Calculation of leptonic decay widths

[luis_m]

Homework Statement

Halzen & Martin, problem 2.25

Homework Equations

The ρ and ω wavefunctions are u$\overline{u}$-d$\overline{d}$ and u$\overline{u}$+d$\overline{d}$ except for a normalization factor.

The Attempt at a Solution

In this problem one has to evaluate the expectation value of the charge operator for each of the mesons listed using their quarks wavefunctions and then square them but I get the same value of that expectation value for both ρ and ω so their squares will never be in the ratio 9:1.

Any ideas of what I'm doing wrong?

Thanks!

 

[mfb]

You have to add the individual contributions and square the sum, not the other way round. This should give a ratio of 3:1 for the sum (as 2/3+1/3 != 2/3-1/3) and 9:1 for its square.

 

[luis_m]

Thank you for your reply, mfb.

I appreciate the point that one has to square afterwards.

I think that my problem lies in the evaluation of the expectation value of the charge operator for each meson. What I'm doing for the ρ meson, for instance, is

$\langle u \overline{u}-d \overline{d}|e_1+e_2|u \overline{u}-d \overline{d} \rangle=\frac{2}{3}-\frac{1}{3}-\frac{2}{3}+\frac{1}{3}=0$,

where $e_1$ and $e_2$ represent the charge operators for each of the quark/antiquarks.

Obviously conceptually there is something wrong but I'm not sure what.

Thank you for your attention.

 

[mfb]

u can go to d + positive lepton only, u-bar can go to d-bar + negative lepton only - they are two different processes and do not add.

 

[luis_m]

I do not quite understand your reply.

So, what I'm doing for calculating the expectation value of $e_1$, for the sake of the argument, is

$\langle u\overline{u}-d\overline{d} | e_1|u\overline{u}-d\overline{d}\rangle=\langle u\overline{u}|e_1|u\overline{u}\rangle + \langle d\overline{d}|e_1|d\overline{d}\rangle = \frac{2}{3}-\frac{1}{3}=\frac{1}{3}$

and analogously for $e_2$.

 

[mfb]

Calculate the decay widths for positive and negative leptons separately - they are two different processes, the amplitudes do not add.