By Polyakov A.

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**Sample text**

S s h h0j(ay ; a)(a + ay)j0i hp(0)x(0)i = i m2h! 2m! t: C (t) = 2m! 33) 2m! 2m! 6 Consider a one-dimensional simple harmonic oscillator. Do the following algebraically, that is, without using wave functions. (a) Construct a linear combination of j0i and j1i such that hxi is as large as possible. (b) Suppose the oscillator is in the state constructed in (a) at t = 0. What is the state vector for t > 0 in the Schrodinger picture? Evaluate the expectation value hxi as a function of time for t > 0 using (i) the Schrodinger picture and (ii) the Heisenberg picture.

QUANTUM DYNAMICS by 43 The average hxi in a one-dimensional simple harmonic oscillator is given hxi = h jxj i = (c0h0j + c1h1j) x (c0j0i + c1j1i) = jc0j2hs0jxj0i + c0c1h0jxj1i + c1cs0h1jxj0i + jc1j2h1jxj1i h h0ja + ayj0i + c c h h0ja + ayj1i = jc0j2 2m! 0 1 2m! s h h h1ja + ayj1i y 2 +c1c0 2m! h1ja + a j0i + jc1j 2m! s s h h <(c c ) = 2m! (c0c1 + c1c0) = 2 2m! 0 1 s q h = 2 2m! 36) q h where we have used that x = 2m! (a + ay). What we need is to nd the values of jc0j and 1 ; 0 that make the average hxi as large as possible.

K=1 n=1 n! 53) = n=0 n! n=1 (n ; 1)! 54) which means that j i is a coherent state. If it is normalized, it should satisfy also h j i = 1. m! jmi] n! m! ( )n mhnjmi = e;j j2 X 1 (j j2)n n! m! n n! 56) where x x ; h jxj i, p p ; h jpj i and c is a purely imaginary number. Since j i is a coherent state we have aj i = j i ) h jay = h j : Using this relation we can write s s h h ( + ay)j i xj i = 2m! (a + ay)j i = 2m! 58) s h h (h jaj i + h jayj i) hxi = h jxj i = 2m! h j(a + ay)j i = 2m! s h ( + ) = 2m!