Proton Mixed A and S wavefunctions charge

  • Thread starter ChrisVer
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  • #1
ChrisVer
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I have the proton wavefunctions of mixed symmetry:

[itex] |MA> = \frac{1}{\sqrt{2}} ( u_1 d_2 u_3 - d_1 u_2 u_3 ) [/itex]

and

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

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

[itex] <MA| q_3 | MA> ~~,~~ <MS|q_3|MS> [/itex]

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

For MA it's easy to see:

[itex] 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> [/itex]

So that: [itex] <MA|q_3 |MA> = \frac{2}{3} [/itex]

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
 

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  • #2
Orodruin
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The charge of the proton is not ##q_3##, it is ##Q = q_1 + q_2 + q_3## ...
 

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