aa r X i v : . [ m a t h . OA ] F e b Eigenstates of C*-Algebras
Luther RinehartFebruary 16, 2021
Abstract
We introduce the notion of eigenstate of an operator in an ab-stract C*-algebra, and prove several properties. Most significantly, ifthe operator is self-adjoint, then every element of its spectrum has acorresponding eigenstate.
In quantum mechanics an important role is played by eigenvectors of opera-tors, namely those vectors v such that x ( v ) = λv for some λ ∈ C . Such vec-tors represent physical states in which the operator x has the definite value λ . The algebraic formulation of quantum mechanics seeks to develop thetheory using only a C ∗ -algebra of operators, without reference to a Hilbertspace. The purpose of this paper is to examine the analogous concept toeigenvector in the purely algebraic setting.A key advantage of considering algebraic states is shown by theorem 5.Typically, not every point in an operator’s spectrum has a correspondingeigenvector. Indeed, an operator might have no eigenvectors at all. However,theorem 5 shows that, as long as the operator is self-adjoint, every point inthe spectrum has an algebraic eigenstate.In the following, A will always be a C ∗ -algebra with 1. Its space of states S ( A ) consists of the normalized positive linear functionals. Definition 1.
Let x ∈ A . A state E ∈ S ( A ) is an eigenstate of x with eigenvalue λ ∈ C if ∀ y ∈ A, E ( yx ) = λE ( y ). Theorem 1.
Every eigenvalue of x is contained in the spectrum of x . Proof.
Let λ be an eigenvalue of x . For contradiction, suppose y = ( x − λ ) − .Let E be the associated eigenstate for λ . By definition, E ( y ( x − λ )) = 0.But this says E (1) = 0, a contradiction. So x − λ is not invertible.1 heorem 2. A state E is an eigenstate of x with eigenvalue λ if and only if E (( x − λ ) ∗ ( x − λ )) = 0. Proof.
The implication ⇒ follows by direct computation or by noting thatsince ∀ y, E ( y ( x − λ )) = 0, then this certainly holds for y = ( x − λ ) ∗ .To show the converse, apply the Cauchy-Schwartz inequality for states: ∀ y | E ( y ( x − λ )) | ≤ E ( y ∗ y ) E (( x − λ ) ∗ ( x − λ )) = 0The next theorem shows that the notion of eigenstate coincides with thestandard notion of eigenvector if we fix a representation and focus only onvector states. Theorem 3.
Let π : A → B ( H ) be a *-representation of A and let v ∈ H .Let E v be the vector state associated with v , namely, E v ( x ) = h v, π ( x ) v i . E v is an eigenstate of x with eigenvalue λ if and only if π ( x ) v = λv . Proof.
Suppose π ( x ) v = λv . ∀ y, E v ( yx ) = h v, π ( yx ) v i = h v, π ( y ) π ( x ) v i = λ h v, π ( y ) v i = λE v ( y ).Conversely, suppose E v is an eigenstate of x with eigenvalue λ . By theorem2, 0 = E v (( x − λ ) ∗ ( x − λ )) = h v, π (( x − λ ) ∗ ( x − λ )) v i = h π ( x − λ ) v, π ( x − λ ) v i .So π ( x − λ ) v = 0, that is π ( x ) v = λv .For the following theorem, recall that if x = x ∗ , the C*-algebra generatedby x can be canonically identified with C ( σ ( x )), the continuous functions onthe spectrum of x . Theorem 4.
Let x = x ∗ . If E is an eigenstate of x with eigenvalue λ , then ∀ f ∈ C ( σ ( x )), E is an eigenstate of f with eigenvalue f ( λ ). Proof.
By induction on n , the theorem holds for f = x n . Indeed, ∀ y, E ( yx n ) = E ( yx n − x ) = λE ( yx n − ) = λλ n − E ( y ) = λ n E ( y ).By linearity, for all polynomials p , E is an eigenstate of p with eigenvalue p ( λ ). Now let f ∈ C ( σ ( x )). By the Stone-Weierstrass theorem, there existsa sequence of polynomials { p n } converging uniformly to f . Using continu-ity of multiplication and of E , ∀ y, E ( yf ) = lim E ( yp n ) = lim p n ( λ ) E ( y ) = f ( λ ) E ( y ). 2f x is self-adjoint, every point of its spectrum has a corresponding eigen-state. The proof is analogous to that of theorem 1.3.1 in [1]. Theorem 5.
Let x = x ∗ . If λ ∈ σ ( x ), then λ is an eigenvalue of x. Proof.
Let x = x ∗ and let λ ∈ σ ( x ). Observe that finding an eigenstate witheigenvalue λ is equivalent to finding a state which vanishes on the left idealgenerated by ( x − λ ). Let J = A ( x − λ ) be this ideal. Since ( x − λ ) is self-adjoint and not invertible, it is not left-invertible, so 1 / ∈ J . Furthermore, noelement of J can be invertible, as this would imply y ∈ J ⇒ y − y = 1 ∈ J .Thus ∀ y ∈ J, ∈ σ ( y ) ⇒ ∈ σ ( y − ⇒ ≤ k y − k So 1 / ∈ ¯ J . By the Hahn-Banach theorem, we can find a linear functional E ∈ A ∗ satisfying E | ¯ J = 0 and E (1) = 1 = k E k . Such E will be aneigenstate of x with eigenvalue λ . Theorem 6.
Any set of eigenstates of x all having distinct eigenvalues, arelinearly independent. Proof.
Let { E α } be a collection of eigenstates of x with eigenvalues { λ α } ,and { λ α } all distinct. By induction on n , if P nα =1 c α E α = 0, then c = ... = c n = 0.The case n = 1 is clear.Make the induction hypothesis that the conclusion holds for n = k , andsuppose that P k +1 α =1 c α E α = 0. Apply this sum to the element y ( x − λ k +1 ),with y arbitrary: 0 = k +1 X α =1 c α E α ! ( y ( x − λ k +1 ))= k X α =1 c α ( λ α − λ k +1 ) E α ( y )By the induction hypothesis, c α ( λ α − λ k +1 ) = 0 for α = 1 ...k . But the { λ α } are all distinct, so λ α − λ k +1 = 0 for α = 1 ...k . So c = ... = c k = 0. Finally, c k +1 = 0 by the n = 1 case. 3 efinition 2. A pair of positive linear functionals ω, φ ∈ A ∗ + are orthogonal if k ω − φ k = k ω k + k φ k . (See definition 3.2.3 in [2]) Theorem 7. If x = x ∗ , then any two eigenstates of x with different eigen-values are orthogonal. Proof.
Let E and E be eigenstates of x with respective eigenvalues λ and λ , with λ = λ . Then there exists a continuous function f ∈ C ( σ ( x )) suchthat k f k = 1, f ( λ ) = 1, and f ( λ ) = −
1. Then ( E − E )( f ) = 1 − ( −
1) =2. Consequently, 2 ≤ k E − E k ≤ k E k + k E k = 2, so E and E areorthogonal.Finally, eigenstates of projection operators are precisely those obtainedby projection. Theorem 8.
Let p be a self-adjoint projection and E a state. Let pE be thestate given by pE ( x ) = E ( pxp ) E ( p ) . The following are equivalent:1. pE = E E is an eigenstate of p with eigenvalue 13. E ( p ) = 1 Proof. (1) ⇒ (2): ∀ y ∈ A, pE ( yp ) = E ( pypp ) /E ( p ) = E ( pyp ) /E ( p ) = pE ( y ), so pE is an eigenstate of p with eigenvalue 1.(2) ⇒ (3): Obvious.(3) ⇒ (2): Let E ( p ) = 1. E (( p − ∗ ( p − E (( p − ) = E (1 − p ) =1 − E ( p ) = 0 and apply theorem 2.(2) ⇒ (1): Suppose E is an eigenstate of p with eigenvalue 1, that is ∀ y ∈ A, E ( yp ) = E ( y ). By taking adjoint, this also gives E ( py ∗ ) = E ( y ∗ ).Since ∗ is an involution, we have ∀ y E ( py ) = E ( y ). Now replacing y with yp , E ( pyp ) = E ( yp ) = E ( y ). And since E ( p ) = 1, altogether this shows that ∀ y pE ( y ) = E ( y ), so E = pE . 4 eferences [1] G. J. Murphy, C ∗ -algebras and operator theory , Academic Press, Boston,1990.[2] G. K. Pedersen, C ∗ -algebras and their automorphism groups , LondonMathematical Society Monographs14