Kevin Alan Reiss

Suppose that we have two operators $A$ and $B$ such that: $$ AB=-BA. $$In other words, they anti-commute. What can we say about their eigenbasis? It is well known that for commuting operators, they share an eigenbasis, i.e. their eigenkets are identical. First, let’s look at the representation of $B$ in the basis defined by the eigenbasis of $A$. Let $A=\sum_{i} a_i\ket{a_i}\bra{a_i}$ be the spectral decomposition of $A$. Then, $$ \bra{a_i}(AB+BA)\ket{a_j} = (a_i+a_j)\bra{a_i}B\ket{a_j}=0 \tag{1} $$This tells us immediately that if $a_i\neq-a_j$, then $\bra{a_i}B\ket{a_j}=0$. One immediate consequence is that if all of the eigenvalues of $A$ are nonzero, then the diagonal of $B$ in the $A$ basis is 0, since the diagonal is where $a_i=a_j$ and $a_i=-a_j$ is only met if $a_i=0$. Then sum of the diagonal elements of $B$ is 0 as well – and since the trace is basis-independent, this is the trace of $B$. That is to say, if $A$ is invertible (has no zero-eigenvalues) then any operator which anti-commutes with $A$ will have to have 0 trace. ...

In quantum mechanics, if two observables do not commute ($[A,B]\neq0$), we generally say that they are not simultaneously observable and do not share an eigenbasis. However, this is not necessarily true for all states! For example, consider L$^2$ and $L_z$. They commute, and we assign separate quantum numbers to their eigenvalues: $l$ for L$^2$ and $m$ for $L_z$. However, $L_z$ does not commute with $L_x$ or $L_y$. In particular: $$ [L_p,L_q]=i\hbar\epsilon_{pqr}L_r. $$However, the state with $l=0$ immediately tells us the value of the quantum number $m$ – it’s also 0. The interesting thing about this state is that it also has definite $L_x=L_y=L_z=0$. There is a subset of the whole Hilbert space which may be a simultaneous eigenbasis for two non-commuting observables. That is to say, there are exceptional states $\ket\psi$ such that: ...

Recently, a friend asked me if I believed in the concept of free will. I responded instantly “yes”, as is my wont. He probed deeper, asking “what makes you think that?” Immediately, something welled within me, as I delayed my immediate response of launching into a monologue about the ineffable nature of consciousness and how contradictory it would be to lack free will but feel as if one had some degree of control. The feeling was that of knowing I had an opportunity to be succinct and humorous, and grasping for the words which called me. Without missing much of a beat, I responded to my friend’s “what makes you think that” with, “I chose to”. My friend chuckled and, well, that settled it. In some cases, less is certainly more. ...

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