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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?

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    $\begingroup$ This is the basic assumption of bohr's model.the lower energies fall in forbidden zones. $\endgroup$ – Lelouch Nov 19 '16 at 8:41
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    $\begingroup$ While I imagine you'll find this unsatisfactory, if the electron can transition to a lower energy state, it's not in the ground state. In other words, the ground state is essentially the state of lowest energy. Is you question more like why there is a state of lowest energy? $\endgroup$ – Alfred Centauri Nov 19 '16 at 13:01
  • $\begingroup$ Are you looking for a conceptual or semi-classical explanation? If so I think I might be able to help. $\endgroup$ – David Elm Nov 19 '16 at 19:08
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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, enter image description here

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.

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  • $\begingroup$ I don't understand how the uncertainty principle enters the assumption of an existence of a lowest energy state. Also, any eigenstate of the energy has, by definition, $\sigma_E = 0$. $\endgroup$ – gented Nov 19 '16 at 11:36
  • $\begingroup$ @GennaroTedesco That's not a question - all I can answer, read again. If you have some concrete problems, formulate them properly. The existence of a ground state is not assumption, it's a property of the specific Hamiltonian that can be seen also from $\langle\psi|\hat{H}|\psi\rangle$ having minimum. I know that any eigenstate has $\sigma_E=0$ but the problem I seriously doubt that OP knows what eigenstate is and don't think it's appropriate to answer his question by writing quantum mechanics textbook. $\endgroup$ – OON Nov 19 '16 at 11:49
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The energy of bound particles (in quantum systems) has two characteristics:

  1. The energy is quantized.
  2. 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.

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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.

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The existence of a minimum energy follows from the wave-like aspects of matter.

The allowed values of energy are those corresponding to stationary wave states, so the trite answer to your question is that where you have a set of allowed energy values one of them has to be a minimum.

To get some physical insight into why the minimum level ends up where it is, without resorting to maths, you can picture a semi-classical explanation as follows. The classical treatment would say that the electron could lose potential energy by spiralling closer to the nucleus. But the electron has an associated wavelength, and the lower its energy becomes the longer its wavelength becomes. If the electron is going to be in some standing state its orbit needs to be at least one wavelength long, so the lower its energy, the longer its wavelength and the more widely spaced its orbit. The elongation of the wavelength, through the loss of energy, ensures that the electron can't get any close to the nucleus and thus can't lose any more energy, so you arrive at a natural minimum.

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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?

If you are asking if the energy level could be different, the answer is that the physical constants entering in the calculation of energy levels match the measured energy level for hydrogen. This is an experimental fact, a photon of the specific energy is needed to ionize hydrogen. So no, there cannot exist a lower energy level for the hydrogen atom.

If you are asking why there is a lower energy level, a limiting level for the electrons in the atom, the answer is given in the fact that if there were not a stable lowest energy level,atoms as we know them would not exist because the electron would fall on the proton . It is the basic argument for the development of the Bohr model of the atom and the subsequent theory of quantum mechanics based on the Schrodinger equation and its solutions, as described in other answers.

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    $\begingroup$ 1st sentence paragraph 2: should it be "why there isn't a lower energy level." $\endgroup$ – Bill N Nov 20 '16 at 2:16
  • $\begingroup$ @BillN Well, I think the paragraph above covers that, "different" means either higher or lower. $\endgroup$ – anna v Nov 20 '16 at 4:14
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In the case at hand of the hydrogen atom the Hamiltonian can be written as sum of two Casimir operators in the algebra $so(4)$, in particular one can see that it can be of the form $M^2 + N^2$, where both $M, N$ are some sort of angular momentum operators, hence their representations are characterised by positive eigenvalues. One can show that because of this the spectrum must be of the form $$ E_n \sim -\frac{1}{n^2} $$ where the lowest $n$ may be is $n=1$. That would correspond to the groundest of the states. In the relativistic correction one shows that an additional contribution to the energy in terms of the total angular momentum $l$ is required, but also in that case the representations are all positive-values. For many atoms systems the argument goes along the same lines, where one basically constructs (anti)-symmetric one-particle states from the above.

Another example can be the harmonic oscillator, where the Hamiltonian can be re-written as a positive operator $N^{\dagger}N$ with spectrum $\sigma_H \subseteq [0, +\infty]$ (see also this answer).

In general Hamiltonians in quantum mechanics are constructed to be bounded from below, namely their spectrum is assumed to allow a lowest minimum weight, for reasons of boundness of physical states (see spectral theory) and stability (especially in case of many-particles interactions).

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    $\begingroup$ I see. Please don't take it as offense... but if your aim is not showing off but better understanding by original poster (look at his other questions), I recommend you to explain almost every word in your answer))) At least you could expand your first paragraph to concrete eqs. $\endgroup$ – OON Nov 19 '16 at 12:45
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    $\begingroup$ @OON Answers aren't necessarily directed at the OP $\endgroup$ – innisfree Nov 21 '16 at 5:44

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