The wave function must be normalized which implies 1 Z \u03c8 x 2 dx A 2 Z e m\u03c9 x 2

# The wave function must be normalized which implies 1

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is a complex constant in general. The wave function must be normalized, which implies 1 = Z -∞ | ψ 0 ( x ) | 2 dx = | A | 2 Z -∞ e - 0 x 2 / ~ dx = | A | 2 r π ~ 0 (86) = A = 0 π ~ 1 / 4 (87) where we used the integral R -∞ e - au 2 du = p π a and chose A to be real. Therefore, the ground state wave function is ψ 0 ( x ) = 0 π ~ 1 / 4 e - 0 x 2 / 2 ~ (88) b) The expectation value x 2 is x 2 = Z -∞ x 2 | ψ 0 ( x ) | 2 dx = r 0 π ~ Z -∞ x 2 e - 0 x 2 / ~ dx = r 0 π ~ · ~ 2 0 r π ~ 0 = ~ 2 0 (89) where we used the integral R -∞ u 2 e - au 2 du = 1 2 a p π a . c) In the momentum representation, this is r 1 2 m ~ ω 0 0 i ~ d dp + ip φ 0 ( p ) = 0 (90) 0 dp + 1 ~ 0 0 ( p ) = 0 (91) This is identical to the differential equation we found in part a), with 0 1 0 , so the solution is φ 0 ( p ) = 1 π ~ 0 1 / 4 e - p 2 / 2 ~ 0 (92) 6
d) The action of ˆ x 2 on φ 0 ( p ) in the momentum representation is ˆ x 2 φ 0 ( p ) = - ~ 2 d 2 φ 0 dp 2 = - ~ 2 1 π ~ 0 1 / 4 d 2 dp 2 e - p 2 / 2 ~ 0 (93) = - ~ 2 1 π ~ 0 1 / 4 - 1 ~ 0 d dp pe - p 2 / 2 ~ 0 (94) = ~ 0 1 π ~ 0 1 / 4 e - p 2 / 2 ~ 0 - 1 ~ 0 p 2 e - p 2 / 2 ~ 0 (95) The expectation value x 2 is x 2 = Z -∞ ψ * 0 ( p x 2 ψ 0 ( p ) dp (96) = ~ 0 r 1 π ~ 0 Z -∞ e - p 2 / ~ 0 dp - 1 ~ 0 Z -∞ p 2 e - p 2 / ~ 0 dp (97) = ~ 0 r 1 π ~ 0 p π ~ 0 - 1 ~ 0 · ~ 0 2 p π ~ 0 (98) = ~ 2 0 (99) where we used the integrals R -∞ e - au 2 du = p π a and R -∞ u 2 e - au 2 du = 1 2 a p π a . e) The Fourier transform of ψ 0 ( x ) is φ 0 ( p ) = 1 2 π ~ Z -∞ ψ 0 ( x ) e - ipx/ ~ dx (100) = 0 π ~ 1 / 4 1 2 π ~ Z -∞ e - 0 x 2 / 2 ~ e - ipx/ ~ dx (101) = 0 π ~ 1 / 4 1 2 π ~ Z -∞ e - ( 0 / 2 ~ ) x 2 - ( ip/ ~ ) x dx (102) = 0 π ~ 1 / 4 1 2 π ~ · r 2 π ~ 0 e - p 2 / 2 ~ 0 (103) = 1 π ~ 0 1 / 4 e - p 2 / 2 ~ 0 (104) where we used the integral R -∞ e - au 2 + bu du = p π a e b 2 / 4 a . f) The matrix representation of ˆ a + = (ˆ a - ) in the energy eigenbasis is ˆ a + = (ˆ a - ) 0 0 0 0 . 1 0 0 0 . 0 2 0 0 . 0 0 3 0 . . . . . . (105) Therefore, the matrix representation of ˆ x in the energy eigenbasis is ˆ x = r ~ 2 0 a + + ˆ a - ) r ~ 2 0 0 1 0 0 . 1 0 2 0 . 0 2 0 3 . 0 0 3 0 . . . . . . (106) Note that this is Hermitian. 7
g) The expectation value x 2 in bracket notation is x 2 = h 0 | ˆ x 2 | 0 i = ( h 0 | ˆ x )(ˆ x | 0 i ). The matrix represen- tation of ˆ x | 0 i in the energy eigenbasis is ˆ x | 0 i → r ~ 2 0 0 1 0 0 . 1 0 2 0 . 0 2 0 3 . 0 0 3 0 . . . . . . 1 0 0 0 . = r ~ 2 0 0 1 0 0 . (107) The norm squared of this state is ( h 0 | ˆ x )(ˆ x | 0 i ) = ~ 2 0 . Therefore, the expectation value is x 2 = ~ 2 0 . 8