This is a little appendix to my post about the consequences of the fermionic CCRs. The results described in the appendix are very well-known – they are taught in any undergrad quantum course – but I’m rather fond of the little proof given, and so am indulging myself by including it here. The results are used in the previous post.
Note: This post is one in a series describing fermi algebras, and a powerful tool known as the Jordan-Wigner transform, which allows one to move back and forth between describing a system as a collection of qubits, and as a collection of fermions. The posts assume familiarity with elementary quantum mechanics, comfort with elementary linear algebra (but not advanced techniques), and a little familiarity with the basic nomenclature of quantum information science (qubits, the Pauli matrices).
Appendix on mutually commuting observables
Any undergraduate quantum mechanics course covers the fact that a mutually commuting set of Hermitian operators possesses a common eigenbasis. Unfortunately, in my experience this fact is usually proved rather early on, and suffers from being presented in a slightly too elementary fashion, with inductive constructions of explicit basis sets and so on. The following proof is still elementary, but from a slightly more sophisticated perspective. It is, I like to imagine, rather more like what would be given in an advanced course in linear algebra, were linear algebraists to actually cover this kind of material. (They don’t, so far as I know, having other fish to fry.)
Suppose [tex]H_1,\ldots,H_n[/tex] are commuting Hermitian (indeed, normal suffices) operators with spectral decompositions:
[tex] H_j = \sum_{jk} E_{jk} P_{jk}, [/tex]
where [tex]E_{jk}[/tex] are the eigenvalues of [tex]H_j[/tex], and [tex]P_{jk}[/tex] are the corresponding projectors. Since the [tex]H_j[/tex] commute, it is not difficult to verify that for any quadruple [tex]j,k,l,m[/tex] the operators [tex]P_{jk}[/tex] and [tex]P_{lm}[/tex] also commute. For a vector [tex]\vec k = (k_1,\ldots,k_n)[/tex] define the operator
[tex] P_{\vec k} \equiv P_{1 k_1} P_{2 k_2} \ldots P_{n k_n}. [/tex]
Note that the order of the operators on the right-hand side does not matter, since they all commute with one another. The following equations all follow easily by direct computation, the mutual commutativity of the [tex]P_{jk}[/tex] operators, and standard properties of the spectral decomposition:
[tex] P_{\vec k}^\dagger = P_{\vec k}; \,\,\,\, \sum_{\vec k} P_{\vec k} = I; \,\,\,\, P_{\vec k} P_{\vec l} = \delta_{\vec k \vec l} P_{\vec k}. [/tex]
Thus, the operators [tex]P_{\vec k}[/tex] form a complete set of orthonormal projectors. Furthermore, suppose we have [tex]P_{\vec k} |\psi\rangle = |\psi\rangle[/tex]. Then we will show that for any [tex]j[/tex] we have [tex]P_{jk_j} |\psi\rangle = |\psi\rangle[/tex], so [tex]|\psi\rangle[/tex] is an eigenstate of [tex]H_j[/tex] with eigenvalue [tex]k_j[/tex]. This shows that the operators [tex]P_{\vec k}[/tex] project onto a complete orthonormal set of simultaneous eigenspaces for the [tex]H_j[/tex], and will complete the proof.
Our goal is to show that if [tex]P_{\vec k} |\psi\rangle = |\psi\rangle[/tex] then for any [tex]j[/tex] we have [tex]P_{jk_j} |\psi\rangle = |\psi\rangle[/tex]. To see this, simply multiply both sides of [tex]P_{\vec k} |\psi\rangle = |\psi\rangle[/tex] by [tex]P_{jk_j}[/tex], and observe that [tex]P_{jk_j} P_{\vec k} = P_{\vec k}[/tex]. This gives [tex]P_{\vec k}|\psi\rangle = P_{jk_j}|\psi\rangle[/tex]. But [tex]P_{\vec k}|\psi\rangle = |\psi\rangle[/tex], so we obtain [tex]|\psi\rangle = P_{j k_j}|\psi\rangle[/tex], which completes the proof.