum_{polarizations} \epsilon_{\mu}(k) \epsilon_{\nu}^*(k) \\
= \frac{g^2}{2} \text{Tr} [ \gamma^{\mu} (\gamma^{\sigma}q_{\sigma} + m_1) \gamma^{\nu} (\gamma^{\rho} p_{\rho}+m_2) ] T_{\mu \nu} = \frac{g^2}{2}[q_{\sigma} p_{\rho} \text{Tr}(\gamma^{\mu} \gamma^{\sigma} \gamma^{\nu} \gamma^{\rho}) + m_1 m_2 \text{Tr}(\gamma^{\mu} \gamma^{\nu})] T_{\mu \nu} \\
= 2g^2 [ q_{\sigma} p_{\rho} (g^{\mu \sigma}g^{\nu \rho} - g^{\mu \nu}g^{\sigma \rho} + g^{\mu \rho}g^{\sigma \nu}) + m_1 m_2 g^{\mu \nu}] T_{\mu \nu} \\
= 2g^2 (p^{\nu} q^{\mu} + p^{\mu} q^{\nu} - p_{\rho} q^{\rho} g^{\mu \nu} +m_1 m_2 g^{\mu \nu} ) T_{\mu \nu}##

I have made the replacement ## T_{\mu \nu} = \displaystyle\sum_{polarizations} \epsilon_{\mu}(k) \epsilon_{\nu}^*(k) ## to make it look a bit nicer.

Now the solution splits into two, depending on if ## m_A = 0 ## or not.

Assume ## k^2 = m_A^2 ## such that ## T_{\mu \nu} = -g_{\mu \nu} + \frac{k_{\mu} k_{\nu}}{m_A^2} ##. I then get

## |\bar{\mathcal{M}}|^2 = -2g^2 (p^{\nu} q^{\mu} + p^{\mu} q^{\nu} - p_{\rho} q^{\rho} g^{\mu \nu} +m_1 m_2 g^{\mu \nu} ) (g_{\mu \nu} - \frac{k_{\mu} k_{\nu}}{m^2}) \\
= -2g^2 (- p \cdot q + 3 m_1 m_2 - 2 \frac{(p \cdot k)(q \cdot k)}{m_A^2}) \\
= 2g^2 ( p \cdot q - 3 m_1 m_2 + 2 \frac{(p \cdot k)(q \cdot k)}{m_A^2}) \geq 0 ##
This quantity is positive or zero and could be ok, even though some issues arise in the massless limit ## m_A \rightarrow 0##

The really strange stuff happens if I assume that ## k^2 = 0 ## when computing the polarization sum.

Assume ## k^2 = 0 ## such that ## T_{\mu \nu} = -g_{\mu \nu}##. I then get.

## |\bar{\mathcal{M}}|^2 = -2g^2 (p^{\nu} q^{\mu} + p^{\mu} q^{\nu} - p_{\rho} q^{\rho} g^{\mu \nu} +m_1 m_2 g^{\mu \nu} ) g_{\mu \nu} \\
= -2 g^2 (2 p \cdot q - 4 p \cdot q + 4 m_1 m_2) \\
= 4 g^2 (p \cdot q - 2 m_1 m_2) < 0 ##
This does not make any sense at all. I've computed the absolute square of a complex number and obtained something which is negative. I have tried to figure out my error for quite a while now but I cannot seem to find it. I would be very grateful if anyone could point me in the right direction.

Most of the calculations in Peskin and Schroeder are quite advanced=I have the book, but I am hardly competent with any of it. In any case, I think I see an error that you made. When you wrote out ## |\bar{M} |^2 ##, in two places, you have separate ## \mu ## and ## \nu ## superscripts and subscripts (on the ## \gamma ## terms and on the ## \epsilon ## terms), when these need to be the same (e.g. both ## \mu ## ). ## \\ ## Editing: And then again, I'm not so sure about what I just mentioned, because the ## \mu ## looks like it gets repeated for both ## M ## and ## M^* ##. Each term has the form: ## \gamma^{\mu} \epsilon_{\mu} ##, (with a sum over ## \mu ##). Instead, I think, because of this, you need to take care in how the ## \gamma^{\mu} \gamma^{\nu} ## and ## \epsilon_{\mu} \epsilon_{\nu} ## get combined, and I'm not sure that they can get separated and grouped in the manner that you did. ## \\ ## Additional editing: On further study, I think what you have for this ## \mu ## and ## \nu ## is correct. I'll keep looking to see if something else might appear... ## \\ ## Suggestion is to review the first couple of pages of Chapter 5 which are found on pp.131-133 of my copy of the book. There they discuss how to compute ## |M|^2 ## and a couple of tricks that simplify the calculation.

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What is a negative square amplitude in a decay process?

A negative square amplitude refers to a mathematical quantity that represents the probability of a decay process occurring. It is a complex number with both a real and an imaginary component.

Why does a decay process have a negative square amplitude?

In quantum mechanics, probabilities are calculated by taking the absolute value squared of the amplitude. In some cases, the amplitude may have a negative value, which results in a negative square amplitude. This does not affect the final probability, as the negative sign cancels out when taking the absolute value squared.

Does a negative square amplitude mean the decay process is impossible?

No, a negative square amplitude does not necessarily mean that the decay process is impossible. It is simply a mathematical representation of the probability of the decay occurring. The final probability will always be a positive value, regardless of the sign of the square amplitude.

Can the negative square amplitude be used to predict the decay rate?

No, the negative square amplitude cannot be used to predict the decay rate. The decay rate is determined by other factors such as the energy and mass of the particles involved in the process.

How is the negative square amplitude related to the uncertainty principle?

The uncertainty principle states that the position and momentum of a particle cannot be precisely known at the same time. In the case of a decay process, the negative square amplitude can be thought of as the uncertainty in the energy of the decaying particle. This uncertainty in energy is related to the uncertainty in time, which is a fundamental aspect of the uncertainty principle.

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