Question

3. The quantum harmonic oscillator is the quantum-mechanical analog of the classical harmonic oscillator. Because an arbitrary potential can usually be approximated as a harmonic potential at the vicinity of a stable equilibrium point, it is one of the most important model systems in quantum mechanics. Consider an electron trapped by a one-dimensional harmonic potential V(x)=-5 mo?x” (where m is the electron mass, o is a constant angular frequency). In this case, the Schrödinger equation takes the following form, ** dºv(x) + 5 merrºv (x) = Ey (x). 2m dar The electron is initially trapped at the ground level. After absorbing a photon, it transits to an excited level. The wave functions of the ground and excited levels take the following forms, respectively, . (x) = exp mox? 2h 2mox? h -1 exp mox 2h Determine the energy of the electron at the ground and excited levels, respectively, and therefore express the wavelength of the incident photon in terms of . (30 marks)

Answer 1

83. Giver Potential V(x) 4more? The wave function of ground state mux? e ati let st we can write where E- -h² & +v(x) 2m da? سه dyo da da da² 2 dk 봉 To find the energy of the ground state of electon. sing schreckigex equation + = EP = energy of electron HopHonian (222) = -fxe danse (**) he****{*}} kter 2²e (**) 44=2(231) 4 - +2 2² (288²-1) 4. ti? (1-xx2) 4. **ių. - *)(-x) 4. temir" uz - 444 =(-+ - hex nư c ) + 1 tw Y. = EU Å 46 = 1 how to so The energy of ground state E = 1 h -ħw da? dx? 2.m 2m d x2 2m -K? 0 am da? hyo So

4 = /2 mw x² • The wave function of excited state (223-1) -mw x2 2 + te dh da I d dre? dre a = (22-1){ $") a 434-1) (yd?x-o) } ANF e* (*** *** -sXw") - *(sex - 244} (5** – 24"x2) e(_dix) (54X – 24'ko) te 29 (542-64–) (-52'x? +246x3 +52?6**) *(-*°42'** +5 = 6***)<? (24*** +5=1483) (24'*=+) (4+5) =<°(*k-s}4 2-5) 4, + 1 hodina útaf stow His mau589 ) + zaverythin' } {* **** fY, = 5 to 4 so energy of excited state dite dx dn 2e 02? - 4 tville - h² lmw mwx-5 1) ( 2m ☆ am due - ps & = & the

Stre The energy of the incident Photon is equal to the energy difference between ६ and & Ey = E,- € & I kw - ko = 2 ho 2 we that the energy Photon know of wavelength of A is given by Ey =hc ㅢ 2to so C-speed of light ho he 2ħw 2h w - į to h 25 he XIE - che JU date The wavelength of incident Photon for this transition Ps TC