I've been watching Susskind's lectures on Quantum Entanglement, and something he said regarding (non-)commuting projection operators confused me.

Consider two subspaces {$|a\rangle$} and {$|b\rangle$} of Hilbert space, with operators $K$ and $L$ for which:

  • $K |a\rangle = \lambda |a\rangle (1)$
  • $L |b\rangle = \mu |b\rangle (2)$

Now considers operators $P_K $ and $P_L$ that project any vector in Hilbert space onto their respective subspaces, that is:

  • $K (P_K |\psi\rangle) = \lambda (P_K |\psi\rangle) $
  • $L (P_L |\psi\rangle) = \mu (P_L |\psi\rangle) $

We want to find simultaneous eigenstates of both $K$ and $L$. If $P_K$ and $P_L$ commute: $P_K (P_L |\psi\rangle) = P_L (P_K |\psi\rangle)$. Now the left-hand satiesfies $(1)$, and the right-hand side satisfies $(2)$, so these are the required states.

In fact, if $P_K$ and $P_L$ operators commute, they share a complete set of eigenstates. The eigenstates of projection operators are those that span the subspace they project onto, so apparently $P_K$ and $P_L$ project onto the same subspace, which means they're the same operator? Then, is the statement: "projection operators commute $\rightarrow$ they're the same" correct, or do they somehow project states onto the same subspace in a different way?

Furthermore, we can imagine the subspaces geometrically as 'planes', and where these planes intersect we can find states that satisfy both $(1)$ and $(2)$. Now, according to Susskind, if $P_K$ and $P_L$ do not commute, finding such states is impossible. If the previous paragraph holds (does it?), then them commuting implies the intersection of their subspaces is the entire subspace. I don't know what non-commuting means geometrically, but shouldn't there be a case where the intersection of their subspaces isn't the entire subspace (for example, imagine two 2D perpendicular planes intersecting each other on a 1D line)? Susskind's comment seems to contradict that, and can't see exactly where I'm going wrongly.


A complete set of eigenstates spans the whole space, not just the subspace the projection operators project on. In this set of eigenstates you also have a basis of the subspace belonging to the eigenvalue 0.

  • $\begingroup$ Right, any vectors orthogonal to the subspace would have eigenvalue 0. Then I suppose it is possible for two commuting projection operators to project onto different subspaces. I now have a hard time imagining two projection operators that don't commute though. $\endgroup$ – Timsey Apr 12 '14 at 22:21
  • $\begingroup$ Think of projection operators onto spin eigenstates; $ \frac{1}{2} ( 1 + \sigma_i)$ $\endgroup$ – taupunkt Apr 12 '14 at 23:16
  • $\begingroup$ Ah, of course! Then I believe that solves it, thank you. $\endgroup$ – Timsey Apr 13 '14 at 18:06

"projection operators commute → they're the same" Are you sure he said this predicate ? or it is your conclusion? Anyway, it is not true ! Consider two dimensional X-Projector And Y-Projector , they commute but they are not the same!

  • $\begingroup$ It's a consequence I arrived at, because I misunderstood what space is spanned by a complete set of vectors. It's been cleared up though! $\endgroup$ – Timsey Apr 13 '14 at 17:49

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