Griffiths' QM book: series solution to harmonic oscillator

[tjkubo]

I'm trying to read through Griffiths' QM book, and right now I'm on the series solution to the harmonic oscillator (ch 2). I'm having a hard time following the math (especially after equation 2.81) in this section, so if anyone has read this book, please help.

My first question is about the recursive formula
a_{j+2}=\frac{2}{j}a_{j}
The text says the approximate solution is
a_{j}=\frac{C}{(j/2)!}
How was this derived?

Second, how was the following derived?
\displaystyle\sum \frac{C}{(j/2)!} \xi^j \approx \displaystyle\sum \frac{1}{j!} \xi^{2j}

Third, the first paragraph on page 54 is totally baffling to me. Why must the power series terminate? Why must a_n+2 = 0? How does this truncate either h_even or h_odd? I think if I can get through this paragraph the rest will come naturally.

 

[kuruman]

First question
If
a_{j+2}=\frac{a_{j}}{j/2}
Then
a_{j}=\frac{a_{j-2}}{(j-2)/2}= \frac{a_{j-2}}{(j/2)-1}
so that when I replace this in the first equation, I get
a_{j+2}=\frac{a_{j-2}}{(j/2)(j/2-1)}
Now write a_j-2 in terms of a_j-4, replace and keep on going down in indices until you reach a₀ which is a constant, call it C.

Second question

Let k = j/2. Then j = 2k so that

<br /> \displaystyle\sum \frac{C}{(j/2)!} \xi^j \approx \displaystyle\sum \frac{1}{k!} \xi^{2k}<br />

But k is a dummy index and can be replaced by j.

Third question

The recursion relation 2.81 is good for either even or odd j. That is how it truncates either h_even or h_odd. Think what would happen if the series did not terminate and ran all the way out to infinity. You would not be able to match the asymptotic behavior for large ξ (Equation 2.76).

Good luck getting through the rest of the section.

 

[tjkubo]

Thank you, kuruman. That was very helpful!
However, I'm still having trouble understanding the paragraph on page 54. I think I now understand why the series must terminate, but I still don't see why if n is for example even, the odd powers must disappear. Why can't h(ξ) = a₀ + a₁ξ + a₂ξ² for example? Why does h(ξ) have to include only even powers or only odd powers? I hope this makes sense.

 

[kuruman]

The discussion on page 54 shows you how to calculate the coefficients a_j once you have chosen a value of n, i.e. for given energy eigenstate. Note that the maximum value that j can have before the series terminate is n. So if n is even, only the even powers in h(ξ) can and will be terminated while the odd powers will all be present. This means that the wavefunction will blow up for large ξ unless none of the odd powers are present to begin with, i.e. a₁=0. Naturally, if n is odd, then you must set a₀=0 for similar reasons.

Also, when you have a symmetric potential, i.e. V(-x) = V(x), as in this case, the energy eigenstates are either even or odd functions. This is because the Hamiltonian operator commutes with the parity operator. If you don't know what I am talking about (this stuff is in the next chapter of Griffiths), accept the previous explanation and move on. What I just said will be clearer later.

 

[hudeel]

i want the sclution for ( 1-d) harmoio oscillator by serise

 

[hudeel]

Iwant the sclution for ( 1-d) harmonic osillator by bra and ket