Chapter10_full - In the next several weeks we will 1 Solve...

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9/9/13 1 In the next several weeks, we will 1. Solve a one dimensional problem, the particle in a box Chapter 10 2. Solve another one dimensional problem, the simple harmonic oscillator Chapter 11 3. Solve the hydrogen atom multi-electron atoms, and hydrogen molecule Chapter 12, 13 Schedule: Monday September 9 th Finish Chapter 10 Wednesday September 11 th Review Friday September 13 th First exam Approach for solving Schrodinger s equation 1. Determine the coordinate system, for one dimension use just x, for three dimensions use x,y,z or r, Θ , φ 2. For the given problem determine the potential everywhere, that is, write and expression for V(r) 3. Substitute the function V(r) into the equation and solve for the wavefunctions ψ (x) 4. Use the boundary conditions in order to determine the allowed values of any parameters 5. Use the normalization condition so that the total probability is one E " ( r ) = # ! 2 2 m $ 2 " ( r ) + V ( r ) " ( r )
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9/9/13 2 Write Schrodinger’s equation if V(x)=0 b . E ! ( x ) = ! ! 2 2 m d 2 dx 2 ! ( x ) + V ( x ) ! ( x ) d . E ! ( x ) = ! ! 2 2 m d 2 dx 2 ! ( x ) + kx 2 2 ! ( x ) a . E ! ( x ) = ! ! 2 2 m d 2 dx 2 ! ( x ) c . E ! ( x ) = ! ! 2 2 m d 2 dx 2 ! ( x ) + ( Ax ) ! ( x ) Correct answer General formula Answer if V(x)=Ax Answer for harmonic oscillator Chapter 10 One-dimensional particle in the box E " ( x ) = # ! 2 2 m d dx 2 2 " ( x ) + V ( x ) " ( x ) A particle of mass m is confined between two infinite walls at x = 0 and x = L ! ( x ) = 0 at x = 0 and x = L E ! ( x ) = ! ! 2 2 m d dx 2 2 ! ( x ) By convention, the particle cannot be in the region with an infinite potential, so we can constrain the wavefunction at the two edges of the box: The particle is only within the box where V (x)= 0 so the equation simplifies to
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9/9/13 3 E " ( x ) = # ! 2 2 m d dx 2 2 " ( x ) Need to solve the equation For what function is the 2nd derivative proportional to the function? d dx 2 2 " ( x ) # " ( x ) Answer: sin(x), cos(x), and e x E " ( x ) = # ! 2 2 m d dx 2 2 " ( x ) ! ( x ) = A sin kx + B cos kx Assume d dx A sin kx + B cos kx ( ) = A d dx sin kx + B d dx cos kx = A ( k cos kx ) + B ( k ( ! sin kx )) E ! ( x ) = !
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  • Spring '14
  • dx, 2 L, 3 $, 1 nm, 3.0 $

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