## Scattering of a scalar particle on a spin-1/2 particle

Hi,

Following my recent encounters with density matrix formulations of scattering theory, here's a question I faced on yesterday's particle physics exam. I did some mechanical computation, but I am not fully convinced of what I did (basically computed the density matrix and tried to compute the polarizations). The problem is that I am still not comfortable with the entire machinery.

I'd appreciate if someone could discuss this question and explain me the machinery. I would like to work out the solution on my own.

 The $T$ matrix for the scattering of a scalar particle on a spin $\frac{1}{2}$ particle is written as $$T = f + ig \vec{\sigma}\cdot(\vec{k}_i - \vec{k}_f)$$ where $\vec{k}_{i,f}$ refer to initial and final momenta in the centre of mass frame. (a) Determine the polarization of the proton after scattering if it is initially polarized. (b) Determine the asymmetry in the scattering to the $|\uparrow\rangle$ and the $|\downarrow\rangle$ states if it is polarized initially with a positive helicity. The states written above refer to the axis perpendicular to the scattering plane.
Thanks!
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 Anyone?
 Hello, In your prevision question about density matrix I already answered how to do it. The definition of S matrix is $$\Phi(\infty) = (I+iT)\Phi(-\infty)$$. $$\Phi(-\infty)\equiv v$$ Now what is the problem? You know that $$\Phi^{*}_{\alpha}\Phi_{\beta} = \rho^{'}_{\alpha\beta}$$ So you know the final density matrix $$\rho^{'} = \frac{T^{\dag}\rho T}{Sp(T^{\dag}\rho T)}$$ The definition of polarization is $$<\boldsymbol \sigma> = \boldsymbol \xi$$ $$\boldsymbol \xi$$ is a polarization vector. So definition $$= Sp(\rho A)$$ now what you should do is to calculate $$\boldsymbol \xi^{'} = \frac{Sp(\rho T^{\dag}\boldsymbol \sigma T)}{Sp(T^{\dag}\rho T)}$$ Regards.

## Scattering of a scalar particle on a spin-1/2 particle

Thank you Tupos. I'm a bit slow in understanding.

Can you explain me a small thing though. In writing the density matrix for the final state, we typically use define T = S - 1 (please see my note here, or appendix E of Karl Blum's book here). But in QFT books and in your reply, you have written S = 1 + iT. Also, in class, for the density matrix calculation, we wrote S = 1 - iT, although we did not use it explicitly anywhere since we only worked with the T matrix.

But the point is...only if you define T as T = S -1 does the unscattered wave get "subtracted" (as shown on the wiki page I've linked to above). Is this just a matter of convention?
 Hello maverick, First of all you should forget about all notations that you wrote and think what you are doing step by step. I will explain you it now. What is the equation for S matrix in interaction picture $$i\frac{\partial S(t,t_0)}{\partial t} = H_{I}(t)S(t,t_0)$$ what you want to do is to solve this using perturbation method. The sense if S matrix is very important as it connect not abstract in and out states as wrote in your wiki, it connect states in Heisenberg picture,with final states in Interaction picture. This is very important. Because we consider that QED equations initially written in Heisenberg form. It is a consequence of the fact that equation of motion in Heisenberg picture are equal to classical equation. When we quantize our field we know that non quantize electron positron and electromagnetic wave function satisfy to Maxwell and Dirac equations. Thats why after we quantized our field and wave functions became operators we can say that they satisfy the same equations in Heisenberg picture. And only after this fact we consider that initial state vector in interaction picture is equal to state vector in Heisenberg picture. This is really important to understand. Go back to equation for S matrix. When we solve this with perturbation method we get T ordered exponent. The zero order is 1, and it is not interested. The next order 1st one is $$-iT(\int{H_I(t)dt})$$ so as you can see it is just matter of convention what we call T matrix. I called it $$\boldsymbol{\hat T}\equiv T(\int{H_I(t)dt}) + ...$$ because Ahiezer and Beresteckij use this convention, I got used to it. Regards.
 Thank you for the detailed description tupos.

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