SU(3) Generators & Physical Quantity Corresp.

[fresh_42]

If we consider the generators of SU(3) in the standard model. Is there a direct correspondence between them and a physical quantity, esp. if we only consider $T_+=\frac{1}{2}(\lambda_1+i\lambda_2)\; , \;U_+=\frac{1}{2}(\lambda_6+i\lambda_7)\; , \;V_+=\frac{1}{2}(\lambda_4+i\lambda_5)$, do they directly represent a quantity? And what does $V_+$ stand for?

 

[stevendaryl]

The idea of the SU(3) operators can be illustrated by the quark model, where we consider the three lightest quarks, U, D, S (up, down, strange). They aren't actually an example of SU(3) symmetry, since the strange quark is so much more massive than the up or down quarks, but it's an approximate application of SU(3). You view the S, U, and D quarks as if they were three different states of the same particle, in the same way that a spin-up or spin-down electron are two different states of the electron. In the same way that for electron spin, you have "raising" and "lowering" operators that convert a spin-up electron into a spin-down electron, or vice-versa, in SU(3), there are similar raising and lowering operators for moving around the space of possible quark states. The operator $I_+$ converts a D quark into a U quark. The operator $V_+$ converges an S quark into a U quark. The operator $U_+$ converts an S quark into a D quark.

(I think what you're calling $T$ is sometimes called $I$ (isospin) when it comes to quark flavors.

In the case of spin-1/2 particles, you can combine more than one spin-1/2 particle to get spins of all different magnitudes. In the case of SU(3), you can consider composite states of two or more quarks to get more complicated representations of SU(3).

 

[fresh_42]

I thought the Isospin (3rd component) refers to an element in the CSA, i.e. an operator leaving something invariant, namely all which I listed. I therefore used the Gell-Mann matrices $\lambda_k$ for clarity. I left out the entire CSA ($\lambda_3 , \lambda_8$) and the negative counterparts here.

The problem I have with your explanation is, that $T_+,U_+,V_+$ doesn't define a SU(2) copy, since $[T_+,U_+]=V_+$ and $V_+$ can only be shifted to zero by the given operators. This means that $V_+$ is of a different quality as the other two, the highest weight of ad so to say, or the outer entry of the upper triangular matrices which I choose. Now how does this circumstance find itself in physics?

 

[stevendaryl]

The problem I have with your explanation is, that $T_+,U_+,V_+$ doesn't define a SU(2) copy, since $[T_+,U_+]=V_+$ and $V_+$ can only be shifted to zero by the given operators. This means that $V_+$ is of a different quality as the other two, the highest weight of ad so to say. Now how does this circumstance find itself in physics?

I'm not sure if I understand your point. The algebra of Isospin is the same as the algebra of SU(2). So if you stick to just $I_+, I_-, I_3$ (where $I_{\pm} = \frac{1}{2} (\lambda_1 \pm i \lambda_2)$ and $I_3 = \frac{1}{2} \lambda_3$), then you have the same algebra as spin-1/2. Similarly, sticking to just $U_{\pm}$ or $V_{\pm}$ will give you something isomorphic to SU(2).

I don't know what you mean by $V_+$can only be shifted to zero...

 

[stevendaryl]

What I was describing is discussed on page 17 of this document:

http://hepwww.rl.ac.uk/Haywood/Group_Theory_Lectures/Lecture_4.pdf

 

[fresh_42]

I'm not sure if I understand your point. The algebra of Isospin is the same as the algebra of SU(2). So if you stick to just $I_+, I_-, I_3$ (where $I_{\pm} = \frac{1}{2} (\lambda_1 \pm i \lambda_2)$ and $I_3 = \frac{1}{2} \lambda_3$), then you have the same algebra as spin-1/2. Similarly, sticking to just $U_{\pm}$ or $V_{\pm}$ will give you something isomorphic to SU(2).

I don't know what you mean by $V_+$can only be shifted to zero...

I meant in your notation just $I_+$ alone, no $I_-$ and no $I_3$, and the same with the other copies of SU(2). Those elements build an algebra on their own, even if we would add the CSA, the diagonal elements. So my question was basically: If I do not consider the entire group SU(3) but single generators instead, what does it mean physically? And especially what does - in your notation - $\langle I_+,U_+,V_+ \rangle$ mean, where $[I_+,U_+] = V_+ \text{ and } [I_+,V_+]=[U_+,V_+]=0$; just the ladder up operators alone

Regarding the lecture you quoted: You cannot raise strange, what does this mean?

 

[samalkhaiat]

If we consider the generators of SU(3) in the standard model. Is there a direct correspondence between them and a physical quantity, esp. if we only consider $T_+=\frac{1}{2}(\lambda_1+i\lambda_2)\; , \;U_+=\frac{1}{2}(\lambda_6+i\lambda_7)\; , \;V_+=\frac{1}{2}(\lambda_4+i\lambda_5)$, do they directly represent a quantity? And what does $V_+$ stand for?

The ladder operators do not correspond (directly) to physical quantities. See
https://www.physicsforums.com/threads/su-3-multiplet.860348/#post-5399569

 

[fresh_42]

The ladder operators do not correspond (directly) to physical quantities. See
https://www.physicsforums.com/threads/su-3-multiplet.860348/#post-5399569

Thanks.

 

[stevendaryl]

I meant in your notation just $I_+$ alone, no $I_-$ and no $I_3$, and the same with the other copies of SU(2). Those elements build an algebra on their own, even if we would add the CSA, the diagonal elements. So my question was basically: If I do not consider the entire group SU(3) but single generators instead, what does it mean physically? And especially what does - in your notation - $\langle I_+,U_+,V_+ \rangle$ mean, where $[I_+,U_+] = V_+ \text{ and } [I_+,V_+]=[U_+,V_+]=0$; just the ladder up operators alone

Regarding the lecture you quoted: You cannot raise strange, what does this mean?

I'm not sure exactly what you mean. We have the nonzero commutation relations:

1. $[I_+, U_+] = V_+$
2. $[I_-, V_+] = U_+$
3. $[V_-, I_+] = U_-$
4. $[V_+, U_-] = I_+$
5. $[U_-, I_-] = V_-$
6. $[U_+, V_-] = I_-$

[edit] There are three more:

1. $[I_+, I_-]$
2. $[U_+, U_-]$
3. $[V_+, V_-]$

So you can go in any direction by a combination of the two other directions. There is no particular meaning to the distinction between "up" ladders and "down" ladders. The ladders take you in one of three possible directions, and there is no particular notion of "up" or "down".

 

[fresh_42]

I'm not sure exactly what you mean. We have the nonzero commutation relations:

1. $[I_+, U_+] = V_+$
2. $[I_-, V_+] = U_+$
3. $[V_-, I_+] = U_-$
4. $[V_+, U_-] = I_+$
5. $[U_-, I_-] = V_-$
6. $[U_+, V_-] = I_-$

[edit] There are three more:

1. $[I_+, I_-]$
2. $[U_+, U_-]$
3. $[V_+, V_-]$

So you can go in any direction by a combination of the two other directions. There is no particular meaning to the distinction between "up" ladders and "down" ladders. The ladders take you in one of three possible directions, and there is no particular notion of "up" or "down".

I tried to understand the role of $V_+$ from where you cannot go up anymore. Since this is the only element with this property, I wondered if it has some meaning; and more general the role of $\operatorname{span}\{\,I_3,Y,I_+,U_+,V_+\,\}$. What does it mean, if we drop half of the group? So far, the answer seems simply to be: nothing. I thought because this is also a Lie group resp. algebra, I'd thought there is a meaning.

 

[stevendaryl]

I tried to understand the role of $V_+$ from where you cannot go up anymore. Since this is the only element with this property,

There is nothing special about $V_+$. I think looking at the "+" and "-" is misleading. There are six directions you can travel in:

1. From s to u.
2. From u to s.
3. From s to d.
4. From d to s.
5. From d to u.
6. From u to d.

The fact that they call #1 $V_+$ and call #2 $V_-$ is completely arbitrary. They are just two operators that are the adjoints of each other.

 

[fresh_42]

Sure, but algebraically $\mathfrak{su}(3)=\mathfrak{N}_{-} \oplus \langle Y,I_3 \rangle \oplus \mathfrak{N}_+$ so I was looking whether this decomposition means anything. Of course you get a completely different situation by restriction to say $\mathfrak{N}_+$ or $\langle Y,I_3 \rangle \oplus \mathfrak{N}_+$, as e.g. the Killing form isn't non-degenerated anymore. I was not expecting to find the entire situation within this smaller range, I just wanted to know, whether this decomposition has a physical meaning.