Complete set of eigenfunctions

[Nikitin]

Hi! So let's say we measured the angular momentum squared of a particle, and got the result $2 \hbar^2$, so $l=1$. Now we have the choice of obtaining a sharp value of either $L_z, L_y$ or $L_x$. Okay, fair enough. But I have two questions:

1) The degeneration degree is $3$ because the eigenvalues squared of $L_z$ must be smaller than or equal to $l^2$. So $m \in [-1,0,1]$.

However, if you were to use $L_x$ or $L_y$ to find the degeneration degree, you'd get the same result because what is defined as $L_x$ and $L_y$ is only dependant on the coordinate system, and the physics doesn't care about your coordinate systems. Is this correct reasoning?

2) The complete set of eigenfunctions for $l=1$ can be found by adding together all the functions that are simultaneously eigenfunctions for both $L_z$ and $L$. So The complete set for $l=1$ are: $Y_{10}, Y_{1-1},Y_{11}$, where $Y_{lm}$ is an eigenfunction to both $L_z$ and $L$.

However, why is 2) true? I mean this means you can write eigenfunctions of both $L_y$ and $L_x$ as linear combinations of those $Y$s. Can somebody post the relevant theorem explaining this? Thanks.

PS: I got my exam tomorrow so pls help

 

[Nikitin]

Also can somebody define "complete set of eigenfunctions"? I'm pretty sure it means something like "all the possible states a system can be in, given it has a specific eigenvalue", but I'm not sure if I'm 100% correct.

 

[bhobba]

Also can somebody define "complete set of eigenfunctions"? I'm pretty sure it means something like "all the possible states a system can be in, given it has a specific eigenvalue", but I'm not sure if I'm 100% correct.

The answer I will give, while correct, will probably leave you saying so. It's simply the basis of the vector space of the system you are dealing with. To find it you find the maximal number of observables that are actually measuring different things, called a complete set commuting observables, and the 'tensor' product of their eigenfunctions is such a basis.

It probably doesn't resolve it for you, but Google complete set of commuting observables and things will likely be clearer.

Thanks
Bill

 

[Nikitin]

But a "complete set of eigenfunctions for an eigenvalue" is different from a "complete set of commutators". Isn't it?

The first is a linear combo of all the degenerate states of an eigenvalue, while the second tells us which measurements may be done on a system.

PS: Sorry if I am unclear. My brain has been beat to a pulp.

 

[msumm21]

I THINK a complete set of eigenfunctions for an eigenvalue is any set of functions which spans the eigenspace -- i.e. all eigenvectors with that eigenvalue can be written as a linear combination of the elements in a complete set. This is different from a basis because a basis must also have the fewest number of elements possible, while a complete set can have extra, redundant, eigenvectors.

I am not 100% sure I remember this right, so you may want to sanity check this with how the phrase was used and maybe someone else will follow up to confirm / deny here.

 

[bhobba]

PS: Sorry if I am unclear. My brain has been beat to a pulp.

This stuff makes my brain hurt to.

But best if I just give you a bit of a pointer to what's going on rather than post the full excruciating detail - and less time consuming on my part as well - its also standard textbook stuff.

Start here and follow your nose:
http://www.glue.umd.edu/afs/glue.umd.edu/department/phys/courses/Phys622/public_html/ji/lecture5.pdf

Thanks
Bill

 

[strangerep]

I got my exam tomorrow so pls help

Geez, you're leaving it a bit late. This is nontrivial stuff.

1) The degeneration degree is $3$ because the eigenvalues squared of $L_z$ must be smaller than or equal to $l^2$. So $m \in [-1,0,1]$.

However, if you were to use $L_x$ or $L_y$ to find the degeneration degree, you'd get the same result because what is defined as $L_x$ and $L_y$ is only dependant on the coordinate system, and the physics doesn't care about your coordinate systems. Is this correct reasoning?

Essentially yes, but here's a little more detail...

What's really going on here is that we're trying to find all "unitary irreducible representations" of the rotation group. In more mundane terms, this means all Hilbert spaces on which elements of the rotation group can be represented as operators on the Hilbert space, such that the Hilbert space's inner product is preserved. (Actually, there's a bit more to it than that since one must consider phase, but I'll skip that for now.)

To find all possible such "unitary irreducible representations", one first looks for a maximal set of mutually commuting elements in the rotation algebra. There is the Casimir operator $J^2$ which commutes with everything, and then we can pick another generator expression arbitrarily, usually just $J_z$. Assuming that both of these, and the other generators $J_x$ and $J_y$ are represented as self-adjoint (Hermitian) operators on some Hilbert space, one can derive the possible eigenvalues of $J^2$ and the associated set of eigenvalues of $J_z$ for each $J^2$ eigenvalue. In performing this derivation, the other generators are used in the forms $J_\pm := J_x \pm iJ_y$ -- called "ladder" operators. We find that the action of $J_\pm$ moves us up and down the list of eigenvalues of $J_z$. The structure of the full eigenvalue set essentially gives the structure of the Hilbert space -- via the associated eigenvectors.

You should definitely study Ballentine section 7.1 in detail to see how this works. Actually you should have studied it a while ago... <insert finger--waggling icon here>

There is an important theorem -- the "spectral theorem" which says that the eigenvectors of any self-adjoint operator span the Hilbert space: any state vector in the Hilbert space can be expressed as a linear combination of those eigenvectors. This covers the case of fixed $l=1$, since it specifies a Hilbert subspace of the larger Hilbert space, hence the spectral theorem applies there too.

More generally, there are extensions of the theorem that cover sets of mutually-commuting self-adjoint operator. That's what we work with in the general theory of quantum angular momentum.

HTH.