Time-independent Schrödinger-EQ in 2D rectangular box is separable in spatial coordinates

let's consider a particle in a 2D-Box of length $$L_x, L_y$$.

Hamiltonian: $$\hat{H}=-\frac{\hbar^2}{2m}(\partial_x^2 + \partial_y^2)$$

Potential: $$V(x,y)=\begin{cases} 0, & 0\leq x \leq L_x, \ 0\leq y \leq L_y\\ \infty ,& \text{else}\end{cases}$$

Schrödinger-EQ: $$\hat{H}\Psi(x,y)=E\Psi(x,y)$$

We solve the Schrödinger-EQ using the ansatz $$\Psi(x,y)=\psi_{n_x}(x)\psi_{n_y}(y)$$

Inside of the box, we have:

$$\frac{-\hbar^2}{2m}\big( \psi_{n_y}(y)\partial_x^2\psi_{n_x}(x) + \psi_{n_x}(x)\partial_y^2\psi_{n_y}(y) \big) = E\psi_{n_x}(x)\psi_{n_y}(y) \tag{1}$$

We divide by $$\psi_{n_x}(x)\psi_{n_y}(y)$$ and end up with

$$\frac{-\hbar^2}{2m}\frac{1}{\psi_{n_x}(x)}\partial_x^2\psi_{n_x}(x) - \frac{-\hbar^2}{2m}\frac{1}{\psi_{n_y}(y)}\partial_y^2\psi_{n_y}(y) = E \tag{2}$$

Now apparently we can argue like this:

In (2) the first term on the LHS depends only on $$x$$ while the second term only depends on $$y$$. Becuase of that, both terms have to be constant and we can look at each other separately:

$$\frac{-\hbar^2}{2m}\frac{1}{\psi_{n_x}(x)}\partial_x^2\psi_{n_x}(x)= E_{n_x} \tag{3}$$ $$\frac{-\hbar^2}{2m}\frac{1}{\psi_{n_y}(y)}\partial_y^2\psi_{n_y}(y)= E_{n_y} \tag{4}$$

whereas $$E_{n_x}+E_{n_y}=E$$

Now my question is about the quote above. How exactly do we know that both terms are constant on their own? It's more a mathematical question. I'd like to prove this but I fail.

Here's my attempt:

Let

$$f: \mathbb R^2 \to \mathbb R, \ \vec{v}\mapsto \vec{v}\cdot\begin{pmatrix}1\\ 0\end{pmatrix}$$

$$g: \mathbb R^2 \to \mathbb R, \ \vec{v}\mapsto \vec{v}\cdot\begin{pmatrix}0\\ 1\end{pmatrix}$$

whereas $$\vec{v}=(v_x, v_y)$$

Claim: $$f(\vec{v}) + g(\vec{v}) = E$$ for a fixed $$E\in\mathbb R \vec{5}$$ implies $$f,g$$ are constant Proof: Choose $$f,g$$ as above. Then

$$g(\vec{v})+f(\vec{v})=E \tag{5.1}$$

$$\vec{v}\cdot\begin{pmatrix}1\\ 0\end{pmatrix} + \vec{v}\cdot\begin{pmatrix}0\\ 1\end{pmatrix} = E \tag{5.2}$$

Let's identify $$g(\vec{v})$$ with $$\hat{g}(y)$$ and $$f(\vec{v})$$ whereas obviusly $$\hat{f}(x)$$ whereas $$\hat{g},\hat{f}: \mathbb R \to \mathbb R$$

Now we can write

$$\hat{f}(x) + \hat{g}(y) = E \tag{5.3}$$

so now we have the situation in the quote. (I did this, just so we have two one-dimensional functions and don't have to use vectrs. I'm pedantic.)

Now, what stops me from just saying:

Choose $$\hat{f}(x)$$ as a non-constant function and define $$\hat{g}(y)=\hat{f}(y)+\text{const}$$ with $$\text{const}\in \mathbb R$$

Now furthermore choose $$\text{const}=E$$ and we get $$E=\hat{g}(y)-\hat{f}(y)$$ whereas $$\hat{g}, \hat{f}$$ are both non-constant functions.

Of course, that only works because we now identified implicitly $$x=y$$. But I don't fully see how I can get the claim properly now.

• It is really a question about separation of variables, which is a general mathematical technique. Apr 9 '20 at 13:19
• This technique is a standard procedure in the solution of partial differential equations. It is called "separation of variables"; you can find the information you need in a first-year-university-level mathematics textbook, and on the web too I am sure. Or else just think it through, e.g. by the method proposed here by walber97. Apr 9 '20 at 13:20
• I know that this is separation of variables and I know that you use a separation constant for it usually - I could have taken a different example but I just noticed that I'm not satisfied with my understanding while solving this example. Apr 10 '20 at 8:38

Let me try it this way. $$f(x)+g(y)=E$$ Take the partial derivative with respect to $$x$$ on both sides. Since $$g$$ depends only on $$y$$, and $$E$$ is a constant, we get $$\frac{\partial f(x)}{\partial x}=0$$ But $$f$$ depends only on $$x$$, and given that it's a two dimensional problem. Thus it is not a function of $$y$$ or $$z$$. Hence we get $$f$$ as a constant. Do the similar step by taking partial derivative with respect to $$y$$, you will get $$g$$ an another constant

I think we can have a proof by contradiction.

If we assume $$f(x)$$ is not constant while $$g(y)\equiv Y$$, then we would have $$max\{f(x)\}\neq min\{f(x)\}$$, which means $$max\{f(x)-g(y)\}= max\{f(x)\}-Y \neq min\{f(x)\}-Y = min\{f(x)-g(y)\}$$. Therefore, $$f(x)-g(y)$$ can't be a constant since $$max\{f(x)-g(y)\} \neq min\{f(x)-g(y)\}$$.

Similarly, we can exclude the cases when both $$f(x)$$ and $$g(y)$$ are not constant and when only $$f(x)$$ is constant. As the result, both $$f(x)$$ and $$g(y)$$ must be constants.

What we have is the equation: $$f(x)=E-g(y)$$

Since this equality must hold for any arbitrary $$(x,y)$$, the only way this can happen is if both the functions are constants.

• Yeah, sure I also had that in mind but I think that's too weak of an argumentation. Thanks thought :) Apr 10 '20 at 8:37
• @handy, not sure why you think it’s a weak argument. You may think of it graphically. Any solution to this issue at the end of the day will always lead to this idea. Apr 10 '20 at 9:29