# Physical Interpretation ~Quantum Numbers

1. Dec 8, 2004

### saiarun

:rofl: 1.Can any one explain the physical interpretation of the Quantum numbers?
2. How they are vectors?
3.Are they known before solving the hydrogen atom problem?
4. Are there quantum numbers for protons and neutrons also?

2. Dec 8, 2004

### dextercioby

1.The quantum numbers are nothing but eigenvalues of densly-defined,self adjoint linear operators which have olny discrete spectrum.(in units hbar equal one)
2.They're not.Vectors are eigenvectors (elements of a separable Hilbert space) whiile q numbers are defined above.Don'te be fooled by MR.Dirac's notation.
3.Yes,if u're talking about "l" and "m".They're known from the QM theory of angular momentum.
4.Yes.For example the spin number is 1/2 for neutron,proton and electron.

Daniel.

3. Dec 9, 2004

### saiarun

Can you deifne what is eigen value of densly defined state. Why eigen vector can't have a direction (vector).

4. Dec 9, 2004

### dextercioby

An eigenvalue of an operator is the number from the equality:
$$\hat{A} |\psi> =a |\psi>$$,where a is the eigenvalue of the operator A,if the spectrum of A is discrete and |psi> designate vectors from the Hilbert space on which the operator A acts.
What do you mean by direction...??The Hilbert space of states is abstract,its vectors have no real geometrical meaning (no angles,no cosine directors,no direction,no support,just a modulus,which,by chance,is a still a real number),yet the're assigned by the first postulate of QM to represent the quantum possible states of the system.

5. Dec 10, 2004

### marlon

When a physical formula (like the Lagrangian for example) exhibits a global (ie : independent of time and position) symmetry, the Noether-theorem assures us that the system will have a conserved quantity (like energy, momentum, charge,...)

When we make that symmetry local (ie dependent of time and space), the gauge-bosons will appear. These particles mediate the interactions between matter-particles like protons, electrons... The photon is an example of a gauge-boson that "brings over" the EM-interaction.

Now, it can be shown that the socalled quantum-numbers generate operations that will respect the symmetry of the model. "The quantumnumbers are the generators of operations that leave the Lagrangian invariant..."

Basically this means that the "conserved quantity" due to a global symmetry is expressed in terms of a quantumnumber (like "L" for conservation of angular momentum) and the interactions of particles will obey this conservation-law (L is conserved during all interactions). You could see this conservation law as the referee that sets the rules of the game (ie the interaction) and the language he uses is in terms of quantumnumbers...like charge needs to be conserved or momentum needs to be conserved...

regards
marlon

6. Dec 10, 2004

### dextercioby

Thenx,Marlon,for generalizing the problem to QFT,but i still wonder if the simple QM picture which i strove to describe,ignoring symmetries,is clear to him.If it is,then your more elaborate approach could be seen as comprehendable,up to a point at which u're gettin'fed up with group theory and irreductible representations.

Daniel.

7. Dec 10, 2004

### marlon

Well, the QM-picture does not ignore symmetries...It is because of symmetry that quantumnumbers are born...

For example, look at the way the quantumnumbers (eg : l) are inserted in QM because of the symmetry-properties of spherical harmonics...These properties need to be respected because otherwise these wave-functions are not physical etc...

regards
marlon

8. Dec 10, 2004

### dextercioby

You're right about the relevance of symmetries in QM.A symmetry group for a QM system admtis an irreductible representation whose operators commute with the (time independent) hamiltonian,and by means of the Noether theorem for QM,all these operators describe at QM level conserved quantities.
But i'm afraid your example is not the greatest when talking abot symmetries.Because the notion of angular momentum in QM is much,much more general,than symmetries.I shall prove my assertion.
We refer at QM operators for angular momentum as at a set of 3 operators which are operators of an irreductible representation of the Lie algebra of angular momentum obtained through quantization via the second principle and the classical definiton of the angular momentum.The angular momentum operators Jsquared and Jz act in an irreductible linear space of the representation,where the quantum numbers "l" and "m" have the significance of merely hbar units eigenvalues.These thing are very abstract and mathematized.But that's QM at essence.
So,who said anything about symmetries/Noether theorem wrt to angular momentum,SO FAR??
To resume this necessary mathematical divagation:to state that symmetries give birth to Q numbers for angular momentum,that's bull****. These Q numbers are hbar unit eigenvalues of those operators.Yet symmetries of the Hamiltonian of the H atom are responsible for their values,but not for their fundamental significance.
AAAAAAAAA,when discussing specific exmples (i have a hunch yours refers to H atom),then,yes,symmetries become essential,for 2 reasons:1)the possibilities induced by the Noether theorem;2) the possibility of chosing specific realizations of the irreductible space of the representation.Actually the basis in that space becomes essential.
For the H atom,my friend,u're only given these 3 relations:

$$\hat{H} =\frac{\hat{p}^{2}}{2m} -\frac{e^{2}}{4\pi\epsilon_{0}r}\hat{1}$$

$$\hat{L}^{2} |\psi> =\hbar ^{2} l(l+1) |\psi>$$
$$\hat{L}_{z} |\psi> =\hbar m |\psi>$$

Solving the H atom from these 3,would require a chain of deductions,forrmulas,integrations,to find what u were saying:"l" and "m" have the values they have (for the H atom) because of symmetries.
Sorry for repeating myself,but to state that "l" and "m" are symmetry related in general,just doesn't sound right to me.

Daniel.

PS But you're right about the relevance of symmetries in quantum physics.They guide our steps to knowledge.But mathematical rigurosity,logics and definitons are somethig else.I think so.I might be wrong,though.You're free to say your point of view.

PPS.Sorry,again,Mr.Integral,for putting too much soul in my exposures,but i guess it's just another thing people on PF should become familiar with,as long as i'm still here.When i won't,things would return to normal. :tongue2:

9. Dec 10, 2004

### Mike2

Could you recommend a book or hyperlink on the subject of symmetries, group theory, and Quantum Mechanics/QFT, maybe something with a primere on group theory in general? Thanks.

10. Dec 10, 2004

### marlon

That sure as hell is the case...

Second principle ??? I hope you don't mean second quantization my dear friend because then you are way off...Second quantization gives rise to the creation and annihilation-operators and is the foundation of QFT.

:surprised
Well, errr, the mere fact that you are talking about irreducible representations justifies my approach and answers your question here...

Let me answer by asking you this question : How do you think, irreducible representations are born? What justifies their existence? Take a look at the fundamental representation of QCD, using the three quark-colour as a fundamental SU(3) representation. How do you think these quark-colours are born ?? Perhaps it has something to do with symmetries ?

Let me apply this to QM, the system is the same...What is the motivation for the fact that for example the magnetic quantumnumber has the specific values that it has. I mean why (+/-)l, ...,0 and so on ???

I suggest you go on and study some group-theory...

You are completely right,...and why do you think that is ???
How do you justify the commutation-relations of such operators with the Hamiltonian : THIS is the essential part and nothing else... ???

What do you think [H,Q]= 0 means ??? How is this justified...Please, don't tell me this has nothing to do with symmetry...

Moving on to QFT : ever heard of the Nambu-Goldstone-theorem???

regards
marlon

11. Dec 10, 2004

### marlon

QM : QUANTUMMECHANICS by Bransden and Joachain
QFT : QFT in a nutshell by Anthony Zee

regards
marlon

12. Dec 10, 2004

### dextercioby

Okay,then.

I may look stupid sometimes,BUT I'M NOT AN IDIOT.
I was obiously referring to the SECOND POSTULATE OF "NORMAL" QM.U know,the one that states that to any Hamiltonian observable of a classical dynamical system it is associated a closed densly defined selfadjoint linear operator acting on the separable Hilbert space of states+the rule of quantizing the Dirac paranthesis for a dynamical system+the rule of symmetrizing wrt to all possible distinct products of noncommuting operators.

Let's not mix things...There are symmetry groups in QFT (u've given the example of a gauge group) and symmetry groups in QM.QM is essentially build upon the Halmiltonian and its symmetry groups,while in QFT when speaking of symmetries,we're referring to symmetries of the Lagrangian density,and not to the ones of the Hamilton density.
Quark colors have everything to do with symmetries.But i believe,that's an attempt to change the subject.

.

Those values are only in the happy/fortunate case when u're applying GENERAL THEORY OF ANGULAR MOMENTUM TO A PARTICULAR SYSTEM.As to general theory of angular momentum,"m" could take any real value in the domain "[-l,+l]".That's the only constraint (the boundness of the spectrum of L_z) whiwh results from the theory.GENERAL THEORY.I guess that,it's either that u don't see my point,or u won't.I'll go for the second,as i believe u're a smart guy and i made myself pretty clear so far.Or the least i was trying to.
The reason why those 2 spectrums of the 2 angular momentum operators are discrete is that U HAVE PROVEN THAT USING A PARTICULAR REPRESENTATION FOR THE COORDINATES AND MOMENTA.The coordinate representation,in which it's easy to show that,in order to have both continuous,bounded (hence normalizable) eigenfunctions for the 2 operators,(which obviuously commute,so they share the same subspace of complete orthonormal eigenvectors),u must have discrete spectrum,and more,"l" cannot be negative.
YOU ARE REFUSING TO BELIEVE THAT THE THEORY OF ANGULAR MOMENTUM IN QUANTUM MECHANICS IS SO GENERAL,THAT,IN ORDER TO BE FULLY CONSTRUCTED,IT NEEDS NOT THE CONCEPT OF SYMMETRY,AS LONG AS THAT CONCEPT OF SYMMETRY IN QUANTUM MECHANICS CANNOT BE DEVELOPED WITHOUT A SYSTEM,WITHOUT A HAMILTONIAN.

Believe me,i have studied enough as to make a difference between an irreductible representation of a (symmetry) group and an irreductible representation of an abstract Lie algebra.Apparently u're not willing to make that difference.

My version is:you're right.But u didn't get when i was talking about.

I'd suggest we leave spontaneous symmetry breaking of discrete and continuous global symmetry in field theory aside,as the VEV-s' of the scalar fields (for example) connection with irreductible representations of the angular momentum Lie algebra of QM,if it exists,is not easily comprehendable,not by me,at least.I think u know as well as me,that the notion of vacuum in not of the realm of QM.

Daniel.

PS.We can discuss symmetries of QFT in detail,if u want to,as apparently for the ones of QM we have no different opinions.

13. Dec 10, 2004

### humanino

better look at local symmetries in this context
:tongue:

You belgian guys can not keep a cool head can you ?

14. Dec 10, 2004

### dextercioby

I simply suggested him that his statement of whether or not i'm aware of Nambu-Goldstone theorem,was off context,as i was striving to prove that in QM the general theory of angular momentum does not require the need of a symmetry group of the hamiltonian.That theory can be entirely deduced by looking and analyzing the relation giving the Lie algebra of the angular momentum.Nonetheless,once u have a system and its hamiltonian has rotational symmetry,then u can use group theory ((S)O_{3}) to identify the generators of the irreductible representations of this group to be exactly the angular momentum operators.That' s what he was trying to say.What i was trying to say,was that the eigenstates (and thus the eigenvalues) could be found just by simply searching for the irreductible representations of the Lie algebra of angular momentum,algebra which is deductable exactly from the second principle of QM,FOR ANY SYSTEM,NOT ONLY FOR THE ONES ROTATIONALLY SYMMETRIC.For a rotational symmetric hamiltonian (as it's the case of H atom) it can be shown that the eigenstates and eigevectors of the angular momentum operators can be found by giving a particular representation of the momentum operators.That particular reprtesentations is given by the PE part of Hamiltonian which depends of the classical coordinates of the particle,therefore implying that at QM,when quantizing the system (H atom),we'd better use coordinate representation for each operator and for each eigenstate.

Daniel.

15. Dec 10, 2004

### humanino

Daniel : you perfectly make sens. And I would better not participate this debate, because I do not see any point to it. Except that you are arguing for the pleasure of arguing. Well, once again I will side to Godfather Marlon, because I disagree with :
Indeed, one needs not even in classical mechanics to state that Energy conservation comes from the arbitrariness of the origin (invariance under translation group = another symmetry) but that would be missing a lot of the story.

In fact, I am too fascinated by symmetry principles, and the fact that they allow to derive so many fundamental results. Maybe every statement about the physical world could be, at the end of the day, reduced to a certain invariance of a certain lagrangian. That is one of the common main paradigms in physics, that emerged from 20th century discoveries.

16. Dec 10, 2004

### dextercioby

Let me make the statement that i'm not diminishing the relevance of symmetry to fundamental phyiscs,as the latter cannot be possibly conceived without the notion of symmetry.
And i'm not isinuating that u could neglect the symmetry groups of the Hamiltonian,i'm just saying that the construction of angular momentum theory in QM can be made without taking into consideration the rotational invariance of the Hamiltonian,because the latter could not exist,yet every mechanical system that i know of can have angular momentum and can be quantized by the second principle.Those relations of commutation between certain components of momenta have deep roots in classical dynamics of a phase space seen as a symplectic manifold and the Hamiltonian is nothing but a function on that manifold.Nobody asks that the Hamiltonian be invariant under rotations,but the fundamental Poisson/Dirac brackets remain invariant.
It is just a remarkable (mathematically originated) coincidence that the Lie algebra of angular momentum is isomorphic with the Lie algebra su(2) of the spin group SU(2) which in turn is locally isomorphic with the SO(3) group which is seen to be as the symmetry group for a rotationally invariant Hamiltonian.But that rotational symmetry may not exist (of course it is found that it is determined by the fact whether the PE part of the Hamiltonian depends upon the modulus of the distance between the particle and the center of rotation,like the case of central fields),and yet the algebra of of angular momentum would still exist.

Anyways,i agree with you.Symmetry and geometry rule physics!!

Daniel.

17. Dec 10, 2004

### 1100f

If your hamiltonian does not have rotational invariance, it will not commute with the angular momentum, so that the hamiltonian eigenstates are not eigenstates of the angular momentum.

BTW. In the hydrogen atom, not only l and m come from symmetry but also n, since in the Kepler problem, the hamiltonian is invariant not only under rotations but it is invariant under SO(4) where 3 generators are the angular momentum and three other generators are the three components of the Runge-Lentz vector

18. Dec 10, 2004

### dextercioby

So WHAT??What if those 2 operators do not commute/the angular mometum is not a constant of motion??What does that have to do with the irreductible representations of the angular momentum Lie algebra??Does that mean u cannot find eigenvalues and eigenvalues for the L's???I hope not...

Thank you with providing an information absolutely not necessary to the discussion. The discussion was not about symmetries of the H atom,but about the nature of "l" and "m".They are hbar units eigenvalues for the L squared and L_z.

19. Dec 10, 2004

### humanino

The more I discover about instantons, the more symmetry fascinates me.
Instantons are due to a non-trivial global connection which does not appear in the local structure, in the generators. So this is really the part of the gauge symmetry which we cannot compute "trivially". They are self-dual objects.

And then of course, there are many new types of symmetries in the non-perturbative sector of string theories. They are also called duality symmetries. Yet I am not certain that they can really be interpreted in terms of a duality operation in a Hilbert space, is it a Pontryagin duality (the dual of a group being the set of characters), or is it merely a categorical duality (which to me is essentially involution $$\hat{O}^2=1$$, but in any case this is central in duality). For those duality symmetries which I am aware of (I might forget some, maybe important) :
• S-duality (strong/weak)
• U-duality (generalization of the two previous, gauge/string)
is it always possible to have a simple interpretation in terms of elementary Hilbert space duality, or is it merely the fact that those operations are involutive ?

20. Dec 10, 2004

### RedX

uhhh

Okay I didn't understand what most of you just said. That was beyond me (I'm an engineer... ).

But I overheard you talking about the 2nd postulate of quantum mechanics and angular momentum, and if I recall, the 2nd postulate is quantization of classical variables.

Actually angular momentum can be derived from deeper principles, as the generators of infinitismal rotations, and if you consider a sequence of rotations in this order: e_x about the x-axis, e_y about the y-axis, -e_x about the x-axis, -e_y about the y-axis, then ignoring quadratic terms but keeping cross-terms, it's equivalent to a rotation of -e_x*e_y about the z-axis, just from the concept of a rotation and this has nothing to do with QM.

Anyways, so you write the rotation operator in terms of the generators and they have to satisfy the above identity, and you derive the commutation relation for angular momentum: JxJ=ihJ, from which you can construct the ladder operators and get everything you want.