My question is a duplicate of this.
Please consider the equation
$\nabla^2\psi + (2m/\hbar^2)[E-V]\psi=0$ (1)
Potential of electron revolving hydrogen atom is given as
$V=\frac{-e}{4\pi\epsilon_0r}$ (2)
Equation (1) is a second order differential equation in variable $\psi$. $V$ is a continuous variable and takes all values from -$\propto$ to 0 at different distances. What is intuitive understanding for E and hence $\psi$ being discrete?
Consider the thought process, an electron gets closer to nucleus and it's potential rises then it's kinetic energy has to automatically adjust to one of the chosen discrete energy levels
If I draw a comparison with a similar orbital system, Total energy of a satellite revolving earth is given by $E=-\frac{Gm_1m_2}{2r}$
And it is continuous.
I understand that Schrodinger equation is a wave equation and from linked question I take that confined strings can have quantized number of hops. However an electron in an isolated atom is of course excited at one end by potential supplied by nucleus but what confines electron at other end?
I'd like to draw your attention to another similar analogy. Please consider wave equation of EM wave
$\nabla^2E=\mu_0\epsilon_0\frac{\partial^2 E}{\partial t^2} $ Also has a solution with continuous angular momentum and energy.
We know quantized energy levels are proportional to $\frac{1}{n^2}$ and summation of this series is finite. Is this a reason for energy levels being discrete? (Atom cannot supply infinite energy) If this is so why this doesn't apply to gravitational system?
Please help me with intuitive understanding rather mathematical equations
Answer
What confines the electron “at the other end”, i.e., far from the nucleus, is the requirement that the wavefunction be normalizable... in other words, that the total probability of finding the electron somewhere be 1. For this to happen, the wavefunction must go to zero far from the nucleus; otherwise the probability would be infinite. So this is like clamping down one end of a string.
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