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Complete set of commuting observables
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<h2 id="the-compatibility-theorem">The compatibility theorem</h2> <p>Consider two observables, A {\displaystyle A} and B {\displaystyle B} , represented by the operators A ^ {\displaystyle {\hat {A}}} and B ^ {\displaystyle {\hat {B}}} . Then the following statements are equivalent: </p> <ol><li> A {\displaystyle A} and B {\displaystyle B} are compatible observables.</li> <li> A ^ {\displaystyle {\hat {A}}} and B ^ {\displaystyle {\hat {B}}} have a common eigenbasis.</li> <li>The operators A ^ {\displaystyle {\hat {A}}} and B ^ {\displaystyle {\hat {B}}} <a href="/facts/Commutator/eYxzFmFY">commute</a>, meaning that [ A ^ , B ^ ] = A ^ B ^ − B ^ A ^ = 0 {\displaystyle [{\hat {A}},{\hat {B}}]={\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}=0} .</li></ol> <h3>Proofs</h3> Proof that a common eigenbasis implies commutation <p>Let { | ψ n ⟩ } {\displaystyle \{|\psi _{n}\rangle \}} be a set of orthonormal states (i.e., ⟨ ψ m | ψ n ⟩ = δ m , n {\displaystyle \langle \psi _{m}|\psi _{n}\rangle =\delta _{m,n}^{\,}} ) that form a complete eigenbasis for each of the two compatible observables A {\displaystyle A} and B {\displaystyle B} represented by the self-adjoint operators A ^ {\displaystyle {\hat {A}}} and B ^ {\displaystyle {\hat {B}}} with corresponding (real-valued) eigenvalues { a n } {\displaystyle \{a_{n}\}} and { b n } {\displaystyle \{b_{n}\}} , respectively. This implies that </p> A ^ B ^ | ψ n ⟩ = A ^ b n | ψ n ⟩ = a n b n | ψ n ⟩ = b n a n | ψ n ⟩ = B ^ A ^ | ψ n ⟩ , {\displaystyle {\hat {A}}{\hat {B}}|\psi _{n}\rangle ={\hat {A}}b_{n}|\psi _{n}\rangle =a_{n}b_{n}|\psi _{n}\rangle =b_{n}a_{n}|\psi _{n}\rangle ={\hat {B}}{\hat {A}}|\psi _{n}\rangle ,} <p>for each mutual eigenstate | ψ n ⟩ {\displaystyle |\psi _{n}\rangle } . Because the eigenbasis is complete, we can expand an arbitrary state | Ψ ⟩ {\displaystyle |\Psi \rangle } according to </p> | Ψ ⟩ = ∑ n c n | ψ n ⟩ , {\displaystyle |\Psi \rangle =\sum _{n}c_{n}|\psi _{n}\rangle ,} <p>where c n = ⟨ ψ n | Ψ ⟩ {\displaystyle c_{n}=\langle \psi _{n}|\Psi \rangle } . The above results imply that </p> ( A ^ B ^ − B ^ A ^ ) | Ψ ⟩ = ∑ n c n ( A ^ B ^ − B ^ A ^ ) | ψ n ⟩ = 0 , {\displaystyle ({\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,} <p>for <i>any</i> state | Ψ ⟩ {\displaystyle |\Psi \rangle } . Thus, A ^ B ^ − B ^ A ^ = [ A ^ , B ^ ] = 0 {\displaystyle {\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}}=[{\hat {A}},{\hat {B}}]=0} , meaning that the two operators commute. </p> Proof that commuting observables possess a complete set of common eigenfunctions <p><i>When A {\displaystyle A} has </i>non-degenerate<i> eigenvalues:</i> </p> <p>Let { | ψ n ⟩ } {\displaystyle \{|\psi _{n}\rangle \}} be a complete set of orthonormal eigenkets of the self-adjoint operator A {\displaystyle A} corresponding to the set of real-valued eigenvalues { a n } {\displaystyle \{a_{n}\}} . If the self-adjoint operators A {\displaystyle A} and B {\displaystyle B} commute, we can write </p> A ( B | ψ n ⟩ ) = B A | ψ n ⟩ = a n ( B | ψ n ⟩ ) {\displaystyle A(B|\psi _{n}\rangle )=BA|\psi _{n}\rangle =a_{n}(B|\psi _{n}\rangle )} <p>So, if B | ψ n ⟩ ≠ 0 {\displaystyle B|\psi _{n}\rangle \neq 0} , we can say that B | ψ n ⟩ {\displaystyle B|\psi _{n}\rangle } is an eigenket of A {\displaystyle A} corresponding to the eigenvalue a n {\displaystyle a_{n}} . Since both B | ψ n ⟩ {\displaystyle B|\psi _{n}\rangle } and | ψ n ⟩ {\displaystyle |\psi _{n}\rangle } are eigenkets associated with the same non-degenerate eigenvalue a n {\displaystyle a_{n}} , they can differ at most by a multiplicative constant. We call this constant b n {\displaystyle b_{n}} . So, </p> B | ψ n ⟩ = b n | ψ n ⟩ {\displaystyle B|\psi _{n}\rangle =b_{n}|\psi _{n}\rangle } , <p>which means | ψ n ⟩ {\displaystyle |\psi _{n}\rangle } is an eigenket of B {\displaystyle B} , and thus of A {\displaystyle A} and B {\displaystyle B} <i>simultaneously</i>. In the case of B | ψ n ⟩ = 0 {\displaystyle B|\psi _{n}\rangle =0} , the non-zero vector | ψ n ⟩ {\displaystyle |\psi _{n}\rangle } is an eigenket of B {\displaystyle B} with the eigenvalue b n = 0 {\displaystyle b_{n}=0} . </p> <p><i>When A {\displaystyle A} has </i>degenerate<i> eigenvalues:</i> </p> <p>Suppose each a n {\displaystyle a_{n}} is g {\displaystyle g} -fold degenerate. Let the corresponding orthonormal eigenkets be | ψ n r ⟩ , r ∈ { 1 , 2 , … , g } {\displaystyle |\psi _{nr}\rangle ,r\in \{1,2,\dots ,g\}} . Since [ A , B ] = 0 {\displaystyle [A,B]=0} , we reason as above to find that B | ψ n r ⟩ {\displaystyle B|\psi _{nr}\rangle } is an eigenket of A {\displaystyle A} corresponding to the <i>degenerate</i> eigenvalue a n {\displaystyle a_{n}} . So, we can expand B | ψ n r ⟩ {\displaystyle B|\psi _{nr}\rangle } in the basis of the degenerate eigenkets of a n {\displaystyle a_{n}} : </p> B | ψ n r ⟩ = ∑ s = 1 g c r s | ψ n s ⟩ {\displaystyle B|\psi _{nr}\rangle =\sum _{s=1}^{g}c_{rs}|\psi _{ns}\rangle } <p>The c r s {\displaystyle c_{rs}} are the expansion coefficients. The coefficients c r s {\displaystyle c_{rs}} form a self-adjoint matrix, since ⟨ ψ n s | B | ψ n r ⟩ = c r s {\displaystyle \langle \psi _{ns}|B|\psi _{nr}\rangle =c_{rs}} . Next step would be to diagonalize the matrix c r s {\displaystyle c_{rs}} . To do so, we sum over all r {\displaystyle r} with g {\displaystyle g} constants d r {\displaystyle d_{r}} . So, </p> B ∑ r = 1 g d r | ψ n r ⟩ = ∑ r = 1 g ∑ s = 1 g d r c r s | ψ n s ⟩ {\displaystyle B\sum _{r=1}^{g}d_{r}|\psi _{nr}\rangle =\sum _{r=1}^{g}\sum _{s=1}^{g}d_{r}c_{rs}|\psi _{ns}\rangle } <p>So, ∑ r = 1 g d r | ψ n r ⟩ {\displaystyle \sum _{r=1}^{g}d_{r}|\psi _{nr}\rangle } will be an eigenket of B {\displaystyle B} with the eigenvalue b n {\displaystyle b_{n}} if we have </p> ∑ r = 1 g d r c r s = b n d s , s = 1 , 2 , . . . g {\displaystyle \sum _{r=1}^{g}d_{r}c_{rs}=b_{n}d_{s},s=1,2,...g} <p>This constitutes a system of g {\displaystyle g} linear equations for the constants d r {\displaystyle d_{r}} . A non-trivial solution exists if </p> det [ c r s − b n δ r s ] = 0 {\displaystyle \det[c_{rs}-b_{n}\delta _{rs}]=0} <p>This is an equation of order g {\displaystyle g} in b n {\displaystyle b_{n}} , and has g {\displaystyle g} roots. For each root b n = b n ( k ) , k = 1 , 2 , . . . g {\displaystyle b_{n}=b_{n}^{(k)},k=1,2,...g} we have a non-trivial solution d r {\displaystyle d_{r}} , say, d r ( k ) {\displaystyle d_{r}^{(k)}} . Due to the self-adjoint of c r s {\displaystyle c_{rs}} , all solutions are linearly independent. Therefore they form the new basis </p> | ϕ n ( k ) ⟩ = ∑ r = 1 g d r ( k ) | ψ n r ⟩ {\displaystyle |\phi _{n}^{(k)}\rangle =\sum _{r=1}^{g}d_{r}^{(k)}|\psi _{nr}\rangle } <p> | ϕ n ( k ) ⟩ {\displaystyle |\phi _{n}^{(k)}\rangle } is simultaneously an eigenket of A {\displaystyle A} and B {\displaystyle B} with eigenvalues a n {\displaystyle a_{n}} and b n ( k ) {\displaystyle b_{n}^{(k)}} respectively. </p> <h3>Discussion</h3> <p>We consider the two above observables A {\displaystyle A} and B {\displaystyle B} . Suppose there exists a complete set of kets { | ψ n ⟩ } {\displaystyle \{|\psi _{n}\rangle \}} whose every element is simultaneously an eigenket of A {\displaystyle A} and B {\displaystyle B} . Then we say that A {\displaystyle A} and B {\displaystyle B} are <i>compatible</i>. If we denote the eigenvalues of A {\displaystyle A} and B {\displaystyle B} corresponding to | ψ n ⟩ {\displaystyle |\psi _{n}\rangle } respectively by a n {\displaystyle a_{n}} and b n {\displaystyle b_{n}} , we can write </p> A | ψ n ⟩ = a n | ψ n ⟩ {\displaystyle A|\psi _{n}\rangle =a_{n}|\psi _{n}\rangle } B | ψ n ⟩ = b n | ψ n ⟩ {\displaystyle B|\psi _{n}\rangle =b_{n}|\psi _{n}\rangle } <p>If the system happens to be in one of the eigenstates, say, | ψ n ⟩ {\displaystyle |\psi _{n}\rangle } , then both A {\displaystyle A} and B {\displaystyle B} can be <i>simultaneously</i> measured to any arbitrary level of precision, and we will get the results a n {\displaystyle a_{n}} and b n {\displaystyle b_{n}} respectively. This idea can be extended to more than two observables. </p> <h3>Examples of compatible observables</h3> <p>The Cartesian components of the position operator r {\displaystyle \mathbf {r} } are x {\displaystyle x} , y {\displaystyle y} and z {\displaystyle z} . These components are all compatible. Similarly, the Cartesian components of the momentum operator p {\displaystyle \mathbf {p} } , that is p x {\displaystyle p_{x}} , p y {\displaystyle p_{y}} and p z {\displaystyle p_{z}} are also compatible. </p> <h2 id="formal-definition">Formal definition</h2> <p>A set of observables A , B , C . . . {\displaystyle A,B,C...} is called a CSCO if:<a class="footnote-ref" id="fnref:2" href="#fn:2"><sup>2</sup></a> </p> <ol><li>All the observables commute in pairs.</li> <li>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.</li></ol> <p>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. </p> <h2 id="discussion">Discussion</h2> <p>Let us have an operator A ^ {\displaystyle {\hat {A}}} of an observable A {\displaystyle A} , which has all <i>non-degenerate</i> eigenvalues { a n } {\displaystyle \{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 A ^ {\displaystyle {\hat {A}}} corresponding to the eigenvalue a n {\displaystyle a_{n}} can be labelled as | a n ⟩ {\displaystyle |a_{n}\rangle } . Such an observable is itself a self-sufficient CSCO. </p><p>However, if some of the eigenvalues of a n {\displaystyle a_{n}} are <i>degenerate</i> (such as having <a href="/facts/Degenerate_energy_levels/ulrnMXTR">degenerate energy levels</a>), 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 {\displaystyle B} ), which is compatible with A {\displaystyle A} . The compatibility theorem tells us that a common basis of eigenfunctions of A ^ {\displaystyle {\hat {A}}} and B ^ {\displaystyle {\hat {B}}} can be found. Now if each pair of the eigenvalues ( a n , b n ) {\displaystyle (a_{n},b_{n})} uniquely specifies a state vector of this basis, we claim to have formed a CSCO: the set { A , B } {\displaystyle \{A,B\}} . The degeneracy in A ^ {\displaystyle {\hat {A}}} is completely removed. </p><p>It may so happen, nonetheless, that the degeneracy is not completely lifted. That is, there exists at least one pair ( a n , b n ) {\displaystyle (a_{n},b_{n})} which does not uniquely identify one eigenvector. In this case, we repeat the above process by adding another observable C {\displaystyle C} , which is compatible with both A {\displaystyle A} and B {\displaystyle B} . If the basis of common eigenfunctions of A ^ {\displaystyle {\hat {A}}} , B ^ {\displaystyle {\hat {B}}} and C ^ {\displaystyle {\hat {C}}} is unique, that is, uniquely specified by the set of eigenvalues ( a n , b n , c n ) {\displaystyle (a_{n},b_{n},c_{n})} , then we have formed a CSCO: { A , B , C } {\displaystyle \{A,B,C\}} . If not, we add one more compatible observable and continue the process till a CSCO is obtained. </p><p>The same vector space may have distinct complete sets of commuting operators. </p><p>Suppose we are given a finite CSCO { A , B , C , . . . , } {\displaystyle \{A,B,C,...,\}} . Then we can expand any general state in the Hilbert space as </p> | ψ ⟩ = ∑ i , j , k , . . . C i , j , k , . . . | a i , b j , c k , . . . ⟩ {\displaystyle |\psi \rangle =\sum _{i,j,k,...}{\mathcal {C}}_{i,j,k,...}|a_{i},b_{j},c_{k},...\rangle } <p>where | a i , b j , c k , . . . ⟩ {\displaystyle |a_{i},b_{j},c_{k},...\rangle } are the eigenkets of the operators A ^ , B ^ , C ^ {\displaystyle {\hat {A}},{\hat {B}},{\hat {C}}} , and form a basis space. That is, </p> A ^ | a i , b j , c k , . . . ⟩ = a i | a i , b j , c k , . . . ⟩ {\displaystyle {\hat {A}}|a_{i},b_{j},c_{k},...\rangle =a_{i}|a_{i},b_{j},c_{k},...\rangle } , etc <p>If we measure A , B , C , . . . {\displaystyle A,B,C,...} in the state | ψ ⟩ {\displaystyle |\psi \rangle } then the probability that we simultaneously measure a i , b j , c k , . . . {\displaystyle a_{i},b_{j},c_{k},...} is given by | C i , j , k , . . . | 2 {\displaystyle |{\mathcal {C}}_{i,j,k,...}|^{2}} . </p><p>For a complete set of commuting operators, we can find a unitary transformation which will <a href="/facts/Simultaneously_diagonalize/y3MkyTCy">simultaneously diagonalize</a> all of them. </p> <h2 id="examples">Examples</h2> <h3>The hydrogen atom without electron or proton spin</h3> <p class="note">Main article: <a href="/facts/Hydrogen-like_atom/0rmuVpjT">Hydrogen-like atom</a></p> <p>Two components of the angular momentum operator L {\displaystyle \mathbf {L} } do not commute, but satisfy the commutation relations: </p> [ L i , L j ] = i ℏ ϵ i j k L k {\displaystyle [L_{i},L_{j}]=i\hbar \epsilon _{ijk}L_{k}} <p>So, any CSCO cannot involve more than one component of L {\displaystyle \mathbf {L} } . It can be shown that the square of the angular momentum operator, L 2 {\displaystyle L^{2}} , commutes with L {\displaystyle \mathbf {L} } . </p> [ L x , L 2 ] = 0 , [ L y , L 2 ] = 0 , [ L z , L 2 ] = 0 {\displaystyle [L_{x},L^{2}]=0,[L_{y},L^{2}]=0,[L_{z},L^{2}]=0} <p>Also, the <a href="/facts/Hamiltonian_function/2ReerlNA">Hamiltonian</a> H ^ = − ℏ 2 2 μ ∇ 2 − Z e 2 r {\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2\mu }}\nabla ^{2}-{\frac {Ze^{2}}{r}}} is a function of r {\displaystyle r} only and has rotational invariance, where μ {\displaystyle \mu } is the reduced mass of the system. Since the components of L {\displaystyle \mathbf {L} } are generators of rotation, it can be shown that </p> [ L , H ] = 0 , [ L 2 , H ] = 0 {\displaystyle [\mathbf {L} ,H]=0,[L^{2},H]=0} <p>Therefore, a commuting set consists of L 2 {\displaystyle L^{2}} , one component of L {\displaystyle \mathbf {L} } (which is taken to be L z {\displaystyle L_{z}} ) and H {\displaystyle H} . The solution of the problem tells us that disregarding spin of the electrons, the set { H , L 2 , L z } {\displaystyle \{H,L^{2},L_{z}\}} forms a CSCO. Let | E n , l , m ⟩ {\displaystyle |E_{n},l,m\rangle } be any basis state in the Hilbert space of the hydrogenic atom. Then </p> H | E n , l , m ⟩ = E n | E n , l , m ⟩ {\displaystyle H|E_{n},l,m\rangle =E_{n}|E_{n},l,m\rangle } L 2 | E n , l , m ⟩ = l ( l + 1 ) ℏ 2 | E n , l , m ⟩ {\displaystyle L^{2}|E_{n},l,m\rangle =l(l+1)\hbar ^{2}|E_{n},l,m\rangle } L z | E n , l , m ⟩ = m ℏ | E n , l , m ⟩ {\displaystyle L_{z}|E_{n},l,m\rangle =m\hbar |E_{n},l,m\rangle } <p>That is, the set of eigenvalues { E n , l , m } {\displaystyle \{E_{n},l,m\}} or more simply, { n , l , m } {\displaystyle \{n,l,m\}} completely specifies a unique eigenstate of the Hydrogenic atom. </p> <h3>The free particle</h3> <p class="note">Main article: <a href="/facts/Free_particle/Fjbx90yW">Free particle § Non-Relativistic Quantum Free Particle</a></p> <p>For a <a href="/facts/Free_particle/Fjbx90yW">free particle</a>, the Hamiltonian H = − ℏ 2 2 m ∇ 2 {\displaystyle H=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}} is invariant under translations. Translation commutes with the Hamiltonian: [ H , T ^ ] = 0 {\displaystyle [H,\mathbf {\hat {T}} ]=0} . However, if we express the Hamiltonian in the basis of the translation operator, we will find that H {\displaystyle H} has doubly degenerate eigenvalues. It can be shown that to make the CSCO in this case, we need another operator called the <a href="/facts/Parity_(physics)/emSbJ4hD">parity</a> operator Π {\displaystyle \Pi } , such that [ H , Π ] = 0 {\displaystyle [H,\Pi ]=0} . { H , Π } {\displaystyle \{H,\Pi \}} forms a CSCO. </p><p>Again, let | k ⟩ {\displaystyle |k\rangle } and | − k ⟩ {\displaystyle |-k\rangle } be the degenerate eigenstates of H {\displaystyle H} corresponding the eigenvalue H k = ℏ 2 k 2 2 m {\displaystyle H_{k}={\frac {{\hbar ^{2}}{k^{2}}}{2m}}} , i.e. </p> H | k ⟩ = ℏ 2 k 2 2 m | k ⟩ {\displaystyle H|k\rangle ={\frac {{\hbar ^{2}}{k^{2}}}{2m}}|k\rangle } H | − k ⟩ = ℏ 2 k 2 2 m | − k ⟩ {\displaystyle H|-k\rangle ={\frac {{\hbar ^{2}}{k^{2}}}{2m}}|-k\rangle } <p>The degeneracy in H {\displaystyle H} is removed by the momentum operator p ^ {\displaystyle \mathbf {\hat {p}} } . </p> p ^ | k ⟩ = k | k ⟩ {\displaystyle \mathbf {\hat {p}} |k\rangle =k|k\rangle } p ^ | − k ⟩ = − k | − k ⟩ {\displaystyle \mathbf {\hat {p}} |-k\rangle =-k|-k\rangle } <p>So, { p ^ , H } {\displaystyle \{\mathbf {\hat {p}} ,H\}} forms a CSCO. </p> <h3>Addition of angular momenta</h3> <p>We consider the case of two systems, 1 and 2, with respective angular momentum operators J 1 {\displaystyle \mathbf {J_{1}} } and J 2 {\displaystyle \mathbf {J_{2}} } . We can write the eigenstates of J 1 2 {\displaystyle J_{1}^{2}} and J 1 z {\displaystyle J_{1z}} as | j 1 m 1 ⟩ {\displaystyle |j_{1}m_{1}\rangle } and of J 2 2 {\displaystyle J_{2}^{2}} and J 2 z {\displaystyle J_{2z}} as | j 2 m 2 ⟩ {\displaystyle |j_{2}m_{2}\rangle } . </p> J 1 2 | j 1 m 1 ⟩ = j 1 ( j 1 + 1 ) ℏ 2 | j 1 m 1 ⟩ {\displaystyle J_{1}^{2}|j_{1}m_{1}\rangle =j_{1}(j_{1}+1)\hbar ^{2}|j_{1}m_{1}\rangle } J 1 z | j 1 m 1 ⟩ = m 1 ℏ | j 1 m 1 ⟩ {\displaystyle J_{1z}|j_{1}m_{1}\rangle =m_{1}\hbar |j_{1}m_{1}\rangle } J 2 2 | j 2 m 2 ⟩ = j 2 ( j 2 + 1 ) ℏ 2 | j 2 m 2 ⟩ {\displaystyle J_{2}^{2}|j_{2}m_{2}\rangle =j_{2}(j_{2}+1)\hbar ^{2}|j_{2}m_{2}\rangle } J 2 z | j 2 m 2 ⟩ = m 2 ℏ | j 2 m 2 ⟩ {\displaystyle J_{2z}|j_{2}m_{2}\rangle =m_{2}\hbar |j_{2}m_{2}\rangle } <p>Then the basis states of the complete system are | j 1 m 1 ; j 2 m 2 ⟩ {\displaystyle |j_{1}m_{1};j_{2}m_{2}\rangle } given by </p> | j 1 m 1 ; j 2 m 2 ⟩ = | j 1 m 1 ⟩ ⊗ | j 2 m 2 ⟩ {\displaystyle |j_{1}m_{1};j_{2}m_{2}\rangle =|j_{1}m_{1}\rangle \otimes |j_{2}m_{2}\rangle } <p>Therefore, for the complete system, the set of eigenvalues { j 1 , m 1 , j 2 , m 2 } {\displaystyle \{j_{1},m_{1},j_{2},m_{2}\}} completely specifies a unique basis state, and { J 1 2 , J 1 z , J 2 2 , J 2 z } {\displaystyle \{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 J = J 1 + J 2 {\displaystyle \mathbf {J} =\mathbf {J_{1}} +\mathbf {J_{2}} } . The eigenvalues of J 2 {\displaystyle J^{2}} are j ( j + 1 ) ℏ 2 {\displaystyle j(j+1)\hbar ^{2}} where j {\displaystyle j} takes on the values j 1 + j 2 , j 1 + j 2 − 1 , . . . , | j 1 − j 2 | {\displaystyle j_{1}+j_{2},j_{1}+j_{2}-1,...,|j_{1}-j_{2}|} , and those of J z {\displaystyle J_{z}} are m {\displaystyle m} where m = − j , − j + 1 , . . . j − 1 , j {\displaystyle m=-j,-j+1,...j-1,j} . The basis states of the operators J 2 {\displaystyle J^{2}} and J z {\displaystyle J_{z}} are | j 1 j 2 ; j m ⟩ {\displaystyle |j_{1}j_{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 } {\displaystyle \{j_{1},j_{2},j,m\}} , and the corresponding CSCO is { J 1 2 , J 2 2 , J 2 , J z } {\displaystyle \{J_{1}^{2},J_{2}^{2},J^{2},J_{z}\}} . </p> <h2 id="see-also">See also</h2> <ul><li><a href="/facts/Quantum_number/AByP6F0y">Quantum number</a> <ul><li><a href="/facts/Good_quantum_number/HuBnjFsl">Good quantum number</a></li></ul></li> <li><a href="/facts/Degenerate_energy_levels/ulrnMXTR">Degenerate energy levels</a></li> <li><a href="/facts/Mathematical_formulation_of_quantum_mechanics/fRLw4sFU">Mathematical structure of quantum mechanics</a></li> <li><a href="/facts/Operator_(physics)/rCJw8UFz">Operators in Quantum Mechanics</a></li> <li><a href="/facts/Canonical_commutation_relation/UgUW6TuT">Canonical commutation relation</a></li> <li><a href="/facts/Measurement_in_quantum_mechanics/VY7jPfP0">Measurement in quantum mechanics</a></li> <li><a href="/facts/Collapse_of_the_wavefunction/f7KvbBHK">Collapse of the wavefunction</a></li> <li><a href="/facts/Angular_momentum/7TuSVFxq">Angular Momentum (Quantum Mechanics)</a></li></ul> <h2 id="further-reading">Further reading</h2> <ul><li><a href="/facts/Stephen_Gasiorowicz/egj4Dx9u">Gasiorowicz, Stephen</a> (1974), <i>Quantum Physics</i>, New York: <a href="/facts/John_Wiley_%26_Sons/53g3LqJc">John Wiley & Sons</a>, <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-471-29281-4.</li> <li><a href="/facts/Claude_Cohen-Tannoudji/fjs27Pvx">Cohen-Tannoudji, Claude</a>; Diu, Bernard; Laloë, Franck (1977). <i>Quantum mechanics</i>. Vol. 1. New York: Wiley. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-471-16433-3. <a href="/facts/OCLC_(identifier)/Yqad8waq">OCLC</a> <a href="https://search.worldcat.org/oclc/2089460">2089460</a>.</li> <li><a href="/facts/Claude_Cohen-Tannoudji/fjs27Pvx">Cohen-Tannoudji, Claude</a>; Diu, Bernard; Laloë, Franck (1977). <i>Quantum mechanics</i>. Vol. 2. New York: Wiley. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-471-16435-7. <a href="/facts/OCLC_(identifier)/Yqad8waq">OCLC</a> <a href="https://search.worldcat.org/oclc/45727993">45727993</a>.</li> <li><a href="/facts/Paul_Dirac/6bt5yRdk">Dirac, P.A.M.</a> (1958). <i><a href="/facts/The_Principles_of_Quantum_Mechanics/ebaerFCG">The Principles of Quantum Mechanics</a></i>. Oxford: Clarendon Press. <a href="/facts/ISBN_(identifier)/15AdSPa9">ISBN</a> 978-0-19-851208-0. <a href="/facts/OCLC_(identifier)/Yqad8waq">OCLC</a> <a href="https://search.worldcat.org/oclc/534829">534829</a>.</li> <li>R.P. Feynman, R.B. Leighton and M. Sands: <i>The Feynman Lectures on Physics</i>, Addison-Wesley, 1965</li> <li>R Shankar, <i>Principles of Quantum Mechanics</i>, Second Edition, Springer (1994).</li> <li>J J Sakurai, <i><a href="/facts/Modern_Quantum_Mechanics/boVu6kP4">Modern Quantum Mechanics</a></i>, Revised Edition, Pearson (1994).</li> <li>B. H. Bransden and C. J. Joachain, <i>Quantum Mechanics</i>, Second Edition, Pearson Education Limited, 2000.</li> <li>For a discussion on the Compatibility Theorem, Lecture Notes of <i>School of Physics and Astronomy</i> of The University of Edinburgh. <a href="http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect2.pdf">http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect2.pdf</a>.</li> <li>A slide on CSCO in the lecture notes of Prof. S Gupta, Tata Institute of Fundamental Research, Mumbai. <a href="http://theory.tifr.res.in/~sgupta/courses/qm2013/hand3.pdf">http://theory.tifr.res.in/~sgupta/courses/qm2013/hand3.pdf</a></li> <li>A section on the Free Particle in the lecture notes of Prof. S Gupta, Tata Institute of Fundamental Research, Mumbai. <a href="http://theory.tifr.res.in/~sgupta/courses/qm2013/hand6.pdf">http://theory.tifr.res.in/~sgupta/courses/qm2013/hand6.pdf</a></li></ul> <h2 id="references">References</h2> <ol> <li id="fn:1"><p>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. <a href="978-0262366892" target="_blank">978-0262366892</a> <a href="#fnref:1" class="footnote-back-ref">↩</a></p></li> <li id="fn:2"><p>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. <a href="978-0-471-16433-3" target="_blank">978-0-471-16433-3</a> <a href="#fnref:2" class="footnote-back-ref">↩</a></p></li> </ol>