Download Quantum Mechanics of Harmonic Oscillator: Deriving Energy Levels and Wave Functions and more Slides Chemistry in PDF only on Docsity! Use the result for ψ(1) to derive an expression for Eq(2) ∑== k kkqqqq aHHE )0()1()0()1()1()0()2( |ˆ||ˆ| ψψψψ )1()0()1()0( |ˆ| qk k kkq k k HaHa ∑∑ == ψψ But )0()0( )1( qk kq k EE H a − −=∴ ∑ − ⋅ −=∴ k qk qkkq q EE HH E )0()0( )1()1( )2( docsity.com ( )*)1()1( qkkq HH = since H(1) is self-adjoint ∑ ≠ − −=∴ qk qk kq q EE H E )0()0( 2)1( )2( Note: if q is the ground state then Eq(2) < 0 always. docsity.com • The quantized energy levels for the harmonic oscillator are equidistant from one another. • The lowest possible energy, corresponding v = 0, E0 = 0.5 hνo – In classical theory a molecule could be undergoing no vibration, but in quantum mechanics, this is not allowed – Even at 0 K, vibration still occurs with energy ½ hνo . This is the zero-point energy. – Consistent with Heisenberg Uncertainty principle; if no vibration occurred, the position and momentum of the atoms would both have precise values and this is not allowed! ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ +⋅= 2 1vv ohνE where v = 0,1,2,3,… and m kν ho 2π 1 = • The quantum oscillator can penetrate into classically forbidden regions - tunneling. • The relative percentages of ψ in the classical forbidden regions suggests that the tunneling probability decreases as the total vibrational energy increases. ψn (x) ψn*(x) ψn (x) docsity.com Recall: Operator Method for Solving Harmonic Oscillator Problem We defined two operators: )ˆ( 2 1ˆ piqa += )ˆ( 2 1ˆ piqa −=+ Some relationships between a, a+ and H. 1]ˆ,ˆ[ =+aa aHa ˆ]ˆ,ˆ[ = ++ −= aHa ˆ]ˆ,ˆ[ 2 1ˆˆˆ +=+ Haa 2 1ˆˆˆ −=+ Haa )ˆˆˆˆ( 2 1ˆ aaaaH ++ += Note: hhh kmmppxq x ==== ωα α α ˆ1ˆ docsity.com Most relevant operations for nth wave function >++>= >−>= + 1|1|ˆ 1||ˆ nnna nnna docsity.com Next: )0()1( k qk kq a ψψ ∑ ≠ = where ( ))0()0( )1( kq kq k EE H a − = In this problem, q =1 ( ) 2,0, )1()1( 1, 2 1|ˆˆ|1|ˆ| kk k aakHkH δδ += >+>=<=<∴ + Therefore: Hk,1(1) = 0 unless k = 0 or 2 >+>=+=∴ 2|0| 20 )0( 22 )0( 00 )1( aaaa ψψψ )0( 2 )0( 1 2)0( 0 )0( 1 0 2;1 EE a EE a − = − = νν ννν hhEE hhEEnhEn −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=−⇒ =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=−⇒+= 2 5 2 3 2 1 2 3) 2 1( )0( 2 )0( 1 )0( 0 )0( 1 )0(Q docsity.com >−>=∴ 2|20|1)1(1 νν ψ hh Lastly: ∑ ≠ − = qk kq kq q EE H E )0()0( 2)1( )2( Like the first order correction to the wave function Eq(2) is non-zero for k = 0 and 2 ( ) ν νννν h hhhh EE H EE H E 1 2121 2 )0( 2 )0( 1 2)1( 1,2 )0( 0 )0( 1 2)1( 1,0)2( 1 −= −=−= − + − =∴ Summary: >−>−>= −=−+=++= 2|20|11| 1 2 310 2 3 1 )2( 1 )1( 1 )0( 11 νν ψ ν ν ν ν hh h h h hEEEE docsity.com
