If an electron is in ground state, why can't it lose any more energy?
You might as well have asked: "Why can't the string on my guitar produce a lower sound?" The short answer is: the lowest sound depends on the quality of the string, on the tension of the string and on the restricted length of the string. So it's just a matter of configuration! It's the same with an electron in a hydrogen atom. The mass of the electon, the Coulomb-force between it and the nucleus, the distance between these two and the restricted space determine it's lowest energylevel. There is nothing deeper than that. It's because of the way (we think) the atom is built. And if you want to know what the exact energy is for that level, then you will need Bohr's model of the atom or quantummechanics to calculate it.
The deeper reason is quantum behaviour of electrons.The proper description of the electron in the atom is given by quantum wavefunction satisfying Schrödinger equation which is second-order partial differential equation.
Let's restrict ourselves to one-dimensional case for better understanding. The Schrödinger equation for particle moving in one dimension can be written as, \begin{equation} \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+U(x)\right]\psi_t(x)=i\hbar\frac{\partial \psi_t(x)}{\partial t} \end{equation} The wavefunction is not observable by itself. Rather given some observable physical quantity $A$ it describes the probabilities $P(A=a)$ to measure some value $a$ of it and their expectation values $\langle A\rangle$ (i.e. the average value you get if you perform the same experiment again and again) For example, \begin{aligned} \langle x\rangle=\int_{-\infty}^{+\infty}dx\,x|\psi(x)|^2,\quad \langle p_x\rangle=\int_{-\infty}^{+\infty}dx\,\psi(x)^\ast\left(-i\hbar\frac{\partial\psi(x)}{\partial x}\right),\\ \langle E\rangle = \langle \frac{p_x^2}{2m}+U(x)\rangle =\int_{-\infty}^{+\infty}dx\,\psi(x)^\ast\left[-\frac{\hbar^2}{2m}\frac{\partial^2\psi(x)}{\partial x^2}+U(x)\psi(x)\right] \end{aligned}
Unless the state is very special the measurement results are NOT necessarily equal to expectation value but simply has probability distribution centered around expectation value. We can introduce uncertainty \begin{equation} \sigma_A=\sqrt{\langle A^2\rangle-\langle A\rangle^2} \end{equation} that characterizes how much this probability distribution is spread.
One of the properties of quantum wavefunction is Heisenberg uncertainty relation that restricts the uncertainties of coordinate and corresponding momentum. \begin{equation} \sigma_x\sigma_{p_x}=\sqrt{\langle x^2\rangle-\langle x\rangle^2}\sqrt{\langle p_x^2\rangle-\langle p_x\rangle^2}\geq \frac{\hbar}{2} \end{equation} That means that if you consider very localized state in coordinates it has huge uncertainty in momentum. Remember that kinetic energy is $p_x^2/2m$ then for the state for which average momentum is zero $\langle p_x\rangle=0$ we can write, \begin{equation} \langle \frac{p_x^2}{2m}\rangle=\frac{1}{2m}\sigma_{p_x}^2\geq \frac{\hbar^2}{8m\sigma_x^2} \end{equation}
Now consider potential unbounded from below (with $U(x)\rightarrow-\infty$ as $x\rightarrow 0$ like this,
In classical physics where the particle is described by point material particle if we can take $p_x=0$ the energy is given only by potential energy $U(x)$ which we can make arbitrarily low putting particle very close to $x=0$.
However in quantum physics the average energy for state with $\langle p_x\rangle=0$, \begin{equation} \langle E\rangle\geq \frac{\hbar^2}{8m\sigma_x^2}+\int_{-\infty}^{+\infty}dx\,U(x)|\psi(x)|^2 \end{equation} Let's assume also $\langle x\rangle$ is close to zero. If we lower $\sigma_x$ the wavefunction will concentrate near that point deep down the pit and we can lower second term as much as we wish. But doing so we are raising the first term arbitrarily high. What happens with total average energy is determined by the competition of these two terms.
As result if the pit is not wide enough and not steep enough the average energy will never go below some finite value $E_0$. In fact the stronger claim is true - the lowest measurable value of total energy equals $E_0$. The ground state is a state that have that lowest energy $E_0$ and no uncertainty $\sigma_E=0$.
The same happens in three-dimensional case with Coulomb potential. You have this infinitely deep pit near $r=0$ however it's not wide enough and not steep enough and because of that electron being quantum particle never can have total energy below the ground state energy.
The energy of bound particles (in quantum systems) has two characteristics:
- The energy is quantized.
- The lowest allowable energy level is non-zero.
This is true of all bound particle systems, whether atoms, quantum oscillators or other.
Let us have a look at the simplest quantum system of all: the single particle in a 1D box, with zero potential. In this system a particle is trapped in a 1D box (or tube), confined by infinitely high potential energy walls but with zero potential energy within the confines of the box, also known as an infinite potential well. As per the link above the wave function is given by:
$$\psi_n(x)=\sqrt{\frac2a}\sin\Big(\frac{n\pi x}{a}\Big)$$
Where $a$ is the length of the box. The derivation shows that the only allowable values for $n$ are positive integers:
$$n=1,2,3,...$$
This makes sense, as for $n=0\implies\psi_0(x)=0$. As the probability density is given by:
$$\rho(x)=[\psi(x)]^2,$$
for $n=0$, the probability density becomes zero and this would mean the particle has escaped, which is impossible due to the infinitely high potential walls.
Also per the link above the only allowed energy levels are given by:
$$E_n=\frac{\pi^2\hbar^2}{2ma^2}n^2$$
The lowest allowable energy (the ground state) is for $n=1$:
$$E_1=\frac{\pi^2\hbar^2}{2ma^2}$$
For any 'lower' (or non-integer values) of $n$ the Schrödinger equation of the 1D box:
$$-\frac{\hbar^2}{2m}\frac{\mathrm{d^2}}{\mathrm{d}x^2}\psi_n(x)=E_n\psi_n(x),$$
is no longer satisfied.
For a hydrogen atom the situation is remarkably similar (the potential well is shaped as shown in one of the answers above). The energy is quantised as:
$$E_n=\frac{-13.6\mathrm{eV}}{n^2},$$
for $n=1,2,3,...$. From the hydrogen wave functions can be seen that $n=0$ would correspond to the electron residing in the nucleus! Obviously that does not constitute an atom.
The lowest allowable energy level (or ground state) is $-13.6\mathrm{eV}$ and the atom cannot lose any more energy.