# Proton Mixed A and S wavefunctions charge

1. Dec 15, 2014

### ChrisVer

I have the proton wavefunctions of mixed symmetry:

$|MA> = \frac{1}{\sqrt{2}} ( u_1 d_2 u_3 - d_1 u_2 u_3 )$

and

$|MS> = \frac{1}{\sqrt{6}} (2u_1 u_2 d_3 - u_1 d_2 u_3 - d_1u_2u_3 )$

If the charge defined as: $qu =\frac{2}{3} u$ and $qd= -\frac{1}{3} d$ , I need to show what are the expectation values of $q_3$ acting on the third particle:

$<MA| q_3 | MA> ~~,~~ <MS|q_3|MS>$

However I get $q_{MA}= \frac{2}{3}$ and $q_{MS}=0$. Is that rational? None of these are the proton's charge...

For MA it's easy to see:

$q_3 |MA> = \frac{1}{\sqrt{2}} \Big[ q_3 (u_1d_2u_3) - q_3 (d_1u_2u_3 ) \Big] = \frac{1}{\sqrt{2}} \Big[ \frac{2}{3} (u_1d_2u_3) - \frac{2}{3} (d_1u_2u_3 ) \Big] = \frac{2}{3} |MA>$

So that: $<MA|q_3 |MA> = \frac{2}{3}$

In a similar but a bit more complicated way (since |MS> doesn't appear to be an eigenstate) I calculated that $q_{MS}= \frac{1}{6} [ -\frac{4}{3} + \frac{2}{3} + \frac{2}{3}]=0$ ....
However I don't understand why I don't get =+1

2. Dec 15, 2014

### Orodruin

Staff Emeritus
The charge of the proton is not $q_3$, it is $Q = q_1 + q_2 + q_3$ ...