ch05 - CHAPTER 5. 1. We are given dx(A (x) * (x) dx (x) * A...

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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 ) ( 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 ))* ( ( x ) + λψ ( x )) -∞ = dx ( A )* + λ ( A )* + λ *( A )* + | | 2 ( A )* [ ] -∞ On the r.h.s. we get dx ( ( x ) + λψ ( x ))*( A ( x ) + λ A ( x )) -∞ = dx * A + λ * * A + λφ * A + | | 2 * A [ ] -∞ Because of the hermiticity of A , the first and fourth terms on each side are equal. For the rest, sine is an arbitrary complex number, the coefficients of and * are independent , and we may therefore identify these on the two sides of the equation. If we consider , for example, we get dx ( A ( x ))* ( x ) = -∞ dx ( x )* A ( 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 * ( x ) A 2 -∞ ( x ) Now define A ( x ) = ( x ). Then the above relation can be rewritten as
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A 2 ⟩= dx ψ ( x ) A φ ( x ) = dx -∞ -∞ ( A ( x ))* ( x ) = dx -∞ ( A ( x ))* A ( 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
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ch05 - CHAPTER 5. 1. We are given dx(A (x) * (x) dx (x) * A...

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