Complete set of commuting observables

Proof that compatible observables commute.

Let ${\displaystyle \{|\psi _{n}\rangle \}}$ be a complete set of common eigenkets of the two compatible observables ${\displaystyle A}$ and ${\displaystyle B}$, corresponding to the sets ${\displaystyle \{a_{n}\}}$ and ${\displaystyle \{b_{n}\}}$ respectively. Then we can write ${\displaystyle AB|\psi _{n}\rangle =Ab_{n}|\psi _{n}\rangle =a_{n}b_{n}|\psi _{n}\rangle =b_{n}a_{n}|\psi _{n}\rangle =BA|\psi _{n}\rangle }$ Now, we can expand any arbitrary state ket ${\displaystyle |\Psi \rangle }$ in the complete set ${\displaystyle \{|\psi _{n}\rangle \}}$ as ${\displaystyle |\Psi \rangle =\sum _{n}c_{n}|\psi _{n}\rangle }$ So, using the above result, we can see that ${\displaystyle (AB-BA)|\Psi \rangle =\sum _{n}c_{n}(AB-BA)|\psi _{n}\rangle =0}$ This implies ${\displaystyle [A,B]=0}$, which means that the two operators commute.

Proof that commuting observables possess a complete set of common eigenfunctions.

When ${\displaystyle A}$ has non-degenerate eigenvalues: Let ${\displaystyle \{|\psi _{n}\rangle \}}$ be a complete set of eigenkets of ${\displaystyle A}$ corresponding to the set of eigenvalues ${\displaystyle \{a_{n}\}}$. If the operators ${\displaystyle A}$ and ${\displaystyle B}$ commute, we can write ${\displaystyle A(B|\psi _{n}\rangle )=BA|\psi _{n}\rangle =a_{n}(B|\psi _{n}\rangle )}$ So, we can say that ${\displaystyle B|\psi _{n}\rangle }$ is an eigenket of ${\displaystyle A}$ corresponding to the eigenvalue ${\displaystyle a_{n}}$. Since both ${\displaystyle B|\psi _{n}\rangle }$ and ${\displaystyle |\psi _{n}\rangle }$ are eigenkets associated with the same non-degenerate eigenvalue ${\displaystyle a_{n}}$, they can differ at most by a multiplicative constant. We call this constant ${\displaystyle b_{n}}$. So, ${\displaystyle B|\psi _{n}\rangle =b_{n}|\psi _{n}\rangle }$ , which means ${\displaystyle |\psi _{n}\rangle }$ is an eigenket of ${\displaystyle B}$, and thus of ${\displaystyle A}$ and ${\displaystyle B}$ simultaneously. When ${\displaystyle A}$ has degenerate eigenvalues: We suppose ${\displaystyle a_{n}}$ is ${\displaystyle g}$ -fold degenerate. Let the corresponding linearly independent eigenkets be ${\displaystyle |\psi _{nr}\rangle ,(r=1,2,...g)}$ Since ${\displaystyle [A,B]=0}$, we reason as above to find that ${\displaystyle B|\psi _{nr}\rangle }$ is an eigenket of ${\displaystyle A}$ corresponding to the degenerate eigenvalue ${\displaystyle a_{n}}$. So, we can expand ${\displaystyle B|\psi _{nr}\rangle }$ in the basis of the degenerate eigenkets of ${\displaystyle a_{n}}$: ${\displaystyle B|\psi _{nr}\rangle =\sum _{s=1}^{g}c_{rs}|\psi _{ns}\rangle }$ The ${\displaystyle c_{rs}}$ are the expansion coefficients. We now sum over all ${\displaystyle r}$ with ${\displaystyle g}$ constants ${\displaystyle d_{r}}$. So, ${\displaystyle B\sum _{r=1}^{g}d_{r}|\psi _{nr}\rangle =\sum _{r=1}^{g}\sum _{s=1}^{g}d_{r}c_{rs}|\psi _{ns}\rangle }$ So, ${\displaystyle \sum _{r=1}^{g}d_{r}|\psi _{nr}\rangle }$ will be an eigenket of ${\displaystyle B}$ with the eigenvalue ${\displaystyle b_{n}}$ if we have ${\displaystyle \sum _{r=1}^{g}d_{r}c_{rs}=b_{n}d_{s},s=1,2,...g}$ This constitutes a system of ${\displaystyle g}$ linear equations for the constants ${\displaystyle d_{r}}$. A non-trivial solution exists if ${\displaystyle \det[c_{rs}-b_{n}\delta _{rs}]=0}$ This is an equation of order ${\displaystyle g}$ in ${\displaystyle b_{n}}$, and has ${\displaystyle g}$ roots. For each root ${\displaystyle b_{n}=b_{n}^{(k)},k=1,2,...g}$ we have a value of ${\displaystyle d_{r}}$, say, ${\displaystyle d_{r}^{(k)}}$. Now, the ket ${\displaystyle |\phi _{n}^{(k)}\rangle =\sum _{r=1}^{g}d_{r}^{(k)}|\psi _{nr}\rangle }$ is simultaneously an eigenket of ${\displaystyle A}$ and ${\displaystyle B}$ with eigenvalues ${\displaystyle a_{n}}$ and ${\displaystyle b_{n}^{(k)}}$ respectively.

We consider the two above observables ${\displaystyle A}$ and ${\displaystyle B}$. Suppose there exists a complete set of kets ${\displaystyle \{|\psi _{n}\rangle \}}$ whose every element is simultaneously an eigenket of ${\displaystyle A}$ and ${\displaystyle B}$. Then we say that ${\displaystyle A}$ and ${\displaystyle B}$ are compatible. If we denote the eigenvalues of ${\displaystyle A}$ and ${\displaystyle B}$ corresponding to ${\displaystyle |\psi _{n}\rangle }$ respectively by ${\displaystyle a_{n}}$ and ${\displaystyle b_{n}}$, we can write

${\displaystyle A|\psi _{n}\rangle =a_{n}|\psi _{n}\rangle }$

${\displaystyle B|\psi _{n}\rangle =b_{n}|\psi _{n}\rangle }$

If the system happens to be in one of the eigenstates, say, ${\displaystyle |\psi _{n}\rangle }$, then both ${\displaystyle A}$ and ${\displaystyle B}$ can be simultaneously measured to any arbitrary level of precision, and we will get the results ${\displaystyle a_{n}}$ and ${\displaystyle b_{n}}$ respectively. This idea can be extended to more than two observables.

The Cartesian components of the position operator ${\displaystyle \mathbf {r} }$ are ${\displaystyle x}$, ${\displaystyle y}$ and ${\displaystyle z}$. These components are all compatible. Similarly, the Cartesian components of the momentum operator ${\displaystyle \mathbf {p} }$, that is ${\displaystyle p_{x}}$, ${\displaystyle p_{y}}$ and ${\displaystyle p_{z}}$ are also compatible.

Two components of the angular momentum operator ${\displaystyle \mathbf {L} }$ do not commute, but satisfy the commutation relations:

${\displaystyle [L_{i},L_{j}]=i\hbar \epsilon _{ijk}L_{k}}$

So, any CSCO cannot involve more than one component of ${\displaystyle \mathbf {L} }$. It can be shown that the square of the angular momentum operator, ${\displaystyle L^{2}}$, commutes with ${\displaystyle \mathbf {L} }$.

${\displaystyle [L_{x},L^{2}]=0,[L_{y},L^{2}]=0,[L_{z},L^{2}]=0}$

Also, the Hamiltonian ${\displaystyle {\hat {H}}=-{\frac {\hbar ^{2}}{2\mu }}\nabla ^{2}-{\frac {Ze^{2}}{r}}}$ is a function of ${\displaystyle r}$ only and has rotational invariance, where ${\displaystyle \mu }$ is the reduced mass of the system. Since the components of ${\displaystyle \mathbf {L} }$ are generators of rotation, it can be shown that

${\displaystyle [\mathbf {L} ,H]=0,[L^{2},H]=0}$

Therefore, a commuting set consists of ${\displaystyle L^{2}}$, one component of ${\displaystyle \mathbf {L} }$ (which is taken to be ${\displaystyle L_{z}}$) and ${\displaystyle H}$. The solution of the problem tells us that disregarding spin of the electrons, the set ${\displaystyle \{H,L^{2},L_{z}\}}$ forms a CSCO. Let ${\displaystyle |E_{n},l,m\rangle }$ be any basis state in the Hilbert space of the hydrogenic atom. Then

${\displaystyle H|E_{n},l,m\rangle =E_{n}|E_{n},l,m\rangle }$

${\displaystyle L^{2}|E_{n},l,m\rangle =l(l+1)\hbar ^{2}|E_{n},l,m\rangle }$

${\displaystyle L_{z}|E_{n},l,m\rangle =m\hbar |E_{n},l,m\rangle }$

That is, the set of eigenvalues ${\displaystyle \{E_{n},l,m\}}$ or more simply, ${\displaystyle \{n,l,m\}}$ completely specifies a unique eigenstate of the Hydrogenic atom.

For a free particle, the Hamiltonian ${\displaystyle H=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}}$ is invariant under translations. Translation commutes with the Hamiltonian: ${\displaystyle [H,\mathbf {\hat {T}} ]=0}$. However, if we express the Hamiltonian in the basis of the translation operator, we will find that ${\displaystyle H}$ has doubly degenerate eigenvalues. It can be shown that to make the CSCO in this case, we need another operator called the parity operator ${\displaystyle \Pi }$, such that ${\displaystyle [H,\Pi ]=0}$.${\displaystyle \{H,\Pi \}}$ forms a CSCO.

Again, let ${\displaystyle |k\rangle }$ and ${\displaystyle |-k\rangle }$ be the degenerate eigenstates of ${\displaystyle H}$corresponding the eigenvalue ${\displaystyle H_{k}={\frac {{\hbar ^{2}}{k^{2}}}{2m}}}$, i.e.

${\displaystyle H|k\rangle ={\frac {{\hbar ^{2}}{k^{2}}}{2m}}|k\rangle }$

${\displaystyle H|-k\rangle ={\frac {{\hbar ^{2}}{k^{2}}}{2m}}|-k\rangle }$

The degeneracy in ${\displaystyle H}$ is removed by the momentum operator ${\displaystyle \mathbf {\hat {p}} }$.

${\displaystyle \mathbf {\hat {p}} |k\rangle =k|k\rangle }$

${\displaystyle \mathbf {\hat {p}} |-k\rangle =-k|-k\rangle }$

So, ${\displaystyle \{\mathbf {\hat {p}} ,H\}}$ forms a CSCO.

We consider the case of two systems, 1 and 2, with respective angular momentum operators ${\displaystyle \mathbf {J_{1}} }$ and ${\displaystyle \mathbf {J_{2}} }$. We can write the eigenstates of ${\displaystyle J_{1}^{2}}$ and ${\displaystyle J_{1z}}$ as ${\displaystyle |j_{1}m_{1}\rangle }$ and of ${\displaystyle J_{2}^{2}}$ and ${\displaystyle J_{2z}}$ as ${\displaystyle |j_{2}m_{2}\rangle }$.

${\displaystyle J_{1}^{2}|j_{1}m_{1}\rangle =j_{1}(j_{1}+1)\hbar ^{2}|j_{1}m_{1}\rangle }$

${\displaystyle J_{1z}|j_{1}m_{1}\rangle =m_{1}\hbar |j_{1}m_{1}\rangle }$

${\displaystyle J_{2}^{2}|j_{2}m_{2}\rangle =j_{2}(j_{2}+1)\hbar ^{2}|j_{2}m_{2}\rangle }$

${\displaystyle J_{2z}|j_{2}m_{2}\rangle =m_{2}\hbar |j_{2}m_{2}\rangle }$

Then the basis states of the complete system are ${\displaystyle |j_{1}m_{1};j_{2}m_{2}\rangle }$ given by

${\displaystyle |j_{1}m_{1};j_{2}m_{2}\rangle =|j_{1}m_{1}\rangle \otimes |j_{2}m_{2}\rangle }$

Therefore, for the complete system, the set of eigenvalues ${\displaystyle \{j_{1},m_{1},j_{2},m_{2}\}}$ completely specifies a unique basis state, and ${\displaystyle \{J_{1}^{2},J_{1z},J_{2}^{2},J_{2z}\}}$ forms a CSCO. Equivalently, there exists another set of basis states for the system, in terms of the total angular momentum operator ${\displaystyle \mathbf {J} =\mathbf {J_{1}} +\mathbf {J_{2}} }$. The eigenvalues of ${\displaystyle J^{2}}$ are ${\displaystyle j(j+1)\hbar ^{2}}$ where ${\displaystyle j}$ takes on the values ${\displaystyle j_{1}+j_{2},j_{1}+j_{2}-1,...,|j_{1}-j_{2}|}$, and those of ${\displaystyle J_{z}}$ are ${\displaystyle m}$ where ${\displaystyle m=-j,-j+1,...j-1,j}$. The basis states of the operators ${\displaystyle J^{2}}$ and ${\displaystyle J_{z}}$ are ${\displaystyle |j_{1}j_{2};jm\rangle }$. Thus we may also specify a unique basis state in the Hilbert space of the complete system by the set of eigenvalues ${\displaystyle \{j_{1},j_{2},j,m\}}$, and the corresponding CSCO is ${\displaystyle \{J_{1}^{2},J_{2}^{2},J^{2},J_{z}\}}$.