Question

a) Use the variational method to estimate the binding energy of a deuteron. Assume that the potential between the proton and neutron is V(r) = Ae-r/ro where A and ro are constants and use as a trial function W(r) = Ce-Br (4) where C is the normalization constant. b) Consider the Hamiltonian of a nonharmonic oscillator d2 (5) H + x2 + x4 dx2 Use the WKB approximation to find the ground state of the system as x .

Answer 1

Solution - Ay given potential between proton and deutson A and ro are Constanta Variational method Problem to estimate bending energy of deupon. V/8)= A e-8/00 trial function 4lo- cefr ces normalization Conilan't Solve rio 2 O <E> - <T> TLV) feist calculate c naimalization Cmitent for Deutron problem nomalisation Condition Y 4 (5) dr=1 e afr(dr41784) - | 4T 1 c ² Jug eaBr dr=1 411 11² Gamma integration 101² = (2313 jamemnon = Cuntin! Sice 3 x2x1 2ß 8 Ida BBB Br 8T rio JOK - Br 879 e- 4/8)= BBB e 3) 116) = JB It Now calculate LT) for formula IT) = -2 1 2 24 7 6 2 + 2 / 3 4 4100 do 3D

le-aßr dr / Brutto رم a dr 3 :-h2018 am OT apri re dr eBay e BY 二一k4182 1 2m <1) = #JOY A LOT TO on [Refr-4bre portparte folder 2e - Brdr Brdr-4ß XIX1 + 2 x x che (2 p2) (20)? :-ħa 1982 -4B x lataya 21 to et () { 2x { !(40) 4B ß -K2 m e LT) = ore or Zero it 2 (1) =-ve formula LT) = 2 am dy 4178² dr dr apr 4r dr er no Jº 89 888=288 400-dr ♡ 2 སྣ x पप . ħa vp5 F am 3 xeyil tha 485 am [let) XAPTA ha X Z am & ß ha (T) = AQ am

+ 8 ABS roʻx3 xaro, 90 V V= Aerlo V y d (41782) dr. (V) = 8AMS oproti Now let us fiel LV) EXY giren * = A 183 1411 -8/00 -28 r pa e BB + 1/2.) 8 dr Z4 AP3 x X2 aß tt ro 13 A (lizenty e dr. 0 4A 83 동 3 Now we got <E> =<T>+<V) 의 - ka p2 am + BAB358 LED- for Emin E (P) -0 find & men, Too late to to + am ka am (QB) dp [aßrotlj3 [aBrot 1]2 =-8 A 83 83 600 & A 23382 [apro +1]? He ten neglect ła Bro 1173 ta this the tum QA 96A B4r +48A BSico constant 288 + multiply by B on both side (T) ha o W am poro t'j3. -8Ar 33827 Beared. B.2 am

k2pQL 2m Now lm 3 + & Apan (aproto3 (9 = 96ABS NE +48 PAB 264 - 8103383 [Q Bro +173 putting the value of hap on the eqno. <E> 48 A 85 865+ 24 ABY TOM GRAN3 B3 (a protiß for sake of semelet since by placing value of ß from ea on eq" @ we get the menemum value of energy. similar calculation will turn eary if Bro giren then the term [abrott] 3 can be negligible. and <E> becomes as follows <E>=h2 p2 + 8 AB3863 -es) =h22ß + 3 B & A 203 =0 dB to let the fay pa ray]XB h2B2 = -12P3Aro Batha putting value of h2p² in ear <E>= -12 B3 A3+ & AB323= -4 AB383 -4 A ri3 t 8 (24,3 m² A3 8 9 7 (24) 3m3 A²56) 1 am dE 2m B xam am SI aum A 83 IM <E> Amin - 4 h 8

b ) clution Hamiltonian giren He=d² anharmonic oscillater is ) non harmonic licillater 4x2+x4. duz continious on both sides So boundry Condition wed, T ħ Spdx= (n + 2) * V potential) = x? +84. E- pe + x2 + x4 am 41 p=0 E = X x?tay aħ P=J2m (€ - x²-x4) Secl Semce-42-44) dx =(ht!) this is quite bit complicate first then JE ak as vi so take sele, x² (n²41)=f x = LJE 1=IJET linct from JE-Ito JE Vi= x2 E p2 tuz am p=0 E = x2 He IJE 0 SE 2x JE p=52m (E-x2) Fran = (n+7) ah 2x 52m (E-1²) dx = (mtf) at 2x JIME consider x = 4 dna Elle at €34 y 252m 1-22 de (nt) at I alami (n+/) th

azot-o. H = JE t=1 =) 2 x JamE YJE n S't - t dt = (1+2) ah say of lit dt = y (Constent) ax Jim Exp = 0 + ? let su El Eu = (hthah 2 Jamy ta4 E p=0 E = a 4 for E2 = p2 Elly Span= (n+1) n the (n+1) .(n+1) ah 2X Ja mEE" Il STI dt Elly ak 2 Jamle -14) dn Elly tah llyf Х. x 4 dn 2x J2ME E (n + 2) let somsider E N20 tao du Elly at 1=Elly of t=1 Yutl/ E = (n +fjñk Ya Jam. 3/4 - (n+1/) th asim Y

E 113 2 t ((n+1) k k 2.52m ? ground state total energy of anhom armonic Orcellatora E = lite2 = n +9) Akt (en dag ܢ Ah