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A little problem with charge operator 
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#1
Jan413, 10:55 AM

P: 3

I have a problem where it's said that the operator Q is likely to be:
[itex]Q=\sum^3_{i=1}[\frac{1}{2}B_i + I_{3,i}][/itex] 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 <pQp>=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. 


#2
Jan413, 11:07 AM

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P: 12,037

You should get 1 both for "u up u down d up" and for "u up u up d down". They have to be the same (and the same as Q for the whole proton), otherwise the proton would not have a welldefined charge.



#3
Jan413, 11:26 AM

P: 3

mmm let me explain how I'm doing this. Let's take the last way of 3.20:
[itex]Q(u\uparrow u\downarrow d\uparrow> 2u\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> 2u\uparrow u\uparrow d\downarrow>)=12=1[/itex] So yes each part is 1 but because of 2 I get 1. Shouldn't be the result 1? 


#4
Jan413, 12:29 PM

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P: 12,037

A little problem with charge operator
You cannot subtract/multiply with prefactors like that.
Your "real" calculation is ##\langle \psiQ\psi \rangle##. If you expand this, each summand is evaluated independent of the others, and their prefactors are squared. As result, you get ##\left(\frac{1}{\sqrt{5}}\cdot 1\right)^2 \cdot 1 + \left(\frac{1}{\sqrt{5}}\cdot 2\right)^2 \cdot 1 = 1## where I corrected ##\sqrt{3} \to \sqrt{5}##. If you know that the proton has a welldefined charge, both components have to have the same charge, so you can ignore all prefactors and evaluate a single component only with the formula in post 1. 


#5
Jan413, 02:37 PM

P: 3

Oh man, I'm feeling really really genius... I think it's time to stop studying. Enought for today. What a big mistake. All for no writing long expressions...
Thanks a lot! 


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