So where does it come from? I know where the first three come from. Any comments are appreciated.

Cheers.

Actually, maybe I should have been a little clearer. The first three can be found by solving the time-independent Schrodinger equation in spherical-polar coordinates and, of course, using the operator \/^2 (del squared). Then separate the variables and solve the differential equation. Anyway, the point is that this leaves no room for a fourth quantum number!

I have an idea that it has something to do with special relativity.

Another quick question that came to mind: Is there a formula which allows me to work out which energy states are nondegenerate? Like 3 and 12 are nondegenerate but 6,9,11,14,... etc are degenerate.

So where does it come from? I know where the first three come from. Any comments are appreciated.

Cheers.
so you know where the angular momentum quantum number comes from, right? the angular momentum quantum number just tells you what happens to the wavefunction's phase when you go around in a revolution around your origin. the spin quantum number tells you what happens to your wavefunction's phase when you rotate the particle in place.

Another quick question that came to mind: Is there a formula which allows me to work out which energy states are nondegenerate? Like 3 and 12 are nondegenerate but 6,9,11,14,... etc are degenerate.
what are these numbers? energy level quantum numbers for hydrogen? if you are neglecting spin-orbital interactions, Lamb shift, and the other fine structures and hyperfine structures, then all energy levels are degenerate.

if you are not neglecting those, then i don't think any energy levels are degenerate.

Actually, maybe I should have been a little clearer. The first three can be found by solving the time-independent Schrodinger equation in spherical-polar coordinates and, of course, using the operator \/^2 (del squared). Then separate the variables and solve the differential equation. Anyway, the point is that this leaves no room for a fourth quantum number!
right, the spin quantum number doesn't arise from a differential equation. but there is a nice way to think about that differential equation which puts the angular momentum quantum number on equal footing with the spin quantum number.

you are looking for representations of the rotation group on the vector space of wavefunctions &psi;(<b>x</b>). a rotation can act on this guy by rotating the components of the wavefunction or by rotating the argument: R&psi;(R^-1<b>x</b>). unless we wanted to consider particles with an infinite number of components, the first part acts on a finite dimensional vector space, so we can just write down the possible representations from our first book on representation theory. in the second case, we have an infinite dimensional vector space of wavefunctions, and to find representations on spaces of this kind usually requires a differential equation.

I have an idea that it has something to do with special relativity.
not really. perhaps you have heard that spin is predicted by special relativity, but this is pretty much a load of bunk. when people say this, i think they are refering to the fact that Dirac, who wanted to get a first order version of the Klein-Gordon equation, discovered that the first order version only worked if you had a 4 component spinor, instead of a single component wavefunction. and thus we say that by incorporating special relativity with quantum mechanics, he predicted spin.

but this is silly, because by requiring the equation to be first order, he was really already adding spin by hand (although he did not know it at the time). and furthermore, his reasons for doing so were wrong. he though that he could restore positive definiteness to probability density of the wavefunction by going to a first order equation, but he did not. instead, what he ended up doing was deciding that there should be an antiparticle.

Thankyou for replying Lethe.

not really. perhaps you have heard that spin is predicted by special relativity, but this is pretty much a load of bunk. when people say this, i think they are refering to the fact that Dirac, who wanted to get a first order version of the Klein-Gordon equation, discovered that the first order version only worked if you had a 4 component spinor, instead of a single component wavefunction. and thus we say that by incorporating special relativity with quantum mechanics, he predicted spin.

but this is silly, because by requiring the equation to be first order, he was really already adding spin by hand (although he did not know it at the time). and furthermore, his reasons for doing so were wrong. he though that he could restore positive definiteness to probability density of the wavefunction by going to a first order equation, but he did not. instead, what he ended up doing was deciding that there should be an antiparticle.

Okay, this makes sense now.

right, the spin quantum number doesn't arise from a differential equation. but there is a nice way to think about that differential equation which puts the angular momentum quantum number on equal footing with the spin quantum number.

you are looking for representations of the rotation group on the vector space of wavefunctions ψ(x). a rotation can act on this guy by rotating the components of the wavefunction or by rotating the argument: Rψ(R-1x). unless we wanted to consider particles with an infinite number of components, the first part acts on a finite dimensional vector space, so we can just write down the possible representations from our first book on representation theory. in the second case, we have an infinite dimensional vector space of wavefunctions, and to find representations on spaces of this kind usually requires a differential equation.

This... a little harder to understand!

Nonrelativistic energy is given by E = p^2 /2m. And relativistic energy is given by E = p^2*c^2 + m^2*c^2.

When you solve the relativistic Schrodinger equation in 3-dimensions (which is a really messy thing to write here but I'm sure you are familiar with it) it has a solution given by [psi] = R(r)f(theta)g(phi), that is by a change of variables (taking 1 3-dimensional DE and turning it into 3 1-dimensional DE's). Each quantum number is involved in the solution to each of these functions.

To make a long story short we are considering the energy E of the particle is taken to be the same as the Bohr energy. But this energy does not take into account the fact that electrons are relativistic particles. Thus by solving the relativistic Schrodinger equation shouldn't the spin number pop out as the solution to this DE?

what are these numbers? energy level quantum numbers for hydrogen? if you are neglecting spin-orbital interactions, Lamb shift, and the other fine structures and hyperfine structures, then all energy levels are degenerate.

Let E = (pi^2h-bar^2)/(2mL^2)
Then...
3(E) corresponds to 1, 1, 1
12(E) corresponds to 2, 2, 2
But 27(E) corresponds to 3, 3, 3 and 5, 1, 1 and 1, 5, 1 and 1, 1, 5.

So 27(E) is 4-fold DEGENERATE
and 12(E) and 3(E) are NONDEGENERATE

I was wondering if given an (E) can I tell if it is degenerate or nondegenerate? That's all.

This... a little harder to understand!

Nonrelativistic energy is given by E = p^2 /2m. And relativistic energy is given by E = p^2*c^2 + m^2*c^2.
you need a square root, and some more c's in that last equation, but so far, I'm with you

When you solve the relativistic Schrodinger equation in 3-dimensions
the relativistic Schrödinger equation? do you mean the nonrelativistic Schrödinger equation with relativistic corrections added in? or do you mean the Klein-Gordon equation or the Dirac equation?

(which is a really messy thing to write here but I'm sure you are familiar with it) it has a solution given by [psi] = R(r)f(theta)g(phi), that is by a change of variables (taking 1 3-dimensional DE and turning it into 3 1-dimensional DE's). Each quantum number is involved in the solution to each of these functions.
mm hmm... of course, that stuff that i was saying before about the representation theory of the rotation group is secretly hidden here in your choice to change to spherical coordinates.

To make a long story short we are considering the energy E of the particle is taken to be the same as the Bohr energy. But this energy does not take into account the fact that electrons are relativistic particles. Thus by solving the relativistic Schrodinger equation shouldn't the spin number pop out as the solution to this DE?
no, of course it shouldn't! as i said before, relativity does not imply spin! or at least, there is no constraint in the Schrödinger equation or the Klein-Gordon equation about the spin of the particle it describes. so solving those equations will never predict a spin quantum number. The Dirac equation, on the other hand.....



Let E = (pi^2h-bar^2)/(2mL^2)
Then...
3(E) corresponds to 1, 1, 1
12(E) corresponds to 2, 2, 2
But 27(E) corresponds to 3, 3, 3 and 5, 1, 1 and 1, 5, 1 and 1, 1, 5.

So 27(E) is 4-fold DEGENERATE
and 12(E) and 3(E) are NONDEGENERATE

I was wondering if given an (E) can I tell if it is degenerate or nondegenerate? That's all.
oh, i see, you are talking about the particle in the 3d box. in other words, you want to know, given any integer n, how many triples (a,b,c) are there such that a²+b²+c²=n. the energy will be degenerate when there is more than one way. the energy level will be forbidden when there are no ways.

i think there probably is a way to tell, but i don't know what it is. there is a story that Ramanujan and Hardy were walking down the street, and Hardy saw a license plate number 1729 (i think that is the number), and said to Ramanujan "what a boring number!", to which Ramanujan replied "but 1729 is not boring at all! it is the first number that can be written as the sum of two cubes in more than one way"

the point of the anecdote is that questions like "can n be written as the sum of three squares in more way than one?" is something that number theorists work on, so i think there may be an answer.

and hell, maybe a number theorist will actually show up and tell us what the answer is. i am, myself, completely ignorant of number theory, so i can't tell you anything about it other than my anecdote about Ramanujan.

Lethe, what is the Klein-Gordon and Dirac equations? An explanation or a link would be great. Thanks.

Lethe, what is the Klein-Gordon and Dirac equations? An explanation or a link would be great. Thanks.
so, the classical nonrelativistic Hamiltonian for a point particle moving in an external potential is

H=p²/2m+V

if you substitute i&part;/&part;t for H, and -i&part;/&part;x for p, you get the Schrödinger equation (which is nonrelativistic)

i&part;&psi;/&part;t=-1/2m&nabla;²&psi;+V&psi;

if you start instead with the Hamiltonian (squared, for various reasons) of a relativistic free point particle

H²=p²+m²

and perform the same substitution as above, you get the Klein-Gordon equation:

(&part;²+m²)&phi;=0

you can factorize the Klein-Gordon operator into the product of two first order operators:

&part;²+m² = (&gamma;^&mu;&part;_&mu;+m)(&gamma;^&nu;&part;_&nu;-m)

Here, &gamma; is a matrix valued vector that Dirac made up to force this factorization to work. the condition that &gamma; must satisfy for this to work is {&gamma;^&mu;,&gamma;^&nu;}=2&eta;^&mu;&nu;. matrices satisfying relations like this in general generate what is known as a Clifford algebra. you might like to sort of think of these as square roots of -1 (like i), which you would use to factorize (a²+b²)=(a+bi)(a-bi). only here, we are factoring operators, instead of numbers. so these two factors give the Dirac equation (and its Hermitian adjoint)

(&gamma;^&mu;&part;_&mu;-m)&psi;=0