It can be shown that the ground sate of harmonic oscillator |0) hits the uncerta

Question

It can be shown that the ground sate of harmonic oscillator |0) hits the uncertainty limit, meaning that in this state Delta x delta rho = k/2.This is only Eigen state of harmonic oscillator which obeys this limit, in all Other states Delta x Delta rho = (2n+1)pi/2 >pi/2, as it follows from problem 6.10 in your homework. However, the certain linear combinations of Eigen states of harmonic oscillator, known as coherent states also minimize the uncertainty product. It turns out that these coherent states |alpha) are eigenvectors of the lowering operator A = a = . so that A|alpha) = alpha|alpha), where Eigen value a is some complex number, since A is not Hermitian operator Calculate (x) in the state |alpha).

Explanation / Answer

since A = 1/sqrt(2 hbar m w) ( i p + mw x)
and A+ = 1/sqrt(2 hbar m w ) ( - ip + m w x)
if we do A + A+ = 1/sqrt(2 hbar m w ) 2 m w x
= sqrt( 2 m w /hbar ) x
x = sqrt( hbar / 2m w) ( A + A+)

so we want to find <|x|>

= sqrt(hbar/ 2 mw) (<|A|>+<|A+|>)

but <|A|>= <|A>=<|>=

and <|A+|>=<A|>=*

so <x>= sqrt(hbar/2mw) ( + *) = sqrt( hbar/2mw) * 2 Re()

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