The Harmonic Oscillator Asymptotic solution?

[patric44]

Homework Statement

i am trying to find the Asymptotic solution of the quantum harmonic oscillator using power series ?

Relevant Equations

y''-x^2 y = 0

hi guys
i am trying to solve the Asymptotic differential equation of the Quantum Harmonic oscillator using power series method and i am kinda stuck :
$$y'' = (x^{2}-ε)y$$
the asymptotic equation becomes :
$$y'' ≈ x^{2}y$$
using the power series method $y(x) = \sum_{0}^{∞} a_{n}x^{n}$ , this leads to :
$$\sum_{n=2}^{∞} n(n-1)a_{n}x^{n-2}+x^{2}\sum_{n=0}^{∞} (-1)a_{n}x^{n} = 0$$
$$\sum_{n=2}^{∞} n(n-1)a_{n}x^{n-2}+\sum_{n=0}^{∞} (-1)a_{n}x^{n+2} = 0$$
$$\sum_{n=2}^{∞} n(n-1)a_{n}x^{n-2}+\sum_{n=0}^{∞} (-1)a_{n}x^{n+2} = 0$$
changing n to n+4 :
$$\sum_{n=-2}^{∞} (n+4)(n+3)a_{n+4}x^{n+2}+\sum_{n=0}^{∞} (-1)a_{n}x^{n+2} = 0$$
expanding the first sum :
$$\sum_{n=0}^{∞} (n+4)(n+3)a_{n+4}x^{n+2}+\sum_{n=0}^{∞} (-1)a_{n}x^{n+2} +2a_{2}+6a_{3}x= 0$$
$$\sum_{n=0}^{∞}[(n+4)(n+3)a_{n+4}-a_{n}]x^{n+2}+2a_{2}+6a_{3}x=0$$
i am stuck here becouse i have an excess "x" term , so is this equation could be solved exactly using power series , or there is another way other than the ansatz $e^{\pm\frac{x^{2}}{2}}$ ?

 

[vela]

Doesn't that imply that $a_2=a_3=0$?

 

[patric44]

Doesn't that imply that $a_2=a_3=0$?

i am not really sure , but suppose that is the case , this will lead me to the recurrence formula of :
$$a_{n} = (n+4)(n+3)a_{n+4}$$
and since $a_{2} = 0$ this leads to :
$$a_{2} = (2+4)(2+3)a_{6}⇒ a_{6} = 0$$
and so on ?! , how to continue from this recurrence relation ? i know the solution already , it suppose to be
approximately $$y = e^{\frac{\pm\;x^{2}}{2}} $$

 

[patric44]

i revised my solution and i am relatively sure that the recurrence formula is
$$a_{n+4} = \frac{a_{n}}{(n+4)(n+3)}$$
but iam not really sure how to procced from here to reach a form of a function that i could identify , the solution should be approximately $e^{\frac{x^{2}}{2}}$ is this form reachable from this recurrence formula , even with some approximations ?!

 

[vanhees71]

It should rather be $\exp(-x^2/2)$, because otherwise you have no square-integrable function. To see that this is an approximate asymptotic solution for $x \rightarrow \infty$ to the same order as made by neglecting $\epsilon$ you get by simply plugging it into the equation.

 

[patric44]

It should rather be $\exp(-x^2/2)$, because otherwise you have no square-integrable function. To see that this is an approximate asymptotic solution for $x \rightarrow \infty$ to the same order as made by neglecting $\epsilon$ you get by simply plugging it into the equation.

thanks vanhees71 i meant $e^{\frac{-x^{2}}{2}}$ but my question is that i want to show this using the power series method for the equation of the harmonic oscillator after neglecting $\epsilon$ :
$$y''≈x^{2}y$$
is it possiple to show by power series method that the solution to this might be $e^{\frac{-x^{2}}{2}}$ ? because every source i saw just put this asymptotic solution as if its an ansatz !? .

 

[vanhees71]

Yes, the trick is to make the ansatz
$$y(x)=z(x) \exp(-x^2/2)$$
and then apply the Frobenius method (generalized power-series ansatz) to $z$. This works better than applying it directly, because you get the recursion relation for $a_{n+4}=F(a_{n})$. What you'd need is a recursion relation of the type $a_{n+2}=F(a_{n})$ so that you have to choose only $a_0$ and $a_1$ as the arbitrary starting values, leading to two linearly independent solutions of your 2nd-order differential equation.

 

[patric44]

Yes, the trick is to make the ansatz
$$y(x)=z(x) \exp(-x^2/2)$$
and then apply the Frobenius method (generalized power-series ansatz) to $z$. This works better than applying it directly, because you get the recursion relation for $a_{n+4}=F(a_{n})$. What you'd need is a recursion relation of the type $a_{n+2}=F(a_{n})$ so that you have to choose only $a_0$ and $a_1$ as the arbitrary starting values, leading to two linearly independent solutions of your 2nd-order differential equation.

how would i know the ansatz "$e^{\frac{-x^{2}}{2}}$" is the right ansatz to begin with as my asymptotic solution ?! , i know how to complete the solution using power series method after the putting
$$y(x)=z(x) \exp(-x^2/2)$$
and get to the Hermite polynomials ans so on , but how to show that the asymptotic solution it self is approximately $e^{\frac{-x^{2}}{2}}$ , what i really want to do is to just solve the asymptotic equation only to convince my self that $\exp(-x^2/2)$ is the right ansatz , i don't know if that make sense

 

[MathematicalPhysicist]

Yes, the trick is to make the ansatz
$$y(x)=z(x) \exp(-x^2/2)$$
and then apply the Frobenius method (generalized power-series ansatz) to $z$. This works better than applying it directly, because you get the recursion relation for $a_{n+4}=F(a_{n})$. What you'd need is a recursion relation of the type $a_{n+2}=F(a_{n})$ so that you have to choose only $a_0$ and $a_1$ as the arbitrary starting values, leading to two linearly independent solutions of your 2nd-order differential equation.

That is what I would call cheating... :-> I mean how did you know to look for such a product without knowing beforehand what the solution will look like.

My thoughts about this is to find some integration factor for the differential form: $d(y'^2)-x^2d(y^2)=C$ where $C$ is some constant.
You get this equation by multiplying $y''= x^2 y$ by $y'$.
But I am clueless about how to find such an integration factor.

 

[patric44]

That is what I would call cheating... :-> I mean how did you know to look for such a product without knowing beforehand what the solution will look like.

My thoughts about this is to find some integration factor for the differential form: $d(y'^2)-x^2d(y^2)=C$ where $C$ is some constant.
You get this equation by multiplying $y''= x^2 y$ by $y'$.
But I am clueless about how to find such an integration factor.

thats why i tried power series method for $y''= x^2 y$ but i can't go further with my solution

 

[vanhees71]

You think way too complicated ;-). Just use
$$\mathrm{d}_x^2 \exp(-x^2/2)=(x^2-1) \exp(-x^2/2) \simeq_{x \rightarrow \infty} x^2 \exp(-x^2/2).$$

 

[MathematicalPhysicist]

You think way too complicated ;-). Just use
$$\mathrm{d}_x^2 \exp(-x^2/2)=(x^2-1) \exp(-x^2/2) \simeq_{x \rightarrow \infty} x^2 \exp(-x^2/2).$$

Seems like you need to know this relation before doing this exercise.
But yes, seems easy enough, you only need to know how to take derivatives... :-D

 

[patric44]

You think way too complicated ;-). Just use
$$\mathrm{d}_x^2 \exp(-x^2/2)=(x^2-1) \exp(-x^2/2) \simeq_{x \rightarrow \infty} x^2 \exp(-x^2/2).$$

what if i came from a different planet and didn't know before that $e^{\frac{-x^{2}}{2}}$ is the solution

 

[vela]

It's an educated guess. It's easy to see that differentiating $\exp(-x^2/2)$ twice will give you something of the form $p(x)\exp(-x^2/2)$ where $p(x)$ is quadratic.

I doubt you could figure this out from the series solution. It's already hard to determine what a series converges to, and here you're trying to show it will converge to an approximate solution, not the actual solution.

 

[mpresic3]

i revised my solution and i am relatively sure that the recurrence formula is
an+4=an(n+4)(n+3)
but iam not really sure how to procced from here to reach a form of a function that i could identify , the solution should be approximately ex22 is this form reachable from this recurrence formula , even with some approximations ?!

This looks OK so far.

( Sorry I spoke too soon. I think the recurrence relation is related to the past two coefficients, not the one earlier coefficient

In any case, you want to know how to proceed. Once you get the correct recurrence relation. First note with the correct recurrence relation, the current coefficient depends on the past coefficients. You want a way to truncate the series. Otherwise, eventually the high order coefficient will be non zero, and you will get a function of the form x^n exp(-x^2/2) for infinite n, which is not normalizable. So you want to find conditions which make a sub n = 0 for n (finitely large). Do this truncation for n gives you the Hermite polynomials.

 

[pasmith]

i revised my solution and i am relatively sure that the recurrence formula is
$$a_{n+4} = \frac{a_{n}}{(n+4)(n+3)}$$
but iam not really sure how to procced from here to reach a form of a function that i could identify , the solution should be approximately $e^{\frac{x^{2}}{2}}$ is this form reachable from this recurrence formula , even with some approximations ?!

Set $n = 4k$. Then you have $$a_{4(k+1)} = \frac{1}{(4k + 4)(4k + 3)}a_{4k} = \frac{(4k + 2)!}{(4k + 4)!} a_{4k}$$ which has the solution $$a_{4k} = a_0\prod_{r=0}^{k-1} \frac{(4r + 2)!}{(4r + 4)!}.$$ Thus $$y(x)= a_0\sum_{k=0}^\infty \left(\prod_{r=0}^{k-1} \frac{(4r + 2)!}{(4r + 4)!}\right)x^{4k}.$$ It's not obvious that this tends asymptotically to $$\exp(\tfrac12 x^2) = \sum_{n=0}^\infty \frac{1}{2^nn!}x^{2n}$$ and that's your issue: convergence of series tells you what happens as $n \to \infty$ for fixed $x$, and asymptoticness tells you what happens as $x \to \infty$ for fixed $n$.

However: starting from $y'' = (1 + x^2) y$ for which $e^{\frac12 x^2}$ is a solution, using the series method yields a recurrence $$a_{m+2} = \frac{a_m + a_{m-2}}{(m+2)(m+1)}$$ together with $2a_2 = a_0$ and $6a_3 = a_1$. You can see that by neglecting $a_m$ we get the same recurrence as $y'' = x^2 y$, which of course does not prove that the solutions are asymptotic (nor is there any particular justification for neglacting $a_m$).

 

[patric44]

Set $n = 4k$. Then you have $$a_{4(k+1)} = \frac{1}{(4k + 4)(4k + 3)}a_{4k} = \frac{(4k + 2)!}{(4k + 4)!} a_{4k}$$ which has the solution $$a_{4k} = a_0\prod_{r=0}^{k-1} \frac{(4r + 2)!}{(4r + 4)!}.$$ Thus $$y(x)= a_0\sum_{k=0}^\infty \left(\prod_{r=0}^{k-1} \frac{(4r + 2)!}{(4r + 4)!}\right)x^{4k}.$$ It's not obvious that this tends asymptotically to $$\exp(\tfrac12 x^2) = \sum_{n=0}^\infty \frac{1}{2^nn!}x^{2n}$$ and that's your issue: convergence of series tells you what happens as $n \to \infty$ for fixed $x$, and asymptoticness tells you what happens as $x \to \infty$ for fixed $n$.

However: starting from $y'' = (1 + x^2) y$ for which $e^{\frac12 x^2}$ is a solution, using the series method yields a recurrence $$a_{m+2} = \frac{a_m + a_{m-2}}{(m+2)(m+1)}$$ together with $2a_2 = a_0$ and $6a_3 = a_1$. You can see that by neglecting $a_m$ we get the same recurrence as $y'' = x^2 y$, which of course does not prove that the solutions are asymptotic (nor is there any particular justification for neglacting $a_m$).

thank you so much its clear now , though its kinda disappointing that the series solution will not work

 

[patric44]

pasmith i tried to plug the series you provided in wolfram alpha to see how it goes , and i just picked the first few terms in the series which gave :
$$x^{4} +\frac{1}{12} (x^{4} + 1) + 1$$
on compairnig this to $e^{\frac{x^{2}}{2}}$ it was something like this :

for physicists i would say that this is a very good approximation
but i have a little question : how would i get the other solution from the series which in this case will be the real physical solution corresponding to $e^{\frac{-x^{2}}{2}}$

 

[vanhees71]

Why are you so obsessed with this application of the Frobenius method? It's well-known that it doesn't work in this case, at least I don't know any textbook which says otherwise. That's why in every textbook they make the ansatz
$$y(x)=z(x) \exp(-x^2/2)$$
and then apply the Frobenius method to $z(x)$, which obeys the equation
$$z''-2x z' +(\epsilon-1) z=0.$$
For this case, which is of the Fuchs class, where the Frobenius method is applicable, you get a recursion relation for the coefficients showing that any solution, which does not stop at a finite power, goes asymptotically like $\exp(x^2)$. So you must choose $\epsilon$ such that it stops, which gives the known eigenvalues for the energy of the harmonic oscillator (among of the most important results of the entire QM 1 lecture!) and the Legendre Polynomials as solutions for $z$.

 

[vanhees71]

pasmith i tried to plug the series you provided in wolfram alpha to see how it goes , and i just picked the first few terms in the series which gave :
$$x^{4} +\frac{1}{12} (x^{4} + 1) + 1$$
on compairnig this to $e^{\frac{x^{2}}{2}}$ it was something like this :
View attachment 271048
for physicists i would say that this is a very good approximation
but i have a little question : how would i get the other solution from the series which in this case will be the real physical solution corresponding to $e^{\frac{-x^{2}}{2}}$

No, it's obviously wrong! You cannot solve the differential equation in this way!

 

[patric44]

Why are you so obsessed with this application of the Frobenius method? It's well-known that it doesn't work in this case, at least I don't know any textbook which says otherwise. That's why in every textbook they make the ansatz
$$y(x)=z(x) \exp(-x^2/2)$$
and then apply the Frobenius method to $z(x)$, which obeys the equation
$$z''-2x z' +(\epsilon-1) z=0.$$
For this case, which is of the Fuchs class, where the Frobenius method is applicable, you get a recursion relation for the coefficients showing that any solution, which does not stop at a finite power, goes asymptotically like $\exp(x^2)$. So you must choose $\epsilon$ such that it stops, which gives the known eigenvalues for the energy of the harmonic oscillator (among of the most important results of the entire QM 1 lecture!) and the Legendre Polynomials as solutions for $z$.

i am not so obsessed i just was trying to find out why the ansatz is in this form , but i found another method to do that instead of the power series method , i just used a simple substitution then applied the asymptoticness :

it feels some how satisfying now i know a way to deduce this ansatz ! .

 

[patric44]

No, it's obviously wrong! You cannot solve the differential equation in this way!

i know that its wrong but i was trying to see were this series will lead me .