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PHYS4221(Fall2010)_Solution_5

PHYS4221(Fall2010)_Solution_5 - 1 PHYS4221 Suggested...

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1 PHYS4221 Suggested Solution of Homework 5 Question 1 (a) The Hamiltonian H = p 2 x + p 2 y 2 m 0 + m 0 ω 2 0 2 ( x 2 + y 2 ) + λxy can be easily diagonized by a change of variable. We let x ± = x ± y 2 , p ± = p x ± p y 2 . This change of variables corresponds to a unitary transformation , because the new variables preserve the commutation relations [ x + , p + ] = [ x - , p - ] = i ¯ h [ x + , p - ] = [ x - , p + ] = 0 [ x - , x + ] = [ p - , p + ] = 0 . As an example, using [ x, p x ] = [ y, p y ] = i ¯ h and [ x, p y ] = [ y, p x ] = 0, [ x + , p + ] = 1 2 ([ x + y, p x + p y ]) = 1 2 ([ x, p x ] + [ y, p y ]) = 1 2 ( i ¯ h + i ¯ h ) = i ¯ h [ x + , p - ] = 1 2 ([ x + y, p x - p y ]) = 1 2 ([ x, p x ] - [ y, p y ]) = 0 . Furthermore, we can write H as H = p 2 + + p 2 - 2 m 0 + m 0 ω 2 0 2 ( x 2 + + x 2 - ) + λ 2 ( x 2 + - x 2 - ) = p 2 + 2 m 0 + 1 2 m 0 ω 2 + x 2 + + p 2 - 2 m 0 + 1 2 m 0 ω 2 - x 2 - H + + H - where H ± = p 2 ± 2 m 0 + 1 2 m 0 ω 2 ± x 2 ± , ω ± = r ω 2 0 ± λ m 0 . To summarize, in normal-mode coordinates { x + , x - } , the Hamiltonian describes two decoupled harmonic oscillators with frequencies ω + and ω - respectively. Hence the eigenstates of H are given by {| n + , n - i} , with eigenvalues H | n + , n - i = ¯ + n + + 1 2 + ¯ - n - + 1 2 | n + , n - i , n ± = 0 , 1 , 2 , . . . . (i) In coordinate space ( x + , x - ), the eigenfunction of H is just the product of the eigenfunctions of H + and H - : Ψ n + ,n - ( x + , x - ) ≡ h x + , x - | n + , n - i = ψ (+) n + ( x + ) * ψ ( - ) n - ( x - ) , where ψ ( ± ) n ( x ) = 1 q 2 n n ! πx ± 0 H n x x ± 0 exp ( - 1 2 x x ± 0 2 ) , x ± 0 = s ¯ h m 0 ω ± . (c.f. lecture notes no.3, eq.(42)). Rewriting the required eigenfunction in terms of ( x, y ), Ψ n + ,n - ( x, y ) = ψ (+) n + x + y 2 ψ ( - ) n - x - y 2 .
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2 (ii) Similarly, the eigenfunction in momentum space is Ψ n + ,n - ( p + , p - ) = φ (+) n + ( p + ) * φ ( - ) n - ( p - ) , where φ ( ± ) n ( p ) = 1 q 2 n n ! πp ± 0 H n p p ± 0 exp ( - 1 2 p p ± 0 2 ) , p ± 0 = p ¯ hm 0 ω ± . In terms of the original momenta ( p x , p y ), Ψ n + ,n - ( p x , p y ) = φ (+) n + p x + p y 2 φ ( - ) n - p x - p y 2 .
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