dx x x = 1 and + dx α x x α x = + dx α ˆx x x α = α ˆx α as required, in the last equality we used completeness relation +

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1 Physics 5 Assignment #5 Solutions Due My 5, Dim Wvefunctions Wvefunctions ψ α nd φp p α re the wvefunctions of some stte α in position-spce nd momentum-spce, or position representtion nd momentum representtion, respectively. The completeness reltions become d nd dp p p, which re the etensions of completeness reltion in finite spce to tht in infinite spce. From the definition nd mening of wvefunction, we know the left hnd side is just the epecttion vlue of position in some properly normlized stte α, becuse d ψ d α α d α ˆ α α ˆ α s required, in the lst equlity we used completeness reltion d. Plugging in completeness reltion dp p p twice, we hve d ψ d dp dp α p p p p α d dp dp α p e i p e i p p α d dp dp α p dp dp α p dp α p i p e i i δp p p α p i p α p dp φ p i φp, p p p p α where we hve used p e ip/ in the second equlity nd d e ip p / δp p in the fourth equlity.. -Dim Wve-Pcket As we lerned in previous homework, is the uncertinty in position nd is defined s. Here 0, d ψ d ψ d π λ e λ d π λ e λ

2 nd d ψ d ψ d λ e λ d λ e λ π// λ 3/ π/λ / λ. Therefore the uncertinty in position is λ. Agin if we ssume the system is in stte α, ψ α nd φp p α. Plugging in completeness reltion, φp d p α, where p e ik, in which K p/. Thus φp d p α d e ik π e λ / e ik λ π + d e λ λ +ik k π / e K k λ. 3 Similr to the uncertinty in position, the uncertinty in momentum is found to be p p p, where p dp φp p dp φp + dp dp k πλ πλ k, e K k λ p e K k λ K k dk e λ K K k dk e λ

3 nd p dp φp p dp φp dp dp + e K k λ p e K k λ K k dk e λ K K k dk e λ λ + k πλ πλ Hence p λ. λ + k. 4 Finlly p which is certinly stisfying Heisenberg s position-momentum uncertinty reltion with n euqlity reltion rther thn ineuqlity. Therefore the wvefunction in this problem is ctully Gussin wvepcket which is often clled minimum uncertinty wvepcket. 3. -Dim Infinite Squre-Well Since it is n infinite squre well, the probbility for the prticle being outside of the well, >, is zero. Hence the complete form of the wvefunction for the prticle in its lowest energy-eigenstte is { π ψ 0 >. Menwhile it is esy to verify tht ψ is lredy normlized. Keeping this in mind nd following the method we used in problem, the momentum-spce wvefunction is φp d d e i p ψ e i p + d e i p + d e i p π π π π i e p/ π/ p/. 3 + e i π

4 If we plot in unit of, then ψ π in unit of. If we plot p in unit of, then φp p in unit of π π/ p. Both plots re shown below in order It is hrd to directly use momentum-spce wvefunction to clculte the verge of momentum squred for momentum uncertinty, lthough it is not impossible. To mke the world complete, in the following we first prove the counterprt of eqution. If system is in some properly normlized stte α, the epecttion vlue of momentum is α ˆp α dp α ˆp p p α dp α p p p α dp φp p s we lredy used in problem. Therefore p dp φp p dp d d α p p p α dp d d α e i p p e dp d d α d d α d α i d ψ i i i α ψ. p i α p ei α δ α Armed with this eqution, the verges re 0, d ψ d π 4

5 nd Hence 3. 3 d ψ d π, nd p p 0, 0 d ψ i d iπ π d ψ i d ψ d d π. ψ sin π ψ ψ π π π + π Then p π. Therefore p 3,which is perfectly consistent with the uncertinty principle. 4 When width of the infinite well increses from to 4, the system chnges from stte α, its lowest energy-eigenstte or ground-stte, to stte β which is lso the ground-stte but when the well width is 4. Therefore its wvefunction becomes, { β ψ π 4 0 >. 5

6 The probbility of this is P β α, where β α 8 3π. d β α d ψ ψ d Then the probbility is P 8 3π 0.7. π π 4 In following prts let s cll the two energy-eigensttes ψ nd ψ respectively, where ψ π nd ψ sin π. We lredy know tht ψ is the ground-stte with energy eigenvlue E π which could be clculted by simply plugging energy-eigenstte wvefunctions into Schrödinger s eqution 8M m ψ Eψ. We find tht ψ is ctully the first ecited stte with eigenvlue E π 4E M. Now the following prts re strightforwrd fter these preprtions. d ψ, d [ d [ So this stte is properly normlized. π + sin π π sin 3π The time evolution of n energy-eigenstte is + sin ψ, t e iht/ ψ e iht/ ψ + ψ + sin π ] π π ] π e iet/ ψ + e iet/ ψ. Therefore its squre is ψ, t e iet/ ψ + e iet/ ψ e iet/ ψ + e iet/ ψ ψ + ψ E E t + ψ ψ [ π + sin π E E t π π ] + sin. 6

7 The plots shown below hs in unit of nd ψ, t in unit of, for π π 5π t 0, t E E 8,t E E 4 t E E 8, from top to bot tom nd left to right. ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ ÈΨHLÈ Here we clim tht Z + hˆ it d ψ, t Z + E E t ψ + ψ + ψ ψ d Z π π 3E t + sin d Z π 3E t + 3π sin d sin π t, 9π 8M which is obviously periodic in time. 7

8 4 The period is T 3E 8M 3 6M 3π s, where we used m, M kg nd J s. 8