What is the significance of commuting operators and CSCO in quantum mechanics?

[fog37]

Hello,

Today I am studying complete set of commuting observables (CSCO) which is a set of commuting operators, pair by pair, whose eigenvalues completely specify the state of a system. For example, given 4 different commuting observables, there is a set of eigenstates which are eigenstates for all the four commuting operators but the eigenvalues are clearly different.

It is said that the state $$|\Psi>$$ is completely specified by those commuting observables. But what about those observables that do not commute? I am sure that, for a particular system, there may be other observables that do not commute.
Isn't the truly complete state $$|\Psi>$$ of the system represented by a unit vector state in the Hilbert vector space that is the tensor product of all the Hilbert vector subspaces with each subspace associated to a different observable? It seems that all observables would be truly needed to give the most complete description...

I am sure there is something to tweak/correct in my understanding since I do not see how a set containing a finite number of commuting observables could give the most complete description of the state...

Thanks!

 

[facenian]

Hello,

Today I am studying complete set of commuting observables (CSCO) which is a set of commuting operators, pair by pair, whose eigenvalues completely specify the state of a system. For example, given 4 different commuting observables, there is a set of eigenstates which are eigenstates for all the four commuting operators but the eigenvalues are clearly different.

It is said that the state $$|\Psi>$$ is completely specified by those commuting observables. But what about those observables that do not commute? I am sure that, for a particular system, there may be other observables that do not commute.

A certain state is completely specified by CSCO simply means that if given the four eigenstates there is only one vector (except for a complex constant) that is a simultaneous eigentate with those given eigenvalues. Of course there may be other observables that do not comute with all those four or neither of them but is not relevant for the concept of CSCO.

 

[fog37]

Thank Facenian.

I think I see your point. There is a set of eigenstates that are simultaneously eigenstates of all four commuting operators. And a certain state $$|\Psi>$$ which is a superposition of a number of those eigenstates is completely specified by that superposition. But that would seem to apply to any situation in which a particular state is expressed, using the basis of a certain and arbitrary operator, as a superposition of its eigenstates. I don't see how the superposition in terms of the common eigenstates of CSCO specifies the state even more...I am missing something.

 

[facenian]

. And a certain state $$|\Psi>$$ which is a superposition of a number of those eigenstates is completely specified by that superposition. But that would seem to apply to any situation in which a particular state is expressed, using the basis of a certain and arbitrary operator, as a superposition of its eigenstates. I don't see how the superposition in terms of the common eigenstates of CSCO specifies the state even more...I am missing something.

Yes, in fact you are misunderstanding the issue. It is not that an arbitrary state is more specified because you expressed it in the basis of the CSCO.
The states that are completely specified by the CSCO are the common eigenstates the this system. This means that once you fix four eigenvalues threre is only one state that is a simultaneous eigenvector with those eigenvalues. For instance if you only have three of the four observables you can fix three eigenvalues but threre may be many eigenvectors with the same three eigenvalues.

 

[vanhees71]

I would put it even a bit differently to make the physics more clear:

If you prepare a system such that a complete set of compatible observables are determined, the system is prepared uniquely in the state, described by the uniquely determined common eigenvector of the (self-adjoint) operators, representing these observables.

This is one of the fundamental postulates of quantum theory.

 

[fog37]

Ok, I think I am honing into this:

For the simple case of just two commuting observables $A$ and $B$, if the system is in a generic state $|\Psi>$ and we measure the observable $A$, the system will change state and drop (we can only know the probability) into a specific eigenstates $\phi_1$ which and give the eigenvalue $\alpha_1$ as the numerical measurement result. If we measure observable $B$ right after, the system will remain in the eigenstate $\phi_1$ with eigenvalue $\beta_1$. So, from the first measurement, we automatically lock and know the value of the eigenvalue $\beta_1$ from the second measurement.

This generalizes to CSCO when there are more than two commuting observables so we get to automatically know the other operators' eigenvalues even after just measuring the first observable in the set...

 

[facenian]

No. You don't get the other eigenvalues by measuring just the first. This is because the there is degeneracy if the CSCO has more than one observable.

 

[fog37]

Ha! Thanks for the patience.

Ok. So the other eigenvalues, for the other commuting observables, after measuring the first observable, pop out also based on probability.

However, say that, when measuring observable A, the system drops into eigenstate $|phi_1$ with eigenvalue $\alpha_1$. If we measure observable $B$, the eigenstate remains in state $|phi_1$. Isn't there a specific and different eigenvalue $\beta_1$ associated to $|phi_1$ from the perspective of $B$? Both operators share the same eigenstate but with different eigenvalues. I understand that term degeneracy means that one eigenstate has multiple associated eigenvalues...

 

[facenian]

Both operators share the same eigenstate but with different eigenvalues. I understand that term degeneracy means that one eigenstate has multiple associated eigenvalues...

It is the other way around. Degeneracy means different eigenvectors with the same eigenvalue.

 

[kith]

If for your observable $A$, there's only a single eigenstate to every eigenvalue, $A$ by itself is a CSCO because then, a single measurement fixes the state (up to normalization).

But if there are multiple eigenstates $|a_i \rangle$ to an eigenvalue $a$**, getting $a$ as an outcome in a measurement isn't enough to determine a unique state. Any linear combination of the $|a_i \rangle$ is an eigenstate to the eigenvalue $a$. So you only narrowed the state down to a certain subspace and need to perform additional measurements until you arrive at a one-dimensional subspace.

**: as facenian has remarked, this is what's called degeneracy

 

[fog37]

Hi kith,
Thank for your help.

You mention that any linear combination of eigenstates $|a_i>$ is an eigenstate to the eigenvalue $a$. If this true only if there is degeneracy? In general, correct me if I am wrong, the superposition of two (or more) eigenstates of a certain operator is a state that is never an eigenstate of the operator itself, correct?

 

[kith]

Superpositions of eigenstates associated with different eigenvalues are not eigenstates, yes.

You can think about it the following way (picture it in the 2D plane): the eigenvectors of a linear transformation are precisely the vectors for which its action is a simple scaling. A vector which can be written as the (weighted) sum of two eigenvectors gets scaled in two different directions. Only when the scaling factor (=eigenvalue) is the same for both directions do you get an overall scaling of the vector.