Question about the solution to the Harmonic wave equation

[SemM]

Hi, I have been looking in various text about how to find an admissible solution to the Schrödinger eqn in one dim. in the harmonic oscillator model. As in MQM, the solutions to this are said to be $Ae^{ikx}+Be^{-ikx}$, which are then said to be not admissible. The book then goes straigtht to the Hermite polynomials as solutions.

In Bohms Quantum Theory, he splits up the Schrödinger eqn into its factorial form, and uses the two factorial forms to show how the raising and lowering operators are used to develop functions of higher order (and lower order).. Here he goes straight to show an example, $\psi=e^{-x^2}$. He elaborates further on the operators, the generating function and so forth.

In Pauling and Wilsons Quatum Mechanics, on p 69, they write, similar to MQM, the Schrödinger eqn for the harmonic oscillator, and then write:

"This equation is satisfied asymptoticallly by the exponential functions:

\begin{equation}
\psi = e^{\pm \alpha/2 x^2}
\end{equation}

and no way is given on how this is derived, which is surprising, because in all Math books of graduate level, some method for showing how to derive the solution from the original is always given, i.e such as the method of $Ae^{\lambda_1x}+Be^{\lambda_2x}$ for the ODE form $ay''+by'+cy=0$, where $\lambda = [-b\pm\sqrt{b^2-4ac}]/2$.

As one who has recently studied ODEs in both Greenbergs Math, and Kreyszig Adv Eng Math, I was expecting that physicians in all these books, would list up a method for deriving a solution, and I am surprised that always the same example recurs (the form given above and in Bohm and Pauling), and not other versions of it or of similar admissible functions. Especially, when it is known that ODEs have many solutions, and in quantum chemistry knowing several solutions is a great start for DFT methods to calculate the properties of electrons (through the cycling method).

1. Are there really so few admissible examples in literature, and surprisingly, no mathematical method to derive these solns. (i.e. the Bernoulli eqn. method for solving ODEs, or the method for solving the Riccati eqn. etc. etc.)?

2. Can someone provide a method for solving the S.eq. and get the soln. given above, (Pauling and Wilson , and Bohm)?Note that another similar post has been posted previously, but this post has more references, the same example, and even more important, an uttering on whether someone can provide a mathematic method to solve S.Eq. and get the admissible form given above?

This is a question about method, not about why the given soln. results.

Thanks

 

[dRic2]

Try "Introduction to Quantum Mechanics" by David J. Griffiths. It says the same things you did in your post but I really enjoyed the explanation. In fact he shows why the Hermite polynomials are the solution. He actually builds up the solution step by step and when he reaches the final formula he tells you that it is know as Hermite polynomials. In the book it is also explained why the Schrödinger's equation is satisfied asymptoticallly by $\psi = e^{±\alpha x^2}$.

 

[SemM]

Try "Introduction to Quantum Mechanics" by David J. Griffiths. It says the same things you did in your post but I really enjoyed the explanation. In fact he shows why the Hermite polynomials are the solution. He actually builds up the solution step by step and when he reaches the final formula he tells you that it is know as Hermite polynomials. In the book it is also explained why the Schrödinger's equation is satisfied asymptoticallly by $\psi = e^{±\alpha x^2}$.

Fantastic! Thanks for the suggestion, I will get the book from the library.

All the best

 

[dRic2]

If you have any problems finding the book I can send you some pictures of this part. I have this book right know, but I have to return it today or tomorrow.

 

[stevendaryl]

Solving differential equations is often a matter of guess and check. I'm not sure if there is any definitive way to solve them, in general.

However, we can guess the asymptotic form for the wave function pretty easily:

Write \psi(x) = e^{W}. Then in terms of W, the Schrödinger equation becomes:

\frac{-\hbar^2}{2m} (\frac{d^2 W}{dx^2} + (\frac{dW}{dx})^2) + V(x) = E

Let's assume that asymptotically, V(x) \approx A x^l and W(x) \approx B x^n. Then our equation becomes:

\frac{-\hbar^2}{2m} (n(n-1) B x^{n-2}+ B^2 n^2 x^{2n-2}) + Ax^l = E

So there are 4 different exponents in this equation: x^0, x^{n-2}, x^{2n-2}, x^l. Asymptotically, only the largest power is important. Let's assume that:

l > 0, n > 0

Then the equation is only solvable asymptotically if 2n-2 = l or n = \frac{l}{2} + 1. Dividing through by x^l, our equation becomes:

\frac{-\hbar^2}{2m} ((\frac{l}{2} + 1)\frac{l}{2} B x^{-\frac{l}{2} - 1}+ B^2 (\frac{l}{2} + 1)^2) + A = Ex^{-l}

As x \rightarrow \infty, the terms with negative powers of x go to zero, and the equation becomes:

\frac{-\hbar^2}{2m} B^2 (\frac{l}{2} + 1)^2) + A = 0

So B = \pm \frac{\sqrt{2mA}{\hbar}}{\frac{l}{2} + 1}

So, asymptotically, \psi(x) \approx e^{B x^{\frac{l}{2} + 1}}. We can thus use the guess that the full solution is of the form:

\psi(x) = e^{B x^{\frac{l}{2} + 1}} f(x), where f(x) grows more gently for large x than e^{B x^{\frac{l}{2} + 1}}.

In the case of the Harmonic oscillator, l = 2, so \psi(x) is asymptotically e^{B x^2}. (We have to choose the negative square root for B to get a normalizable solution).

If l < 0, then the potential term is unimportant at large x, so we can ignore that. The equation becomes
\frac{-\hbar^2}{2m} (n(n-1) B x^{n-2}+ B^2 n^2 x^{2n-2}) = E

This is solvable asymptotically only when n=1. In that case, we have:
-\frac{\hbar^2}{2m} B^2 = E

So asymptotically, \psi(x) = e^{B x}. If B is real, then we have to choose the negative sign of B to get a normalizable solution.

 

[SemM]

Solving differential equations is often a matter of guess and check. I'm not sure if there is any definitive way to solve them, in general.

Thanks Steven, this was very pedagogically friendly, and is probably an excellet resource for others than me to understand how to solve the S.eq by hand! I will print it out and study it.

I am disappointed however that there is no method to treat the S.Eq as for the Riccati Eqn, or the Bernoulli form. It would indeed be opportune to have. Guessing solutions is really something one would avoid, in other to have reproducibility, because, take a Russian mathematician, like G Perelman, his guess would confuse the best of mathematicians, but be far ahead correct. So I think guessing is a sad part of physical mathematics, and is demoralizing, because I must admit, I would never have guess what you wrote here! Cheers

 

[SemM]

If you have any problems finding the book I can send you some pictures of this part. I have this book right know, but I have to return it today or tomorrow.

Please do! That book will not arrive for several weeks!

 

[vanhees71]

The most simple way to understand the harmonic oscillator is to use purely algebraic methods with annihilation and creation operators (which you need to understand the most important single theory of physics anyway, namely QFT ;-))). To get the energy eigenfunctions you can apply this approach simply within the position representation, and you can easily derive all properties of the Hermite polynomes in a very elegant way. It's much simpler than the standard treatment in older texts. You find this very important derivation in

J. J. Sakurai, Modern Quantum Mechanics, Revised Edition (1994)

 

[SemM]

The most simple way to understand the harmonic oscillator is to use purely algebraic methods with annihilation and creation operators (which you need to understand the most important single theory of physics anyway, namely QFT ;-))). To get the energy eigenfunctions you can apply this approach simply within the position representation, and you can easily derive all properties of the Hermite polynomes in a very elegant way. It's much simpler than the standard treatment in older texts. You find this very important derivation in

J. J. Sakurai, Modern Quantum Mechanics, Revised Edition (1994)

Thanks, now that you mentioned it, do you mean:

Raising operator : $L^{+}$ on position psi

\begin{equation}
\psi_n = \bigg(\frac{d}{dx}-x\bigg)^n \langle \psi, x, \psi^{*} \rangle
\end{equation}

Lowering operator, $L^{-}$ on position psi:\begin{equation}
\psi_n = \bigg(\frac{d}{dx}+x\bigg)^n \langle \psi, x, \psi^{*} \rangle
\end{equation}

?

Probably not, because the Dirac brackets yield a simple number. Can you elaborate on what you mean "apply this approach simply within the position representation"?

Did you alternatively express the S.eq. by the $L^{+}$ and $L^{-}$ and then use their commutation to replace either of them and thereby get another form of the eqn which is directly related to the Hermite polynomials?

P:S: Everyone calls it annhillation and creation operators, but isn't raising and lowering more correct, because if it was creation, it would create something out of zero, but that doesn't work for this creation operator, as it would give zero for any eigenstate.