# hws03 - Homework 3 Solutions TA Kevin Chiou(Original...

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Homework 3 Solutions TA: Kevin Chiou (Original Solutions by Matt Block) Problem 1 Let’s start by stating what we know. We have from Griffiths Eq. (1.33) that: h ˆ p i = - i ¯ h Z Ψ * Ψ ∂x ! d x (1) The limits of integration are implicitly from negative infinity to positive infinity. We have Schr¨ odinger’s Equation (and its complex conjugate): i ¯ h Ψ ∂t = H Ψ (2) - i ¯ h Ψ * ∂t = H Ψ * (3) Here, H = H ( x ) because we are working in position space. Recall that ˆ H is Hermitian so the complex conjugate of H is itself. Lastly, we also have the generic form of the Hamiltonian (in a position space representation): ˆ H . = H ( x ) = - ¯ h 2 2 m 2 ∂x 2 + V ( x ) (4) We’ll be using integration by parts a bit in this problem and I won’t be writing out all the steps. You can look at the footnote on page 16 of Griffiths to recall the details. In particular, every time you integrate by parts, you get two terms; one still has an integral in it and the other has no integral and is being evaluated at the original limits of integration. This second term is called the surface term and since Ψ (and Ψ * ) is (are) zero at ±∞ (for normalizable wave functions), these surface terms will always evaluate to zero throughout this problem. Bear in mind that this is not universally true; I don’t want you to forget how to do integration by parts, but if you notice that I only write one term after doing it, it’s because the surface term is zero. Additionally, the (generally) nonzero term with the integral obeys a simple property if the original integral had a derivative in the integrand; that is, often we integrate by parts to “move” a derivative from one term to another. The price we pay is a minus sign. Consider the following example: assume f ( x ) and g ( x ) go to zero at ±∞ . Then: Z -∞ f ( x ) d g ( x ) d x d x = - Z -∞ d f ( x ) d x g ( x )d x (5) 1

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Notice the minus sign. This technique is very useful so become comfortable with it. Moving on, we start by taking the time derivative of Eq. (22) above: d h ˆ p i d t = - i ¯ h Z Ψ * ∂t Ψ ∂x + Ψ * ∂x Ψ ∂t ! d x (6) Important notational clarification: here we are assuming that x and t are independent variables; that is, x does not depend on t . Therefore, it is legitimate to bring the time derivative inside the integral (as a partial derivative) and to interchange differentiation with respect to x and with respect to t . Now we do our first integration by parts on the second term: d h ˆ p i d t = - i ¯ h Z Ψ * ∂t Ψ ∂x - Ψ * ∂x d t ! d x (7) d h ˆ p i d t = Z " - i ¯ h Ψ * ∂t ! Ψ ∂x + Ψ * ∂x i ¯ h Ψ ∂t !# d x (8) Now we use Eqs. (23) and (24) above: d h ˆ p i d t = Z " H Ψ * Ψ ∂x + Ψ * ∂x H Ψ # d x (9) Now we can insert Eq. (25) for H . I will collect the two terms coming from the kinetic energy part of the Hamiltonian and separately collect the two terms with the potential energy, V ( x ), in them: d h ˆ p i d t = - ¯ h 2 2 m Z " 2 Ψ * ∂x 2 Ψ ∂x + Ψ * ∂x 2 Ψ ∂x 2 # d x + Z " V ( x * Ψ ∂x + Ψ * ∂x V ( x # d x (10) d h ˆ p i d t = - ¯ h 2 2 m Z ∂x " Ψ * ∂x
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