Energy quantization of the particle in a box

[bodyscripter]

I'm self-learning quantum mechanics and I'm reading the famous Shankar book (Principles of Quantum Mechanics - second edittion). At page 161 of the book, I don't understand the following part of this page:

"Let us restate the origin of energy quantization in another way. Consider the search for acceptable energy eigen-functions, taking the finite well as an example. If we start with some arbitrary values $\psi(x_0)$ and $\psi'(x_0)$, at some point $x_0$ to the right of the well, we can integrate Schrodinger's equation numerically. (Recall the analogy with the problem of finding the trajectory of a particle given its initial position and velocity and the force on it.) As we integrate out to $x \to \infty$, $\psi$ will surely blow up since $\psi_{III}$ contains a growing exponential. Since $\psi(x_0)$ merely fixes the overall scale, we vary $\psi'(x_0)$ until the growing exponential is killed. [Since we can solve problem analytically in region III, we can even say what the desired value of $\psi'(x_0)$ is: it is given by $\psi'(x_0) = -\kappa \psi(x_0)$. Verify, starting with Eq. (5.2.4), that this implies $B=0$.] We are now out of the fix as $x \to \infty$, but we are committed to whatever comes out as we integrate to the left of $x_0$. We will find that $\psi$ grows exponential till we reach the well, whereupon it will oscillate. When we cross the well, $\psi$ will again start to grow exponentially, for $\psi_I$ also contains a growing exponentially in general. Thus there will be no acceptable solution at some randomly chosen energy. It can, however, happen that for certain values of energy, $\psi$ will be exponentially damped in both regions I and III. [At any point $x_0'$ in region I, there is a ratio $\psi'(x_0')/\psi(x_0')$ for which only the damped exponential survives. The $\psi$ we get integrating from region III will not generally have this feature. At special energies, however, this can happen.] These are the allowed energies and the corresponding functions are the allowed eigen-functions. Having found them, we can choose $\psi(x_0)$ such that they are normalized to unity. For a nice numerical analysis of this problem see the book by Eisberg and Resnick.$".

I need the formula details to understand it. Another question: "overall scale", what does it mean?

 

[bhobba]

I can't follow it either - maybe with work but not easily.

However I can point you to a series of lectures that does carefully explain what's going on and much more besides:


Thanks
Bill

 

[Nugatory]

I need the formula details to understand it. Another question: "overall scale", what does it mean?

The "formula" is just the time-independent Schrodinger's equation for the infinite square well potential. It will have been covered somewhere in previous 160-odd pages. There is also an assumption here that you know what numerical integration is; you don't have to actually do it, but you have to know what it is.

The bit about "overall scale" is just referring to the fact that if $\psi$ is a solution to Schrodinger's equation, then any constant multiple of $\psi$ is also a solution. Thus we can choose $\psi$ to have any value we please at any single point without losing any generality. We then work from there to find the $\psi$ and $\psi'$ that when numerically integrated give us the desired declining exponential on both sides of the well. That's an unnormalized eigenfunction, but as a final step we can multiply it by whatever constant is necessary to normalize it.

 

[stevendaryl]

I need the formula details to understand it. Another question: "overall scale", what does it mean?

To answer the last question first, Schrodinger's equation for an energy eigenstate is:

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

Obviously, if $\psi(x)$ solves this equation, then so does $\tilde{\psi}(x) \equiv C \psi(x)$ for any constant $C$. So we can choose $\psi(x_0)$ to be anything at all, and solve the equation, then at the end, we can multiply by a constant $C$ to make $\int |\psi(x)|^2 dx = 1$. So he's just saying it doesn't matter what choice we make for $\psi(x_0)$.

As far as the details for his claim, I'm not sure what extra details you want. Mathematically, Schrodinger's equation implicitly gives a function $\psi(x) = F(x,E,x_0, \psi_0,\psi'_0)$ that depends on 5 quantities: (1) $x$, of course, (2) the energy $E$, (3) $x_0$, your starting point, (4) $\psi_0 \equiv \psi(x_0)$, the value at $x=x_0$, and (5) $\psi'_0$, the value of $\psi'$ at $x=x_0$. Let's hold $\psi_0$ and $x_0$ fixed, so we get a function of three arguments: $\psi(x) = F(x,E,\psi'_0)$. There are various ways to compute $F$, either numerically or in a power series.

Now, the first claim being made is that for almost all values of $\psi'_0$, $lim_{x \rightarrow \infty} F(x,E,\psi'_0) = \infty$. Only for one very specific value of $\psi'_0$ will it be the case that $lim_{x \rightarrow \infty} F(x,E,\psi'_0) = 0$. The author claims that that value is $\psi'_0 = -\kappa \psi_0$. So let's pick that value for $\psi'_0$

The second claim being made is that for almost all values of $E$, if $\psi'_0 = -\kappa \psi_0$, then $lim_{x \rightarrow -\infty} F(x,E,\psi'_0) = \infty$. An energy eigenvalue is some value of $E$ such that $lim_{x \rightarrow -\infty} F(x,E,\psi'_0) = 0$ when $\psi'_0 = -\kappa \psi_0$.

So if you are solving the Schrodinger equation numerically, then you could do it this way:

1. Fix $x_0$ and $\psi_0$
2. Make a guess for a starting value for $E$
3. Make a guess for a starting value for $\psi'_0$
   
4. Check if $F(x,E,\psi'_0)$ blows up for $x$ large and positive.
5. If so, adjust $\psi'_0$ and go back to 4.
6. If not, then check if $F(x,E,\psi'_0)$ blows up for $x$ large and negative.
7. If so, adjust $E$ and go back to 3.
8. If not, then you've found an approximate eigenvalue.