Wigner ’ s Theorem in s * and sn ( H ) Spaces

Wigner theorem is the cornerstone of the mathematical formula of quantum mechanics, it has promoted the research of basic theory of quantum mechanics. In this article, we give a certain pair of functional equations between two real spaces s or two real spaces ( ) n s H , that we called “phase isometry”. It is obtained that all such solutions are phase equivalent to real linear isometries in the space s and the space ( ) n s H .


Introduction
Mazur and Ulam in [1] proved that every surjective isometry U between X and Y is a affine, also states that the mapping with ( ) is a linear isometry.we can say the mapping : V X Y → is phase equivalent to J.
If the two spaces are Hilbert spaces, Rätz proved that the phase isometries : V X Y → are precisely the solutions of functional equation in [7].If the two spaces are not inner product spaces, Huang and Tan [8] gave a partial answer about the real atomic p L spaces with 0 p > .Jia and Tan [9] get the conclusion about the  -type spaces.In [6], xiaohong Fu proved the problem of isometry extension in the s space detailedly.
In this artical, we mainly discuss that all mappings s H s H → also have the properties, that are solutions of the functional American Journal of Computational Mathematics equation All metric spaces mentioned in this artical are assumed to be real.

Results about s
First, let us introduction some concepts.The s space in [10], which consists of all scalar sequences and for each elements x y ⊥ .We can suppose And we also have It is easy to know ( ) , similarty to the above ( ) and we have as the same x y x y − = + .It follows that Similarly to the proof of necessity, we get x y ⊥ . Lemma Proof: We proof first that for any ( ) ≤ ≤ , there is a unique ( ) Hence, by the "pigeon nest principle" (or Pigeonhole principle) there must exist ( ) a contradiction which implies l p = .From this 1 2 θ θ = − follows.Finally, there is a unique θ with

=
, by the result in the last step, we have ( ) So, we get and we also have through the two equalities of above In the end, The proof is complete.
Note that the defination of 0 V , we can easily get ( ) , where x e e e e e r r λ λ On the other hand, we have Combiniing the two equations, we obtain that Analysis of the equation, according to the monotony of the function, that is The proof is complete. The next result shows that a mapping satisfying functional Equation (1) has a property close to linearity.
Lemma 2. 4. Let surjective mapping satisfying Equation (1).there exist two real numbers α and β with absolute 1 such that for all nonzero vectors x and y in X, x and y are orthogonal.Proof: Let x and y be nonzero orthogonal vectors in X, we write where . We infer from Equation (1) that Through the above equation we can get 0 , and similarly ( ) mapping satisfying Equation (1).Then V is injective and ( ) ( ) Proof: Suppose that V is surjective and ( ) ( ) Putting y x = in the Equation ( 1), this yields , using the Equation (1) for , , x y z , we obtain This lead to the contradiction that ( ) 0 surjective mapping satisfying Equation (1).Then V is phase equivalent to a linear isometry J.
Proof: Fix 0 γ ∈ Γ , and let . By Lemma 2.4 we can write for any z Z ∈ .Then, we can define a mapping for 0 λ ∀ ≠ ∈ .The J is phase equivalent to V.So it is easily to know that J satisfies functional Equation (1).For any z Z ∈ , and for any z Z ∈ , and 0 λ ∀ ≠ ∈ .
That yields . On the other hand, x y X ∈ , by assumed conditions, so J is a surjective isometry.Theorem 2.7.Let X s = and Y s surjective mapping satisfying Equation (1).Then V is phase equivalent to a linear isometry J.
Proof: According to [10] Theorem 1, Theorem 2 the author presents some results of extension from some spheres in the finite dimensional spaces ( ) n s .And also we have the above Theorem 2.6, so we can get the result easily.

( )
n s H denote the set of all elements of the form Some notations used: Specially, when ( ) 0 x k = , we have Next, we study the phase isometry between the space ( ) if V is a surjective phase isometry, then V is phase equivalent to a linear isometry J.
Proof: "⇒" Take any two elements . Then we have at the same time, we have That means The proof of sufficiency is similar to the Lemma 2.1.
Finally, we assert that, there exists ( ) , by the result in the last step, we have e e e e x l x l x l x l x l x l x l x l , where ( ) ( ) ( ) ( ) We infer from Equation (1) that , Through the above equation we can get ( ) ( ) The proof is complete.Lemma 3.5.Let American Journal of Computational Mathematics for any z Z ∈ .Then, we can define a mapping ( ) ( ) for 0 λ ∀ ≠ ∈ .The J is phase equivalent to V.So it is easily to know that J satisfies functional Equation (1).For any z Z ∈ , and 0 λ ∀ ≠ ∈ , That means ( ) ( ) . On the other hand, , z z Z ∀ ∈ , It follows that ( ) ( ) J x J y x y − = − for all , x y X ∈ , by assumed conditions, so J is a surjective isometry.

Conclusion
Through the analysis of this article, we can get the conclusion that if a surjective mapping satisfying phase-isometry, then it can phase equivalent to a linear isometry in the space s and the space ( ) s H .
then U is linear.Let X and Y be normed spaces, if the mapping : It was called isometry.About it's main properties in sequences spaces, Tingley, D, Ding Guanggui, Fu Xiaohong in [2] [3] [4] [5] [6] proved.So, we give a new definition that if there is a function which is a contradiction.So we obtain z x = − , and we must have y x = .For otherwise we get y z x = = − and In this part, we mainly introduce the space ( ) n s H , where H is a Hilbert space.In [11] mainly discussed the isometric extension in the space , the F-norm of x is defined by

So, we can get k
Equation(1).there exist two real numbers α and β with absolute 1 such that for all nonzero vectors x and y in X, x and y are orthogonal.Proof: Let