*1. Introduction*

Quantum mechanics achieved an outstanding success during the 20th century. But in the 1960s, particle physicists decided that the standard form of quantum mechanics contradicts experimental results, regarding the states of the particles Δ^{++}, Δ^{–} and Ω^{–}. They were so sure about it, that they invented a substantial extension of the theory that depends on a new kind of force with three colors. The new force has properties that were never known before, like particles that can be detected only if they have equal amount of colors, forces that become stronger when particles move away from one another, and other new concepts that aren’t similar to anything in ordinary quantum mechanics (see QCD history).

In 1987 the European Muon Collaboration found what is called “the second EMC effect” or the proton spin crisis. This effect remained unexplained until today, and it is considered as one of the most important problems in particle physics.

The current discussion shows that the experimental results of the 1960s can be explained quite easily by the laws of ordinary quantum mechanics, and that the origin of the proton spin crisis is that particle physicists continue using the same erroneous assumptions as they did in the 1960s.

Although the proton spin crisis is considered as an important unexplained phenomenon, I believe that the issue of the Δ^{++}, Δ^{–} and Ω^{–} is far more important. The reason is simple: if these particles can be explained easily by quantum mechanics, then there was no need to invent a completely new theory in the 1960s, the theory of Quantum Chromodynamics, a central component of the standard model. Would you believe in a theory which is based on fantastic new assumptions that was invented just in order to explain something that can be explained by a well established theory?

The article is divided into several sections in order to make it easier for physicists and popular science readers. The second section describes well known terms and physicists may skip it. The last sections explain the proton spin crisis and what went wrong during the 1960s.

*2. Quantum mechanical principles of bound states*

**Pauli exclusion principle and electronic shells**

There are four properties, called four quantum numbers, which are assigned to an electron state in an atomic shell. These properties are the following:

– Radial quantum number (n_{r}). It is a positive integer which denotes the radial excitation of the electron.

– Orbital quantum number (l) or “azimuthal quantum number”, denotes the electron’s spatial angular momentum. It takes a non-negative integral number. It is denoted by the symbols l=s (or l=0), l=p (or l=1), l=d (or l=2), l=f (or l=3), l=g (or l=4), l=h (or l=5), etc. The “principal quantum number” (n) is the sum of the radial and angular quantum numbers (n_{r}+l). This is a positive integer.

– Magnetic quantum number (m). It denotes the projection of the quantum number l on the z-axis. It can take 2l+1 integral values –l <= m <= l. The l=s shell has only one m-value. The l=p shell has 3 m values (-1,0,1). l=d has 5 values (-2,-1,0,1,2) etc.

– Spin projection (m_{s}). The electron spin is 1/2. Its z-axis projection can be +1/2 or -1/2.

Examples of possible electronic states:

n=1 l=s m=0 m_{s}=+1/2

n=1 l=s m=0 m_{s}=-1/2

n=2 l=s m=0 m_{s}=+1/2

n=2 l=s m=0 m_{s}=-1/2

n=2 l=p m=1 m_{s}=+1/2

n=2 l=p m=1 m_{s}=-1/2

n=2 l=p m=0 m_{s}=+1/2

n=2 l=p m=0 m_{s}=-1/2

n=2 l=p m=-1 m_{s}=+1/2

n=2 l=p m=-1 m_{s}=-1/2

n=3 l=d m=2 m_{s}=-1/2

…

According to Pauli exclusion principle it is impossible that two electrons in an atom would have the same quantum state; therefore, they cannot have the same four quantum numbers.

In the case of n-electrons, an acceptable set of single particle angular momenta of these electrons is called a configuration.

**Angular momentum and spin**

One of the important quantities in physics is the angular momentum. The angular momentum of a particle (composite or elementary) is called the particle’s spin. The angular momentum of a composite particle is the sum of the angular momenta of its components.

When we calculate the contribution of an electron to the total angular momentum of the atom, we sum-up two quantities:

– The electron spin. This number can be -1/2 or +1/2. This quantity is related to the electron itself.

– The spatial angular momentum. This is the quantum number “m” above. This quantity is related to the movement of the electron in the atomic shell.

Angular momentum is a vectorial quantity, therefore summing up the quantities is not straightforward. But we can say that for electron with l=0 (such state is often called “s-wave”) the contribution to the angular momentum is exactly the electron spin, because m=0.

Each electronic state is related to the energy level of the electron. For example, electrons with a higher principal quantum number (n) have a higher energy because of the electron’s radial and/or orbital excitation.

The lowest energy level of an atom is called the ground state.

*3. QCD motivation and the proton spin crisis*

**The situation inside the proton**

The proton has typically three quarks of the uud flavor and its total spin is 1/2. The proton has sometimes an additional pair of a quark and an antiquark. Particle physicists assume the following assumptions about the proton:

(1) There is only one configuration with exactly 3 quarks (u,u,d). There are additional configurations that have additional quark-antiquark pairs.

(2) All the proton’s quarks have s-wave, meaning that the quarks do not have angular momentum, except for their spin.

In the experiment that caused the proton spin crisis [1] physicists measured the instantaneous spin direction of the proton’s quarks, which is the m_{s} quantum number (+1/2 or -1/2), and counted how many times it is parallel to the spin of the proton (+1/2), and how many times it has an opposite direction (-1/2). Assuming the two assumptions above, physicists expected that in configurations that have 3 quarks without the quark-antiquark pair, two quarks will have +1/2 spin and one will have -1/2 spin. This is mandatory because the total spin of the proton is +1/2.

However, in that experiment, the number of quarks with spin projection +1/2 was found to be almost equal to the number of quarks with spin projection -1/2.

**Δ ^{++}, Δ^{–} and Ω^{–}: the motivation for QCD foundation**

As already mentioned, properties of the baryons Δ

^{++}, Δ

^{–}and Ω

^{–}provided a primary motivation for developing QCD as an extension of ordinary quantum mechanics.

Let’s talk about Δ^{++}. Analogous discussion may be applied to the other two baryons.

The Δ^{++} baryon contains three u-quarks, and its spin and parity are 3/2+. Particle physicists thought that this is impossible: all the three u-quarks, they thought, must have the same quantum numbers:

u quark, n=1 l=0 m=0 m_{s}=+1/2

u quark, n=1 l=0 m=0 m_{s}=+1/2

u quark, n=1 l=0 m=0 m_{s}=+1/2

And this is obviously impossible by the Pauli exclusion principle.

Notice that the scientists assume that there is only one configuration which must be an l=0 s-wave.

Another possibility could be to use higher principal quantum numbers:

u quark, n=1 l=0 m=0 m_{s}=+1/2

u quark, n=2 l=0 m=0 m_{s}=+1/2

u quark, n=3 l=0 m=0 m_{s}=+1/2

In this case the energy level of Δ^{++} should be much higher than that of the proton (because this configuration consists of much higher principal quantum numbers). But this is not inline with the measured energy levels: the proton’s mass is 938 MeV and the Δ^{++} mass is only 1232 MeV.

This is why they concluded that there must be a new quantum number. This is why they have invented the color.

*4. A naïve explanation according to Comay model*

Comay model assumes that the proton (and all other baryons) has a core that attracts the quarks, and there are inner closed quark shells having the u,d flavor. The uud valence quarks that we observe belong to the outer quark shell.

Therefore, due to the Pauli exclusion principle, the three uud quarks in the proton’s outer shell cannot have the lowest energy state, which is already occupied by the quarks in inner shells. Possible configurations of the proton and the corresponding m-values may be:

u quark, n=1 l=0 m=0 m_{s}=+1/2 (inner quark)

u quark, n=1 l=0 m=0 m_{s}=-1/2 (inner quark)

d quark, n=1 l=0 m=0 m_{s}=+1/2 (inner quark)

d quark, n=1 l=0 m=0 m_{s}=-1/2 (inner quark)

u quark, n=2 l=0 m=0 m_{s}=+1/2

d quark, n=2 l=0 m=0 m_{s}=+1/2

u quark, n=2 l=0 m=0 m_{s}=-1/2

and:

u quark, n=1 l=0 m=0 m_{s}=+1/2 (inner quark)

u quark, n=1 l=0 m=0 m_{s}=-1/2 (inner quark)

d quark, n=1 l=0 m=0 m_{s}=+1/2 (inner quark)

d quark, n=1 l=0 m=0 m_{s}=-1/2 (inner quark)

u quark, n=2 l=0 m=0 m_{s}=+1/2

d quark, n=2 l=1 m=0 m_{s}=-1/2

u quark, n=2 l=1 m=1 m_{s}=-1/2

The two configurations above have similar stability (because they have the same principal quantum numbers), and the difference is that two quark functions are p-wave instead of s-wave. The total of m_{s} of the first configuration is +1/2 and the total of m_{s} of the second is -1/2. If the two states above appear in reality in a similar probability then a cancellation of quarks’ spin arises. This fact alone explains the proton spin crisis.

Similar ideas work in order to explain the situation of Δ^{++} which has 3 u valence quarks and total angular momentum of 3/2. A possible configuration is:

u quark, n=1 l=0 m=0 m_{s}=+1/2 (inner quark)

u quark, n=1 l=0 m=0 m_{s}=-1/2 (inner quark)

d quark, n=1 l=0 m=0 m_{s}=+1/2 (inner quark)

d quark, n=1 l=0 m=0 m_{s}=-1/2 (inner quark)

u quark, n=2 l=0 m=0 m_{s}=+1/2

u quark, n=2 l=1 m=0 m_{s}=+1/2

u quark, n=2 l=1 m=1 m_{s}=-1/2

This configuration does not violate the Pauli exclusion principle and can explain the Δ^{++} quantum numbers and its low energy as well.

*5. The full explanation – multiple configurations*

The electron in the hydrogen atom takes a quite unique state. For example, in the hydrogen ground state, this electron has the following quantum numbers:

n=1 l=s m=0 m_{s}=+1/2 or n=1 l=s m=0 m_{s}=-1/2.

However, if there is more than one electron in the atomic shell the situation becomes much more complicated.

When there is more than one electron in the atom, there are electromagnetic forces between the electrons, and the electrons are not stored exactly like in the “clean” case where one electron is attracted to the nuclear charge. The reason is that here, for each electron, the electric field of all other electrons doesn’t have a well defined center like in the case of the hydrogen atom. The impact of this fact is that these electrons are not stored in a unique configuration but in a multiple configuration state where each configuration has an appropriate statistical weight. The “multiple configurations” scenario happens even in the simplest case of the spherically symmetric spin=0 ground state of the helium atom. A fortiori, it happens in atoms with more electrons as well.

This fact is known since the 1940s and 1950s [2] and a special mathematical tool, called “angular momentum algebra” or “Wigner and Racah calculus” has been developed in order to calculate the probability (weight) of each configuration. In the beginning of the 1960s these configurations were calculated by early computers, and it was found that their impact is tremendous and cannot be ignored: even for the simplest case of the ground state of the helium atom, several configurations have two digit probabilities and no configuration dominates the state. The paper from 1961 [3] considers 35 configurations (!!) of the helium ground state, each has a non-negligible probability. Other atoms, with more electrons and/or a higher spin, have a much more complex situation with many more possible configurations.

As a consequence, even in its ground state each of the two electrons of the helium atom has a significant probability to be in p-wave or d-wave and not only in s-wave.

The “multiple configurations” concept has not attracted an adequate attention from physicists. There are many educational web pages which ignore it [4]. In these web pages the helium ground state is described as if it has one configuration only: each electron has n=1,l=s,m=0 quantum numbers and the two electrons have opposite spins. There are two related errors in this description:

(1) The helium ground state contains 1 configuration only. This is incorrect.

(2) The state of these electrons is l=0 (s-wave), which means that they do not have spatial angular momentum. This is also a mistake.

It can be concluded that the incorrect idea stating that any atomic ground state can be described properly by a unique configuration prevails in contemporary literature.

**The proton spin crisis explained**

Obviously, the concept of the quantum numbers is applicable to composite particles which contain particles that obey the Pauli exclusion principle, and apply forces on one another. Therefore, it would be natural to assume that this concept applies to the quarks inside protons and neutrons.

Here, there are three quarks which apply strong forces on one another, meaning that the number of non-negligible configurations is even much higher than in the spin=0 ground state of the helium atom.

Notice that the proton spin crisis experiment measured only the instantaneous value of the spin of each quark, the m_{s} quantum number, and not its spatial angular momentum. The spatial angular momentum is very significant in a situation of multiple configurations because many configurations contain quarks in p-wave or higher orbitals. In this case, the quark’s spin is sometimes parallel and sometimes anti-parallel to the proton’s spin. Therefore, a measurement of the instantaneous quark’s spin ignores the spatial angular momentum of the proton’s quarks and certainly does not measure properly the total spin of the proton.

In fact, according to Comay model there are inner closed quark shells inside the proton. This makes the number of configurations much higher, and the relative contribution of the spatial angular momentum even higher.

**Δ ^{++} explained**

We already know the answer to this question, right? Δ

^{++}has many configurations like the proton. It is easy to build many possible configurations for this particle, if we use not only quarks in s-wave, but also quarks in p-wave or even higher orbital values like d-wave etc.

The only question that remains is why the Δ^{++} energy is so close to that of the proton. First, they both have multiple configurations. Second, according to Hund rule which applies to electrons in atomic shells, lower energy levels are obtained if the electrons’ spin are in the same direction. This is explained by the fact that electrons repel each other, and a lower energy state is achieved when their mean distance is larger. Thus, if the electrons’ spin are parallel (symmetric) then, by virtue of the Pauli principle, the spatial components are anti-symmetric. Evidently, the mean distance between a spatially anti-symmetric state is larger than that of the corresponding symmetric state.

The same applies here. And here, like in the electrons, the quarks try to achieve states in which they are far from each other. This is the explanation why the energy of the Δ^{++} is not too far from the proton, and why the energy of Δ^{++}(1750) which has spin 1/2 (unlike the Δ^{++}(1232) which has spin 3/2) has higher energy.

By the way, one may wonder how QCD explains why Δ^{++} with spin 1/2 has a much higher energy than that of the Δ^{++} with spin 3/2. This is the outcome of applying Hund rule to quarks. But in atoms the Hund rule is justified by the fact that electrons repel each other. This is the case in Comay model as well. But this is not the case in QCD and therefore there is no justification to apply Hund rule to the quarks according to QCD.

**Few more words about multiple configurations**

We see here that the reason for the fundamental error of QCD is the wrong perception of the role of multiple configurations in atoms. This mistake is starring in many academic web sites [4].

It is interesting to see how the issue is treated in the Wikipedia topic Electron Configurations (as of September 16, 2011). One can see in the Discussion tab of this topic that a paragraph called “Why electron configurations occur” was removed in 2005 and the author who removed this paragraph says:

“… any system has only one stable electron configuration. If left in equilibrium, it will always have this configuration (called the ground state), though the electrons may be temporarily “excited” to other configurations.”

It seems that the author is aware to the fact that multiple configurations occur, but unaware to the fact that this is not a negligible issue – calculations show that atoms are mostly not in a single configuration, and hence there is no unique “stable electron configuration” which is mentioned above.

[1] J. Ashman et al, *A measurement of the spin asymmetry and determination of the structure function g _{1} in deep inelastic muon-proton scattering*, Physics Letters B, Vol

**206**, (1988). p.364-370

[2] G.R. Taylor and R.G. Parr,

*Superposition of configurations: The helium atom*, Proc., Natl.Acad. Sci. USA

**38**, (1952). p.154-160

[3] AW Weiss,

*Configuration Interaction in Simple Atomic Systems*, Phys. Rev.

**122**, (1961). p.1826–1836

[4] hyperphysics.phy-astr.gsu.edu/hbase/quantum/helium.html

wiki.brown.edu/confluence/download/attachments/29133/Helium+and+Calcium.pdf

quantummechanics.ucsd.edu/ph130a/130_notes/node35.html