Separatibility of wave-function in Schrodinger equation describing motion of particle in spherical shell [closed]

Question

Question. Why wave-function describing rotation of particle around spherical surface is separable? Why solution obtained assuming function is separable is general solution? I am interested in proof of that (see P.S.).

Explanation of the question. For thous who are not familiar with the topic. I don't need the solution of this equation, It can be found in 301 page of Atkins textbook of physical chemistry, 8th edition. The solutions are known as spherical harmonics.

Here is this wave-function. $$\frac{-\hbar^2}{2m}\nabla^2 \psi(x,y,z)=E \psi(x,y,z)$$ It can be expressed like this in spherical polar coordinates, assuming that radius is constant. $$\Lambda^2\psi(\theta,\phi)=\epsilon \psi(\theta,\phi)$$, where $\Lambda^2$ is the Legendrian: $$ \Lambda^2 = \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} + \frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\sin\theta\frac{\partial}{\partial \theta} $$ and $\epsilon=\frac{2IE}{\hbar^2}$. And I need to prove that for any wave-function staisfying the diffferential equation there are $\Theta(\theta)$ and $\Phi(\phi)$ $\forall \theta \forall \phi[\psi(\theta,\phi)=\Theta(\theta)\Phi(\phi)]$. That is: $$\forall \psi [\forall \theta \forall \phi \Lambda^2\psi(\theta,\phi)=\epsilon \psi(\theta,\phi) \implies \exists \Theta \exists \Phi \forall \theta \forall \phi[\psi(\theta,\phi)=\Theta(\theta)\Phi(\phi)]] $$

Relevance. Wave-function describing rotation of particle in spherical shell is used to construct wave-function that describes electron motion in hydrogen atom. Similar arguments are popular in textbooks about quantum chemistry. In several similar cases it is assumed that these wave-function are separable.

My attempt. I have tried to find the solution for several days with no success. For some people this might seem like homework-like (too simple to answer) question. In these case please suggest few classical comprehensible for chemist textbooks where this question is discussed.

My research. In the Atkins textbook of Physical chemistry is 8th edition is justification of wave-function separability. There are several other similar examples. Author stares that separability can be inferred by plugging in differential equation and seeing that it is possible to obtain two single variable differential equations. I am guessing that author see that it should be self-evident that wave-function is separable. So I hove answer should be simple.

P.S. About self-evidence. I am a chemist. Mathematicians and physicians often find some things that other people don't understand as self-evident. They are self evident for them probably because they work with them frequently. This why proofs are needed.

What proofs are. So I need list of statements that starts from things I believe (axioms, assumptions or well-known theorems) and than by known inference rules conclusion is derived. Each step must be executed formally. It is good if proof comes from book or publication. I would like to have proof in predicate logic. I am not interested in the algorithm that is used to solve this kind of problems on the exam. I want to know why it works. At least this algorithm should have a name (e.g. Theorem lalala). Or this algorithm may be composed of several theorems.

About the answer I have accepted. It looked like something I wanted, but it is a sketch of proof. It would be good to have more details. As I have asked I would like to have proof written in predicate logic.

Answer 1

It is possible to prove this although the proof is a two-step proof.

First, separation of variables in an assumption on the form of your solution. In other words, assume the solution is separable and verify that a separable solution can be found.

It’s not the only way to deal with PDE (you could try the method of characteristics) but separation of variables has the advantage of transforming the PDE to ODEs, for which there are multiple techniques.

What is less trivial is to show that all solutions must be linear combinations of the separable ones. This follows from Sturm-Liouville theory. In other words, even if you specialized your search to solutions of a specific (separable) form, you lost nothing because all solutions are superpositions of separable solutions.

Answer 2

Consider the angular momentum operator $L_z={\hbar \over i}(x{\partial \over \partial y}-y{\partial \over \partial x})=-i\hbar{\partial \over \partial \phi}$. Show that $[L_z,H]=0$ where $H=\Lambda^2$ is the Hamiltonian. Since they commute, the eigenfunctions of $L_z$ make the eigenfunctions of $H$, and since $L_z$ only depends on the variable $\phi$, the solutions of $H$ must be separable.

Answer 3

where $\Lambda^2$ is the Legendrian and $\epsilon=\frac{2IE}{\hbar^2}$.

You did not provide an explicit equation for your Legendrian operator. But I assume you mean: $$ \Lambda^2 = \frac{1}{\sin^2\theta}\frac{\partial^2}{\partial \phi^2} + \frac{1}{\sin\theta}\frac{\partial}{\partial \theta}\sin\theta\frac{\partial}{\partial \theta} $$

And I need to prove that $\psi(\theta,\phi)=\Theta(\theta)\Phi(\phi)$.

You already know that the solutions have the form of Spherical Harmonics. So, basically, it seems like you are trying to re-derive the form of the Spherical Harmonics.

What you do is just plug in the form $$ \psi(\theta,\phi)=\Theta(\theta)\Phi(\phi) $$ as an ansatz and show that it works.

You need to plug in a solution of this form and show that you can completely separate the variables. You will see that you end up with one term that only depends on $\theta$ and one that only depends on $\phi$.

You will introduce a "separation constant" called "$-m^2$" and you will see that: $$ \frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2} = -m^2\;, $$ etc.

Answer 4

Like @hft said, the method of separation of variables uses an ansatz. An ansatz is an assumption of the form that your solution takes on.

In this case we simply assume that the wave function is a product of functions of each variable:

$$\psi(\theta,\phi) = \Theta(\theta)\Phi(\phi). $$

It is not a fact. However, this assumption allows us to figure out genuine solutions of the 3D Schrödinger equation$^2$. That this method produces genuine solutions to the differential equations is enough to prove that the ansatz is valid.

Crucially, the genuine solutions we obtain via separation of variables can form (via linear combination) any solution to the given Schrödinger equation. In this case, we say that the solutions form a basis for the solution space. So via separation of variables, we have found a way to construct all solutions of the given differential equation. The fact that the solutions we found form a basis for the solution space is a very special property of the given differential equation$^1$.

Note: the equation you call "the wave-function" is actually the 3D Schrödinger equation. Wave functions are normalizable solutions to the Schrödinger equation (i.e. wave functions are the $\psi$s).

In general, it seems like differential equations are called "equations" (e.g. the Legendre equation) and their solutions are called "functions".

[1] Under "Applicability", there is an outline of an argument as to why separation of variables works to find a basis for the solution space in cases like the one you're interested in: https://en.wikipedia.org/wiki/Separation_of_variables.

[2] Via separation of variables, you split the original Schrödinger equation (a partial differential equation because it depends on more than one variable) into two second order ordinary differential equations (order meaning the highest derivative present, which is the second derivative in this case). There is a theorem somewhere that says an N-order differential equation has N linearly independent solutions. Thus, we should expect two linearly independent solutions from $\Theta(\theta)$ and $\Phi(\phi)$. There ends up being only one from $\Theta$ because the other linearly independent solution is physically unacceptable. And, we get two for $\Phi$.

Answer 5

It is not really a proof (with all mathematical rigour), but if you see that the Hamiltonian is a sum of terms that only contains one of the variables, you can solve the differential equation by separating variables.

In your case, the differential equation is $$ \left[\frac{1}{\sin^{2}\theta}\frac{\partial^{2}}{\partial\phi^{2}} + \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)\right]\psi(\theta,\phi) = E\psi(\theta,\phi) $$ By inspection, you see that there are a $\theta$ term together with a partial derivative with respect to $\phi$. However, you can also verify that using a wavefunction as $\psi(\theta,\phi)=\Theta(\theta)\Phi(\phi)$ your problem becomes $$ \frac{1}{\Phi(\phi)}\frac{d^{2}\Phi(\phi)}{d\phi^{2}} + \frac{\sin\theta}{\Theta}\frac{d}{d\theta}\left(\sin\theta\frac{d\Theta}{d\theta}\right) - E\sin^{2}\theta = 0 $$ You see, the first term now is only depending on $\phi$ and the second and third terms only contains the $\theta$ variable. From this, you can solve two simpler differential equations for each variable.

It might not be really clear why you have to try a product wavefunction here, but it is usually the case when you have $N$-dimensional problems with $N>1$. In other words, you think of each dimension of space as independent of the other, and the problem will tell you if that assumption will give you the correct result.