12b-soln1-10

# 12b-soln1-10 - PH12b 2010 Solutions HW#1 1 a hi = 0 = hi...

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PH12b 2010 Solutions HW#1 1. a) h i =0= h i imply that ( ) 2 = ­ 2 ® and ( ) 2 = ­ 2 ® Using this we get h i = ­ 2 ® 2 + 1 2  2 ­ 2 ® = ( ) 2 2 + 1 2  2 ( ) 2 Now, from the uncertainty relation we know that ¯  2 , therefore h i ¯ 2 8 ( ) 2 + 1 2  2 ( ) 2 b) h i ¯ ¯ ¯ ¯ = 2 8 ( ) 3 +  2 ( )=0 solving for we get = r ¯ 2  This imply that h i ¯  2 Then, the value of the energy lower bound for any quantum state is the same as the ground state energy of the one-dimensional harmonic oscillator. 2. a) First we de f ne 0 0 h 0 i 0 0 h 0 i Because the position and momentum of the particle are "uncorrelated" then h 0 0 i + h 0 0 i =0 Notice that 0 and 0 are operators that do not commute, therefore h 0 0 i 6 = h 0 0 i We know that = 0 + 0  then ( ) 2 = D ( h i ) 2 E = * μ 0 + 0 2 + =( 0 ) 2 + 2 2 ( 0 ) 2 + [ h 0 0 i + h 0 0 i ] 0 ) 2 + 2 2 ( 0 ) 2 where we used h 0 0 i + h 0 0 i From the uncertainty principle we know that 0 ¯  2 0 ,so f nally we get ( ) 2 ( 0 ) 2 + ¯ 2 2 4 2 ( 0 ) 2 b) Because 0 s (

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12b-soln1-10 - PH12b 2010 Solutions HW#1 1 a hi = 0 = hi...

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