Why does an electron not emit energy when it is in a stationary state?

Question

The Bohr's postulate states that an electron does not emit energy when it is in a stationary state. My question is, is it only a postulate or does it have proof? Also on what basis did Bohr come to this conclusion that there are stationary states where an electron does not emit energy?

Answer 1

To begin with, I would suggest not to attach too much importance to Bohr's approach. It had the great merit of suggesting a novel idea and to give the first explanation to a great problem: the one of atomic spectra, as explained by anna v. And it was not a vague idea: his formula for the energy of hydrogen levels $$E_n = -{m\,e^4 \over 2\,\hbar^2 n^2}$$ (Gauss units) fitted experimental data up to at least 6 significant digits, as far as I can remember. It should also be remarked that Bohr's formula is made entirely of known fundamental constants: there was no room for adjustable parameters, either it fitted or not. It did.

But it never grew to a real theory, and its "old quantum mechanics" was short lived: from 1913 to 1925 at most.

Bohr's idea of stationary states was necessary to overcome the absurdity following from Rutherford's planetary model together with classical electromagnetism: if both were true, atoms would not exist. Bohr was the first to know that his postulate was inconsistent with Maxwell's electromagnetism - which is the basis, I suppose, of your question.

However the concept of stationary state survived old q.m. and passed to the "new" q.m. founded by Heisenberg and Schrödinger (not forgetting de Broglie's suggestion of waves associated to particles: his famous $\lambda=h/p$). But here some clarification is in point.

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First, as far as experimental facts are concerned, each atom has only one stationary state: its ground state, i.e. the one of the least energy. If you push the atom, in any way, to an excited state, it sooner or later decays, emitting one ore more photons. But what does theory say?

If Schrödinger equation is applied to hydrogen atom it gives an infinity of stationary states, with energies exactly given by Bohr's formula. And according to q.m. these are true stationary states, i.e. states which do not evolve in time, remaining the same forever. But this is patently wrong, given the experimental facts. Energies are right, as is proven by measured energies of photons emitted or absorbed by the atom. But Schrödinger's stationary states do not exist in nature, other than as an approximate description of what are only transient states (with the only exception of the ground state).

Actually theory was soon able to do a step forward. If an electromagnetic wave hits an atom, calculations show that two things can happen, according to the initial state of the atom and the frequency of the wave:

1. The atom absorbs energy from the wave, jumping to a higher energy state. This is called absorption.
2. The atom loses energy, jumping to a lower energy state. This is called stimulated emission.

In both cases there is a relation between the atom's energy change and the wave's frequency. It is the second famous Bohr's formula: $$|\Delta E| = h\nu.$$

It is almost certain that whoever read this formula thought of photons - and this is right, as Bohr himself deduced the formula thinking of photons. But it is important to note that the theory I alluded above knows of no photons: it is, in physicists jargon, a semi-classical theory. This means that electromagnetic field is treated following Maxwell, whereas the atom is a Schrödinger atom.

So far so good, but what about spontaneous emission? It exists, as is proven by experimental facts (an excited state spontaneously decays via photon emission). Yet semi-classical theory is unable to deal with it. There was however another seminal paper by Einstein (1917) where he showed on general grounds that all three processes must exist (absorption, stimulated emission, spontaneous emission) and gave simple formulas relating the rates of those processes. But a theory of spontaneous emission had to wait for the birth of QED (quantum electrodynamics). In this theory electromagnetic field is treated according the prescriptions of quantum mechanics and its quanta - photons - naturally arise.

Needless to say, QED calculations exactly reproduce Einstein's predictions about the ratios between rates of photon emission and absorption.

Answer 2

Strictly speaking, electrons typically emit energy when they are in stationary states (they transition to other stationary states with lower energy) due to spontaneous emission, so if this postulate can be proven it can only be done within some specific model that does not describe the reality precisely.

Answer 3

As others have already pointed out, the question is a bit vague as to what exactly is being asked. Nevertheless, I can show a slightly mathematical version of something that might be related to your question. If you know about the Dirac formulation of QM, it will be easier to understand. $\newcommand{\ket}[1]{|#1\rangle}$ $\newcommand{\bra}[1]{\langle #1|}$ $\newcommand{\bracket}[1]{\langle #1|#1\rangle}$

One can show that the general solution to the Time-dependant Schrödinger Equation(TSE) has a general solution of the form $ \ket {\psi(t)} = U(t) \ket {\psi(0)}$ where $U(t)$ is the time evolution operator , which is necessarily unitary in this case. Stationery states are the solutions to the TISE, $\hat{H} \ket {\psi} = E \ket \psi$.

Suppose now that $\ket{\psi(0)}$ is a solution of the TISE, with energy eigenvalue $E_0$. Now let's look at the expectation of $\psi(t)$ given by $\bra{\psi(t)} \hat{H} \ket{\psi(t)}= \bra {\psi(0)}U(t)^\dagger\hat{H}U(t)\ket {\psi(0)} $.

Now if you know what is meant by 'symmetries of a quantum system', you can show that $U(t)$ is just that. One can show that for such symmetries, $[\hat{H},U(t)] = 0$. Given the commutativity of $U(t)$ and $\hat{H}$, we infer that, $\bra {\psi(t)}\hat{H}\ket{\psi(t)} = \bra {\psi(0)}\hat{H}\ket{\psi(0)}$, since $U(t)$ is unitary . But we know from the stationary property of $\ket {\psi(0)}$, that $\hat{H} \ket{\psi(0) }= E_0\ket{\psi(0)}$. Thus, assuming orthonormalilty of $\psi(0)$, we have, $ \bra {\psi(0)}\hat{H}\ket{\psi(0)} = E_0$. Hence, we have shown that the energy eigenvalue of a particle in a stationary state does not change.

P.S I am new to these stuff. Pleased point out any mistakes if any.

Answer 4

Of course, Anna V's answer is correct, let me add a few things.

First, you are asking about the ground state of a bound electron. Only bound electrons (that exist around a nucleus) emit real photons, when moving to a lower energy level as per QM.

Other then that, there is the case when accelerated electrons emit EM waves (photons), that is how radio antennas work. In that case it is a common misconception that those are free electrons. They are loosely bound to the nuclei of the metal (of the antenna). They gain kinetic energy from an external EM field (mediated by virtual photons) and thus move along to the next nucleus. But let's disregard that case now.

You are asking about the ground level of an electron that is bound to a nucleus. There are three forces acting on the electron:

1. Kinetic energy of the electron keeps the electron away from the nucleus.

2. EM attraction keeps the electron close to the nucleus.

3. Heisenberg uncertainty principle keeps the electron away from the nucleus.

At the ground state level, these three forces equal out, and the electron is in a stable energy level as per QM.

In this case, for the electron to emit a photon, it would need to lose kinetic energy, and move to a lower energy level, closer to the nucleus. Why can't it do that? It is because at any lower (closer to the nucleus) energy level:

1. these three forces would not equal out

2. as you try to confine the electron to a smaller space, the Heisenberg uncertainty principle would cause the electron to gain more momentum (kinetic energy), moving the electron away from the nucleus

3. at a region smaller then the ground level (what space the electron is confined to at the ground level) there simply is no force that would be strong enough to oppose the Heisenberg uncertainty principle, even the EM attraction is not strong enough to oppose it

Now it is not true that the electron cannot exist for a fraction of time where the nucleus is, or very close to it. As per QM, when the electron is at the ground level, the probability distribution (wavefunction) of the electron describes the position of the electron for all of space. The electron is with a high probability at the ground level, and with small probability at other places in space, including the nucleus.

But this (for the electron to get very close to the nucleus) can only happen with very little probability, as the electron is not stable that near the nucleus, the Heisenberg uncertainty principle will not let it get that close in a stable state.

But for your question, let's disregard this, and say the electron would emit a photon at the ground level. The electron would:

1. lose kinetic energy

2. move closer to the nucleus (because EM attraction would become dominant over kinetic energy's effect to keep the electron away from the nucleus)

3. the Heisenberg uncertainty principle would cause the electron to move back more away from the nucleus until these three forces equal out again, at the ground level.

Answer 5

The answer is that it is an experimental observation that stumped classical electromagnetic theory, because there is no solution within it, and this experimental fact, the light spectrum of atoms led to postulating stable states. That is what postulates do, they are a distillate from experimental observations which are common to all observations and thus become the "axioms" of a physical theory and called postulates to distinguish them from the mathematical axioms of the mathematics used.

In classical EM theory atoms cannot exist, because an electron attracted by a positive nucleus, according to classical electromagnetism would emit continuous radiation , losing momentum and finally falling on the positive nucleus by this loss of momentum. So classical EM cannot explain the atomic spectra.

If you read the link you will see that these lines are fitted with various mathematical series and that is what the Bohr model managed to fit using the postulate of stationary states.

This observation together with the black body radiation and the photoelectric effect lead to the general theory of quantum mechanics with its postulates that fits and predict very well the behavior of elementary particle physics, nuclear physics and solid state etc. when dimensions are very small, commensurate with the Heisenberg uncertainty principle..

The Bohr model was a useful model which lead to the present day mainstream theories.

See also my answer here.