It can be shown that the ground state of harmonic oscillator |0) hits the uncert

Question

It can be shown that the ground state of harmonic oscillator |0) hits the uncertainty limit. meaning that in this state Delata x Delta rho = h/2.This is only Eigen state of harmonic oscillator which obeys this limit, in all other states Delta x Delta rho = (2n+1)h/2>h/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 = (ipn +maxxnp),so that A|alpha) = alpha|alpha).where Eigen value alpha is some complex number, since A is not Hermitian operator (5pts.)Calculate (x) in the state |alpha). (5 pts.) Calculate (rho) in the state |alpha}. (5 pts.) Calculate (x2) in the state |alpha). (5 pts.) Calculate (p2) in the state |alpha). Hint: Represent operators xn, and rho n in terms of operators a and . (5 pts.) Find Delta x and Delta rho and show that Delta x Delta = h/2.

Explanation / Answer

given that (|lpha>=sum_{n=0}^{infty}c_n|n>)

Since it is an eigenstate of the operator a with eigenvalue alpha, so,

(a|lpha>=lpha|lpha>)

thus,

(sum_{n=0}^{infty}c_na|n>=lphasum_{n=0}^{infty}c_n|n>)

we know that (a|n>=sqrt{n}|n-1>)

thus,

(sum_{n=0}^{infty}c_nsqrt{n}|n-1>=sum_{n=0}^{infty}lpha c_n|n>)

putting (n-1 ightarrow n)    in left side, we get

(sum_{n=0}^{infty}c_{n+1}sqrt{n+1}|n>=sum_{n=0}^{infty}lpha c_n|n>)

Taking scalar product with (<n|)    both sides, we will get

(c_{n+1}=rac{lpha c_n}{sqrt{n+1}}=rac{lpha^2 c_{n-1}}{sqrt{(n+1)n}}=rac{lpha^{n+1}}{sqrt{(n+1)!}}c_0)

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