Chemistry 460 Fall 2017 Dr. Jean M. Standard October 2, 2017 The Sommerfeld Polynomial Method: Harmonic Oscillator Example Scaling the Harmonic Oscillator Equation Recall the basic definitions of the harmonic oscillator. The potential energy is defined by V (x) = 1 2 kx 2, (1) where k is the force constant. The Hamiltonian operator therefore has the following form H ˆ =!2 2m d 2 dx 2 + 1 2 kx 2. (2) In order to make the equations easier to work with, we defined dimensionless operators P ˆ and ˆ, which are defined as follows ˆ " = mk % 1/ 4 $ #! 2 ' x ˆ, (3) & P ˆ d = i d. (4) These scaled operators have the property that [ ˆ P, ˆ ] = i. In these new variables, the Schrödinger equation has the form hν 0 2 P ˆ ( 2 + ˆ 2 ) ψ ( ) = E ψ ( ), (5) or hν 0 2 $ & % d 2 d 2 + ˆ 2 ' ) ( ψ ( ) = E ψ ( ), (6) where the harmonic frequency ν 0 is defined as ν 0 = 1/ 2 1 $ k ' & ) 2π % m(. (7) Dividing by hν 0, and defining the dimensionless energy ε asε = E /hν 0, the equation becomes 1 # 2 d 2 % d 2 + ˆ $ 2 & ( ψ ' ( ) = ε ψ( ). (8)
2 Rearranging leads to the following second-order ordinary differential equation, ψ ""( ) + ( 2ε 2 ) ψ ( ) = 0. (9) The Sommerfeld Polynomial Method To solve the differential equation, we apply a general technique called the Sommerfeld polynomial method. This method consists of four steps: 1. An asymptotic solution is obtained. 2. A differential equation for the remainder is determined. 3. The remainder is expressed as a power series, and a recursion formula is developed for the coefficients. 4. Finally, the power series is truncated to obtain quantized eigenvalues. These steps will be applied to the scaled harmonic oscillator equation. Step 1. Asymptotic Solution Since the range of coordinate of the harmonic oscillator is x, the range of the coordinate is the same. Thus, we seek an asymptotic solution to the equation as ±. The equation takes the form ψ ""( ) + ( 2ε 2 ) ψ ( ) = 0 (10) ψ ""( ) 2 ψ( ) = 0 (11) in the limit as ± since 2 >> 2ε in the factor 2ε 2. Rearranging the equation yields This equation has solutions of the form d 2 d 2 ψ ( ) = 2 ψ ( ). (12) in the limit as the absolute value of becomes large. ψ( ) e ± 2 / 2 (13) Since the wavefunction must remain finite as ±, we have to throw out the solution with the positive exponent since it blows up in the limit. Therefore, the asymptotic solution has the form ψ( ) e 2 / 2 in the limit as ±. (14)
3 Step 2. Equation for Remainder Now that we have the asymptotic form, we seek an equation for the remainder. To do this, we write the full solution of the form ψ( ) = H( ) e 2 / 2, (15) where the asymptotic form has been explicitly included. The function H() is called the remainder function. To obtain a differential equation governing the remainder function, the full solution is substituted back into the original equation, ψ ""( ) + ( 2ε 2 ) ψ ( ) = 0. (16) Before substitution, the second derivative should be evaluated. The first derivative is and the second derivative is ( ) = H "( ) e 2 / 2 H( ) e 2 / 2, (17) ψ " ( ) = H ""( ) e 2 / 2 H " ( ) e 2 / 2 H ψ "" ( )e 2 / 2 H " = e 2 / 2[ H ""( ) 2 H "( ) H( ) + 2 H( ) ]. ( ) e 2 / 2 + 2 H ( )e 2 / 2 (18) Substituting this into the differential equation yields e 2 / 2 The terms involving 2 cancel to give ( ) H [ H ##( ) 2 H #( ) H( ) + 2 H( ) ] + 2ε 2 ( )e 2 / 2 = 0. (19) e 2 / 2 [ H ##( ) 2 H #( ) + 2ε H( ) H( ) ] = 0. (20) Dividing both sides by the factor e 2 / 2 leads to the differential equation for the remainder, ( ) 2 H "( ) + ( 2ε 1) H( ) = 0. (21) H ""
4 Step 3. Polynomial Solution for Remainder To solve the differential equation for the remainder function H(), we assume that the function can be written in terms of a power series, ( ) = a n n H. (22) The coefficients a n are unknown at this point. The object is to find the coefficients a n such that the function H() satisfies the differential equation for the remainder. To obtain the coefficients a n, we must first compute the derivatives of H() and substitute them into the differential equation. The first derivative is given by and the second derivative is given by H "" ( ) = n a n n 1 H " Substitution into the equation for the remainder, H "", (23) n=1, (24) n=2 ( ) = n ( n 1) a n n 2 ( ) 2 H "( ) + ( 2ε 1) H( ) = 0, yields n ( n 1) a n n 2 2 n a n n 1 + 2ε 1 n=2 n=1 = 0. (25) ( ) a n n In order to combine these power series, the limits must match. To make them match, we can rewrite the first power series for the second derivative in the form ( ) = ( n +1) ( n + 2) a n+2 n H "". (26) You can verify by expanding the series that it is identical to the original form for the second derivative. The second power series in the differential equation, the one for the first derivative, can be rewritten by adding in a term that is zero, ( ) = n a n n 1 H " Using these two forms in the differential equation, we get. (27) ( n +1) ( n + 2) a n+2 n 2 n a n n 1 + ( 2ε 1) a n n = 0. (28)
5 Inserting the factor of in the second term into the summation gives ( n +1) ( n + 2) a n+2 n 2 n a n n + 2ε 1 = 0. (29) ( ) a n n Notice that now all the power series have the same range and also the same power of, so we can collect terms and add the power series, { ( n +1) ( n + 2) a n+2 2n a n + ( 2ε 1) a n } n = 0, or { ( n +1) ( n + 2) a n+2 ( 2n + 1 2ε) a n } n = 0. (30) The functions 0, 1, 2, n in the summation form a complete, linearly independent set. The only way for the equation to be satisfied is if each factor multiplying the powers is independently set equal to zero. Thus, we must have Solving for the coefficient a n+2 yields ( n +1) ( n + 2) a n+2 ( 2n + 1 2ε) a n = 0. (31) a n+2 = 2n + 1 2ε n +1 ( )( n + 2) a n. (32) This equation is called a recursion formula or recursion relation. Given the value of one coefficient, such as a 0, it allows us to determine the values of all the others. This particular formula is called a two-term recursion relation because the coefficients are spaced two units apart (n and n+2). An interesting feature of the two-term recursion formula is that the coefficients generated are either all even or all odd. If we start with a 0, then the coefficients generated from the recursion formula are a 2, a 4, a 6, a 8, etc. On the other hand, if we start with a 1, then the coefficients generated from the recursion formula are a 3, a 5, a 7, a 9, etc. This means that there are two types of solutions, even and odd, such that ( ) = e 2 / 2 ψ ( ) = e 2 / 2 ψ a 2l 2l EVEN Solutions l=0 a 2l+1 2l+1 ODD Solutions. l=0 (33) The fact that we got solutions that are either even or odd is a reflection of the symmetry of the harmonic oscillator potential. In general, for an arbitrary differential equation, this will not be the case.
6 Step 4. Truncation of the Series The even and odd solutions shown above involve infinite power series. They are valid solutions of the differential equation. However, they do not satisfy all the criteria of a wavefunction. One of the requirements for a wavefunction is that it remains finite so that its square is integrable (in order that it represent probability). Because they contain infinite power series, ψ( ) as ±. The only way this can be alleviated is by truncating the infinite power series so that the wavefunction remains finite. Suppose the truncation of the power series occurs after v terms in the power series. Then all the higher coefficients are zero, a v+2 = a v+4 = a v+6 = = 0. (34) Using the recursion formula for a v+2, this means that 0 = a v+2 = 2v + 1 2ε v +1 ( )( v + 2) a v. (35) Since we truncated the series at v terms, we know that the coefficient a v is not zero. Therefore, we can divide both sides of the equation by a v, 2v + 1 2ε v +1 ( )( v + 2) = 0. (36) Multiplying by the factor in the denominator gives 2v + 1 2ε = 0. (37) Solving for the dimensionless energy ε, ε = v + 1 2, v = 0, 1, 2, 3, (38) This is the quantization condition for the energy. Note that the quantum number v can be any integer since truncation of the power series can occur at any term less than infinity. To get the quantization condition for the original energy, note that the dimensionless energy ε was defined as ε = E /hν 0. Solving for E gives E v = hν 0 ε = hν 0 ( v + 1 2), v = 0, 1, 2, 3, (39)