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Let`s say we have a physical system — an atom, a complicated nucleus, or a molecule or something like that — and it makes no difference if we take the whole system and move it to another place. So we have a Hamilton that has the property that, in a sense, it depends only on internal coordinates and not on absolute position in space. In these circumstances, there is a special symmetry operation that we can perform, which is a translation in space. Let`s define $Dop_x(a)$ as the operation of moving the distance $a$ along the $x$ axis. Then we can do this for each state and get a new state. But again, there may be very special states that have the property that if you move them $a$ along the $$x axis, you get the same state except for a phase factor. It is also possible to prove, as we did above, that when this happens, the phase must be proportional to $$a. So for these special states, we can write $ket{psi_0}$ begin{equation} label{Eq:III:17:23} Dop_x(a),ket{psi_0}=e^{ika},ket{psi_0}. end{equation} The $k$ coefficient is called the $x$ component of momentum when multiplied by $hbar$. And the reason it`s called that is because this number is numerically equal to the classical momentum $p_x$ if we have a large system.

The general statement is that if the Hamilton is unchanged when the system is moved, and if the state starts with some momentum in the direction $x$, then the momentum in the direction $x $ will remain the same over time. The total amount of movement a system before and after collisions – or after explosions or whatever – will be the same. So we start by looking at the symmetries of systems. A very simple example is the molecular hydrogen ion – as much to take the ammonia molecule – in which there are two states. For the molecular hydrogen ion, we took as base states one where the electron was close to the number of protons $1$, and another where the electron was close to the number of protons $2. The two states – which we have called $ketsl{slOne}$ and $ketsl{slTwo}$ – are again shown in Fig. 17–1(a). Well, as long as the two nuclei are exactly the same, there is some symmetry in this physical system.

That is, if we reflected the system in the plane halfway between the two protons – meaning that everything on one side of the plane is moved to the symmetric position on the other side – we would get the situations in Fig. 17-1 (b). Since the protons are identical, the reflection operation changes $ketsl{slOne}$ to $ketsl{slTwo}$ and $ketsl{slTwo}$ to $ketsl{slOne}$. We call this reflection operation $Pop$ and write begin{equation} label{Eq:III:17:1} Pop,ketsl{slOne}=ketsl{slTwo},quad Pop,ketsl{slTwo}=ketsl{slOne}. end{equation} So our $Pop$ is an operator in the sense that it “does something” to a state to create a new state. What is interesting is that $Pop$, which runs in any state, creates another state of the system. In ordinary space, there is not the smallest step by which we can translate a quark or an electron, an atom or a planet in space to the tiny distance scales we can see. We therefore hypothesize that space does not have the smallest scale of distance.

A translation in the continuum of space cannot be considered as an integer number of smaller, discrete steps, since there are no smaller steps. In a continuum, the absence of the smallest step implies an infinite number of possible translational symmetry operations. — Ibid. (p. 83). However, experimental results show that there is an asymmetry in the decomposition. The measured angular distribution goes as $costheta$, as we predict – not as $cos^2theta$ or any other power. Since the angular distribution has this form, we can deduce from these measures that the spin of $Lambda^0$ is $1/2$. We also note that parity is not maintained. In fact, the coefficient $alpha$ is experimentally found as $-0.62pm0.05$, so $b$ is about twice as large as $$a.

The lack of symmetry under a reflection is quite clear. See also this article in the American Journal of Physics (Vol. 72, No. 4, pp. 428-435, April 2004) “Symmetries and Conservation Laws: Consequences of Noether`s Theorem”. The same thing comes from our mathematics. Our definition of symmetry is equation (17.10) or equation (17.11) (good for any state $psi$), begin{equation} label{Eq:III:17:20} hopQop,ket{psi}=Qophop,ket{psi}. end{equation} But we only consider a state $ket{psi_0}$, which is a certain state of energy, so $Hop,ket{psi_0}=E,ket{psi_0}$. Since $E$ is just a floating number through $Qop$ if we want, we have begin{equation*} QopHop,ket{psi_0}=Qop E,ket{psi_0}=EQop,ket{psi_0}. end{equation*} Also begin{equation} label{Eq:III:17:21} Hop{Qop,ket{psi_0}}=E{Qop,ket{psi_0}}. end{equation} So $ket{psi_0`}=Qop,ket{psi_0}$ is also a certain energy state of $Hop$ – and with the same $E$. But according to our hypothesis, there is only one such state; It must be $ket{psi_0`}=E^{Idelta},ket{psi_0}$.

Before applying the result we just found, we would like to discuss the idea of symmetry a little more.