A little problem with charge operator

  • Thread starter Thread starter idontkonw
  • Start date Start date
  • Tags Tags
    Charge Operator
idontkonw
Messages
3
Reaction score
0
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=...mYGwBQ&usg=AFQjCNFeF1HkxpOSTjbWuly9_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 <p|Q|p>=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.
 
Physics news on Phys.org
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 well-defined charge.
 
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&gt; -2|u\uparrow u\uparrow d\downarrow&gt;)= \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&gt; -2|u\uparrow u\uparrow d\downarrow&gt;)=1-2=-1

So yes each part is 1 but because of -2 I get -1. Shouldn't be the result 1?
 
You cannot subtract/multiply with prefactors like that.

Your "real" calculation is ##\langle \psi|Q|\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 well-defined 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.
 
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!
 
Thread 'Why is there such a difference between the total cross-section data? (simulation vs. experiment)'
Well, I'm simulating a neutron-proton scattering phase shift. The equation that I solve numerically is the Phase function method and is $$ \frac{d}{dr}[\delta_{i+1}] = \frac{2\mu}{\hbar^2}\frac{V(r)}{k^2}\sin(kr + \delta_i)$$ ##\delta_i## is the phase shift for triplet and singlet state, ##\mu## is the reduced mass for neutron-proton, ##k=\sqrt{2\mu E_{cm}/\hbar^2}## is the wave number and ##V(r)## is the potential of interaction like Yukawa, Wood-Saxon, Square well potential, etc. I first...
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
Back
Top