PIB1DResonance

PIB1DResonance - Physical Chemistry Course Number: C362 6...

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Physical Chemistry Course Number: C362 6 Particle-in-a-box (PIB) 1. Consider a linear poly-ene. 2. The electrons are completely delocalized inside the poly-ene, but cannot leave the molecular framework. 3. Let us approximate this system by a one-dimensional box, of length L. The potential energy of the electrons inside the polyenes can be approximated by the figure on the board. 4. We will assume that the electron feels the same potential at all points inside the molecular framework. (An approximation that actually works pretty well and explains results in many physical systems.) 5. The particle-in-a-box is “toy” problem, but it is the starting point for many important modern-day ideas such as: (a) A fundamental understanding of resonance on polyenes is possible through PIB. (b) An understanding of band structure in crystals arises from PIB. (c) Frontier molecular orbital theory, which is the starting point for the Woodward Hoffman rules in organic chemistry, is easily explained using PIB. (d) Quantum dots, wells, and wires become very much accessible. 6. So PIB is of fundamental importance. 7. The box edges have infinite repulsive potentials to keep the electrons inside the molecular framework. But inside the molecular framework the electrons are completely free to move as they should be on account of resonance in poly-enes. Chemistry, Indiana University 29 c c 2011, Srinivasan S. Iyengar (instructor)
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Physical Chemistry Course Number: C362 So, V ( x ) = 0 , 0 x L V ( x ) = , x < 0 and x > L (6.1) 8. In region II, the Hamiltonian for the system is: H = p 2 2 m + V = ¯ h 2 2 m 2 ∂x 2 (6.2) (note that the potential is zero). 9. We would like to solve the time-independent Schr¨odinger Equation to obtain the wavefunction for the system inside the box (in region II): H II ψ ( x ) = ( x ) ¯ h 2 2 m 2 ∂x 2 ψ ( x ) = ( x ) 2 ∂x 2 ψ ( x ) + 2 mE ¯ h 2 ψ ( x ) = 0 (6.3) If we make the substitution 2 mE ¯ h 2 = k 2 , 2 ∂x 2 ψ ( x ) + k 2 ψ ( x ) = 0 (6.4) To obtain the solution to this equation, we need to solve for ψ ( x ) . The func- tion should have a form such that, when differentiated twice, it gives the function back multiplied by a constant. The exponential function has this property and hence lets guess the solution to this equation as exp { sx } . Sub- stituting this solution into the equation above leads to: s 2 ψ ( x ) + k 2 ψ ( x ) = 0 (6.5) Chemistry, Indiana University 30 c c 2011, Srinivasan S. Iyengar (instructor)
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Physical Chemistry Course Number: C362 which leads to: s = ± ık (6.6) which gives two solutions: exp { ıkx } and exp {− ıkx } . 10. Hence the general solution has the form:
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This note was uploaded on 01/17/2012 for the course C 362 taught by Professor Amarflood during the Winter '11 term at Indiana.

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PIB1DResonance - Physical Chemistry Course Number: C362 6...

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