Griffith's QM, Harmonic Oscillator approximate solution eq

[Sparky_]

Homework Statement

on page 51 (of my book, probably not current) section 2.3.2 equation 2.74 and 2.75

d²ψ / dξ² ≈ ψξ²

​

Homework Equations

This is an approximation of the Schrodinger equation with a variable introduced ξ = √(mω/h)

The solution is given: ψ(ξ) = Ae^-ξ2/2 +Be^ξ2/2

The Attempt at a Solution

[/B]
I am attempting to reproduce this solution with no luck

I have tried:
integrating once to get rid of the second order derivative and try an integrating factor approach.

∫ d² ψ / dξ² = ∫ ξ² ψ

dψ / dξ = ψ ξ³ / 3

dψ/dξ - ψ ξ³ /3 = 0

I treated this as a 1st order and did the integrating factor and so forth ... I stopped when I saw I am not approaching the solution Ae^-ξ2/2 +Be^ξ2/2

I am getting e^-ξ4/12 terms within the integrating factor

Using separable variables approach I likewise stopped when I wasn't seeing the solution in the book showing up.

For what it's worth, I am very rusty with differential equation (and some math in general) I am not a student nor physicist. I am chasing this as a bucket list item to understand some quantum mechanics.

Bottom line how do I solve the equation to get the solution A e^ξ2/2 + Be^-ξ2

I assume it's a differential equation method I am missing??

Thanks
Sparky_

 

[PeroK]

Try the substitution $z = \xi^2/2$. It should reduce to one of these:

https://en.wikipedia.org/wiki/Cauchy–Euler_equation

 

[PeroK]

I have tried:
integrating once to get rid of the second order derivative and try an integrating factor approach.

∫ d2 ψ / dξ2 = ∫ ξ2 ψ

dψ / dξ = ψ ξ3 / 3

This is not right, as $\psi$ is a function of $\xi$. You've treated it like a constant.

 

[vela]

Homework Statement

on page 51 (of my book, probably not current) section 2.3.2 equation 2.74 and 2.75

d²ψ / dξ² ≈ ψξ²

​

Homework Equations

This is an approximation of the Schrodinger equation with a variable introduced ξ = √(mω/h)

The solution is given: ψ(ξ) = Ae^-ξ2/2 +Be^ξ2/2

That should be $\psi(\xi) = A e^{-\xi^2/2} + B e^{\xi^2/2}$. The factors of 1/2 are in the exponent.

Note that the function isn't a solution to the differential equation. It's only an approximate solution. To verify it works, try just plugging $\psi$ into the differential equation.

 

[Sparky_]

Vela, Perok, and others ,

Well, I have spent some time reviewing the Cauchy- Euler equations.

Perok - I do not follow you (yet) - do I make you suggested substitution within the differential equation : d² ψ/ dξ² ≈ ξ² ψ?

Vela -
Should I not get hung up on trying to solve the differential equation (equation 2.74 in my book) : d² / dξ² ≈ ξ² ψ?

Like you suggested I plugged the approximate solution back in (instead of trying to solve the original diff equation) - I get
ψ = Ae^-ξ2/2 + Be^ξ/2
ψ^'' = -Ae^-ξ2/2 + Aξ²e^-ξ2/2 + Be^ξ2/2 + Bξ²e^ξ2/2

plugging in ξ²ψ = Aξ²e^-ξ2/2 + Bξ²e^ξ2/2

Granted the book clearly shows "approximate" sign not equal sign, I get

-Ae^-ξ2/2 + Aξ²e^-ξ2/2 + Be^ξ2/2 + Bξ²e^ξ2/2 = Aξ²e^-ξ2/2 + Bξ²e^ξ2/2

end up with 2 extra terms -Ae^-ξ2/2 + B e^ξ2/2

I see in the book (or at least I hope I am seeing this correctly) - "The B term is clearly not normalizable" , then there is work on the part with the negative expoent

so some clarifying questions:

1)
Is the main point here not to solve the differential equation but rather show where to focus (ignore the B term) - there is the approximation symbol not equal? Meaning is he (the author) is showing the general shape or form of the solution?

2) If I wanted to really solve the differential equation : d² y/ dx² = x² y? (I change to dy/dx - to make it general and change it to an equal sign) - how do I solve this? It's not a Cauchy - Euler (yet) is it - if I divide by x² then I have an x^-2 as a coefficient on my second derivative - not in the correct form.

sorry for getting a little long here
Thanks

 

[PeroK]

Rather than answer your questions directly, let me simply explain what Griffiths is doing:

First, he derives an exact differential equation of the form:

$\frac{d^2 \psi}{d \xi^2} = (\xi^2 - K)\psi$

Now, one thing he could have done here is simply to say that (based on a prior knowledge of these sorts of differential equations), let's look for a solution of the form:

$\psi(\xi) = h(\xi)e^{-\xi^2 /2} \$ (equation 2.77)

Note that any function $\psi(\xi)$ can be expressed as $\psi(\xi) = h(\xi)e^{-\xi^2 /2}$. Simply let $h(\xi) = \psi(\xi)e^{\xi^2 /2}$, so you're not losing any generality here.

In fact, you could argue that taking this "prior knowldege" approach would have been perfectly satisfactory.

Instead, however, he wants to motivate this approach by giving you a bit more background. So, he looks for an approximate solution to the equation for large $\xi$. First, he looks at an approximation to the equation:

$\frac{d^2 \psi}{d \xi^2} \approx \xi^2 \psi \$ (for large $\xi$)

Then, he has an approximate solution to this approximate equation:

$\psi(\xi) = Ae^{-\xi^2 /2} + Be^{\xi^2 /2}$

If you check this by differentiating it, you get:

$\psi''(\xi) = A(\xi^2 -1)e^{-\xi^2 /2} + B(\xi^2 +1)e^{\xi^2 /2} \approx \xi^2 \psi \$ (for large $\xi$)

Then, he observes that the positive exponential solution is not normalisable, so he drops that. This completes the motivation for his approach, equation 2.77.

Now that I've read the relevant section in Griffith's book, I think there is no need to get bogged down in the details of solving these differential equations - especially as it was only to motivate his approach.

Personally, I would move on at this point, as the power-series solution to equation 2.78 is really what it's all about.

Finally, what I would note here is this: if we dive in with a power series solution to equation 2.72, then it must get messy and we'll have to identify and extract the power series for $e^{-\xi^2 /2}$ and see what's left. Griffith's whole preamble, therefore, is to justify taking out this factor first.

So, it's probably more important to see this big picture of what's going on than getting bogged down in the details. The details of solving equation 2.78 are, however, worth taking note of.

 

[Sparky_]

Thank You!