Lecture15Notes - Chem 120A 02/22/06 Spring 2006 READING:...

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Chem 120A Free Particle and a Particle Meeting a Potential Barrier 02/22/06 Spring 2006 Lecture 15 READING: Engel: Chapters 4 and 5 In the last few lectures, we have been dealing with particle in a potential well. Now we will look at the the solutions to the Schrodinger equation for a free particle. That is a particle which is not confined by any potential ( V=0 everywhere). The Schrodinger equation for a particle in 1D is: - ¯ h 2 2 m d 2 dx 2 Ψ ( x ) = E Ψ ( x ) (1) The general solution to this 2nd-order differential equation is: Ψ ( x ) = A 1 e ikx + A 2 e - ikx (2) where k 2 = 2 mE ¯ h 2 There are several interesting comparisons between the free particle and the particle in a box. First of all, the energy eigenvalue for the free particle solution is E = ¯ h 2 k 2 2 m , so the energies are not quantized. Without ap- plying constraints (boundary conditions) on Ψ , the energy can take on continuous values. Secondly, there is no zero-point energy (which means that the lowest energy is E = 0). Finally, we can think of the free particle as a particle in a box of length a , where a Remember that the uncertainty in momentum is inversely propotinal to a as σ p = n π ¯ h a . Thus, σ p 0 as a 0. On the otherhand, the uncertainty in position is directly proportional to a , thus σ x as a . For a free particle, the momentum is completely defined, while the position is completely uncertain. a)For the bound solutions when E < 0, k will be a complex number. For convience, we can set k = i κ to get rid of the complexity (remember i 2 = - 1), thus Ψ ( x ) = A 1 e - κ x + A 2 e κ x (3) You can easily show that since the bound solutions are real, the probability current, j ( x , t ) x is zero, where j ( x , t ) = i ¯ h 2 m [ Ψ * ( x , t ) Ψ ( x , t ) x - Ψ * ( x , t ) x Ψ ( x , t )] (4) b) Now we want to look at the probability current for the unbound solutions, E > 0. We know that the Chem 120A, Spring 2006, Lecture 15 1
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eigenfunctions of the free-particle Hamiltonian are also eigenfunctions of the momentum operator because ˆ H = ˆ p 2 2 m (5) Thus the unbound free particle state, Ψ 1 ( x ) = A 1 e ikx is also an eigenfunction of momentum ˆ p Ψ 1 = p Ψ 1 (6) where ˆ p = - i ¯ h d dx and p = ¯ hk Ψ 1 represents states with momentum going towards the right. We also have states with momentum to the
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This note was uploaded on 09/29/2009 for the course CHEM 120A taught by Professor Whaley during the Spring '07 term at University of California, Berkeley.

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Lecture15Notes - Chem 120A 02/22/06 Spring 2006 READING:...

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