0
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ It is really a question about separation of variables, which is a general mathematical technique. $\endgroup$ Apr 9 '20 at 13:19
  • $\begingroup$ 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. $\endgroup$ Apr 9 '20 at 13:20
  • $\begingroup$ 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. $\endgroup$
    – handy
    Apr 10 '20 at 8:38
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Yeah, sure I also had that in mind but I think that's too weak of an argumentation. Thanks thought :) $\endgroup$
    – handy
    Apr 10 '20 at 8:37
  • $\begingroup$ @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. $\endgroup$ Apr 10 '20 at 9:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.