Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Usually, in Quantum Mechanics, an observable is an operator on the space of the possible quantum states (labelled as $|\psi\rangle$). If this quantity is conserved, in the meaning that the associated operator $\hat D$ is constant: $$\frac{d\hat D}{dt}=0$$ it has to commute with the Hamiltonian operator $\hat H$. To prove this, in the Heisenberg picture of Classical QM, we use the Schroedinger Equation $$i\hslash\frac{\partial}{\partial t}|\psi\rangle =\hat H|\psi\rangle $$

On the other hand, the Noether theorem states that if a system has a continuous symmetry property, then there are corresponding quantities whose values are conserved in time. And this is true either for classical, relativistic and quantum systems.

Therefore, first question:

  • Is it true, even in the relativistic case, when the Schroedinger eq. doesn't apply, that a conserved quantity has to commute with the hamiltonian to be conserved during the motion?

Then, one usually studies the transformations generated by $\hat D$, for instance if $\hat D$ is the angular momentum, the transformation $U(\theta)=e^{-i\hat D \theta}$ (a rotation of $|\psi\rangle$ with angle $\theta$). And assuming the angular momentum conserved, or verifying this by means of the commutation properties, we can construct the space of the eigenstates of $\hat H$ knowing that they are also eigenstates of $\hat D$.

For instance we construct the eigenstates of a spin-1/2 particle using $\hat D = \hat S$, and knowing that $[S^2, S_z]=0$. Since it exists only two states: spin up and spin down, $|\psi\rangle$ will be a two-components spinor, and $\hat S$ will be a 2x2 matrix. We see that the Lie Group of the $\hat U$ transformation is generated by the Pauli's matrices throught the expressions $$\hat S_i=\hslash\sigma_i/s$$. So, I understand that, wanting to impose the conservation of $S$, we get into the $SU(2)$ Lie-group, because of the commutation relations satisfied by the $S_i$, and the infinitesimal decomposition of each $\hat U$.

Final questions:

  • If we want to conserve something else than the spin, as it happens in the strong interaction study, when we have the Isospin and the Strangeness to be conserved, we use $SU(3)$. Why?
  • Who is the corresponding Casimir operator? (the analogous of $S^2$)
  • Why in this case the generators should be 8, and not 3?
  • What is the relationship between a normal rotation in the physical space, for which the generator is $\hat L$, and the rotation in the total space $\hat J$?
  • Why in the case of the strong interaction we don't consider any spin or orbital angular momentum in the labeling and classification of the eigenstates?

  • At last, is there any good textbook you would suggest to have a global view of this subject? Because I've been studying this problems in several courses during my undergrad courses, in QM classes, RQM classes, Nuclear Physics, and so on, but every time they've just mentioned some concepts without getting deeper, and this causes me a lot of confusion...

share|improve this question
1  
By assuming a Schroedinger equation, you are working in the Schroedinger picture, not in the Heisenberg picture. –  Arnold Neumaier Sep 22 '12 at 16:27
    
The Schroedinger equation continues to work in the relativistic case, only the Hamiltonian is a bit more hidden. –  Arnold Neumaier Sep 22 '12 at 16:28
2  
Don't ask too many question in a single entry. It makes answering very difficult. –  Arnold Neumaier Sep 22 '12 at 16:29
add comment

1 Answer

up vote 5 down vote accepted

Yes, an observable is conserved iff it commutes with the Hamiltonian. This is very clear in the Heisenberg picture with the equations of motion $$ i\hbar \frac{d}{dt} \hat L = [\hat L, \hat H].$$ It's true in QM in general, whether it is relativistic or not.

$SU(3)$ is just a particular symmetry group. It appears in the science of quarks – at two places, in fact. $SU(3)_f$ rotates the flavors because there are three relatively light flavors or types of quarks, up, down, and strange, and there is an approximate symmetry between them. They're "complex" and the symmetry rotating three things is $SU(3)$. Well, the $U(1)$ out of the $U(3)$ has a different fate, I don't want to go into these details.

$SU(3)_c$ is also the gauge group of QCD because there are three colors. There have to be three colors because a proton and a neutron is composed of three quarks and their colors are multiplied by $\epsilon_{abc}$ and they have to be because the resulting object must be color-neutral. There are lots of things to explain here but I can't teach you all of physics in one question.

There are many other groups besides $SU(2)$ and $SU(3)$ and they're everywhere in physics, including $SU(N)$, $E_6$, $E_7$, $USp(2k)$, and so on, and so on.

A Casimir operator is an operator that commutes with all the generators of a group, such as $J^2$ for $SU(2)$. There is an obvious generalization of this quadratic Casimir to any compact Lie group, not only $SU(3)$, it is $\sum J_i J_i$ summed over all generators. But there are also higher-order invariants.

$SU(N)$ has $N^2-1$ generators which is $9-1=8$ for $N=3$. That's because the generators may be viewed as $N\times N$ Hermitian traceless complex matrices. The Hermiticity reduces the $2N^2$ real parameters to one-half and the tracelessness (trace is zero) removes one more real parameter.

For rotations in space, it is $\vec J$ that generates the rotations of the space and it's only this total $\vec J$ that is a generator of a symmetry and that is conserved. When there are no particles with spin, it may be reduced to $\vec J=\vec L$, but more generally, $\vec J = \vec L+\vec S+\dots$ and the spins can't be neglected. One may say that the generator $\vec L$ is responsible for the right "relocation" of the wave functions and fields. However, if the fields have some extra indices, like vector indices, the rotation must also shift one component into another, to make the rotation full-fledged, and that's $\vec S$ does.

Quarks and gluons surely have spin much like electrons. In some notation, those indices or extra quantum numbers may be suppressed. Orbital quantum numbers aren't considered in the Hydrogen-like way because the bound states of quarks and gluons aren't quite the same as the Hydrogen atom and one needs different quantum numbers and labels.

There are lots of books introducing readers to similar questions but the composition of the questions suggests that Georgi's book may be the most relevant one for you to learn these things:

http://www.amazon.com/Lie-Algebras-Particle-Physics-Frontiers/dp/0738202339/ref=sr_1_1?s=books&ie=UTF8&qid=1348329960&sr=1-1&keywords=Georgi

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.