Conversation about the uncertainty principle

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M. Ya. Azbel

on the uncertainty principle

In the second issue of our magazine for 1971, an article by M. Ya. Azbel “Dialogue on Temperature” was published. Stamy attracted many open readers. All the letters said that the article was integran. However, there were many typos in the article, for which the editors and the author apologize. The letters from readers raised very integrated questions. Therefore, we decided not to simply give a list of typos, but to ask the author to continue the conversation with the reader.

Author. I never expected that the editors would help me turn an imaginary dialogue with the reader into a real one, and not even a dialogue, but a conversation. This is what reader Volodya Margulis from Saransk writes.

V. Margulis. It seems to me that your article “Dialogue on Temperature” contains many typos. The Bolshman and Planck constants are overestimated by two orders of magnitude, the electron charge by an order of magnitude, and so on.

Author. You are absolutely right. Let me tell you a professional secret: like many theoretical physicists, I remember only one system of units: CGSE. All estimates were carried out in it and the constants in my article were given: Planck’s constant $h \approx 6.5 \times 10^{-27}$ ere·sec, Bolshman’s constant $k \approx 10^{-16}$ ere/erad, electron charge $e \approx 5 \cdot 10^{-10}$ units. CGSE. (Remember that we are only interested in the order of magnitude!)

However, in an article published in the journal, all calculations turned out to be in the SI system and, moreover, in a “deformed form.” This, of course, is unpleasant, but there is no big problem here. Every student reader who benefits from knowing systems of units will benefit from continuing the calculations and returning them to the correct form. Helpful and respectful. The fact is that typos and typos in scientific articles are not uncommon, and one must learn to treat them without worry, correcting them as necessary.

V. Margulis. I did just that, but the correct calculation for electrons is that the potential energy is two orders of magnitude less than the kinetic energy. And for the gel, the potential energy turned out to be much greater than the kinetic energy! *)

*) For those who have not read the article “Dialogue about temperature”, I will explain what we are talking about. From the so-called uncertainty principle it follows that the smallest possible value of the momentum $p_{\min}$ of a particle concentrated in a region whose dimensions are of the order of $a$, $p_{\min} \sim \frac{h}{a}$. Hence the lowest possible kinetic energy is $E_{\min} \sim \frac{p_{\min}^2}{2m} \sim \frac{h^2}{2ma^2}$, where $m$ is the mass of the particle.

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Author. Another typo has failed you. In all formulas of the article, it is more correct to write Planck’s constant $h$ rather than $h = \frac{h}{2\pi} \cdot \text{Позетому кинетическая энергия нулевых колебаний в гелини порядка}$

$$\frac{h^2}{2ma^2} \sim 2 \cdot 10^{-15} \text{эр}e \sim 20^\circ\text{K}$$

(since $m \sim 10^{-23} e$, $a \sim 3 \cdot 10^{-8} cm$). This value is only 5 times higher than the actual value, despite the roughness of the estimate.

I did not freely calculate the potential energy of interaction between the atoms of the helium, since this cannot be done based on general considerations.

V. Margulis. It is clear to me that this also applies to electron gas. According to the formula $E_{\text{min}} \sim \frac{h^2}{2ma^2}$ with $a \sim 3 \cdot 10^{-8} cm$ it turns out about $2 \cdot 10^{5^\circ}\text{K}$. And since $a$ can change, the estimate only makes sense in order of magnitude. Electronic gas really has an energy of tens, even hundreds of thousands of degrees.

Author. Absolutely right. I must conclude by saying that I really enjoyed your letter. I wish you success in your progress towards science. And one more thing - I apologize to both you and all the other participants in our conversation for turning their letters into replicas.

Reader M. Mokhnatkin (Saratov). But you haven’t convinced me yet. After all, according to the uncertainty principle

$$\Delta p_x \cdot \Delta x = h$$

and not $\Delta p \cdot \Delta x = h$, as written in the “Dialogue”. This means that since $p^2 = p_x^2 + p_y^2 + p_z^2$, the given value of the minimum energy should be tripled!

Author. If uncertainty corresponded with an equals sign, you would be absolutely right. But it is very important to remember: we are talking only about the order of magnitude. Let us assume, for definiteness, that the most important of $\Delta p_x$, $\Delta p_y$, $\Delta p_z$ is $\Delta p_x$. Then for $\Delta p =$

$$\sqrt{(\Delta p_x)^2 + (\Delta p_y)^2 + (\Delta p_z)^2}$$

we get $\Delta p_x \leqslant \Delta p \leqslant \Delta p_x \sqrt{3}$, so $\Delta p \sim \Delta p_x$ and $\Delta p \Delta x \sim h$ are complete.

When calculating orders of magnitude, it is important not to exceed the accuracy of the calculation, “refining” the coefficients - this can only worsen the matter. Such calculations are very useful, but they cannot give more than what they are designed for - an error of ten times is natural.

In particular, for a free particle “locked” in a spherical box of diameter $a$, the exact calculation gives just

$$E_{\text{min}} = h^2/2ma^2$$

Reader V. Rubin (Moscow). Why do you talk about Coulomb interaction in the case of electron gas? After all, electrons by themselves could not form a crystal - they repel each other!

Author. Certainly. That’s why I talked about electrons in metal. There they are not allowed to scatter positive nonsense. To interact with them and relative calculation. And the Coulomb interaction was taken for the following reason. The electrostatic interaction itself is classical. Let there be one free electron for each atom, so that the only characteristic distance of the problem is the mechanical (more precisely, mechanical) distance $a$, and we are only interested in distances of this order. Then the only quantity with the dimension of energy that can be constructed from $a$ and the electron charge $e$ is $\frac{e^2}{a}$.

Of course, such a notation is correct only in order of magnitude and only at such distances. At large distances, different charges shield each other, and the interaction quickly kills with distance.

V. Rubin. One more question. Why, if you have a distance for an electron in a box that exceeds the speed of light, do you still use non-relativistic formulas?

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Author. Because I wanted to show: in non-relativistic quantum mechanics, the presence of “locked” electrons in the nucleus would lead to troubles. And this suggests that, they believe, they cannot be locked there. However, you are, of course, right: such reasoning is not proof.

Since we are talking about this, let’s make a more accurate calculation. Using the uncertainty principles again, we obtain the order of the minimum impulse $p_{\min}$ of an electron “locked” in a nucleus of radius $R$, $p_{\min} \sim \frac{h}{R}$. According to the theory of relativity, $p = mv = \frac{m_0 v}{\sqrt{1 - v^2/c^2}}$ ($m_0$ is the rest mass of the particle, $v$ is its speed). “Our” impulse $p_{\min} \gg m_0 c$. This means that $v \approx c$ and $p \approx mc$ (m is the mass of the relativistic electron). This means that for an electron “locked” in the nucleus, $m \sim \frac{h}{R_c}$, which gives values ​​on the order of the mass of a proton at rest. Consequently, the proximity of the mass of the nucleus to the sum of the masses of only protons and neutrons means the absence of electrons “locked” in the nucleus: once they get into the nucleus, they are easily pulled out of there.

You can see the size of the region in which the particles are concentrated. This region can either strictly limit their movement, like the nucleus - the movement of protons and neutrons or atom - the movement of electrons, or be characteristic only from the point of view of the uncertainty principle.

Reader A. Petrov (Leningrad). If the average distance between particles is $a$, then $\Delta x \approx a$ and possibly even $\Delta x = L$, where $L$ is the size of the system. So, $p_{\min} \sim \frac{h}{L}$!

Author. Absolutely right. We estimate the characteristic energy, and it is determined by the average mass $a$. The smallest possible energy is $\frac{h^2}{2mL^2}$, and intermediate energy values ​​corresponding to $a < \Delta x < L$ are also possible.

Let me explain this remark in more detail. The average distance between electrons in a metal $a$ determines the characteristic distance in the uncertainty principle and the characteristic energy of zero-point vibrations, which we estimated. However, in principle, individual electrons can be in rare states, being at a distance from each other on the order of the size of the system $L$. In accordance with this, the minimum kinetic energy (zero-point energy), which has a small number of electrons, is of the order of $\frac{h^2}{2mL^2}$, that is, extremely small.

And here the cardial difference between Fermi and Bohemian particles is already evident.

Since basic particles can accumulate in unlimited quantities in any state, at very low temperatures they begin to accumulate precisely in a state with low energy - the so-called Bose-Einstein condensation begins in energy space.

What has been said is enough to obtain the basic characteristics of the Bose gas. Let’s start with the distance of thermal radiation concentrated in the cavity of a heated body (and in equilibrium with this body). Since thermal radiation consists of quanta of electromagnetic waves (photons), the number of which per unit volume is determined by the temperature of the heated body emitting them, the only energy characteristic is temperature. Therefore, the average photon energy is on the order of $kT$. Here we are not given a distance stretching, but a warm stretching of energy. Let us adapt the uncertainty principle in a slightly different form by introducing the characteristic time $\Delta t$:

$$h \sim \Delta p \Delta x \sim \frac{\Delta p}{\Delta t} \Delta x \Delta t.$$

But the change in momentum over a given time is the product of the force and the time of its action: $\Delta p = F \Delta t$. This means that $\frac{\Delta p}{\Delta t} \Delta x = F \Delta x$ is the work of force on the path $\Delta x$, that is, according to the law of conservation of energy, the change in energy $\Delta E$. Thus,

$$\Delta E \Delta t \sim h$$

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(a very important form of the uncertainty principle).

For photons, $\Delta E \sim \varepsilon$, and the value $\Delta t$ is a puree of time $\tau$, which characterizes the period of photon oscillations. Therefore $\varepsilon \sim hv\left(v = \frac{1}{\tau}\right)$.

From the resulting equality, taking into account that $\varepsilon \sim kT$, we find the oscillation period $\tau = \frac{1}{v} \sim \frac{h}{kT}$ and the wavelength $\lambda$ (the distance traveled by a wave moving at the speed of light $c$ during the oscillation period $\tau$):

$$\lambda = c\tau \sim \frac{ch}{kT}$$

The wavelength determines the characteristic distance “arriving at the quantum” $^*$). The constituent volume is $\lambda^3$, so the number of photons per unit volume is of the order of

$$\lambda^{-3} \sim \left(\frac{kT}{ch}\right)^3$$

Their total energy per unit volume, that is, the energy density of thermal radiation,

$$E_1 \sim kT \cdot \lambda^3 \sim \frac{(kT)^4}{(ch)^3}$$

We have obtained the significant Stefan-Boltzmann formula! The exact formula differs by the coefficient $-\frac{\pi^2}{20} \approx \frac{1}{2}$.

Now let’s move from the macrocosm of gases to the microscopic world of nuclei. Let’s try to estimate the order of magnitude of nuclear forces, knowing nothing about them, except that the radius of the nucleus is $R \sim 10^{-12} \sim 10^{-13}$ cm, and it consists of protons and neutrons interacting with each other. Kinetic energy of order $\frac{h^2}{MR^2}$. And since the nucleus does not fall apart and is not removed, but is kept in balance by nuclear forces, the potential energy of interaction must be on the order of kinetic!

So we have added to the energy of thousands of billions ($10^{12}$ degrees. Just don’t forget that this is energy measured in degrees,

*) Of course, this statement does not look very convincing, but it turns out to be correct, so just take it on faith.

but not the temperature of the part in the core!

And the difference in the masses of the nuclear parts - proton and neutron - is of the order of the electron mass $m_e$. (This is not accidental, since a neutron can turn into a proton and an electron.) This means that the energy associated with the charge is of the order of $m_e c^2$, that is, “only” of the order of $10^{10}$ degrees and significantly less than the nuclear interaction energy. Therefore, nuclear forces practically do not depend on charge, and it makes no sense to talk about neutrons and protons in the nucleus as separate parts, and they are simply called “nucleons.”

What is the speed of the core parts?

In non-relativistic mechanics $\frac{Mv^2}{2} \sim \frac{h^2}{MR^2}$ and $v \sim \frac{h}{MR}$. The speed is on the order of the speed of light! A more accurate estimate leads to $v \sim 0,1$ $c$, so in many cases for the part in the nucleus, in a rough approximation, it is possible to restrict ourselves to a nonrelativistic theory.

As you can see, understanding functional physical laws gives functional physical consequences, of which we have touched only a small part.

Exercises

1. Determine the dependence of pressure on the density of a part (their number per unit volume) in an ideal Fermi gas $^*$ at low temperatures. Obtain the equation of state of such a gas (up to a numerical factor). Compare it with the equation of state of a classical gas.

2. Find a criterion for the quantumness of an ideal Fermi gas.

3. Determine the temperature dependences of the heat capacity of a weightless gas at low temperatures.

4. Find the criterion for a classical gas.

5. Estimate the size of hydrogen atoms.

6. Given that the connection of atoms in liquids and crystals is determined by the basic electrical system interaction, estimate the distance between atoms.

*) See “Dialogue about temperature”.