# ch05 - CHAPTER 5 1 We are given dx(A(x(x dx(x A(x Now...

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CHAPTER 5 . 1. We are given dx ( A Ψ ( x ))* Ψ ( x ) = -∞ dx Ψ ( x )* A Ψ ( x ) -∞ Now let Ψ ( x ) = φ ( x ) + λψ ( x ) , where λ is an arbitrary complex number. Substitution into the above equation yields, on the l.h.s. dx ( A φ ( x ) + λ A ψ ( x ))* ( φ ( x ) + λψ ( x )) -∞ = dx ( A φ )* φ + λ ( A φ )* ψ + λ *( A ψ )* φ + | λ | 2 ( A ψ )* ψ [ ] -∞ On the r.h.s. we get dx ( φ ( x ) + λψ ( x ))*( A φ ( x ) + λ A ψ ( x )) -∞ = dx φ * A φ + λ * ψ * A φ + λφ * A ψ + | λ | 2 ψ * A ψ [ ] -∞ Because of the hermiticity of A , the first and fourth terms on each side are equal. For the rest, sine λ is an arbitrary complex number, the coefficients of λ and λ * are independent , and we may therefore identify these on the two sides of the equation. If we consider λ , for example, we get dx ( A φ ( x ))* ψ ( x ) = -∞ dx φ ( x )* A ψ ( x ) -∞ the desired result. 2. We have A + = A and B + = B , therefore ( A + B ) + = ( A + B ). Let us call ( A + B ) = X . We have shown that X is hermitian. Consider now ( X + ) n = X + X + X + X + = X X X X = ( X ) n which was to be proved. 3. We have A 2 ⟩= dx ψ * ( x ) A 2 -∞ ψ ( x ) Now define A ψ ( x ) = φ ( x ). Then the above relation can be rewritten as

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A 2 ⟩= dx ψ ( x ) A φ ( x ) = dx -∞ -∞ ( A ψ ( x ))* φ ( x ) = dx -∞ ( A ψ ( x ))* A ψ ( x ) 0 4. Let U = e iH = i n H n n ! n = 0 . Then U + = ( - i ) n ( H n ) + n ! n = 0 = ( - i ) n ( H n ) n ! n = 0 = e - iH , and thus the hermitian conjugate of e iH is e -iH provided H = H +. . 5. We need to show that e iH e - iH = i n n ! n = 0 H n ( - i ) m m ! m = o H m = 1 Let us pick a particular coefficient in the series, say k = m + n and calculate its coefficient. We get, with m = k n , the coefficient of H k is i n n ! n = 0 k ( - i ) k - n ( k - n )!
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