Understanding a math step in solving Schrödinger's equation for single particle in a box

[zenterix]

Homework Statement

I am following the calculations in solving Schrödinger's time-independent equation for a single particle in a box.

Relevant Equations

$$-\frac{\hbar^2}{2m}\frac{\partial^2\psi(x)}{\partial x^2}=\psi(x)(E-V(x))\tag{1}$$

I have a question about one of the steps in the calculations.

For context about where in the calculations I am, we assume that a particle is subjected to a potential energy function that is infinite everywhere along the $x$-axis except for a line segment of length $l$ where the potential energy is zero.

Thus, the $x$-axis is divided into three regions where we will solve Schrödinger's equation

It can be shown that in regions I and III, the solution to (1) is just $\psi(x)=0$.

Consider region II.

$$-\frac{\hbar^2}{2m}\frac{\partial^2\psi_{II}(x)}{\partial x^2}=\psi_{II}(x)E$$

$$\frac{\partial^2\psi_{II}(x)}{\partial x^2}=-\frac{2m}{\hbar^2}E\psi_{II}(x)$$

$$\frac{\partial^2\psi_{II}(x)}{\partial x^2}+\frac{2mE}{\hbar^2}\psi_{II}(x)=0$$

$$\frac{\partial^2\psi_{II}(x)}{\partial x^2}+k^2\psi_{II}(x)=0$$

This is a 2nd-order, linear, homogeneous ODE with

$$k^2=\frac{2mE}{\hbar^2}$$

A general (complex) solution is

$$\psi_{II}=c_1e^{x\frac{i\sqrt{2mE}}{\hbar}}+c_2e^{\frac{-ix\sqrt{2mE}}{\hbar}}$$

$$=c_1e^{ikx}+c_2e^{-ikx}$$

$$=c_1(\cos(kx)+i\sin{kx})+c_2(\cos{kx}-i\sin{kx})$$

$$=(c_1+c_2)\cos{kx}+i(c_1-c_2)\sin{kx}$$

$$=A\cos{kx}+B\sin{kx}$$

where $c_1, c_2, A,$ and $B$ are complex constants.

We find $A$ and $B$ by applying boundary conditions.

Our boundary conditions arise from assuming that $\psi(x)$ is continuous at $x=0$ and $x=l$.

To be continuous at $x=0$ it must be that the limits of the wavefunction approaching $0$ from left and right are the same.

$$\lim\limits_{x\to 0}\psi_{I}=\lim\limits_{x\to 0}\psi_{II}$$

$$0=\lim\limits_{x\to 0}\left [A\cos{\left ( \frac{\sqrt{2mE}x}{\hbar} \right )}+B\sin{\left ( \frac{\sqrt{2mE}x}{\hbar} \right )}\right ]$$

$$=A$$

Thus

$$\psi_{II}=B\sin{\left ( \frac{\sqrt{2mE}x}{\hbar} \right )}\tag{2}$$

My question is about the calculations involving the second boundary condition, namely that the wavefunction is continuous at $x=l$.

$$\lim\limits_{x\to l}\psi_{III}=\lim\limits_{x\to l}\psi_{II}$$

$$0=\lim\limits_{x\to l}\left [B\sin{\left ( \frac{\sqrt{2mE}x}{\hbar} \right )}\right ]$$

$$B\sin{\left ( kl \right )}=0$$

$$\implies kl = n\pi, n\in \mathbb{N}\tag{3}$$

Okay, notice the $n\in\mathbb{N}$ in (3). Why do we not say $n\in\mathbb{Z}$?

Well, if $n<0$ then we get essentially the same solutions (and I think the only difference is the sign on the energies that we get).

My question is, however, about the case of $n=0$.

If $n=0$ then at first glance it would seem that $k=0$ and from (2) we would have $\psi_{II}=0$.

Can we just conclude this directly like this?

Another way to reach this conclusion is to look at the original differential equation. If $n=0$ then

$$\frac{\partial^2\psi_{II}(x)}{\partial x^2}+k^2\psi_{II}(x)=0$$

$$\frac{\partial^2\psi_{II}(x)}{\partial x^2}=0$$

$$\implies \psi(x)=ax+b$$

which is a solution that only obeys both boundary conditions if $a=b=0$.

Thus, it must be that $\psi(x)=0$.

Is it ok to simply insert $n=0$ into (2) and from there conclude that $\psi(x)=0$?

 

[BvU]

Is it ok to simply insert $n=0$ into (2) and from there conclude that $\psi(x)=0$?

There is one other condition: the particle has to be somewhere, so $\int \psi^* \psi =1$

$\$

 

[zenterix]

Here is why I think I am asking this.

We had a 2nd-order linear ODE.

The characteristic equation was

$$p(s)=s^2+k^2=0$$

with discriminant

$$\Delta=-4k^2$$

and with roots

$$s=\pm ik$$

But if $n=0$ and so $k=0$ then the discriminant is zero and we have only the single root $s=0$.

When the discriminant is zero, the general solution is not $c_1e^{s_1x}+c_2e^{s_2x}$.

Rather it is $c_1+c_2x$.

This isn't physics, this is just the math of solving a 2nd-order ODE with constant coefficients (I am recalling now the exact theorem that I saw in Apostol for this).

So, if we assume $n=0$ then it must be that $k=0$. But if $k=0$ in the original ODE then the discriminant of the characteristic equation is zero and the solution is a line $c_1+c_2x$.

To satisfy the boundary conditions it must then be that $c_1=c_2=0$.

The question I asked in the OP is whether it is mathematically ok to conclude this from (2).

Inserting $k=0$ into (2) tells us that $B$ can be anything since the term involving it is always zero (and in addition, we had previously concluded that $A=0$).

Is it just an accident that we arrived at this solution of $\psi(x)=0$ from this route?

Suppose the boundary condition at $x=0$ were that $\psi(0)=1$. Now, just to be clear, this doesn't seem to be possible with the potential function we chose.

If somehow it were possible then the first boundary condition would tell us that $A=1$ but the second boundary condition would have

$$\psi(l)=A=0$$

$$\implies 1=0$$

In other words, this route would tell us there is no solution.

If, on the other hand, we had considered the actual general solution $\psi(x)=c_1+c_2x$ then the boundary conditions would mean

$$\psi(0)=1=c_1$$

$$\psi(l)=0=1+c_2l\implies c_2=-\frac{1}{l}$$

$$\implies \psi(x)=1-\frac{1}{l}x$$

So it seems the answer to the question in the OP is that no, we can't use (2) since it is not the general solution to our 2nd order equation and so imposing boundary conditions won't necessarily give us a solution at all.

 

[zenterix]

There is one other condition: the particle has to be somewhere, so $\int \psi^* \psi =1$

$\$

Right, so this is what effectively prohibits the solution $\psi(x)=0$.

 

[Albertus Magnus]

There is no problem with the solution $\psi(x)=0$ for the Schrodinger equation in the purely mathematical sense, it is trivial however. On the other hand, physical solutions must always be *normalizable*, i.e. $$\int^{+\infty}_{-\infty}\psi^{*}(x)\psi(x)\;dx.$$
Some kinds of non-normalizable states are particularly repugnant, e.g. states like: $$\int^{+\infty}_{-\infty}\psi^*(x)\psi(x)\; dx\gt\infty,$$
however, states like the one under consideration $\langle\psi(x)|\psi(x)\rangle=0$ are not so incorrigible, i.e. one can think of them as solutions referring to particles that do not exist. But this is after all physically meaningless in that it makes no sense to talk about things that do not exist. In the spirit of the great Henri Poincare, it is good scientific philosophy to pursue only those things which are useful.

Note: As per posts below adressing the typographical error, the above equation should read: $\int^{+\infty}_{-\infty}\psi^*(x)\psi(x)\; dx=\infty.$

 

[renormalize]

$\int^{+\infty}_{-\infty}\psi^*(x)\psi(x)\; dx\gt\infty,$

Can you give an example of a function $\psi(x)$ that makes the integral $\int^{+\infty}_{-\infty}\psi^*(x)\psi(x)dx$ bigger than infinity?

 

[Albertus Magnus]

Can you give an example of a function $\psi(x)$ that makes the integral $\int^{+\infty}_{-\infty}\psi^*(x)\psi(x)dx$ bigger than infinity?

Excellent point, I really meant: $\int^{+\infty}_{-\infty}\psi^*(x)\psi(x)\;dx=\infty.$

 

[Albertus Magnus]

Excellent point, I really meant: $\int^{+\infty}_{-\infty}\psi^*(x)\psi(x)\;dx=\infty.$

So would the correct thing be to edit the original post, or let it slide?

 

[haruspex]

So would the correct thing be to edit the original post, or let it slide?

I favour editing it in a way that makes it clear that it is a correction.