Complete set of commuting observables

Consider two observables, A and B, represented by the operators \hat{A} and \hat{B}. Then the following statements are equivalent: • A and B are compatible observables. • \hat{A} and \hat{B} have a common eigenbasis. • The operators \hat{A} and \hat{B} commute, meaning that [\hat{A}, \hat{B}] = \hat{A}\hat{B}-\hat{B}\hat{A}=0. Proofs {{math proof|title = Proof that a common eigenbasis implies commutation | proof = Let \{|\psi_n \rangle\} be a set of orthonormal states (i.e., \langle \psi_m | \psi_n \rangle = \delta^{\,}_{m,n}) that form a complete eigenbasis for each of the two compatible observables A and B represented by the self-adjoint operators \hat{A} and \hat{B} with corresponding (real-valued) eigenvalues \{a_n\} and \{b_n\}, respectively. This implies that :\hat{A}\hat{B}|\psi_n\rangle=\hat{A}b_n|\psi_n\rangle =a_nb_n|\psi_n\rangle =b_na_n|\psi_n\rangle =\hat{B}\hat{A}|\psi_n\rangle , for each mutual eigenstate |\psi_n \rangle. Because the eigenbasis is complete, we can expand an arbitrary state |\Psi\rangle according to :|\Psi\rangle=\sum_n c_n|\psi_n\rangle, where c_n = \langle \psi_n | \Psi \rangle. The above results imply that :(\hat{A}\hat{B}-\hat{B}\hat{A})|\Psi\rangle=\sum_{n}c_n(\hat{A}\hat{B}-\hat{B}\hat{A})|\psi_n\rangle=0, for any state |\Psi\rangle. Thus, \hat{A}\hat{B}-\hat{B}\hat{A}=[\hat{A},\hat{B}]=0, meaning that the two operators commute. }} {{math proof| title = Proof that commuting observables possess a complete set of common eigenfunctions | proof = ---- '''When A has non-degenerate eigenvalues:''' ---- Let \{|\psi_n\rangle\} be a complete set of orthonormal eigenkets of the self-adjoint operator A corresponding to the set of real-valued eigenvalues \{a_n\}. If the self-adjoint operators A and B commute, we can write :A(B|\psi_n\rangle)=BA|\psi_n\rangle=a_n(B|\psi_n\rangle) So, if B|\psi_n\rangle \neq 0, we can say that B|\psi_n\rangle is an eigenket of A corresponding to the eigenvalue a_n. Since both B|\psi_n\rangle and |\psi_n\rangle are eigenkets associated with the same non-degenerate eigenvalue a_n, they can differ at most by a multiplicative constant. We call this constant b_n. So, :B|\psi_n\rangle=b_n|\psi_n\rangle , which means |\psi_n\rangle is an eigenket of B, and thus of A and B simultaneously. In the case of B|\psi_n\rangle = 0, the non-zero vector |\psi_n\rangle is an eigenket of B with the eigenvalue b_n=0. ---- '''When A has degenerate eigenvalues:''' ---- Suppose each a_n is g -fold degenerate. Let the corresponding orthonormal eigenkets be |\psi_{nr}\rangle, r \in \{1,2,\dots,g\}. Since [A,B]=0, we reason as above to find that B|\psi_{nr}\rangle is an eigenket of A corresponding to the degenerate eigenvalue a_n. So, we can expand B|\psi_{nr}\rangle in the basis of the degenerate eigenkets of a_n: :B|\psi_{nr}\rangle=\sum_{s=1}^g c_{rs}|\psi_{ns}\rangle The c_{rs} are the expansion coefficients. The coefficients c_{rs} form a self-adjoint matrix, since \langle \psi_{ns} |B|\psi_{nr}\rangle = c_{rs}. Next step would be to diagonalize the matrix c_{rs}. To do so, we sum over all r with g constants d_r. So, :B\sum_{r=1}^gd_r|\psi_{nr}\rangle=\sum_{r=1}^g\sum_{s=1}^g d_rc_{rs}|\psi_{ns}\rangle So, \sum_{r=1}^gd_r|\psi_{nr}\rangle will be an eigenket of B with the eigenvalue b_n if we have :\sum_{r=1}^g d_rc_{rs}=b_nd_s, s=1,2,...g This constitutes a system of g linear equations for the constants d_r. A non-trivial solution exists if :\det[c_{rs}-b_n\delta_{rs}]=0 This is an equation of order g in b_n, and has g roots. For each root b_n=b_n^{(k)}, k=1,2,...g we have a non-trivial solution d_r, say, d_r^{(k)}. Due to the self-adjoint of c_{rs}, all solutions are linearly independent. Therefore they form the new basis :|\phi_n^{(k)}\rangle=\sum_{r=1}^g d_r^{(k)}|\psi_{nr}\rangle |\phi_n^{(k)}\rangle is simultaneously an eigenket of A and B with eigenvalues a_n and b_n^{(k)} respectively. }} Discussion We consider the two above observables A and B. Suppose there exists a complete set of kets \{|\psi_n\rangle\} whose every element is simultaneously an eigenket of A and B. Then we say that A and B are compatible. If we denote the eigenvalues of A and B corresponding to |\psi_n\rangle respectively by a_n and b_n, we can write :A|\psi_n\rangle=a_n|\psi_n\rangle :B|\psi_n\rangle=b_n|\psi_n\rangle If the system happens to be in one of the eigenstates, say, |\psi_n\rangle, then both A and B can be simultaneously measured to any arbitrary level of precision, and we will get the results a_n and b_n respectively. This idea can be extended to more than two observables. Examples of compatible observables The Cartesian components of the position operator \mathbf{r} are x, y and z. These components are all compatible. Similarly, the Cartesian components of the momentum operator \mathbf{p}, that is p_x, p_y and p_z are also compatible. ==Formal definition==

A set of observables A, B, C... is called a CSCO if: • All the observables commute in pairs. • 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. ==Discussion==

Let us have an operator \hat{A} of an observable A, which has all non-degenerate eigenvalues \{a_n\}. 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 \hat{A} corresponding to the eigenvalue a_n can be labelled as |a_n\rangle. Such an observable is itself a self-sufficient CSCO. However, if some of the eigenvalues of a_n 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 B), which is compatible with A. The compatibility theorem tells us that a common basis of eigenfunctions of \hat{A} and \hat{B} can be found. Now if each pair of the eigenvalues (a_n, b_n) uniquely specifies a state vector of this basis, we claim to have formed a CSCO: the set \{A, B\}. The degeneracy in \hat{A} is completely removed. It may so happen, nonetheless, that the degeneracy is not completely lifted. That is, there exists at least one pair (a_n, b_n) which does not uniquely identify one eigenvector. In this case, we repeat the above process by adding another observable C, which is compatible with both A and B. If the basis of common eigenfunctions of \hat{A}, \hat{B} and \hat{C} is unique, that is, uniquely specified by the set of eigenvalues (a_n, b_n, c_n), then we have formed a CSCO: \{A, B, C\}. 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 \{A, B, C,...,\}. Then we can expand any general state in the Hilbert space as :|\psi\rangle = \sum_{i,j,k,...} \mathcal{C}_{i,j,k,...} |a_i, b_j, c_k,...\rangle where |a_i, b_j, c_k,...\rangle are the eigenkets of the operators \hat{A}, \hat{B}, \hat{C}, and form a basis space. That is, :\hat{A}|a_i, b_j, c_k,...\rangle = a_i|a_i, b_j, c_k,...\rangle, etc If we measure A, B, C,... in the state |\psi\rangle then the probability that we simultaneously measure a_i, b_j, c_k,... is given by |\mathcal{C}_{i,j,k,...}|^2. For a complete set of commuting operators, we can find a unitary transformation which will simultaneously diagonalize all of them. ==Examples==

The hydrogen atom without electron or proton spin Two components of the angular momentum operator \mathbf{L} do not commute, but satisfy the commutation relations: :[L_i, L_j]=i\hbar\epsilon _{ijk} L_k So, any CSCO cannot involve more than one component of \mathbf{L}. It can be shown that the square of the angular momentum operator, L^2, commutes with \mathbf{L}. :[L_x, L^2] = 0, [L_y, L^2] = 0, [L_z, L^2] = 0 Also, the Hamiltonian \hat{H} = -\frac{\hbar^2}{2\mu}\nabla^2 - \frac{Ze^2}{r} is a function of r only and has rotational invariance, where \mu is the reduced mass of the system. Since the components of \mathbf{L} are generators of rotation, it can be shown that :[\mathbf{L}, H] = 0, [L^2, H] = 0 Therefore, a commuting set consists of L^2, one component of \mathbf{L} (which is taken to be L_z) and H. The solution of the problem tells us that disregarding spin of the electrons, the set \{H, L^2, L_z\} forms a CSCO. Let |E_n, l, m\rangle be any basis state in the Hilbert space of the hydrogenic atom. Then :H|E_n, l, m\rangle=E_n|E_n, l, m\rangle : :L^2|E_n, l, m\rangle=l(l+1)\hbar^2|E_n, l, m\rangle : :L_z|E_n, l, m\rangle=m\hbar|E_n, l, m\rangle That is, the set of eigenvalues \{E_n, l, m\} or more simply, \{n, l, m\} completely specifies a unique eigenstate of the Hydrogenic atom. The free particle For a free particle, the Hamiltonian H = -\frac{\hbar^2}{2m} \nabla^2 is invariant under translations. Translation commutes with the Hamiltonian: [H,\mathbf{\hat T}]=0. However, if we express the Hamiltonian in the basis of the translation operator, we will find that H has doubly degenerate eigenvalues. It can be shown that to make the CSCO in this case, we need another operator called the parity operator \Pi, such that [H,\Pi]=0.\{H,\Pi\} forms a CSCO. Again, let |k\rangle and |-k\rangle be the degenerate eigenstates of Hcorresponding the eigenvalue H_k=\frac{{\hbar^2}{k^2}}{2m}, i.e. :H|k\rangle=\frac{{\hbar^2}{k^2}}{2m}|k\rangle :H|-k\rangle=\frac{{\hbar^2}{k^2}}{2m}|-k\rangle The degeneracy in H is removed by the momentum operator \mathbf{\hat p}. :\mathbf{\hat p}|k\rangle=k|k\rangle :\mathbf{\hat p}|-k\rangle=-k|-k\rangle So, \{\mathbf{\hat p}, H\} forms a CSCO. Addition of angular momenta We consider the case of two systems, 1 and 2, with respective angular momentum operators \mathbf{J_1} and \mathbf{J_2}. We can write the eigenstates of J_1^2 and J_{1z} as |j_1m_1\rangle and of J_2^2 and J_{2z} as |j_2m_2\rangle. :J_1^2|j_1m_1\rangle=j_1(j_1+1)\hbar^2|j_1m_1\rangle :J_{1z}|j_1m_1\rangle=m_1\hbar|j_1m_1\rangle :J_2^2|j_2m_2\rangle=j_2(j_2+1)\hbar^2|j_2m_2\rangle :J_{2z}|j_2m_2\rangle=m_2\hbar|j_2m_2\rangle Then the basis states of the complete system are |j_1m_1;j_2m_2\rangle given by :|j_1m_1;j_2m_2\rangle=|j_1m_1\rangle \otimes |j_2m_2\rangle Therefore, for the complete system, the set of eigenvalues \{j_1,m_1,j_2,m_2\} completely specifies a unique basis state, and \{J_1^2, J_{1z}, J_2^2, J_{2z}\} forms a CSCO. Equivalently, there exists another set of basis states for the system, in terms of the total angular momentum operator \mathbf{J} = \mathbf{J_1}+\mathbf{J_2}. The eigenvalues of J^2 are j(j+1)\hbar^2 where j takes on the values j_1+j_2, j_1+j_2-1,...,|j_1-j_2|, and those of J_z are m where m=-j, -j+1,...j-1,j. The basis states of the operators J^2 and J_z are |j_1j_2;jm\rangle. Thus we may also specify a unique basis state in the Hilbert space of the complete system by the set of eigenvalues \{j_1,j_2,j,m\}, and the corresponding CSCO is \{J_1^2, J_2^2,J^2,J_z\}. ==See also==