Lecture13.pptx

# Lecture13.pptx - 2/9/11 PHYS360QuantumMechanics...

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2/9/11 1 PHYS 360 Quantum Mechanics Wed Feb 9, 2011 Lecture 13: How can you get quanfzed energy levels ±rom an harmonic oscillator, which has no boundary condifons? x ( t ) = A sin( ω t ) + B cos( t ) The Harmonic Oscillator Consider the ±orce exerted by a spring: The solufon has the ±orm: k m . where In quantum mechanics we rarely deal with springs, but many systems in equilibrium have restoring ±orces that look like Hooke’s Law (above). F = ma = m d 2 x dt 2 = kx 2 The potenfal energy associated with a spring is…. F = kx = dV dx V ( x ) = 1 2 kx 2 = 1 2 m 2 x 2 -2 -1 0 1 2 1 2 3 Put this potenfal into the Schrödinger Equafon…. . 3 V ( x ) = 1 2 m 2 x 2 The harmonic potenfal: 2 2 m d 2 ψ n dx 2 + 1 2 m 2 x 2 n = E n n We need to solve the fme‐independent eigenvalue equafon: This is not as simple as it may look. There are two completely di²erent approaches: using operator algebra, or using power series expansions. We will ±ocus now on the operator algebra method…. 4 Raising and Lowering operators: 1 2 m [ ˆ p 2 + ( m ˆ x ) 2 ] n = E n n Rewrite the Schrödinger equafon as: We emphasize the operators: ˆ p i d dx , ˆ x x ˆ H = 1 2 m [ ˆ p 2 + ( m ˆ x ) 2 ]. And the Hamiltonian is:

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## This note was uploaded on 04/23/2011 for the course PHYS 360 taught by Professor Durbin,stephen during the Spring '11 term at Purdue University-West Lafayette.

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Lecture13.pptx - 2/9/11 PHYS360QuantumMechanics...

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