A little problem with charge operator

by idontkonw
Tags: charge, isospin, proton, q operator
 P: 3 I have a problem where it's said that the operator Q is likely to be: $Q=\sum^3_{i=1}[\frac{1}{2}B_i + I_{3,i}]$ I have to apply this to the proton wave function which is the same as you can see in equation (3.20) here: https://www.google.es/url?sa=t&rct=j...uly9_EJcuTEWeQ I have only the formula for the first equal. If I apply it because B number is the same for all of them this contribution is 0 and applying the isospin I get also 0 so I have that =0 which I assume it's wrong because Q=1. What do I do wrong? Also, in this formula (3.20) how do they get the second equal? I mean the sum of this terms. I suppose all is about transpose operator but I'm getting quite confused with this.
 P: 3 mmm let me explain how I'm doing this. Let's take the last way of 3.20: $Q(|u\uparrow u\downarrow d\uparrow> -2|u\uparrow u\uparrow d\downarrow>)= \left[ \left(\frac{1}{2}\frac{1}{3} + \frac{1}{2}\right)+\left(\frac{1}{2}\frac{1}{3} + \frac{1}{2}\right)+\left(\frac{1}{2}\frac{1}{3} - \frac{1}{2}\right)\right] -2 \left[ \left(\frac{1}{2}\frac{1}{3} + \frac{1}{2}\right)+\left(\frac{1}{2}\frac{1}{3} + \frac{1}{2}\right)+\left(\frac{1}{2}\frac{1}{3} - \frac{1}{2}\right)\right](|u\uparrow u\downarrow d\uparrow> -2|u\uparrow u\uparrow d\downarrow>)=1-2=-1$ So yes each part is 1 but because of -2 I get -1. Shouldn't be the result 1?