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

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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.