1. Consider a one-particle with mass m, one-dimensional system with length l and potential energy V=V0for41lx43l,V=0for0x41land43lxl and V= elsewhere, where V0=ml22 as shown in Figure 1. (a) Use the trial variation function 1=(l2)1/2sin(lx) for 0xl to estimate the ground-state energy. Let your answer be in ,m and l. Compare with the true ground-state energy E=5.750345ml22 and comment. Hint: To save time in evaluation integrals, note that nH^n=nT^n+nV^n and nT^n equals the particle-in-a-box ground-state energy nT^n=8ml2n2h2. (b) Consider the above one-particle system in Figure 1. Treat the system as a perturbed particle in a box. Find the first-order energy correction for the groundstate energy E(1) and then E(0)+E(1). Let your answer be in ,m and l. (c) Explain why E(0)+E(1) is the same as obtained by the variational treatment in 1( a). xsinbxdx=b21sinhxbxcosbx sin2bxdx=2x4b1sin(2bx) xsin2bxdx4x24bxsin(2bx)8b21cos(2bx) x2sin2bxdx6x3(4bx28h41)sin(2bx)4b2xcos(2bx) sinaxsinbxdx=sin(ab)xsin(ab)x,a2=b2 xeh+dx(bxb21)em x2ebxdx=ebx(bx2b22x+b32) 0nxnequdxqnn:,n>1,q>0 0e5x2dx21(b)1/2,b>0 0x2nee2dx=22n11n!(2n)!(b2n1)1/2,b>0,n=1,2,3, tznendzan11n!ent(1+at+2!a2t2++n!antn),n0,1,2,,a>0