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

by Spinnor
Tags: components, currents, force, formula, strong
P: 1,376
In the ψ below, there are 4 components for the Dirac spinnor times three possible color states for a total of 12 components for ψ?

 Quote by ChrisVer ... Nevermind, to get the color current, you need the interactive Lagrangian: $L_{int}= -g_{3} \bar{ψ} γ^{μ}λ^{a} ψ A_{μ}^{a}/2$ the corresponding conserved current (if you remember from the Dirac's current case) is: $J_{SU(3)}^{μa}= g_{3} \bar{ψ} γ^{μ}(λ^{a}/2) ψ$ What can we see from that? That we have 8 conserved currents. Each of them is individually conserved. The continuity relation for the currents, is given by their conservation, thus you have again 8 different continuity relations: $∂_{μ}J_{SU(3)}^{μa}= 0$ and the color charge is: $Q_{c}=\int{d^{3}x J_{SU(3)}^{0a}}$ If they also carry electric charge, you'll get also another current, corresponding to $U(1)_{Q}$ interaction... [/itex] ...
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!
 Thanks P: 1,948 No, color confinement prevents anything like a classical color current.
P: 883
 Quote by Spinnor In the ψ below, there are 4 components for the Dirac spinnor times three possible color states for a total of 12 components for ψ?
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.

P: 1,376
ψ in formula for strong force currents, how many components?

 Quote by dauto No, color confinement prevents anything like a classical color current.
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!
P: 1,376
 Quote by 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.

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

P: 883
 Quote by Spinnor Can the above sum be simply realized as a single multicomponent matrix sandwiched between two multicomponent vectors? Thanks for your help!
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.
 P: 1,058 Is that true Duck? I mean it'd be like breaking the tensor product of representations into tensor sum
Thanks
P: 1,948
 Quote by 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.
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).
P: 883
 Quote by ChrisVer Is that true Duck? I mean it'd be like breaking the tensor product of representations into tensor sum
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.
 P: 1 what happens to electrons in nuclear reaction ?
P: 1,376
 Quote by crisami what happens to electrons in nuclear reaction ?
You mean like when a neutron turns into a proton?

http://en.wikipedia.org/wiki/File:Be...tive_Decay.svg

http://en.wikipedia.org/wiki/W_and_Z..._nuclear_force

 Related Discussions Quantum Physics 1 Introductory Physics Homework 4 Electrical Engineering 41 Chemistry 3 General Physics 11