10 Compatibility

As said, the sta­bility of “ordi­nary” mate­rial objects rests on the sta­bility of atoms and mol­e­cules. This in turn rests on the fuzzi­ness of the rel­a­tive posi­tions and momenta of their con­stituents, and it implies an inequality of the form (2.4.3): the product of the “uncer­tain­ties” asso­ci­ated with a rel­a­tive posi­tion and the cor­re­sponding rel­a­tive momentum must have a pos­i­tive lower limit. This makes it impos­sible to simul­ta­ne­ously mea­sure both quan­ti­ties with arbi­trary pre­ci­sion. In other words, the sta­bility of matter requires, via the sta­bility of atoms, that mea­sure­ments of the two quan­ti­ties be incom­pat­ible.

The World According to Quantum MechanicsWe now want to deter­mine the con­di­tions under which two ele­men­tary tests are compatible.

Let the two pos­sible out­comes of an ele­men­tary test T1 be rep­re­sented by the sub­spaces M and M, and let the two pos­sible out­comes of another ele­men­tary test T2 be rep­re­sented by the sub­spaces N and N. Sup­pose that when­ever T1 yields M, a sub­se­quent per­for­mance of T2 yields N. (This means that M is con­tained in N.) It fol­lows that when­ever T2 yields N, the out­come of the first test cannot have been M; it must have been M. (This means that N is con­tained in M.)

There are three pos­sible com­bi­na­tions of out­comes: (i) M and N, (ii) M and N, and (iii) M and N. If the two tests are to be com­pat­ible, then for each of these com­bi­na­tions there must be an algo­rithm assigning to it a prob­a­bility equal to 1. In other words, there must be

  • a line (1-​​dimensional sub­space) L1 con­tained in both M and N,
  • a line L2 con­tained in both M and N,
  • a line L3 con­tained in both M and N.

It is easy to see that these require­ments are sat­is­fied: M is con­tained in N, so L1 exists. Because M is not equal to N (oth­er­wise the two tests would be iden­tical), N con­tains a line orthog­onal M, so L2 exists. And N is con­tained in M, so L3 exists. This war­rants the fol­lowing con­clu­sion: if either out­come of the first test implies either out­come of the second test, the two tests are compatible.

If nei­ther out­come of the first test implies either out­come of the second test, then com­pat­i­bility requires, in addi­tion, the exis­tence of

  • a line L4 con­tained in both M and N,

which assigns prob­a­bility 1 to both M and N.

It is readily shown that these four require­ments are sat­is­fied if and only if

the entire vector space V is spanned by four inter­sec­tions: that of M with N, that of M with N, that of M with N, and that of M with N.

The span of two sub­spaces A, B is the smallest sub­space con­taining both A and B. The inter­sec­tion of two sub­spaces A, B is the sub­space that con­tains all vector that are con­tained in both A and B.

Our next order of busi­ness is to trans­late this result into the lan­guage of projectors.

Some def­i­n­i­tions, to begin with:

  • The norm of a vector a is the pos­i­tive square root of <a|a>.
  • The norm of a unit vector equals unity.
  • A set of mutu­ally orthog­onal vec­tors a1, a2, a3,… is com­plete if for every vector b in V there are real or com­plex num­bers b1, b2, b3,… such that b = b1a1 + b2a2 + b3a3 + ···
  • A basis is a com­plete set of mutu­ally orthog­onal unit vectors.
  • The num­bers b1, b2, b3,… are the com­po­nents of b with respect to the basis a1, a2, a3,.…

Next, a couple of facts:

For every sub­space A of an n–dimen­sional vector space we can find a basis a1, a2, …, an such that a1, …, am span A, while am+1, …, an span A.

For every m–dimen­sional sub­space A of an n–dimen­sional vector space there is an oper­ator PA that projects vec­tors into A. Plug in the vector b = b1a1 + ··· + bnan, and out pops the vector PAb = b1a1 + ··· + bmam. In other words, the pro­jector PA cuts off the terms orthog­onal to A.

And finally, the trans­la­tion of the above result into the lan­guage of pro­jec­tors: T1 (with out­comes M and M) and T2 (with out­comes N and N) are com­pat­ible if and only if

PM and PN com­mute (that is, if for every vector b, PMPNb = PNPMb).

Be it men­tioned in passing that if PM and PN com­mute, then so do PM and PN, PN and PM, and PM and PN.

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