Complete set of commuting observables

In quantum mechanics, a complete set of commuting observables (CSCO) is a set of commuting operators whose common eigenvectors can be used as a basis to express any quantum state. In the case of operators with discrete spectra, a CSCO is a set of commuting observables whose simultaneous eigenspaces span the Hilbert space, so that the eigenvectors are uniquely specified by the corresponding sets of eigenvalues.

In some simple cases, like bound state problems in one dimension, the energy spectrum is nondegenerate, and energy can be used to uniquely label the eigenstates. In more complicated problems, the energy spectrum is degenerate, and additional observables are needed to distinguish between the eigenstates.[1]

Since each pair of observables in the set commutes, the observables are all compatible so that the measurement of one observable has no effect on the result of measuring another observable in the set. It is therefore not necessary to specify the order in which the different observables are measured. Measurement of the complete set of observables constitutes a complete measurement, in the sense that it projects the quantum state of the system onto a unique and known vector in the basis defined by the set of operators. That is, to prepare the completely specified state, we have to take any state arbitrarily, and then perform a succession of measurements corresponding to all the observables in the set, until it becomes a uniquely specified vector in the Hilbert space (up to a phase).

The compatibility theorem


Consider two observables,   and  , represented by the operators   and  . Then the following statements are equivalent:

  1.   and   are compatible observables.
  2.   and   have a common eigenbasis.
  3. The operators   and   commute, meaning that  .


Proof that a common eigenbasis implies commutation

Let   be a set of orthonormal states (i.e.,  ) that form a complete eigenbasis for each of the two compatible observables   and   represented by the self-adjoint operators   and   with corresponding (real-valued) eigenvalues   and  , respectively. This implies that


for each mutual eigenstate  . Because the eigenbasis is complete, we can expand an arbitrary state   according to


where  . The above results imply that


for any state  . Thus,  , meaning that the two operators commute.

Proof that commuting observables possess a complete set of common eigenfunctions

When   has non-degenerate eigenvalues:

Let   be a complete set of orthonormal eigenkets of the self-adjoint operator   corresponding to the set of real-valued eigenvalues  . If the self-adjoint operators   and   commute, we can write


So, if  , we can say that   is an eigenket of   corresponding to the eigenvalue  . Since both   and   are eigenkets associated with the same non-degenerate eigenvalue  , they can differ at most by a multiplicative constant. We call this constant  . So,


which means   is an eigenket of  , and thus of   and   simultaneously. In the case of  , the non-zero vector   is an eigenket of   with the eigenvalue  .

When   has degenerate eigenvalues:

Suppose each   is   -fold degenerate. Let the corresponding orthonormal eigenkets be  . Since  , we reason as above to find that   is an eigenket of   corresponding to the degenerate eigenvalue  . So, we can expand   in the basis of the degenerate eigenkets of  :


The   are the expansion coefficients. The coefficients   form a self-adjoint matrix, since  . Next step would be to diagonalize the matrix  . To do so, we sum over all   with   constants  . So,


So,   will be an eigenket of   with the eigenvalue   if we have


This constitutes a system of   linear equations for the constants  . A non-trivial solution exists if


This is an equation of order   in  , and has   roots. For each root   we have a non-trivial solution  , say,  . Due to the self-adjoint of  , all solutions are linearly independent. Therefore they form the new basis


  is simultaneously an eigenket of   and   with eigenvalues   and   respectively.



We consider the two above observables   and  . Suppose there exists a complete set of kets   whose every element is simultaneously an eigenket of   and  . Then we say that   and   are compatible. If we denote the eigenvalues of   and   corresponding to   respectively by   and  , we can write


If the system happens to be in one of the eigenstates, say,  , then both   and   can be simultaneously measured to any arbitrary level of precision, and we will get the results   and   respectively. This idea can be extended to more than two observables.

Examples of compatible observables


The Cartesian components of the position operator   are  ,   and  . These components are all compatible. Similarly, the Cartesian components of the momentum operator  , that is  ,   and   are also compatible.

Formal definition


A set of observables   is called a CSCO if:[2]

  1. All the observables commute in pairs.
  2. If we specify the eigenvalues of all the operators in the CSCO, we identify a unique eigenvector (up to a phase) in the Hilbert space of the system.

If we are given a CSCO, we can choose a basis for the space of states made of common eigenvectors of the corresponding operators. We can uniquely identify each eigenvector (up to a phase) by the set of eigenvalues it corresponds to.



Let us have an operator   of an observable  , which has all non-degenerate eigenvalues  . As a result, there is one unique eigenstate corresponding to each eigenvalue, allowing us to label these by their respective eigenvalues. For example, the eigenstate of   corresponding to the eigenvalue   can be labelled as  . Such an observable is itself a self-sufficient CSCO.

However, if some of the eigenvalues of   are degenerate (such as having degenerate energy levels), then the above result no longer holds. In such a case, we need to distinguish between the eigenfunctions corresponding to the same eigenvalue. To do this, a second observable is introduced (let us call that  ), which is compatible with  . The compatibility theorem tells us that a common basis of eigenfunctions of   and   can be found. Now if each pair of the eigenvalues   uniquely specifies a state vector of this basis, we claim to have formed a CSCO: the set  . The degeneracy in   is completely removed.

It may so happen, nonetheless, that the degeneracy is not completely lifted. That is, there exists at least one pair   which does not uniquely identify one eigenvector. In this case, we repeat the above process by adding another observable  , which is compatible with both   and  . If the basis of common eigenfunctions of  ,   and   is unique, that is, uniquely specified by the set of eigenvalues  , then we have formed a CSCO:  . If not, we add one more compatible observable and continue the process till a CSCO is obtained.

The same vector space may have distinct complete sets of commuting operators.

Suppose we are given a finite CSCO  . Then we can expand any general state in the Hilbert space as


where   are the eigenkets of the operators  , and form a basis space. That is,

 , etc

If we measure   in the state   then the probability that we simultaneously measure   is given by  .

For a complete set of commuting operators, we can find a unitary transformation which will simultaneously diagonalize all of them.



The hydrogen atom without electron or proton spin


Two components of the angular momentum operator   do not commute, but satisfy the commutation relations:


So, any CSCO cannot involve more than one component of  . It can be shown that the square of the angular momentum operator,  , commutes with  .


Also, the Hamiltonian   is a function of   only and has rotational invariance, where   is the reduced mass of the system. Since the components of   are generators of rotation, it can be shown that


Therefore, a commuting set consists of  , one component of   (which is taken to be  ) and  . The solution of the problem tells us that disregarding spin of the electrons, the set   forms a CSCO. Let   be any basis state in the Hilbert space of the hydrogenic atom. Then


That is, the set of eigenvalues   or more simply,   completely specifies a unique eigenstate of the Hydrogenic atom.

The free particle


For a free particle, the Hamiltonian   is invariant under translations. Translation commutes with the Hamiltonian:  . However, if we express the Hamiltonian in the basis of the translation operator, we will find that   has doubly degenerate eigenvalues. It can be shown that to make the CSCO in this case, we need another operator called the parity operator  , such that  .  forms a CSCO.

Again, let   and   be the degenerate eigenstates of  corresponding the eigenvalue  , i.e.


The degeneracy in   is removed by the momentum operator  .


So,   forms a CSCO.

Addition of angular momenta


We consider the case of two systems, 1 and 2, with respective angular momentum operators   and  . We can write the eigenstates of   and   as   and of   and   as  .


Then the basis states of the complete system are   given by


Therefore, for the complete system, the set of eigenvalues   completely specifies a unique basis state, and   forms a CSCO. Equivalently, there exists another set of basis states for the system, in terms of the total angular momentum operator  . The eigenvalues of   are   where   takes on the values  , and those of   are   where  . The basis states of the operators   and   are  . Thus we may also specify a unique basis state in the Hilbert space of the complete system by the set of eigenvalues  , and the corresponding CSCO is  .

See also



  1. ^ Zwiebach, Barton (2022). "Chapter 15.8: Complete Set of Commuting Observables". Mastering quantum mechanics: essentials, theory, and applications. Cambridge, Mass: The MIT press. ISBN 978-0262366892.
  2. ^ Cohen-Tannoudji, Claude; Diu, Bernard; Laloë, Franck (1977). Quantum mechanics. Vol. 1. New York: Wiley. pp. 143–144. ISBN 978-0-471-16433-3. OCLC 2089460.

Further reading