# 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.

We 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.

→ Next