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# Lect_37_note - Chapter 40-42 Introduction to Quantum...

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Chapter 40-42 Introduction to Quantum Physics

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1D Schrödinger Equation 22 2 2 d ψ U ψ E ψ md x −+ = h U(x) z E is the energy of the particle z ψ ( x ) and d / dx must be continuous
Particle in a Box z A particle is confined to a one-dimensional region of space z The “box” is one- dimensional z The particle is bouncing elastically back and forth between two impenetrable walls separated by L

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Particle in a Box – Classic Solution z It moves at a constant speed v . z v can be any value. z Given its initial position, its position at any later time can be calculated according to v . z E = ½ mv 2 , P = mv
Particle in a Box – Quantum Solution 22 2 2 d ψ U ψ E ψ md x −+ = h U = U =0 U =

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Potential Energy for a Particle in a Box U = U =0 U =
Schrödinger Equation of a Particle in a Box 22 2 2 d ψ U ψ E ψ md x −+ = h 2 2 d ψ U ψ E ψ x = h 2 2 d ψ U ψ E ψ x = h 2 2 d ψ U ψ E ψ x = h 0 ψ (x) = 0 (x) = 0

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Wave Function of a Particle in a Box z The most general solution to the equation is ψ ( x ) = A sin kx + B cos kx 2 2 22 d ψ mE mE ψ k ψ where k dx =− = hh
Wave Function of a Particle in a Box z There is zero probability of finding the particle outside the box z ψ (x) = 0 for x < 0 and x > L z The wave function must also be 0 at the walls z The function must be continuous z (0) = 0 and (L) = 0

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Wave Function of a Particle in a Box z ψ ( x ) = A sin kx + B cos kx z ψ (0) = 0 and ( L ) = 0, So B=0 z ψ (0) = A sin k 0 = 0 z ψ ( L ) = A sin kL = 0, so kL =n π, k =n π / L z ψ ( x ) = A sin( x n π / L )
Wave Function of a Particle in a Box () s in n π x ψ xA L ⎛⎞ = ⎜⎟ ⎝⎠

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Lect_37_note - Chapter 40-42 Introduction to Quantum...

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