Question

qm 09.2

2. (i) In one dimension, the momentum operator is given by d Ô = -ih- dx Determine the x dependence of the (un-normalised) momentum eigenfunction for a particle of momentum p, free to move along the x axis. [4 marks] (ii) A particle that is free to move along the x axis is described by a wavefunction v(x) = 1/ va, 0, |x<a/2 1x1 >a/2. (a) Show that the probability of measuring a momentum between p and p + dp is 14 (p)/2 dp, where sin(pa/2h) (p) = A (pa/2h) and A is a normalisation constant. [6 marks] (b) Sketch the momentum probability distribution and calculate the momentum spread, defined as the full width at half maximum (FWHM) of the central peak. [3 marks] (c) Comment on the relationship between the momentum spread and the spa- tial probability distribution ly(x)/2 with regard to the Heisenberg Uncertainty Principle. [2 marks] (iii) The Apollo 11, 14 and 15 Moon landing missions left lunar reflectors on the surface of the Moon. These "corner-cube" mirrors reflect light back in the direc- tion it came from and so allow the round-trip time of a laser from Earth to be measured. (a) A laser emitting light of wavelength 532 nm routed through a telescope with an exit aperture of 3.5 m is directed at a reflector on the surface of the Moon, which is approximately 400,000 km from the Earth. Neglecting atmospheric effects, estimate the FWHM spread of the laser spot size on the Moon's surface. [3 marks] (b) The Apollo 15 mission left a larger reflector than the previous two. Comment on the possible motivation for this change. [2 marks] [Total 20 marks] You may assume the solution to the equation v2 sin x = x is x = 1.392.

Answer 1

ہے }. dn with p- Let the wave fa be 142 momentum eegen value theim, §147 = plyn Q Likdy =py d de & de Y y=ube ipak Where Vo is a cond, 6) 4(x)= a 1218 1214 = 0 (21212 Now if the momentum ware them fa is & (P) t tipon PP) = Sy wedan q apar tipu 맛 e a da alv tah tipala e F al (EPK e og 2 h لو e = llo (LP/k) ipar ipada 2 h ease Ja 2. at sin (pahan pła Ja sin sin (pa ak (pa (ak sin (phy Y(P/ -14 (pa ako Now probability of finding it p+dp is given by 14 (Pr. dp p and . Re a. sen ah (pa ah dp les lela -Р at maxima 1 YCP) = a then halt maxime 14C PDI = a ola sin (pa а ol Ipa at, Sea (22) = T (2h the rel √2 aim pa per a rering va sina we wrete x=1.392 =n pa at 1.392 P=2.784 i Now 2 y =)) a Then the rela distributton and between probability momentum spread p = 2.784k now we ead writi using uncertainly pribcipal is Apaazh here 40= 2.784 a Ca Aa: then 2.784 h Apan Then Heisenberg worlaunay prinei faal is Salesfied as APAK_2.78947 Nist