Minor mathematical query from Griffiths [QM]

[maverick280857]

Hello all,

I am trying to understand the analytic method for the solution of the (quantum) harmonic oscillator equation.

$$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^{2}x^{2}\psi = E\psi$$

Let

$$\zeta = \sqrt{\frac{m\omega}{\hbar}}x$$

$$K = \frac{2E}{\hbar\omega}$$

We consider

$$\psi(\zeta) = h(\zeta)e^{-\zeta^{2}/2}$$

The equation transforms to,

$$\frac{d^2 h}{d\zeta^2} - 2\zeta\frac{dh}{d\zeta} + (K-1)h = 0$$

Assuming the power series solution,

$$h(\zeta) = \sum_{j=0}^{\infty}a_{j}\zeta^{j}$$

we get the recursion formula,

$$a_{j+2} = \frac{2j+1-K}{(j+1)(j+2)}a_{j}$$

For large j,

$$a_{j+2} \cong \frac{2}{j}a_{j}$$

So,

$$a_{j+2} \cong \frac{2}{j}\frac{2}{(j-1)}\frac{2}{(j-2)}\ldots\frac{2}{1}a_{1}$$

which simplifes to

$$a_{j+2} \cong \frac{2^j}{j!}a_{1}$$

But the solution given is

$$a_{j} = \frac{C}{(j/2)!} = C \frac{1}{(j/2)}}\frac{1}{((j/2)-1)}\ldots\ldots$$

I think I am missing something here. What am I doing wrong? Also, I know that the $(j/2)!$ is just notation but what if the "large j" include some possibly odd values? In that case it wouldn't even be defined. Thanks for your help.

EDIT: 800th post...yay

 

[George Jones]

Because the index of $a$ is $j + 2$, I think

$$a_{j+2} \cong \frac{2}{j}\frac{2}{(j-1)}\frac{2}{(j-2)}\ldots\frac{2}{1}a_{1}$$

should, for odd $j$, actually be something like

$$a_{j+2} \cong \frac{2}{j} \frac{2}{ \left(j - 2 \right)} \frac{2}{\left( j - 4 \right)} \ldots \frac{2}{1} a_{1} \cong \frac{1}{j/2 \left( j/2 - 1 \right) \left( j/2 - 2 \right) \ldots 1/2} a_{1}.$$

For real numbers, the Gamma function generalizes the factorial function.

Regards,
George

 

[maverick280857]

Because the index of $a$m is $j + 2$, I think



should, for odd $j$, actually be something like

$$a_{j+2} \cong \frac{2}{j} \frac{2}{ \left(j - 2 \right)} \frac{2}{\left( j - 4 \right)} \ldots \frac{2}{1} a_{1} \cong \frac{1}{j/2 \left( j/2 - 1 \right) \left( j/2 - 2 \right) \ldots 1/2} a_{1}.$$

For real numbers, the Gamma function generalizes the factorial function.

Regards,
George

Thanks again George! I'll look this up.

 

[Gokul43201]

Assuming the power series solution,

$$h(\zeta) = \sum_{j=0}^{\infty}a_{j}\zeta^{j}$$

we get the recursion formula,

$$a_{j+2} = \frac{2j+1-K}{(j+1)(j+2)}a_{j}$$

For large j,

$$a_{j+2} \cong \frac{2}{j}a_{j}$$

So,

$$a_{j+2} \cong \frac{2}{j}\frac{2}{(j-1)}\frac{2}{(j-2)}\ldots\frac{2}{1}a_{1}$$

Your approximation fails with all the terms near the end of the product (ie: j not large). And these terms are significantly larger than terms in the beginning, so the error gets quite big. Or am I messing up ?

 

[maverick280857]

Your approximation fails with all the terms near the end of the product (ie: j not large). And these terms are significantly larger than terms in the beginning, so the error gets quite big. Or am I messing up ?

Yes I think you're right...I thought about it that way (interestingly this is obvious if you expand the "factorials"). But Griffiths has used the approximation anyway.