# Griffiths' QM book: series solution to harmonic oscillator

1. Aug 11, 2009

### 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 an+2 = 0? How does this truncate either heven or hodd? I think if I can get through this paragraph the rest will come naturally.

2. Aug 11, 2009

### 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 aj-2 in terms of aj-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

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

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 heven or hodd. 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.

3. Aug 11, 2009

### 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.

4. Aug 12, 2009

### kuruman

The discussion on page 54 shows you how to calculate the coefficients aj 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.

5. Nov 3, 2009

### hudeel

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

6. Nov 3, 2009

### hudeel

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

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