# This Quantum World/Implications and applications/Why energy is quantized

## Why energy is quantized

Limiting ourselves again to one spatial dimension, we write the time independent Schrödinger equation in this form:

${\displaystyle {d^{2}\psi (x) \over dx^{2}}=A(x)\,\psi (x),\qquad A(x)={2m \over \hbar ^{2}}{\Big [}V(x)-E{\Big ]}.}$

Since this equation contains no complex numbers except possibly ${\displaystyle \psi }$ itself, it has real solutions, and these are the ones in which we are interested. You will notice that if ${\displaystyle V>E,}$ then ${\displaystyle A}$ is positive and ${\displaystyle \psi (x)}$ has the same sign as its second derivative. This means that the graph of ${\displaystyle \psi (x)}$ curves upward above the ${\displaystyle x}$ axis and downward below it. Thus it cannot cross the axis. On the other hand, if ${\displaystyle V then ${\displaystyle A}$ is negative and ${\displaystyle \psi (x)}$ and its second derivative have opposite signs. In this case the graph of ${\displaystyle \psi (x)}$ curves downward above the ${\displaystyle x}$ axis and upward below it. As a result, the graph of ${\displaystyle \psi (x)}$ keeps crossing the axis — it is a wave. Moreover, the larger the difference ${\displaystyle E-V,}$ the larger the curvature of the graph; and the larger the curvature, the smaller the wavelength. In particle terms, the higher the kinetic energy, the higher the momentum.

Let us now find the solutions that describe a particle "trapped" in a potential well — a bound state. Consider this potential:

Observe, to begin with, that at ${\displaystyle x_{1}}$ and ${\displaystyle x_{2},}$ where ${\displaystyle E=V,}$ the slope of ${\displaystyle \psi (x)}$ does not change since ${\displaystyle d^{2}\psi (x)/dx^{2}=0}$ at these points. This tells us that the probability of finding the particle cannot suddenly drop to zero at these points. It will therefore be possible to find the particle to the left of ${\displaystyle x_{1}}$ or to the right of ${\displaystyle x_{2},}$ where classically it could not be. (A classical particle would oscillates back and forth between these points.)

Next, take into account that the probability distributions defined by ${\displaystyle \psi (x)}$ must be normalizable. For the graph of ${\displaystyle \psi (x)}$ this means that it must approach the ${\displaystyle x}$ axis asymptotically as ${\displaystyle x\rightarrow \pm \infty .}$

Suppose that we have a normalized solution for a particular value ${\displaystyle E.}$ If we increase or decrease the value of ${\displaystyle E,}$ the curvature of the graph of ${\displaystyle \psi (x)}$ between ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ increases or decreases. A small increase or decrease won't give us another solution: ${\displaystyle \psi (x)}$ won't vanish asymptotically for both positive and negative ${\displaystyle x.}$ To obtain another solution, we must increase ${\displaystyle E}$ by just the right amount to increase or decrease by one the number of wave nodes between the "classical" turning points ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$ and to make ${\displaystyle \psi (x)}$ again vanish asymptotically in both directions.

The bottom line is that the energy of a bound particle — a particle "trapped" in a potential well — is quantized: only certain values ${\displaystyle E_{k}}$ yield solutions ${\displaystyle \psi _{k}(x)}$ of the time-independent Schrödinger equation: