• Support PF! Buy your school textbooks, materials and every day products Here!

A question about a certain math argument in Griffiths

  • Thread starter Coffee_
  • Start date
  • #1
259
2
1. In the part of the analytical solution of the harmonic oscillator, p38 in my version of the book a certain mathematical trick is used. The position part of the wave function in function of a dimensionless variable z, is given by

##\frac{d^{2}\Psi(z)}{dz^{2}}=(z^{2}-K) \Psi##

In the limit of z going to +- infinity the solution to this ODE (bearing in mind that we want the function to be normalizable)

##\Psi(z)=Ae^{\frac{-z^{2}}{2}}##

Then he says that for any z now the function can be written as

##\Psi(z)=h(z)e^{\frac{-z^{2}}{2}}##

My question. How certain can I be, only from this reasoning that ##h(z)## does converge? A informal reasoning along the lines:

Well as z goes to the limit of +- infinity the behaviour of ##\Psi(z)## has to be that of the earlier mentioned exponential. This can only be true if ##h(z)## doesn't diverge because even if ##h(z)=z## for example, it would change the behaviour in the limit to ##\Psi(z)=ze^{\frac{-z^{2}}{2}}##

This was an informal reasoning that can have gaps in it. Can I be sure that this is 100% true in a formal way as well?
 
Last edited:

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,057
3,027
How certain can I be, only from this reasoning that ##h(z)## does converge?
There is no certainty whatsoever (If you mean ##\displaystyle \lim_{z\rightarrow \infty} h(z) = 0##. None of the h that are found in a later stage do this).

Usable wave functions must be normalizable, so basically all that is required for that is that the h diverge "less fast" than ##e^{+z^2/2}##.

And the behaviour of ##\Psi(z)## for ##{z\rightarrow \infty}## doesn't have to be exactly as ##e^{-z^2/2}## (that, by the way would mean ##h\rightarrow 1##), only "like" ##e^{-z^2/2}##. As long as ##\Psi## satisfies the Schroedinger equation, everything is hunky dory.

Basically what's done here is the factoring out of the asymptotic behaviour and then the treatment continues with looking at what's left over and what equation and conditions it has to fulfil.
 
Last edited:
  • #3
259
2
There is no certainty whatsoever (If you mean ##\displaystyle \lim_{z\rightarrow \infty} h(z) = 0##. None of the h that are found in a later stage do this).
By converging I meant going to a constant and not zero necessarily.

And the behaviour of ##\Psi(z)## for ##{z\rightarrow \infty}## doesn't have to be exactly as ##e^{-z^2/2}## (that, by the way would mean ##h\rightarrow 1##), only "like" ##e^{-z^2/2}##.
Yeah but the constant A is still in front of the exponential so all that requires is that h(z) has an existing limit towards -+ infinity.

I think you are spot on onto my problem upon underlining exactly. Is it allowed for example in theory that ##h(z)=z## from these facts we have here?

Thanks for the reply.
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,800
5,057
Is it allowed for example in theory that h(z)=z
No, that would not be allowed.
Let's be clear about what this means: "in the limit, ##\Psi(z)=Ae^{\frac{-z^{2}}{2}}##" (which should really be written ##\Psi(z)##~##Ae^{\frac{-z^{2}}{2}}##).
It means ##\lim_{z\rightarrow \infty}\Psi(z)e^{\frac{z^{2}}{2}}=A##. From that you can conclude ##\lim_{z\rightarrow \infty}h(z) = A##.
 
  • #5
BvU
Science Advisor
Homework Helper
2019 Award
13,057
3,027
By converging I meant going to a constant and not zero necessarily.
In that interpretation there is one eigenstate that converges. You can verify that h=const satisfies the Schroedinger equation. The h in all other eigenstates then "don't converge".

Yeah but the constant A is still in front of the exponential so all that requires is that h(z) has an existing limit towards -+ infinity.
No. All that is requireed is that ##\Psi## satisfies the Schroedinger equation and is normalizable. If the former is satisfied, then all that is required for h is that the h diverge "less fast" than ##e^{ +z^2 /2}##. Any finite power of z is acceptable.

Is it allowed for example in theory that ##h(z)=z## from these facts we have here?
Yes. In the wavefunction of the first excited state h is proportional to z. Your book top of p 41.
 
  • #6
259
2
Thanks a lot, this really helped me. One final small thing so that I reall understand it.

Consider a function ##f(x)=(x^{2}+x^{3}-2)e^{-x}##

It is correct to say that this function behaves like ##e^{-x}## in the limit?
 
  • #7
259
2
No, that would not be allowed.
Let's be clear about what this means: "in the limit, ##\Psi(z)=Ae^{\frac{-z^{2}}{2}}##" (which should really be written ##\Psi(z)##~##Ae^{\frac{-z^{2}}{2}}##).
It means ##\lim_{z\rightarrow \infty}\Psi(z)e^{\frac{z^{2}}{2}}=A##. From that you can conclude ##\lim_{z\rightarrow \infty}h(z) = A##.
What about what the other poster pointed out below you, that indeed ##h(z)=z## is a form of one of the accepted solutions for ##h(z)## in this problem later on?
 
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,800
5,057
What about what the other poster pointed out below you, that indeed ##h(z)=z## is a form of one of the accepted solutions for ##h(z)## in this problem later on?
We seem to be answering different questions. I based my answer on the limiting behaviour condition stated ##\Psi(z)=Ae^{-\frac{z^2}{2}}##, where, presumably, A is a constant. BvU,I believe, is allowing the more permissive condition that ##\Psi(z)## converges to 0 as |z| tends to infinity. But I'm approaching this based purely on the algebra. I have little knowledge of quantum theory.
 
  • #9
BvU
Science Advisor
Homework Helper
2019 Award
13,057
3,027
I sympathize with the nay sayers, but it doesn't matter: if it satisfies the Schroedinger equation and it is normalizable, then it's a suitable steady state solution.

When in doubt, make a drawing:

times_exp-x2.jpg


Of course a logarithmic y axis mercilessly reveals an ever increasing ratio between the two, but that really doesn't matter.

A bit later on, when you have the steady state solutions, the probabilities to find the particle outside the classical region drop off exponentially, that's the main thing. It doesn't matter all that much (understatement!) whether the actual probability density is 10-44 or 10-38 (your function, x=100 - where classically x ##\approx# 3) .
 
  • #10
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
32,800
5,057
Thanks a lot, this really helped me. One final small thing so that I reall understand it.

Consider a function ##f(x)=(x^{2}+x^{3}-2)e^{-x}##

It is correct to say that this function behaves like ##e^{-x}## in the limit?
The usual definition of functions having the same asymptotic behaviour can be found at http://www.math.uiuc.edu/~hildebr/595ama/ama-ch2.pdf. As I posted, f(x)~g(x) as ##x \rightarrow x_0## means ##\lim _{x \rightarrow x_0} f(x)/g(x) =1##. Your f(x) above does not behave like e-x in that sense. But ln(f(x)) ~ -x.
 
  • #11
BvU
Science Advisor
Homework Helper
2019 Award
13,057
3,027
We seem to be answering different questions. I based my answer on the limiting behaviour condition stated ##\Psi(z)=Ae^{-\frac{z^2}{2}}##, where, presumably, A is a constant. BvU,I believe, is allowing the more permissive condition that ##\Psi(z)## converges to 0 as |z| tends to infinity. But I'm approaching this based purely on the algebra. I have little knowledge of quantum theory.
Indeed we are. And we all agree that the answer to the original question -- with the given meaning of "convergence" -- is that you can not be sure at all that h converges (except for the ground state, which happens to have h = constant :) ). In fact, for all other solutions, you can be quite sure h diverges, in the sense that ##h\uparrow \pm \infty## for ##x\uparrow\pm\infty##. (do I have to draw an updownarrow in these to accomodate the - sign too?)

What I try to bring across is that the question itself is not relevant in the context of this approach to solve the Schroedinger equation analytically.

[commercial]:
The harmonic oscillator example is incredibly important in quantum mechanics and way beyond, in the physics sense (it is the first order approach to any non-trivial potential function) and also in the didactical sense. So, dear coffee and all others: pay attention, invest time and effort -- and ask (e.g. in PF) when in doubt or stuck ;)
 
  • #12
259
2
I start to get it a little bit more. Thanks a lot for your replies both of you.

As you seem to encourage me to post about confusions about the harmonic oscillator there's one more thing that I is troubling me, related to this in some way. That deserves a different thread though, which I'll make when I'll find some time. Thank again.
 

Related Threads on A question about a certain math argument in Griffiths

  • Last Post
Replies
4
Views
898
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
15
Views
6K
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
4
Views
790
Replies
4
Views
12K
Replies
1
Views
577
Replies
1
Views
3K
Top