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Put two non-interacting particles of mass π in a one-dimensional harmonic oscillator potential of frequency π. The two particle wave function is: Ξ¨(π₯1 , π₯2 ) = ππ (π₯1 ) β ππ (π₯2 ) Where ππ and ππ are the nβth and mβth Harmonic oscillator eigenfunction and π₯1 and π₯2 are positions for the two particles. We can write the wave functions as the following. 1 4 1 πππ₯1 ππ β ππ (π₯1 ) = β( ) β π 2β β π»π (π¦) π πβ β2 β π! 1 4 1 πππ₯2 ππ β ππ (π₯2 ) = β( ) β π 2β β π»π (π¦) π πβ β2 β π! Where π¦ = β ππ π β π₯ for π₯ = π₯1 , π₯2 We need to first find the expectation value β©π₯1 β π₯2 βͺ: β©π₯1 β π₯2 βͺ = β©π₯1 βͺ β β©π₯2 βͺ β©π₯1 βͺ πππ β©π₯2 βͺ can be represented as: β β©π₯1 βͺ = β« (ππβ (π₯1 ) β π₯1 β ππ (π₯1 ))ππ₯1 ββ β β ( ) β©π₯2 βͺ = β« (ππ π₯2 β π₯2 β ππ (π₯2 ))ππ₯2 ββ β β β ( ) β©π₯1 β π₯2 βͺ = β« (ππβ (π₯1 ) β π₯1 β ππ (π₯1 ))ππ₯1 β β« (ππ π₯2 β π₯2 β ππ (π₯2 ))ππ₯2 ββ ββ Putting in the value we find out that these integrals become 0, so: β©π₯1 β π₯2 βͺ = 0 Since the expectation value is finding the probability of the particle around a point, and in this case, the probability of finding a particle to the right of the other particle is equal to the probability of finding it to the left, so it becomes zero at the exact point. Now we will need to find β©(π₯1 β π₯2 )2 βͺ: β©(π₯1 β π₯2 )2 βͺ = β©π₯12 β 2π₯1 π₯2 + π₯22 βͺ = β©π₯12 βͺ β β©2π₯1 π₯2 βͺ + β©π₯22 βͺ = β©π₯12 βͺ β β©2π₯1 βͺβ©π₯2 βͺ + β©π₯22 βͺ So we can do this in steps. For not crowding this document we can take: (ππβ (π₯1 ) β π₯1 β ππ (π₯1 )) = π β ( ) (ππ π₯2 β π₯2 β ππ (π₯2 )) = π First letβs find β©π₯12 βͺ: β β©π₯12 βͺ =β« ββ (ππβ (π₯1 ) 2 β β π₯1 β ππ (π₯1 )) ππ₯1 = β« π 2 ππ₯1 ββ β©π₯12 βͺ = 5 β β2βπ 2 2 β ππ β π»π (π¦)4 3 ππ 2 βπ β ( β ) β β β (π!)2 Now for β©2π₯1 βͺβ©π₯2 βͺ: β β β©2π₯1 βͺβ©π₯2 βͺ = β« (2π)ππ₯1 β β« (π)ππ₯2 = 0 ββ ββ 2 And finally for β©π₯2 βͺ: β β©π₯22 βͺ =β« β β ( ) (ππ π₯2 β π₯2 β ππ (π₯2 ))ππ₯2 = β« (π)ππ₯2 ββ ββ 5 β©π₯22 βͺ = 2β2β2βπ β ππ β π»π (π¦)4 3 2 ππ βπ β ( β ) β β β (π!)2 When we add all these together then we get: 5 β©π₯12 βͺ β β©2π₯1 βͺβ©π₯2 βͺ + β©π₯22 βͺ = 2β2β2βπ β ππ β π»π (π¦)4 3 2 5 β0+ ππ βπ β ( β ) β β β (π!)2 2β2β2βπ β ππ β π»π (π¦)4 3 ππ 2 βπ β ( β ) β β β (π!)2 Or simply: β©(π₯1 β π₯2 )2 βͺ = β©π₯12 βͺ + β©π₯22 βͺ = 2 β5β2βπ 4 2 β ππ β π»π (π¦) 3 2 2 ππ ) β β β (π!) βπ β ( + β β©(π₯1 β π₯2 )2 βͺ = β©π₯12 βͺ + β©π₯22 βͺ = ππ( 2 β5β2βπ 4 2 β π»π (π¦) 3 2 ππ) β β β (π!)2 β βπ β ( 2 β5β2βπ 4 2 β ππ β π»π(π¦) 3 ππ)2 β β β (π!)2 βπ β ( β + 2 β5β2βπ 4 2 β π»π (π¦) 3 2 ππ) β β β (π!)2 β βπ β ( )