# Question about the proof that compatible obverables commute.

First, a preliminary question about Bra-ket notation. On a Hilbert space $$H$$, the inner product $$(\cdot,\cdot):H\times H\to\mathbb C$$ gives an identification of a vector ket $$\varphi = \vert \varphi\rangle$$ with a functional bra $$f_\varphi = \langle \varphi\vert$$, as $$(\varphi,\psi) = (\vert\varphi\rangle,\vert\psi\rangle) := f_\varphi(\psi) = \langle \varphi\vert (\vert\psi\rangle)=\langle\varphi\vert\psi\rangle.$$ Here we are not taking the inner product of the ket and the bra, because they lie in different vector spaces (in particular the bra lies in the dual space of $$H$$). Instead we are defining a bra by $$f_\varphi(\psi) = (\varphi,\psi)$$ for $$\psi\in H$$. What exactly is the meaning of $$\langle\varphi\vert\psi\rangle$$? Just a way of denoting the $$(\mathrm{bra},\mathrm{ket})$$ pair?

Now let $$A$$ and $$B$$ be observables. Suppose there exists a complete set of kets $$\{\vert\psi_n\rangle\}$$ whose every element is simultaneously an eigenket of $$A$$ and $$B$$. Then we says that $$A$$ and $$B$$ are compatible. Let $$\{a_n\}$$ be a sequence of eigenkets of $$A$$ and $$\{b_n\}$$ a sequence of eigenkets of $$B$$. Then we can write $$AB\vert\psi_n\rangle = Ab_n\vert\psi_n\rangle = a_nb_n\vert\psi_n\rangle = b_na_n\vert\psi_n\rangle = BA\vert\psi_n\rangle.$$ Now, we can expand any arbitrary state ket $$\vert\Psi\rangle$$ in the complete set $$\{\vert\psi_n\rangle\}$$ as $$\vert\Psi\rangle = \sum_{n=1}^\infty c_n\vert\psi_n\rangle,$$ where $$c_n\in\mathbb C$$. So we can see that $$(AB - BA)\vert\Psi\rangle = \sum_{n=1}^\infty c_n(AB-BA)\vert\psi_n\rangle = 0.$$ This implies that $$[A,B]=0$$, which means that the operators commute.

My question: what is the meaning of $$[A,B]$$ in the above line? Is it the commutator of $$A$$ and $$B$$? If so, isn't it immediately apparent that $$AB-BA=0$$ from $$AB\vert\psi_n\rangle = BA\vert\psi_n\rangle$$? I suppose not, since we went to the effort of showing this implies $$AB\vert\Psi\rangle = BA\vert\Psi\rangle$$ for an arbitrary $$\Psi\in H$$. Am I missing anything here?

• I don't understand your comment about the bra-ket notation. If you understand $\mathbf{v}\cdot\mathbf{w}:=\mathbf{v}^T\mathbf{w}$ then you shouldn't have too much trouble understanding bra-ket. Dec 29 '19 at 8:35
• All the previous line shows is that $|\psi_n\rangle$ is in the null space of $[A,B]$, not that $[A,B] = 0$. In order to show the latter, one must show all vectors are in the null space of $[A,B]$. But since $|\psi_n \rangle$ form a complete basis for the vector space, any vector is a linear combination of these $|\psi_n\rangle$, and then applying $[A,B]$ to an arbitrary vector gives a weighted sum of zeros, otherwise known as zero. Dec 29 '19 at 8:47
• @JackyChong The notation $\langle\varphi\vert\psi\rangle$ is confusing me, since normally $\langle\cdot,\cdot\rangle$ is the notation for the inner product on $H$. Moreover, what is the significance of the $\vert$? Is it just standard notation? Dec 31 '19 at 21:49
• Well, there are two ways of interpreting it. The first is that (by merely replacing , with |) it is the inner product of $\phi, \psi \in H$. Or secondly, by the bijection $f$, we can view it as the number obtained by evaluating $f_{\phi} \equiv \langle \phi| \in H^*$ on the vector $\psi \equiv | \psi \rangle \in H$ ($\equiv$ meaning its just same thing written with different notation) So, which perspective you choose to adopt is up to you. Jan 2 '20 at 1:01
• and yes, replacing $,$ with $|$ is merely notation. Don't read too much into it. I could just as well ask you why do you use a comma rather than a smiley face as a "spacer" $\langle \phi \,\, \ddot{\smile} \,\, \psi \rangle$? The answer is of course that "it is just notation". Jan 2 '20 at 1:03

• $$[A, B]$$ is the commutator
• It is immediately obvious that $$[A, B] =0$$ since $$\{|\psi_n\rangle\}$$ is complete.