Ch03S13ThmsHermOps - Section 3.13: Theorems for Hermitian...

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3.71 Section 3.13: Theorems for Hermitian Operators (under construction) This section discussions some theorems for operators (refer to T. D. Lee’s Book). The “common theorems” show when two operators are equal and also discusses alternate methods for determining when an operator is Hermition. Previous sections in this book provide the basic definition of a Hermitian operator. Most important, these earlier sections show that a Hermitian operator has real eigenvalues and orthonormal eigenvectors. For quantum mechanics where a Hermitian operator represents an observable, we must also know the eigenvectors form a basis set for the Hilbert space. In particular, this means that the set of eigenvectors must be complete. Every fundamental motion of the physical situation can then be represented by one of the basis vectors. Topic 3.13.1: Common theorems Theorem 3.13.1: If ˆ O is a linear operator on a Hilbert space then ˆ O0 = iff ˆˆ xOy 0 == for all x,y in the Hilbert Space Proof: () O0 xO y x 0 0 =⇒ = = If ˆ 0 = for all x,y in the Hilbert space then take ˆ xO y = to get Oy Oy 0 = . Therefore, by the definition of inner product from Topic 3.2.3, we must have ˆ Oy 0 = for every “y”. Therefore, by definition of the zero operator, we must have ˆ = . Theorem 3.13.2: If ˆ H is a Hermitian linear operator in a Hilbert Space, then H0 xH x 0 =⇔ = for every vector x in the Hilbert space. Proof: x x = = We will show two results that hold for all vectors x,y in the Hilbert space, namely (a) xHy yHx 0 += and (b) 0 −= For then, by addition, ˆ 0 = for all x,y in the Hilbert space. Therefore by Thm 3.13.1, we have ˆ = To show result (a), we will require the starting assumption ˆ xHx 0 = for x in the Hilbert space. Not that if x,y are in the Hilbert space, then so is x+y. Therefore, by assumption, we must have ˆ ˆ ˆ 0 x yH x y yHy =+ + = + + + .
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3.72 Also note by assumption ˆˆ xHx yHy 0 = = , so that xHy yHx 0 + = as required for (a). To show (b), replace the vector “y” with the complex vector “iy” in part (a) to get x Hiy iy Hx 0 += . Factoring out the complex “i” using the complex conjugate implicity in the bra, we find 0 = . We replaced “y” with “iy” and “iy” is not in a “real” Hilbert space. So to show 0 H 0 =→ = for a real Hilbert space, we use result (a) as follows: ˆ ˆ 0 x Hy y Hx x Hy H y x + =+=+ where the last step follows from the definition of adjoint. Next, using the definition of Hermitian and the fact that the adjoint of the inner product reverses the order and includes a complex conjugate, we find * ˆ ˆ ˆ ˆ 0 x Hy Hy x x Hy x Hy x Hy x Hy =+ where the last step follows for real Hilbert spaces. Therefore, as a result, we have ˆ 0 = for all x,y in the Hilbert Space. Now use Theorem 5.14.1 to conclude ˆ H0 = . Theorem 3.13.3: A linear operator ˆ O on a Hilbert space is Hermitian provided ˆ xOx Rea l = for all vectors x in the Hilbert space.
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Ch03S13ThmsHermOps - Section 3.13: Theorems for Hermitian...

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