12-15 Lecture Supplement: Harmonic Oscillator Energies and Wavefunctions via Raising and Lowering Operators

Page 1 12-15 Lecture Supplement Harmonic Oscillator Energies and Wavefunctions via Raising and Lowering Operators We can rearrange the Schrödinger equation for the HO into an interesting form ... 2 1 ⎡⎛ ! d ⎞ 1 2⎤ ⎡ p 2 + ( mω x )2 ⎤ ψ = Eψ + ( mω x ) ⎥ ψ = ⎢⎜ ⎟ ⎦ 2m ⎢⎣⎝ i dx ⎠ 2m ⎣ ⎥⎦ with H= which has the same form as 1 ⎡ p 2 + ( mω x )2 ⎤ ⎦ 2m ⎣ u 2 + v 2 = ( iu + v )( −iu + v ) . We now define two operators 1 a± ≡ 2!mω ( ∓ip + mω x ) that operate on the test function f(x) to yield ( a a ) f ( x ) = ⎛⎝ − = + 1 2!mω 1 ( ip + mω x )( −ip + mω x )⎞ f ( x ) ⎠ ⎡ p 2 + ( mω x ) − imω (xp − px) ⎤ f (x) ⎦ 2!mω ⎣ = { 2 ( } ) 1 i 2 p 2 + ( mω x ) − [ x, p ] f ( x ) 2!mω 2! a− a+ = ( ) 1 1 1 1 2 p 2 + ( mω x ) + = H+ 2!mω 2 !ω 2 Which leads to a new form of the Schrödinger equation in terms of a+ and a- … 1⎞ ⎛ H ψ = !ω ⎜ a− a+ − ⎟ ψ ⎝ 2⎠ If we reverse the order of the operators-- a− a+ ⇒ a+ a− -- we obtain … Page 2 12-15 Lecture Supplement 1⎞ ⎛ H ψ = !ω ⎜ a+ a− + ⎟ ψ ⎝ 2⎠ or 1⎞ ⎛ !ω ⎜ a± a∓ ± ⎟ ψ = Eψ ⎝ 2⎠ and the interesting relation a− a+ − a+ a = [ a a ] = 1 − − + A CLAIM: If ψ satisfies the Schrödinger equation with energy E, then a+ψ satisfies it with energy (E+!ω ) ! 1⎞ 1 ⎞ ⎛ ⎛ H ( a + ψ ) = ! ω ⎜ a + a − + ⎟ ( a + ψ ) = !ω ⎜ a + a − a + + a + ⎟ ψ ⎝ ⎝ 2⎠ 2 ⎠ 1⎞ ⎧ ⎛ ⎛ = !ω a+ ⎜ a− a+ + ⎟ ψ = a+ ⎨!ω ⎜ a+ a− + 1 + ⎝ 2⎠ ⎩ ⎝ 1⎞ ⎫ 1⎞ ⎧ ⎛ ⎫ ⎟⎠ ψ ⎬ = a+ ⎨!ω ⎜⎝ a+ a− + ⎟⎠ + !ω ⎬ψ 2 2 ⎭ ⎩ ⎭ = a+ ( H + !ω )ψ = ( E + !ω )( a+ψ ) H ( a+ψ ) = (E + !ω ) ( a+ψ ) Likewise, a-ψ satisfies the Schrödinger equation with energy (E-!ω ) … 1⎞ 1 ⎞ 1⎞ ⎛ ⎛ ⎛ H ( a−ψ ) = !ω ⎜ a− a+ − ⎟ ( a−ψ ) = !ω ⎜ a− a+ a− − a− ⎟ ψ = a !ω ⎜ a+ a− − ⎟ ψ ⎝ ⎝ ⎝ 2⎠ 2 ⎠ 2⎠ − 1⎞ ⎫ ⎧ ⎛ = a− ⎨!ω ⎜ a− a+ − 1 − ⎟ ψ ⎬ = a− ( H − !ω )ψ = a− ( E − !ω )ψ 2⎠ ⎭ ⎩ ⎝ H ( a−ψ ) = ( E − !ω ) ( a−ψ ) So, these are operators connecting states and if we can find one state then we can use them to generate other wavefunctions and energies. In the parlance of the trade the a± are known as LADDER operators or a+ = RAISING and a- = LOWERING operators. We know there is a bottom rung on the ladder ψ0 so that a−ψ 0 = 0 Page 3 12-15 Lecture Supplement 1 2!mω Integrating this equation yields dψ ∫ ψ 00 = − ⎛ d ⎞ ⎜⎝ ! + mω x ⎟⎠ ψ 0 = 0 dx dψ 0 mω =− xψ 0 dx ! mω xdx ! ∫ ⇒ ψ ( x) = A e 0 E0 comes from plugging below. − mω x2 2! and E0 = 0 ψ 0 into Hψ lnψ 0 = − mω 2 x + A0 2h 1 !ω 2 = Eψ . We will perform the normalization Now that we are firmly planted on the bottom rung of the ladder, we can utilize a+ repeatedly to obtain other wavefunctions, ψn , and energies, En. That is, mω n − 2 ! x2 e , ⎛ ψ n ( x ) = An ( a+ ) with En = ⎜ n + ⎝ 1⎞ ⎟ !ω 2⎠ Thus, for ψ1 we obtain 1 1 ⎛ mω ⎞ 4 ⎛ 2mω ⎞ 2 − m2!ω x2 ψ 1 ( x ) = A1 ⎜ xe ⎝ π ! ⎠⎟ ⎜⎝ ! ⎠⎟ where you still have to determine the normalization constant A1. Page 4 12-15 Lecture Supplement Algebraic Normalization of the Wavefunctions: We can perform the normalization algebraically. We know that a+ψ n = cnψ n+1 a−ψ n = dnψ n−1 What are the proportionality factors cn and dn ? For any functions f (x) and g(x) ∫ ∞ −∞ ∫ f ∗ (a± )gdx = ∞ −∞ (a∓ f )∗ g dx here a∓ is the Hermitian conjugate of a± Proof: ∫ ∞ −∞ f ∗ (a± g) dx = 1 (2!mω )1 2 d ⎛ ⎞ f ∗ ⎜ ∓! + mω x ⎟ g dx −∞ ⎝ dx ⎠ ∫ ∞ recall that a± = 1 1 ∓ip + mwx ] = 12 [ (2!mω ) (2!mω )1 2 1 ⎡ ⎛! d ⎞ ⎤ ⎢ ∓i ⎜⎝ i dx ⎟⎠ + mwx ⎥ = (2!mω )1 2 ⎣ ⎦ d ⎡ ⎤ ∓! ⎢⎣ dx + mwx ⎥⎦ Integrate by parts ∫ 1 f (a± g) dx = dx = (2!mw) ∗ d ⎡⎛ ⎞ ∫−∞ ⎣⎢⎜⎝ ±! dx + mwx ⎟⎠ ∞ ∗ ⎤ f ⎥ gdx = ⎦ ∞ ∫ ∞ (a∓ f ) g dx − So we can write ∫ ∞ −∞ (a±ψ n ) *(a±ψ n )dx = ∞ ∫ (a∓ a±ψ −∞ n ) * ψ n dx We now use 1⎞ 1⎞ ⎛ ⎛ !ω ⎜ a± a∓ ± ⎟ ψ n = Enψ n and En = ⎜ n + ⎟ !w ⎝ ⎝ 2⎠ 2⎠ 1⎞ 1⎞ ⎛ ⎛ ⎜⎝ a+ a− ± ⎟⎠ ψ n = ⎜⎝ n + ⎟⎠ ψ n 2 2 and therefore a+ a−ψ n = nψ n And 1⎞ 1⎞ ⎛ ⎛ !ω ⎜ a− a+ − ⎟ ψ n = !ω ⎜ n + ⎟ ψ n ⎝ ⎝ 2⎠ 2⎠ ∗ Page 5 12-15 Lecture Supplement a− a+ψ n = ( n + 1)ψ n We can now calculate cn : ∫ ∞ −∞ (a+ψ n ) * (a+ψ n )dx = cn 2 * * ∫−∞ψ n+1ψ n +1dx = ∫−∞ ( a− a+ψ n ) *ψ n dx = (n + 1) ∫−∞ψ nψ n dx ∞ ∞ ∞ cn = n +1 The calculation for dn proceeds in a similar manner : ∫ ∞ −∞ (a−ψ n ) * (a−ψ n )dx = dn 2 ∞ ∞ ∞ * * ∫−∞ψ n −1ψ n −1dx = ∫−∞ ( a+ a−ψ n ) ψ n dx = n ∫−∞ψ nψ n dx * dn = n Thus we obtain the two normalization constants for the a+ψ n = ( n + 1) ψ n+1 12 a± a−ψ n = n1 2 ψ n−1 HO Wavefunctions: Rearranging the equations to a slightly more useful form yields ψ n +1 = 1 ( n + 1) 12 ψ n −1 = a+ψ n 1 a−ψ n n1 2 We can now use these equations to generate other wavefunctions. Thus, if we start with ψ 0 we obtain: n=0 n =1 n=2 n=3 So that ψ n is ψ1 = 1 ( 0 + 1)1 2 a+ψ 0 = a+ψ 0 1 1 a+ψ 1 = ( a+ )2 ψ 0 2 2 1 1 ψ3 = a+ψ 2 = a+3ψ 0 2 +1 3⋅ 2 1 1 ψ4 = a+ψ 3 = a+4ψ 0 3+1 4 ⋅ 3⋅ 2 ψ2 = ψn = n 1 a+ ) ψ 0 ( n! Page 6 12-15 Lecture Supplement Orthogonality of the HO Wavefunctions: Recall the orthogonality condition for two wave functions is ∞ ∫−∞ψ nψ mdx = δ nm Using the * a± operators we can show this condition also holds for the HO wavefunctions. The proof is as follows. ∫ ψ ( a+ a− )ψ * m n dx = n ∫ ψ m* ψ n dx ∫ ( a−ψ ) ( a−ψ ) dx = ∫ ( a+ a−ψ ) ψ ∗ ∗ m m m n dx = m ∫ ψ m* ψ n dx ( n − m ) ∫ ψ m* ψ n dx = 0 The trivial case occurs when n = m ; but when n ≠ m then ∫ψ ψ n dx = 0 * m Potential Energy of the Harmonic Oscillator: We can now use the a± operators to perform some illustrative calculations. Consider the potential energy associated with the HO. V= and therefore V = 1 2 1 kx = mω 2 x 2 2 2 1 1 mω 2 x 2 = mω 2 ∫ ψ n* x̂ 2ψ n dx 2 2 First, we express x̂ and p̂ in terms of ⎛ ! ⎞ x̂ = ⎜ ⎝ 2mω ⎟⎠ 12 ( a+ + a− ) a± operators … ⎛ !mω ⎞ p̂ = i ⎜ ⎝ 2 ⎟⎠ 12 ( a+ − a− ) and 2 ⎛ ! ⎞ ⎛ ! ⎞ x2 = ⎜ a+ + a− ) = ⎜ ( ( a a + a+ a− + a− a+ + a− a− ) ⎟ ⎝ 2mω ⎠ ⎝ 2mω ⎟⎠ + + 12-15 Lecture Supplement Page 7 Dirac Notation: Before we evaluate this expression let’s introduce some new notation that will make life simpler for us on future occasions. Instead of writing the integral between ±∞ we use brackets to denote this integral. The first half is called a “bra” and the second a “ket”. That is, “bra”-c-“ket” notation is bra = and ket = and for the probability density we would have an expression such as where the present of ψ m* ∫ ∞ −∞ ψ m* ψ n dx = mn is understood. Using this notation, integrals for x , p , and x 2 assume the form ∫ ∞ −∞ ψ m* x̂ψ n dx = m x̂ n V = and ∫ ∞ −∞ ψ m* p̂ψ n dx = m p̂ n 1 1 1 mω 2 x 2 = mω 2 ∫ ψ n* x̂ 2ψ n dx = mω 2 n x̂ 2 n 2 2 2 ⎛ ! ⎞1 V =⎜ mω 2 ⎡⎣ n a+ a+ n + n a+ a− n + n a− a+ n + n a− a− n ⎤⎦ ⎝ 2mω ⎟⎠ 2 yielding V = hω !ω ⎛ 1⎞ ⎡⎣ n + n + 1⎤⎦ = n+ ⎟ ⎜ 4 2 ⎝ 2⎠ It is important to not get totally embroiled in the equations and neglect the chemistry and physics. Accordingly, we should ask the question as to the physical significance of this formula ?