As far as I know, an electron can't go below what is known as the ground state, which has an energy of -13.6 eV, but why can't it lose any more energy? is there a deeper explanation or is this supposed to be accepted the way it is?
Answer
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.
No comments:
Post a Comment