The disintegration of the $\boldsymbol{\Lambda}^{\boldsymbol{0}}$
Now we want to give an example of how we use the theorem of conservation of angular momentum in a specifically quantum physical problem. We look at break-up of the lambda particle ($\Lambda^0$), which disintegrates into a proton and a $\pi^-$ meson by a “weak” interaction: \begin{equation*} \Lambda^0\to\text{p}+\pi^-. \end{equation*} Assume we know that the pion has spin zero, that the proton has spin one-half, and that the $\Lambda^0$ has spin one-half. We would like to solve the following problem: Suppose that a $\Lambda^0$ were to be produced in a way that caused it to be completely polarized—by which we mean that its spin is, say “up,” with respect to some suitably chosen $z$-axis—see Fig. 17–6(a). The question is, with what probability will it disintegrate so that the proton goes off at an angle $\theta$ with respect to the $z$-axis—as in Fig. 17–6(b)? In other words, what is the angular distribution of the disintegrations? We will look at the disintegration in the coordinate system in which the $\Lambda^0$ is at rest—we will measure the angles in this rest frame; then they can always be transformed to another frame if we want.  We begin by looking at the special circumstance in which the proton is emitted into a small solid angle $\Delta\Omega$ along the $z$-axis (Fig. 17–7). Before the disintegration we have a $\Lambda^0$ with its spin “up,” as in part (a) of the figure. After a short time—for reasons unknown to this day, except that they are connected with the weak decays—the $\Lambda^0$ explodes into a proton and a pion. Suppose the proton goes up along the $+z$-axis. Then, from the conservation of momentum, the pion must go down. Since the proton is a spin one-half particle, its spin must be either “up” or “down”—there are, in principle, the two possibilities shown in parts (b) and (c) of the figure. The conservation of angular momentum, however, requires that the proton have spin “up.” This is most easily seen from the following argument. A particle moving along the $z$-axis cannot contribute any angular momentum about this axis by virtue of its motion; therefore, only the spins can contribute to $J_z$. The spin angular momentum about the $z$-axis is $+\hbar/2$ before the disintegration, so it must also be $+\hbar/2$ afterward. We can say that since the pion has no spin, the proton spin must be “up.”  If you are worried that arguments of this kind may not be valid in quantum mechanics, we can take a moment to show you that they are. The initial state (before the disintegration), which we can call $\ket{\Lambda^0,\text{spin \(+z\)}}$ has the property that if it is rotated about the $z$-axis by the angle $\phi$, the state vector gets multiplied by the phase factor $e^{i\phi/2}$. (In the rotated system the state vector is $e^{i\phi/2}\,\ket{\Lambda^0,\text{spin \(+z\)}}$.) That’s what we mean by spin “up” for a spin one-half particle. Since nature’s behavior doesn’t depend on our choice of axes, the final state (the proton plus pion) must have the same property. We could write the final state as, say, \begin{equation*} \ket{\text{proton going $+z$, spin $+z$; pion going $-z$}}. \end{equation*} But we really do not need to specify the pion motion, since in the frame we have chosen the pion always moves opposite the proton; we can simplify our description of the final state to \begin{equation*} \ket{\text{proton going $+z$, spin $+z$}}. \end{equation*} Now what happens to this state vector if we rotate the coordinates about the $z$-axis by the angle $\phi$?  Since the proton and pion are moving along the $z$-axis, their motion isn’t changed by the rotation. (That’s why we picked this special case; we couldn’t make the argument otherwise.) Also, nothing happens to the pion, because it is spin zero. The proton, however, has spin one-half. If its spin is “up” it will contribute a phase change of $e^{i\phi/2}$ in response to the rotation. (If its spin were “down” the phase change due to the proton would be $e^{-i\phi/2}$.) But the phase change with rotation before and after the excitement must be the same if angular momentum is to be conserved. (And it will be, since there are no outside influences in the Hamiltonian.) So the only possibility is that the proton spin will be “up.” If the proton goes up, its spin must also be “up.”  We conclude, then, that the conservation of angular momentum permits the process shown in part (b) of Fig. 17–7, but does not permit the process shown in part (c). Since we know that the disintegration occurs, there is some amplitude for process (b)—proton going up with spin “up.” We’ll let $a$ stand for the amplitude that the disintegration occurs in this way in any infinitesimal interval of time.10
 Now let’s see what would happen if the $\Lambda^0$ spin were initially “down.”  Again we ask about the decays in which the proton goes up along the $z$-axis, as shown in Fig. 17–8. You will appreciate that in this case the proton must have spin “down” if angular momentum is conserved. Let’s say that the amplitude for such a disintegration is $b$.  We can’t say anything more about the two amplitudes $a$ and $b$. They depend on the inner machinery of $\Lambda^0$, and the weak decays, and nobody yet knows how to calculate them. We’ll have to get them from experiment. But with just these two amplitudes we can find out all we want to know about the angular distribution of the disintegration. We only have to be careful always to define completely the states we are talking about.  We want to know the probability that the proton will go off at the angle $\theta$ with respect to the $z$-axis (into a small solid angle $\Delta\Omega$) as drawn in Fig. 17–6. Let’s put a new $z$-axis in this direction and call it the $z'$-axis. We know how to analyze what happens along this axis. With respect to this new axis, the $\Lambda^0$ no longer has its spin “up,” but has a certain amplitude to have its spin “up” and another amplitude to have its spin “down.”  We have already worked these out in Chapter 6, and again in Chapter 10, Eq. (10.30). The amplitude to be spin “up” is $\cos\theta/2$, and the amplitude to be spin “down” is11 $-\sin\theta/2$. When the $\Lambda^0$ spin is “up” along the $z'$-axis it will emit a proton in the $+z'$-direction with the amplitude $a$. So the amplitude to find an “up”-spinning proton coming out along the $z'$-direction is \begin{equation} \label{Eq:III:17:33} a\cos\frac{\theta}{2}. \end{equation} Similarly, the amplitude to find a “down”-spinning proton coming along the positive $z'$-axis is \begin{equation} \label{Eq:III:17:34} -b\sin\frac{\theta}{2}. \end{equation} The two processes that these amplitudes refer to are shown in Fig. 17–9.  Let’s now ask the following easy question. If the $\Lambda^0$ has spin up along the $z$-axis, what is the probability that the decay proton will go off at the angle $\theta$? The two spin states (“up” or “down” along $z'$) are distinguishable even though we are not going to look at them. So to get the probability we square the amplitudes and add. The probability $f(\theta)$ of finding a proton in a small solid angle $\Delta\Omega$ at $\theta$ is \begin{equation} \label{Eq:III:17:35} f(\theta)=\abs{a}^2\cos^2\frac{\theta}{2}+ \abs{b}^2\sin^2\frac{\theta}{2}. \end{equation} Remembering that $\sin^2\theta/2=\tfrac{1}{2}(1-\cos\theta)$ and that $\cos^2\theta/2=\tfrac{1}{2}(1+\cos\theta)$, we can write $f(\theta)$ as \begin{equation} \label{Eq:III:17:36} f(\theta)=\!\biggl(\frac{\abs{a}^2\!+\abs{b}^2}{2}\biggr)\!+\! \biggl(\frac{\abs{a}^2\!-\abs{b}^2}{2}\biggr)\!\cos\theta. \end{equation} The angular distribution has the form \begin{equation} \label{Eq:III:17:37} f(\theta)=\beta(1+\alpha\cos\theta). \end{equation} The probability has one part that is independent of $\theta$ and one part that varies linearly with $\cos\theta$. From measuring the angular distribution we can get $\alpha$ and $\beta$, and therefore, $\abs{a}$ and $\abs{b}$.  Now there are many other questions we can answer. Are we interested only in protons with spin “up” along the old $z$-axis? Each of the terms in (17.33) and (17.34) will give an amplitude to find a proton with spin “up” and with spin “down” with respect to the $z'$-axis ($+z'$ and $-z'$). Spin “up” with respect to the old axis $\ket{+z}$ can be expressed in terms of the base states $\ket{+z'}$ and $\ket{-z'}$. We can then combine the two amplitudes (17.33) and (17.34) with the proper coefficients ($\cos\theta/2$ and $-\sin\theta/2$) to get the total amplitude \begin{equation*} \biggl(a\cos^2\frac{\theta}{2}+b\sin^2\frac{\theta}{2}\biggr). \end{equation*} Its square is the probability that the proton comes out at the angle $\theta$ with its spin the same as the $\Lambda^0$ (“up” along the $z$-axis).  If parity were conserved, we could say one more thing. The disintegration of Fig. 17–8 is just the reflection—in the $xy$-plane of the disintegration—of Fig. 17–7.12 If parity were conserved, $b$ would have to be equal to $a$ or to $-a$. Then the coefficient $\alpha$ of (17.37) would be zero, and the disintegration would be equally likely to occur in all directions.  The experimental results show, however, that there is an asymmetry in the disintegration. The measured angular distribution does go as $\cos\theta$ as we predict—and not as $\cos^2\theta$ or any other power. In fact, since the angular distribution has this form, we can deduce from these measurements that the spin of the $\Lambda^0$ is $1/2$. Also, we see that parity is not conserved. In fact, the coefficient $\alpha$ is found experimentally to be $-0.62\pm0.05$, so $b$ is about twice as large as $a$. The lack of symmetry under a reflection is quite clear.  You see how much we can get from the conservation of angular momentum. We will give some more examples in the next chapter.  

Parenthetical note. By the amplitude $a$ in this section we mean the amplitude that the state $\ket{\text{proton going \(+z\), spin \(+z\)}}$ is generated in an infinitesimal time $dt$ from the state $\ket{\text{\(\Lambda\), spin \(+z\)}}$, or, in other words, that \begin{equation} \label{Eq:III:17:38} \bracket{\text{proton going $+z$, spin $+z$}} {H} {\text{$\Lambda$, spin $+z$}} =i\hbar a, \end{equation} where $H$ is the Hamiltonian of the world—or, at least, of whatever is responsible for the $\Lambda$-decay. The conservation of angular momentum means that the Hamiltonian must have the property that \begin{equation} \label{Eq:III:17:39} \bracket{\text{proton going $+z$, spin $-z$}} {H} {\text{$\Lambda$, spin $+z$}} =0. \end{equation} By the amplitude $b$ we mean that \begin{equation} \label{Eq:III:17:40} \bracket{\text{proton going $+z$, spin $-z$}} {H} {\text{$\Lambda$, spin $-z$}} =i\hbar b. \end{equation} Conservation of angular momentum implies that \begin{equation} \label{Eq:III:17:41} \bracket{\text{proton going $+z$, spin $+z$}} {H} {\text{$\Lambda$, spin $-z$}} =0. \end{equation} If the amplitudes written in (17.33) and (17.34) are not clear, we can express them more mathematically as follows. By (17.33) we intend the amplitude that the $\Lambda$ with spin along $+z$ will disintegrate into a proton moving along the $+z'$-direction with its spin also in the $+z'$-direction, namely the amplitude \begin{equation} \label{Eq:III:17:42} \bracket{\text{proton going $+z'$, spin $+z'$}} {H} {\text{$\Lambda$, spin $+z$}}. \end{equation} By the general theorems of quantum mechanics, this amplitude can be written as \begin{equation} \label{Eq:III:17:43} \sum_i \bracket{\text{proton going $+z'$, spin $+z'$}} {H}{\Lambda,i} \braket{\Lambda,i} {\text{$\Lambda$, spin $+z$}}, \end{equation} where the sum is to be taken over the base states $\ket{\Lambda,i}$ of the $\Lambda$-particle at rest. Since the $\Lambda$-particle is spin one-half, there are two such base states which can be in any reference base we wish. If we use for base states spin “up” and spin “down” with respect to $z'$ ($+z'$, $-z'$), the amplitude of (17.43) is equal to the sum \begin{align} &\bracket{\text{proton going $+z'$, spin $+z'$}} {H}{\Lambda,+z'} \braket{\Lambda,+z'}{\Lambda,+z}\notag\\[.5ex] \label{Eq:III:17:44} +\,&\bracket{\text{proton going $+z'$, spin $+z'$}} {H}{\Lambda,-z'} \braket{\Lambda,-z'}{\Lambda,+z}. \end{align} The first factor of the first term is $a$, and the first factor of the second term is zero—from the definition of (17.38), and from (17.41), which in turn follows from angular momentum conservation. The remaining factor $\braket{\Lambda,+z'}{\Lambda,+z}$ of the first term is just the amplitude that a spin one-half particle which has spin “up” along one axis will also have spin “up” along an axis tilted at the angle $\theta$, which is $\cos\theta/2$—see Table 6–2. So (17.44) is just $a\cos\theta/2$, as we wrote in (17.33). The amplitude of (17.34) follows from the same kind of arguments for a spin “down” $\Lambda$-particle. 