Msschusetts Institute of Technology Quntum Mechnics I (8.) Spring 5 Solutions to Problem Set 6 By Kit Mtn. Prctice with delt functions ( points) The Dirc delt function my be defined s such tht () (b) 3 δ(x) { for x for x. dxδ(x)f(x) f() for ny function f(x). Evlute the following integrls: dx ( x 3 3x + x ) δ(x + ) ( ) 3 3( ) + ( ) dx (cos(3x) + ) δ(x π) cos(3π) +. Squre well centered t origin, prity (5 points) Solve the time-independent Schrödinger eqution with pproprite boundry conditions for n infinite squre well of width centered t x, i.e. { for / x / V (x) elsewhere. () ( points) Check tht the llowed energies re consistent with those derived in lecture for infinite well of width centered t x /. Confirm tht the wve functions ψ(x) cn be obtined from those found in lecture if one uses the substitution x x /. Why re there no eigensttes with E <? Solve the following Schrödinger eqution, h ψ(x) m x Eψ(x) () ψ(x) Ae ikx + Be ikx, () where k me. Now, pply the boundry conditions, h ψ( /) ψ(/) (3) Ae ik/ + Be ik/ () Ae ik/ + Be ik/. (5)
Solving for B nd k, we obtin, B A (6) k nπ Therefore the normlized eigenfunctions nd eigenenergies re, for n,,3,... (7) ψ(x) cos(kx) (8) E h k m h π m n (9) The eigenfunctions cn lso be obtined from those found in clss if we substitute x x /, ψ [,] (x /) sin(kx k/) sin(kx π/) cos(kx), () since k π π. All eigenenergies re positive becuse V >. Note tht kinetic energy is lwys positive. (b) ( points) We define prity opertor ˆP by ˆP ψ(x) ψ( x). Verify tht the energy eigensttes for the bove potentil re lso eigensttes of the prity opertor. Wht re the corresponding eigenvlues? Is the sme true for the eigensttes of the well centered t x /? Clculte ˆP ψ(s) where ψ(x) is n eigensttes for the bove potentil, ˆP ψ(x) ψ( x) cos( kx) cos(kx) ψ(x), () so ψ(x) is n eigenstte of the prity opertor with n eigenvlue. This is not true for the eigensttes of the well centered t x /, where ψ(x) for x [, ] nd otherwise, ˆP ψ [,] (x) ψ [,] ( x), () since V (x) for x <. (c) (5 points) Find generl condition either on the potentil V (x) or on the Hmiltonin tht ensures tht the energy eigensttes re simultneously eigensttes of the prity opertor. The energy eigensttes re simultneously eigensttes of the prity opertor if nd only h if Ĥ(x) Ĥ( x) V (x) V ( x). Note tht the kinetic-energy term m x lwys conserve prity.
3. Gussin wvepcket in free spce ( points) A free prticle of mss m hs initil wve function ψ(x, t ) (π) / w / where the initil width of the wvepcket w is rel, positive constnt. e x w, (3) () ( points) Compute the Fourier trnsform of ψ(x, ) nd show tht it equls: φ(k, t ) where k w. Clculte the Fourier trnsform of ψ(x, ), (π) / k / e x w, () φ(k, ) π π (πw) / dxψ(x, )e ikx dxe x w ikx (5) Complete the squre, x w + ikx Substitute Eq. 6 into Eq. 5, nd we obtin, φ(k, ) ( ) x + ikw + k w w. (6) e k w π (πw) / e k w π (πw) / dxe ( πw ) x +ikw w ( w π (πk ) ) / e k w / e k k, (7) where k w. (b) (5 points) Show tht the wvefunction t lter time is given by ( ) w / ψ(x, t) x π χ(t) e [χ(t)], (8) 3
where the complex quntity chrcterizing the wvepcket χ(t) t time t is given by χ(t) w + i ht. (9) mw In free spce, the momentum eigenfucntions re lso energy eigenfunctions, nd the energy of prticle of wvevector, E h k. Therefore, the time evolution of φ(k, t) is: m φ(k, t) φ(k, )e Et h φ(k, )e i hk t m. () Clculte the wvefunction t lter time, ψ(x, t), by using the Fourier trnsform, ψ(x, t) π π (πk) / π dkφ(k, t)e ikx (πk ) / dke k k dke ( k i hk t m e ikx ) + i ht k m +ikx. () We cn see tht this integrl is similr to the one in Prt ). Fist, let χ(t) w + i ht, nd then complete the squre, mw [χ(t)] k + ikx [χ(t)] ( From Eq. nd Eq., we obtin, ψ(x, t) x [χ(t)] e π (πk) / x [χ(t)] k e π (πk) / w (π) / [χ(t)] e ( ) w / e x [χ(t)] π χ(t) (π) / x [χ(t)] e ( w + k + i ht m ix ) x [χ(t)] [χ(t)]. () dke [χ(t)] ( π [χ(t)] x [χ(t)] ) k ix [χ(t)] (3) ) () i ht mw
(c) (5 points) Clculte ψ(x, t), nd show tht it represents Gussin wvepcket. Wht is the time-dependent (rel) width w(t) of the wvepcket if we write the probbility density in the form ψ(t) x (π) / w(t) e [w(t)]? (5) Does the width w(t) t long time depend on the initil width w? clculte ψ(x, t), ψ(x, t) ( w π ( w π ) / e x [χ(t)] χ(t) ) / χ(t)χ (t) e x e [χ (t)] χ (t) ( x [χ(t) + ] [χ (t) ] ). (6) Clculte χ(t)χ (t) nd [χ(t)] + [χ (t)], ( χ(t)χ (t) w + i ht mw w + h t m w [χ(t)] + [χ (t)] + [χ(t)] [χ (t)] [χ(t)χ (t)] w w w ( + h t m w ( + h t m w where w(t) w + h t. Therefore, ψ(x, t) becomes, m w ) ) ( i ht mw ) w w(t), (7) ) [w(t)], (8) ψ(x, t) ( ) w / π π[w(t)] e w w(t) e x x [w(t)] [w(t)]. (9) From Eq. 9, w(t) represents the width of the Gussin wvepcket. w(t) increses with time, which mens the probbility density in the position spce is spreding with time. h For lrge t, i.e. t, w(t) ht m w mw. (d) ( points) Determine the time evolution of the wve function in the wvevector representtion φ(k, t) (see bove). Write down the probbility density in wvevector spce 5
Figure : ψ(x, t) vs x. Figure : φ(k, t) vs k. 6
φ(k, t), nd show tht it does not spred in time. Plot ψ(x, t) nd φ(k, t), nd explin in one or two sentences how it is possible tht the wvefunction spreds in position spce without spreding or shrinking in momentum spce, if one is the Fourier trnsform of the other. From Eq., the time evolution of the wvefunction in the wvevector representtion is, φ(k, t) φ(k, )e i hk t m (πk ) k k e e i hk t / Therefore, the probbility density in wvevector spce is, m. (3) φ(k, t) (πk ) (πk ) k k e e i hk t i hk m / e t m / e k k. (3) We cn see tht φ(k, t) does not chnge with time. Therefore there is no spreding of the probbility density in wvevector spce in time. However, we lerned in Prt c) tht the probbility density in position spce spreds with time. When n rbitrry phse is dded to the wvefunction in wvevector spce (in the cse, the extr phse is hk t), ech m component of plnes wve re no longer in-phse when they evolve with time. This cuses the wvepcket in position spce to spred.. Dibtic (sudden) expnsion of infitnite box. (5 points) A prticle of mss m is prepred in the ground stte of n infinite-potentil box of size extending from x to x. Suddenly, the wll t x is moved to x within time t doubling the box size. You my ssume tht the wvefunction is the sme immeditely fter the chnge, if the chnge hppens fst enough. () (5 points) How fst is fst enough? ( t is mthemticin s nswer, nd not good enough for physicist.) For physicist, the chnge is fst enough when it does not llow the wvefunction to evolve with time. We know tht the time evolution of n eigenstte is proportionl to e ient/ h. Therefore, if the chnge hppens in time scle of t E, where E is the h difference between the ground stte nd n excited stte, we will not immeditely observe the chnge in the wve function. (b) (5 points) Wht is the probbility tht the prticle is in the second (n) stte of the new well, immeditely fter the chnge? (Note tht the wvelength within the well, nd hence the energy, for this stte is the sme s for the initil stte in the old well.) Mke sure tht you use properly normlized wvefunctions for your clcultions. 7
The prticle is prepred in he ground stte of the old well. Therefore, its initil wvefunction is ψ i (x) sin(k ix) for x nd otherwise, (3) where k i π. The second (n) eigenstte of the new well is ψ n (x) sin(k fx), (33) where k f π π k i. The probbility tht the prticle is in the second eigenstte in the new well is P (n ) dxψn(x)ψ i (x) ( ( ) πx dx sin ). (3) (c) (5 points) Wht is the probbility tht the prticle would be found in the ground stte of the sudden expnsion? The ground stte of the new well is ψ(x) ( ) πx sin (35) The probbility tht prticle is in the ground stte is P (n ) dxψn(x)ψ i (x) dx sin ( 3π ( ) πx sin ( ) πx ) 3 9π. (36) (d) (5 points) Wht is the expecttion vlue of the energy of the prticle before nd fter the sudden expnsion? Before the sudden expnsion, the prticle is in the eigenstte (ground stte). Therefore, the energy expecttion is equl to the energy of the ground stte, E E h π m. 8
After the sudden expnsion, E h m dxψ i (x) ( h ) ψ m x i (x) dx π sin ( ) πx h π m h π m. (37) The expecttion vlue of energy remins the sme before nd fter the sudden chnge becuse the chnge is so quick tht the wvefunction stys the sme. (e) (5 points) If you wnted to ensure tht the prticle hs unity probbility to be found in the ground stte of the new potentil, how would you chnge the potentil (time scle)? To ensure tht the prticle hs unity probbility to be found in the ground stte, we need to chnge the potentil slowly enough so tht new wvefunction is lwys in the ground stte of new well. It requires tht the time scle of the chnge is t E, where h E is the difference between the ground stte nd n excited stte. This chnge of the wvefunction is continuous, i.e. the eigenenergy of the ground stte chnges continuously from the old eigenenergy to new eigenenergy. This does not violte energy conservtion becuse the prticle will interct with the wlls nd will lose some of its energy to the wlls. 9