# Ïˆ in formula for strong force currents, how many components?

1. Dec 21, 2013

### Spinnor

In the Ïˆ below, there are 4 components for the Dirac spinnor times three possible color states for a total of 12 components for Ïˆ?

Are there low energy, weak field limits of the above that allow us to consider classical color counterparts of electric current densities and electric charge densities?

Thanks for any help!

Thanks to ChrisVer for the original post!

2. Dec 22, 2013

### dauto

No, color confinement prevents anything like a classical color current.

3. Dec 22, 2013

### The_Duck

Yes. Unfortunately in QFT we are usually swimming in a sea of suppressed indices. Here the Dirac and color indices of the $\psi$ field have been suppressed. If you write out all the indices in the expression for the SU(3) current, you get

$j^{a \mu} = \bar \psi_{\alpha i} \gamma^\mu_{\alpha \beta} \lambda^a_{ij} \psi_{\beta j}$

Here $\alpha$ and $\beta$ are Dirac indices, which take on four possible values; $i$ and $j$ are indices for the fundamental representation of SU(3), and take on three possible values; $a$ is an index for the adjoint representation of SU(3), and takes on eight possible values; and $\mu$ is a Lorentz index and takes on four possible values.

$j^{a0}$ is the color charge density and $j^{a\mu}$, $\mu = 1, 2, 3$ is the color current density.

4. Dec 23, 2013

### Spinnor

I think it was shown to me that asymptotic freedom results because the number of quark types is less then 16. Could we arbitrarily add more quark types to the standard model so as to obtain my classical weak field low energy color physics limit?

At high energies in a quark/gluon plasma does need for colorless combinations of quarks and antiquarks still apply?

Thanks for the help!

5. Dec 23, 2013

### Spinnor

Can the above sum be simply realized as a single multicomponent matrix sandwiched between two multicomponent vectors?

6. Dec 23, 2013

### The_Duck

Sure, if you like you can think of $\psi$ and $\bar \psi$ as 12-component vectors and $\gamma^\mu \lambda^a$ as a $12 \times 12$ matrix.

7. Dec 23, 2013

### ChrisVer

Is that true Duck? I mean it'd be like breaking the tensor product of representations into tensor sum

8. Dec 23, 2013

### dauto

And that's just for a single quark. For the complete current you have to sum over all quarks which means two more indices (one for the isospin and one for the family).

9. Dec 23, 2013

### The_Duck

This isn't about breaking up a tensor product into a tensor sum; it's just about how you can write a member of a tensor product representation as one big matrix. As a concrete example, consider a system of two spin-1/2 particles. It lives in the tensor product representation $\frac{1}{2} \otimes \frac{1}{2}$ of the rotation group. The representation has dimension four and a convenient choice of basis states is $|\uparrow\uparrow\rangle, |\uparrow\downarrow\rangle, |\downarrow\uparrow\rangle, |\downarrow\downarrow\rangle$. Then on this basis we can write the tensor product operators as single $4 \times 4$ matrices, for example

$$\sigma_3 \otimes \sigma_3 = \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)$$

$$\sigma_1 \otimes \sigma_1 = \left(\begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{array}\right)$$

Similarly $\gamma^\mu \lambda^a$ can in principle be thought of as a $12 \times 12$ matrix acting on the dimension-12 tensor product representation that $\psi$ lives in.

10. Dec 24, 2013

### crisami

what happens to electrons in nuclear reaction ?

11. Dec 24, 2013