Particle Phys.; Isospin conservation for cross sections of scattering

[mhsd91]

Homework Statement

What relations can you derive from isospin conservation for the cross sections of the following scattering processes:


K$^{+}$ + p → K$^{+}$ + p

K$^{+}$ + n → K$^{0}$ + p

K$^{0}$ + n → K$^{0}$ + n

The Attempt at a Solution

I do not really have my own solution as I've just started learning this, and fail to understand the question: I do not necessarily need the answer, but a starting tip and an explanation of the question at hand would be much appreciated!

 

[fzero]

You should start by recalling how p and n are related by isospin. Does the form of the reactions, together with isospin conservation, suggest anything about whether or not $K^+$ and $K^0$ should be related by isospin as well? Before asking specifically about the cross-section, you'll next want to ask how the matrix elements describing those processes might depend on the isospin degrees of freedom.

 

[mhsd91]

Thank you very much, this helped a lot, really: was able to finish and get the exercise approved.

Anyways, if you (or any) got the time, I'd like to tidy up some elementary stuff,

Wikipedia, WolframAlpha and other sources state that the proton and the neutron both have isospin 1/2. This must imply that their isospin is ±1/2, right?

From the exercise over, we may write a two-particle system of particle X and Y as,

$|X, Y > = |\alpha_X , \alpha_Y> = (CG)|\alpha , \beta>$,

where (CG) is a clebsch-gordan coefficient, and the isospin, $I = I_1+I_2+I_3$, and $\alpha_X, \alpha_Y, \alpha, \beta$ are defined by,

$I = I_X + I_Y \\ I_3 = I_{X3} + I{Y3}$

$|\alpha_X - \alpha_Y| \leq \alpha \leq (\alpha_X + \alpha_Y)\\ \beta = \beta_X + \beta_Y$

$\hat{I}^2 |\alpha,\beta> = \alpha (\alpha + 1) |\alpha,\beta> \\ \hat{I}_3 |\alpha,\beta> = \beta |\alpha,\beta>\\$

Now that we have some theory and definitions: let particle X and Y be a proton and a neutron with isospins $\alpha_X = \alpha_p,\alpha_Y = \alpha_n,$respectively. What values of isospin will these have, and why? To solve my problems I had to set $\alpha_n = -1/2, \alpha_p = 1/2$, but the math would have come out if I changed these two (it seems so at least...).

 

[fzero]

Now that we have some theory and definitions: let particle X and Y be a proton and a neutron with isospins $\alpha_X = \alpha_p,\alpha_Y = \alpha_n,$respectively. What values of isospin will these have, and why? To solve my problems I had to set $\alpha_n = -1/2, \alpha_p = 1/2$, but the math would have come out if I changed these two (it seems so at least...).

The proton and neutron form an isospin doublet. This is analogous to a spin 1/2 system. The eigenvalue of $\hat{I}^2$ is $\tfrac{1}{2}(\tfrac{1}{2}+1)$ for both particles. By convention, the eigenvalue of $\hat{I}_3$ is $+\tfrac{1}{2}$ for the proton (isospin up) and $-\tfrac{1}{2}$ for the neutron (isospin down). Hopefully this is what you meant.

We could have reversed this choice of labels for the up and down states and no physics would change.

 

[mhsd91]

Yes, this is exactly what I mean, thanks!
I'm really sorry about all these follow-up-questions ... but I find this rather intriguing, and not very trivial:

So by convention, the proton gets isospin up, and the neutron down, in their isospin doublet. If we were to consider another doublet, say swap the proton with a $K^+$ as in the original question. Would the neutron still have isospin down (and the Kaon up)? What if we have $K^0$, or $K^-$ instead?

And lastly before i stop bothering you: when I took my first intro course to QM, a year ago, we used the tables for Clebsch-Gordon Coefficients for 'ordinary' spin-particles. As it seems to me now, the same tables are being used for both (ordinary) spin and isospin, independantly. Is this correct? I will be more than satisfied with a yes/no answer, as I suspect it may have a rather long and complicated answer.

 

[fzero]

So by convention, the proton gets isospin up, and the neutron down, in their isospin doublet. If we were to consider another doublet, say swap the proton with a $K^+$ as in the original question. Would the neutron still have isospin down (and the Kaon up)? What if we have $K^0$, or $K^-$ instead?

The kaons have an additional charge called strangeness. Since the proton and neutron have zero strangeness, we can't swap either of them for a kaon. So the kaons form their own doublets under isospin. From the middle reaction, you can guess at the isospin up and down assignments for $K^+$ and $K^0$.

And lastly before i stop bothering you: when I took my first intro course to QM, a year ago, we used the tables for Clebsch-Gordon Coefficients for 'ordinary' spin-particles. As it seems to me now, the same tables are being used for both (ordinary) spin and isospin, independantly. Is this correct? I will be more than satisfied with a yes/no answer, as I suspect it may have a rather long and complicated answer.

It's not really a long answer. Isospin corresponds to the group $SU(2)$, which is the same group that appears in the discussion of spin under the rotation group. Isospin is not the same as the rotational symmetry, since isospin is considered an "internal" symmetry, but the generators and multiplets follow the same rules.