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ps05soln

# ps05soln - 1 Physics 316 Cornell University Solution for...

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1 Physics 316 Solution for homework 5 Spring 2005 Cornell University I. NOTATION Throughout these exercises we will occasionally make use of a notation known as Dirac’s bra-ket notation. The solutions Ψ( x, t ) to Schr¨ odinger’s time-depentent equation will be denoted by | Ψ i . This is called a “ket”. Similarly, this solution’s complex conjuget Ψ( x, t ) * will be denoted by h Ψ | , called a “bra”. When these objects stand alone, they are to be treated just like ordinary functions. However, when they are facing up against eachother, they represent integration. The products of bras and kets, a “bra-ket”, are defined according to the following rule h Ψ | Ψ i = Z -∞ dx Ψ * Ψ . (1.1) Also, when a bra and ket are facing each other on either side of some other quantity, we are representing an intergral according to the following rule h Ψ | f ( x ) | Ψ i = h f ( x ) i = Z -∞ dx Ψ * f ( x . (1.2) This notation will also be applied to solutions Φ E ( x ) of the time-independent Schr¨ odinger equation. Doing so will result in expressions that are intended to seem familiar from your experiences with vector arithmetic. For example, the rule that eigenstates are orthonormal can be written as h Φ E | Φ E 0 i = δ EE 0 , where δ ab = 1 if a = b , and δ ab = 0 otherwise. II. EXERCISE 1 The oscillation amplitude x of a damped classical harmonic oscillator with the force - Cx - b d dt x and the mass m satisfies the differential equation m d 2 dt 2 + b d dt x = - Cx. (2.1) This is an eigenvalue equation for the operator on the left hand side. Find the solution x ( t ) for the starting conditions x (0) = x 0 and d dt x (0) = ˙ x 0 . Why is there a solution for all eigenvalues - C whereas there is only a set of discrete eigenvalues for the Schr¨ odinger equation of the Harmonic oscillator potential? To solve this linear, ordinary differential equation with a constant coefficient, assume a trial solution of the form x ( t ) e λt . Then the auxiliary equation is: 2 + + C = 0 . (2.2) The two roots of this quadratic equation are λ = - γ ± q γ 2 - ω 2 0 (2.3) where we have defined a damping parameter γ = b 2 m and the characteristic frequency in the absence of damping ω 0 = q C m . We can further define ω = p γ 2 - ω 2 0 . This quantity is either 0, real or purely imaginary. The general solution to the differential equation can be written as follows x ( t ) = ( e - γt ( Ae ωt + Be - ωt ) if ω 6 = 0 e - γt ( A + Bt ) if ω = 0 . (2.4) The first case includes both the under-damped ( ω 2 0 > γ 2 ) and over-damped ( ω 2 0 < γ 2 ) case; while the second, the critically damped case, occurs when the two roots are equal. The integration constants A and B are determined from the initial conditions x (0) = x 0 and x 0 (0) = ˙ x 0 . In the first case, we get: A = ( ω + γ ) x 0 + ˙ x 0 2 ω (2.5) B = ( ω - γ ) x 0 - ˙ x 0 2 ω , (2.6) 2 while for the second, we have A = x 0 (2.7) B = ˙ x 0 + γx 0 . (2.8) The solution exists for any value of - C . The reason for this is, that a linear second order differential equation has two independent solutions x 1 ( t ) and x 2 ( t

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ps05soln - 1 Physics 316 Cornell University Solution for...

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