{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lecture13 - WedFeb9,2011 Lecture13 ,whichhasnoboundary...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
2/9/11 1 PHYS 360 Quantum Mechanics Wed Feb 9, 2011 Lecture 13: How can you get quanFzed energy levels from an harmonic oscillator, which has no boundary condiFons? x ( t ) = A sin( ω t ) + B cos( ω t ) The Harmonic Oscillator Consider the force exerted by a spring: The soluFon has the form: ω k m . where In quantum mechanics we rarely deal with springs, but many systems in equilibrium have restoring forces 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 ω 2 x 2 -2 -1 0 1 2 1 2 3 Put this potenFal into the Schrödinger EquaFon….. 3 V ( x ) = 1 2 m ω 2 x 2 The harmonic potenFal: 2 2 m d 2 ψ n dx 2 + 1 2 m ω 2 x 2 ψ n = E n ψ n We need to solve the Fme‐independent eigenvalue equaFon: This is not as simple as it may look. There are two completely different approaches: using operator algebra, or using power series expansions. We will focus now on the operator algebra method…. 4 Raising and Lowering operators: 1 2 m [ ˆ p 2 + ( m ω ˆ x ) 2 ] ψ n = E n ψ 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 ω ˆ x ) 2 ].
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}