[Q]Question about harmonic oscilator

[good_phy]

Hi, Finally! I reached harmonic oscilator! Congratulation!

Most of all QM textbook introduced this formula :

Time independent energy eigenstate equation is

[tex] ( - \frac{\hbar^2}{2m} \frac{\partial}{\partial x) + \frac{Kx^2}{2} )\varphi = E\varphi [/tex]

(1)[itex] \varphi_{xx} = -k^2 \varphi [/itex]

[itex] \frac{\hbar^2k^2(x)}{2m} = E - \frac{K}{2}x^2 > 0 [/itex]

We focused classically forbidden domain [itex] x^2 > x_{o}^2, E < \frac{Kx^2}{2} [/itex]

In this case, kinetic energy is negative, so [itex] \varphi_{xx} = k'^2 \varphi [/itex] [itex] \frac{\hbar^2k'^2}{2m} = \frac{K}{2}x^2 - E > 0 [/itex]

For asymptotic domain, [itex] Kx^2/2 >> E [/itex]

(2) [itex] \varphi_{xx} = \frac{mK}{\hbar^2}\varphi = \beta^4x^2\varphi [/itex] where subscript means 2nd differential, [itex] \beta^2 = \frac{mw_{o}}{\hbar} [/itex]

We let (3) [itex] \epsilon = \beta x[/itex]

(2) appears as (4) [itex] \varphi_{\epsilon\epsilon} = \epsilon^2 \varphi [/itex]

If [itex] \epsilon >>1 [/itex] then (2) is approximated to

(5) [itex] \varphi \approx Aexp(\pm\frac{\epsilon^2}{2}) = Aexp(\pm\frac{(\beta x)^2}{2}) [/itex]

I have a question. Liboff said (2) become (4) by introducing (3). But If (3) is right, I thought (4) should be [itex] \varphi_{\epsilon\epsilon} = \beta^2\epsilon^2\varphi [/itex]. Is it right?

And I don't know how to derive (5) from (4). Please lead me.

 

[malawi_glenn]

(2) [itex] \varphi_{xx} = \frac{mK}{\hbar^2}\varphi = \beta^4x^2\varphi [/itex] where subscript means 2nd differential, [itex] \beta^2 = \frac{mw_{o}}{\hbar} [/itex]

We let (3) [itex] \epsilon = \beta x[/itex]

(2) appears as (4) [itex] \varphi_{\epsilon\epsilon} = \epsilon^2 \varphi [/itex]

If [itex] \epsilon >>1 [/itex] then (2) is approximated to

(5) [itex] \varphi \approx Aexp(\pm\frac{\epsilon^2}{2}) = Aexp(\pm\frac{(\beta x)^2}{2}) [/itex]

I have a question. Liboff said (2) become (4) by introducing (3). But If (3) is right, I thought (4) should be [itex] \varphi_{\epsilon\epsilon} = \beta^2\epsilon^2\varphi [/itex]. Is it right?

And I don't know how to derive (5) from (4). Please lead me.

I don't have Liboff, but since you make substitution of variables from x to beta*epsilon, the second derivative on phi w.r.t to x will change..

[tex]\frac{d^2\phi}{dx^2} \rightarrow \beta ^2 \frac{d^2\phi}{d\epsilon^2}[/tex] (chain rule of calculus)

so it should be: [tex] \phi_{\epsilon\epsilon} = \epsilon^2 \phi [/tex]

 

[good_phy]

Thank you for your help! I'm very pleased with you. But Could you tell me how to apply chain rule of calculus on that formula? I just subsitute [itex] x = \beta\epsilon [/itex] into the x of dominator to get your formula. is it right procedure?

What is more, Can you give me a answer of second question?

 

[malawi_glenn]

[tex]\frac{d\phi}{dx} = \frac{d\phi}{d\epsilon}\frac{d\epsilon}{dx}[/tex]

what the second question is about is that that "far" away from the potential, (classical allowed region) wave function must go down as an exponential.

http://en.wikipedia.org/wiki/Image:HarmOsziFunktionen.jpg