Printer Friendly


L. Herman [*]

Abstract.--This note defines a scalar measure of the commutativity of two orthogonal projections.

This note is set in the context of a complex Hilbert space (H, [less than]*/*[greater than]). It's purpose is to define a scalar that tells whether or not two projections commute. A linear operator P on H is a projection if it is self adjoint (P = [P.sup.*]) and idempotent (P = [P.sup.2] It is well known that the projections form a lattice where if P and Q are projections, the greatest lower bound P[conjunction]Q is the projection on the intersection of the range of P and the range of Q and the least upper bound P V Q is the projection on the closure of the linear sum of the range of P with the range of Q (Holland 1970). Now the sum of projections need not be a projection, though the sum is still self adjoint. Self adjoint operators can be partially ordered by declaring S [less than or equal to]T if and only if [less than]Sx/x[greater than] [less than or equal to] [less than]Tx/x[greater than] for all x in H. The following fact is needed in the proof of the first theorem.

Let P and Q be projections and P + Q [less than or equal to] I in the ordering on self adjoint operators. Then P [perpendicular to] Q; that is, the range of P is orthogonal to the range of Q. Note that P + Q [less than or equal to] I implies P [less than or equal to] I - Q (Halmos 1957). One last reminder. The operator norm for a self adjoint T is given by [parallel]T[parallel] = LUB {/[less than]Tx/x [greater than]/:[parallel]x[parallel] [less than or equal to] 1 } (Berberian 1961). Now for the first theorem.


Suppose P and Q are projections that are not both the zero operator. Then the following conditions are equivalent:

(1.1) PQ = QP

(1.2) PQP = QPQ

(1.3) P + Q - P [conjunction] Q is a projection

(1.4) [parallel] P + Q - P [lambda]Q[parallel] = 1

(1.5) [parallel] P + Q - P [lambda] Q[parallel] [less than or equal to] 1

(1.6) I - P - Q + P [lambda] Q is positive semidefinite

Proof. (1.1) implies (1.2): Suppose PQ = QP. Then PQP = PPQ = PQ and QPQ = QQP = QP. Thus PQP = PQ = QP = QPQ.

(1.2) implies (1.1): First note PQP = QPQ implies PQP = PQPQ (PQ) [2] so (PQP) [2] = PQPPQP = PQPQP = [(PQ).sup.2]P = PQPP = PQP. This shows PQP is idempotent. It is clearly self adjoint so PQP is a projection. To establish (1.1) the fact that T*T = 0 implies T = 0 for operators on Hubert space is used. Compute (PQP - QP)*(PQP - QP) = (PQP - PQ)(PQP - QP) = [(PQP).sup.2] - PQPQP - PQPQP + PQQP = PQP - (PQP)QP - (PQ)(PQP) + PQP = 2 PQP - [(PQ).sup.2] P - [(PQ).sup.2]P = 2 PQP - (PQP)P - (PQP)P = 0. Thus PQP = QP. By taking adjoints, PQP PQ also so (1.1) follows (Note: the argument given above works in any * -ring i.e. a ring with involution where x*x = 0 implies x = 0).

(1.1) implies (1.3): If PQ = QP, then PQ P [lambda] Q and P + Q - P [lambda] Q is the supremum of P and Q in the lattice of projections (Halmos 1957).

(1.3) implies (1.4): If P + Q - P [lambda] Q is the zero projection then P + Q = P [lambda] Q so PQ + Q = (P + Q)Q = (P [lambda] Q)Q = P [lambda]Q = P + Q which shows PQ = P whence P [less than or equal to]Q. Similarly QP = Q and so Q [less than or equal to]P. But then P = Q so 2P = P + Q = P [lambda] Q = P [lambda] P = P so P = 0 = Q, a contradiction. Thus P + Q - P [lambda] Q is a nonzero projection and the operator norm of any nonzero projection is one.

(1.4) implies (1.5) is trivial.

(1.5) implies (1.6): Next suppose [parallel]P + Q - P [lambda] Q [less than or equal to] 1. Then [less than](P + Q - P [lambda] Q) x x[greater than] [less than or equal to]\ [less than](P + Q - P [lambda] Q)x \ x [greater than] \ [less than or equal to] [parallel] (P + Q - P [lambda] Q)x[parallel] [parallel]x[parallel] [less than or equal to] [parallel] P + Q - P [lambda] Q // [[parallel]x[parallel].sup.2] [less than or equal to] [[parallel]x[parallel].sup.2] which shows that P + Q - P [lambda] Q is positive semidefinite.

(1.6) implies (1.1): Assume the condition. Then P + (Q - P [lambda] Q) [less than or equal to]I so by the remarks above, P ""(Q-P [lambda] Q). Thus PQ-P [lambda] Q = P (Q - P [lambda] Q) = 0 = (Q - P [lambda] Q)P = QP- P [lambda] Q. Therefore PQ = QP.

Note that numerous purely lattice theoretic conditions equivalent to PQ = OP are known. For example,(PV[Q.sup.[perpendicular line]]) [conjunction] Q = P [conjunction] Q and (P [conjunction] Q) [disjunction] (P [conjunction] [Q.sup.[perpendicular line]] = P are two such conditions (Holland 1970).

Now a scalar is introduced that will tell if two projections commute. For P and Q not both the zero operator, define [delta](P,Q) = [parallel]P + Q - P [conjunction] Q[parallel] - 1.

In (Piziak et. al. 1999), explicit formulas are given for P [conjunction] Q in terms of P and Q when the Hubert space is finite dimensional so the formula for [delta] can be made more explicit in this case.


With the notation above

(2.1) [delta](P,Q) = [delta](Q,P)

(2.2) 0 [less than or equal to] [delta] (P,Q) [less than or equal to] 1

(2.3) [delta](P,Q) = 0 if and only if PQ = QP

(2.4) [delta](P,Q) = 1 if and only if [parallel]P + Q - P [conjunction] Q[parallel] = 2

Proof. In view of Theorem 1, only (2.2) is not immediate. If PQ = QP then [delta](P,Q)0 while if PQ [not equal to] QP then [delta](P,Q) [greater than] 0. Also [parallel]P + Q - P [conjunction] Q[parallel] [less than or equal to] [parallel]P[parallel] + [parallel]Q - P [conjunction] Q[parallel] [less than or equal to] 2 since Q - P [conjunction] Q is a projection. Thus [delta](P, Q) [less than or equal to] 1 always.

It is natural to speculate in view of Theorem 2, if some kind of probability interpretation can be given to [delta](P,Q). This is pursued next.


Let x and y be unit vectors in H and P and Q the projections onto the one dimensional subspaces they generate. Then [parallel]P + Q[parallel] = 1 + [perpendicular to] [less than]x [perpendicular to] y [greater than] [perpendicular to].

Proof. There is no loss in generality in assuming the dimension of H is 2 since the subspaces generated by x and y reduce P and Q and each is the zero operator on the orthocomplementary subspaces. First suppose that x and y are dependent. Then P = Q and [parallel]P + Q[parallel] = [parallel]2P[parallel] = 2.

Say y = [beta]x then [perpendicular to][beta][perpendicular to] = [parallel][beta]x[parallel] = [parallel]y[parallel] = 1 and so [perpendicular to] [less than] x [perpendicular to]y[greater than] [perpendicular to] = [perpendicular to] [less than]x [perpendicular to] [beta]x [greater than] [perpendicular to] = [perpendicular to][beta][perpendicular to] [parallel]x[[parallel].sup.2] = [perpendicular to][beta][perpendicular to] = 1. Thus [parallel]P + Q[parallel] = 1 + [perpendicular to] [less than]x [perpendicular to] y [greater than][perpendicular to] in this case. Now assume x and y are linearly independent. Then they form a basis of H. With respect to this basis, the matrix of P is [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and the matrix of Q is [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] where [alpha] = [less than] x \ y [greater than]. Since P + Q is self adjoint, [parallel]P + Q[parallel] is the largest of its eigenvalues (Halmos 1958). Now the matrix of [lambda]I - (P + Q) with respec t to this basis is [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] and so det[[lambda]I - (P + Q)] = [([lambda] - 1).sup.2] - [\[alpha]\.sup.2] = [[lambda] - 1 - \[alpha]\] [[lambda] - 1 + \[alpha]\]. So again [parallel]P + Q[parallel] = 1 + \[alpha]\.

Now the connection to [delta](P, Q) is apparent.


Let x and y be linearly independent unit vectors in our Hubert space and P and Q the projections onto the subspaces they generate. Then [delta](P,Q) = \[less than]x\y[greater than]\.

Proof. Since x and y are linearly independent P [conjunction] Q = 0. Then [delta](P,Q) = [parallel]P + Q[parallel] - 1 = \ [less than]x y [greater than]\ by the previous theorem.

Note that if x and y are linearly dependent then \ [less than]x y[greater than] 1 = 1. However, P = Q and so [delta](P,Q) = [parallel]P + Q - P [conjunction] Q[parallel] - 1 = [parallel]P + Q - Q[parallel] - 1 = [parallel]P[parallel] - 1 = 0.

Finally, in the conventional interpretation of quantum mechanics, each unit vector represents a state. Then [\[less than]x \ y[greater than] \.sup.2] is interpreted as the probability p(x, y) of transition between the states. By Corollary 4, p(x, y) = [delta][(P,Q).sup.2] if x and y are linearly independent. So in particular p(x,y) = 0 if and only if PQ = QP and p(x,y) = 1 if and only if [parallel]P + Q[parallel] = 2, i.e. is as large as possible.

(*.) Deceased


Halmos, p. 1957. An Introduction to Hilbert Space, Chelsea Press, New York, 114 pp.

Halmos, P. 1958. Finite Dimensional Vector Spaces, Van Nostrand, Princeton, 200 pp.

Berberian, S. K. 1961. Introduction to Hubert Space, Oxford University Press, New York, 206 pp.

Holland, Jr., S. 1970. Current interest in orthomodular lattices, Trends in Lattice Theory, Van Nostrand Reinhold, New York, 41-126 pp.

Maezynski, M.J. 1981. A numerical characterization of commuting projections in Hubert Space. Bull. of L'Acad. Polon. des Sci., 29:157-163.

Young, N. 1988. An Introduction to Hilbert Space, Cambridge Mathematical Textbooks, New York, 239 pp.

Piziak, R., P. L. Odell & R. Hahn. 1999. Constructing Projections on Sums and Intersections. Computers and Mathematics with Applications, 37:67-74.
COPYRIGHT 2001 Texas Academy of Science
No portion of this article can be reproduced without the express written permission from the copyright holder.
Copyright 2001 Gale, Cengage Learning. All rights reserved.

Article Details
Printer friendly Cite/link Email Feedback
Title Annotation:orthogonal projections
Author:Herman, L.; Piziak, R.
Publication:The Texas Journal of Science
Geographic Code:1U7TX
Date:Feb 1, 2001

Terms of use | Privacy policy | Copyright © 2019 Farlex, Inc. | Feedback | For webmasters