On fixed points of Lüders operationLiu Weihua and Wu Junde Citation: Journal of Mathematical Physics 50, 103531 (2009); doi: 10.1063/1.3253574 View online: http://dx.doi.org/10.1063/1.3253574 View Table of Contents: http://scitation.aip.org/content/aip/journal/jmp/50/10?ver=pdfcov Published by the AIP Publishing Articles you may be interested in On supremum of bounded quantum observable J. Math. Phys. 50, 083513 (2009); 10.1063/1.3204082 Cumulants and the moment algebra: Tools for analyzing weak measurements J. Math. Phys. 50, 042103 (2009); 10.1063/1.3093270 The Fixed Point Iteration and Newton’s Methods for the Nonlinear Wave Equation AIP Conf. Proc. 1048, 482 (2008); 10.1063/1.2990968 On spectra of Lüders operations J. Math. Phys. 49, 022110 (2008); 10.1063/1.2840472 Invariant analytic orthonormalization procedure with an application to coherent states J. Math. Phys. 48, 043505 (2007); 10.1063/1.2711371

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On fixed points of Lüders operationLiu Weihua and Wu Jundea�

Department of Mathematics, Zhejiang University, Hangzhou 310027, People’s Republic ofChina

�Received 27 July 2009; accepted 16 September 2009; published online 28 October 2009�

In this paper, we give a concrete example of a Lüders operation LA with n=3, suchthat LA�B�=B does not imply that B commutes with all E1, E2, and E3 in A, thisexample answers an open problem of Professor Gudder. © 2009 American Instituteof Physics. �doi:10.1063/1.3253574�

Let H be a complex Hilbert space, B�H� be the bounded linear operator set on H, E�H�= �A :0�A� I�, A= �Ei�i=1

n �E�H�, and �i=1n Ei= I, where 1�n��. The famous Lüders operation

LA is a map which is defined on B�H� by

LA:A → �i=1

n

Ei1/2AEi

1/2.

A question related to a celebrated theorem of Lüders operation is whether LA�A�=A for someA�E�H� implies that A commutes with all Ei for i=1,2 , . . . ,n.1 The answer to this question ispositive for n=2 �Ref. 2� and negative for n=5.1 In this paper it is shown, by using a simplederivation of the example of Arias–Gheondea–Gudder in Ref. 1, that the answer is negative aswell for n=3, a question raised by Gudder in 2005.3

First, we denote B�H�LA= �B�B�H� :LA�B�=B� is the fixed point set of LA, A� is the com-mutant of A.

Lemma 1: �Ref. 1� If B�H�LA=A�, then A� is injective.Lemma 2: �Ref. 1� Let F2 be the free group generated by two generators g1 and g2 with

identity e, C be the complex numbers set, and H= l2�F2� be the separable complex Hilbert space,

H = l2�F2� = � f �f:F2 → C, � �f�x��2 � �� .

For x�F2 define �x :F2→C by �x�y� equals 0 for all y�x and 1 when y=x. Then ��x �x�F2� isan orthonormal basis for H. Define the unitary operators U1 and U2 on H by U1��x�=�g1x andU2��x�=�g2x. Then the von Neumann algebra N which is generated by U1 and U2 and its com-mutant N� are not injective.

Now, we follow Lemma 1 and Lemma 2 to prove our main result:Let C1 be the unit circle in C and h be a Borel function defined on C1 as the following:

h�ei��=� for �� �0,2��. Then A1=h�U1� and A2=h�U2� are two positive operators in N. Takingthe real and imagined parts of U1=V1+ iV2 and U2=V3+ iV4, N is generated by the self-adjointoperators �V1 ,V2 ,V3 ,V4�. Since functions cos and sin are two Borel functions, we have V1

= 12 �U1+U1

��=cos�A1�, V2=sin�A1�, V3=cos�A2�, and V4=sin�A2�. Thus N is contained in the vonNeumann algebra which is generated by A1 and A2.

On the other hand, it is clear that the von Neumann algebra which is generated by A1 and A2

is contained in N. So N is the von Neumann algebra which is generated by A1 and A2. Let E1

=A1 /2A1, E2=A2 /2A2, and E3= I−E1−E2. Then A= �E1 ,E2 ,E3��E�H� and E1+E2+E3= I.Now, we define the Lüders operation on B�H� by

a�Author to whom correspondence should be addressed. Electronic mail: wjd@zju.edu.cn.

JOURNAL OF MATHEMATICAL PHYSICS 50, 103531 �2009�

50, 103531-10022-2488/2009/50�10�/103531/2/$25.00 © 2009 American Institute of Physics

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LA�B� = �i=1

3

Ei1/2BEi

1/2.

It is clear that the von Neumann algebra which is generated by �E1 ,E2 ,E3� is N. It follows fromLemma 1 and Lemma 2 that B�H�LA�A�, thus there exists a D�B�H�LA \A�. The real part or theimaginary part D1 of D also satisfies D1�B�H�LA \A�. If D2= D1I−D1, then D2�0. Thus D3

=D2 / D2�E�H� and D3�B�H�LA \A�. We have proven the following theorem which answeredthe question in Ref. 3.

Theorem 1: Let H= l2�F2�, A= �Ei�i=13 be defined as above. Then there is a B�E�H�, such that

LA�B�=B, but B does not commute with all E1, E2, and E3.

ACKNOWLEDGMENTS

The authors wish to express their thanks to the referee for his �her� important comments andsuggestions. This project is supported by Natural Science Foundation of China �Grant Nos.10771191 and 10471124�.

1 A. Arias, A. Gheondea, and S. Gudder, J. Math. Phys. 43, 5872 �2002�.2 P. Busch and J. Singh, Phys. Lett. A 249, 10 �1998�.3 S. Gudder, Int. J. Theor. Phys. 44, 2199 �2005�.

103531-2 L. Weihua and W. Junde J. Math. Phys. 50, 103531 �2009�

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