Theorem. For any bounded operators A and B on a complex Hilbert space H, it is impossible that AB – BA = I, where I is the identity operator on H.
To establish the theorem, we note some elementary results concerning the spectrum of a bounded operator on H.
Definition. For any bounded operator A on H, the spectrum of A is given by .
Proposition 1. For any bounded operator A, sp(A) is a nonempty closed subset of the closed disk with radius .
Proof: [Next time. The relatively more difficult part is that the spectrum is never empty.]
Proposition 2. For any bounded operators A and B, .
Proof: Suppose c is nonzero such that AB -cI is invertible. It suffices to consider the case c = 1. Let C be the inverse of AB – I. We claim that BCA – I is the inverse of BA – I. Indeed,
(BA – I)(BCA – I) = BABCA – BA – BCA + I = B(AB – I)CA – BA + I = BA – BA + I = I
and similarly (BCA -I)(BA – I) = I.
Proof of Theorem: Suppose AB – BA = I for some bounded operators A and B on H. Then by Proposition 2, for any positive integer n, if both sp(AB) and sp(BA) contain 0 or do not contain 0 then
and if 0 lies in precisely one of sp(AB) and sp(BA), say in sp(BA), then
Therefore sp(AB) or sp(BA) is unbounded, contradicting Proposition 1.
A consequence of the theorem is that the Heisenberg commutation relation in quantum mechanics cannot be implemented using bounded operators.