S117Lecture42011

# S117Lecture42011 - Heisenberg Uncertainty Principle In the...

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Heisenberg Uncertainty Principle If you try to specify/measure the exact position of a particle you cannot simultaneously know its momentum exactly. If you try to specify/measure the exact momentum of a particle, you cannot simultaneously know its position exactly . p x 2 h In the microscopic world, Notice that this is very different from the macroscopic world (like throwing a football around). In the latter case, you CAN specify both the position and the momentum very well. We call the position as a function of time the TRAJECTORY

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Schrodinger’s equation In 1926, Erwin Schrodinger describes how particles behave on a microscopic scale (Quantum Mechanics) E H Hamiltonian operator wavefunction energy wavefunction One particular case (result) in Quantum Mechanics is the behavior of the electron in the hydrogen atom.
Particle (electron) in a box (one dimensional) This is NOT the problem of the Hydrogen atom but will allow us to UNDERSTAND Quantum Mechanics and the Schrodinger equation a little . x m 1. Electron is forced to be inside box. It cannot be outside (Thick repulsive walls). 2. The particle experiences no force acting on it inside the box (until it reaches a wall). 3. Can the electron simply move around the box (Follow a trajectory)? NO! Why not? x=0 x=L V= V= V=0

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E H E V T ) ˆ ˆ ( T and V are the kinetic energy operator and the potential energy operator respectively. However, inside the box V(x) =0 2 2 2 2 ˆ dx d m T E dx d m 2 2 2 2 2 2 2 2 mE dx d Rearranging, Note that m,E, and hbar are all constants. This is called a second order ordinary differential equation. We guess the solution, ) sin( ) ( kx A x
) cos( ) sin( kx Ak kx A dx d ) sin( 2 2 kx A dx d dx d dx d  2 2 2 2 2 )) sin( ( ) sin( ) cos( k kx A k kx Ak kx Ak dx d dx d 2 2 2 2 mE dx d Notice that what we wanted was: By inspection, our trial solution works so 2 2 2 mE k m k E 2 2 2 ) sin( ) ( kx A x and But what are the constants A and k ?

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The role of boundary conditions At the walls of the box, the wavefunction must be zero. Why? So, 0 ) 0 sin( ) 0 ( A x 0 ) sin( ) ( kL A L x 0 ) sin( kL L n k ) sin( ) ( x L n A x What is the value of A? ,... 3 , 2 , 1 n where Consider the fact that the particle MUST be in the box. That is to say, the probability of finding the particle in the box is 100%.
Total probability of finding the Particle in the box 1 ) ( 0 2 L dx x ) sin( ) ( x L n A x 1 ) ( sin 0 2 2 L dx x L n A 1 ) ( sin 0 2 2 L dx x L n A 2 0 2 1 ) ( sin A dx x L n L Value of integral = L/2 2 1 2 A L So, L A 2 )

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## This note was uploaded on 01/17/2012 for the course S 117 taught by Professor Stephenjacobson during the Fall '11 term at Indiana.

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S117Lecture42011 - Heisenberg Uncertainty Principle In the...

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