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Section 5.5 State Representation for Spin

We can now apply the mathematical framework of state vectors from the previous sections to the case of spin states for particles. When considering the spin state of an electron, the component of spin measured along a certain direction has only two possible states, either “spin-up” or “spin-down”; i.e., with a component of either \(+\hbar/2\) or \(-\hbar/2\text{.}\) For the component of spin in the \(z\)-direction, we can denote the “spin-up” and “spin-down” states by the kets \(| +z \rangle\) and \(| -z \rangle\text{,}\) respectively. For example, in Figure 5.4, the electrons exiting the top and bottom of the SG device are in states \(| +z \rangle\) and \(| -z \rangle\text{,}\) respectively. Measurement of the \(z\)-component of spin \(S_z\) for the state \(| +z \rangle\) will always produce the value \(+\hbar/2\) (as indicated by the experiment in Figure 5.5), and the same measurement for the \(| -z \rangle\) state will always produce a value \(S_z = -\hbar/2\text{.}\) Similarly, we can define the states \(| +x \rangle\) and \(| -x \rangle\) as the states with \(x\)-component of spin \(S_x = +\hbar/2\) and \(S_x = -\hbar/2\text{,}\) respectively. Similar definitions apply for the states \(| +y \rangle\) and \(| -y \rangle\text{.}\)

In the experiment depicted in Figure 5.7 we found that a measurement of the \(z\)-component of spin for an electron in the state \(| +x \rangle\) will produce either \(+\hbar/2\) or \(-\hbar/2\text{.}\) This implies that it should be possible to write \(| +x \rangle\) as a superposition of the states \(| +z \rangle\) and \(| -z \rangle\text{:}\)

\begin{equation} | +x \rangle = c_+ | +z \rangle + c_- | -z \rangle\label{eq_superState}\tag{5.14} \end{equation}

where \(c_+\) and \(c_-\) are constants. We interpret this equation in the following way: if an electron is in a state \(| +x \rangle\) and we measure the \(z\)-component of spin \(S_z\text{,}\) then the probabilities that the measurement will yield the result \(+\hbar/2\) or \(-\hbar/2\) are \(|c_+|^2\) and \(|c_-|^2\text{,}\) respectively.

Since the results of the experiment in Figure 5.7 give beams of equal intensity from the third SG device measuring \(S_z\text{,}\) this means that \(|c_+|^2 = |c_-|^2 = 1/2\text{.}\) Therefore the simplest linear superposition of states representing \(| +x \rangle\) consistent with this result is

\begin{equation} | +x \rangle = \sqrt{\frac{1}{2}}| +z \rangle + \sqrt{\frac{1}{2}}| -z \rangle\text{.}\tag{5.15} \end{equation}

The same can be done for the states \(| -x \rangle\text{,}\) \(| +y \rangle\text{,}\) and \(| -y \rangle\text{.}\) The linear superpositions for these states in terms of the states \(| +z \rangle\) and \(| -z \rangle\) are presented below without proof:

\begin{align} | +x \rangle =\mathstrut \amp \sqrt{\frac{1}{2}} \,| +z \rangle + \sqrt{\frac{1}{2}} \,| -z \rangle\label{eq_plusx}\tag{5.16}\\ | -x \rangle =\mathstrut \amp \sqrt{\frac{1}{2}} \,| +z \rangle - \sqrt{\frac{1}{2}} \,| -z \rangle\label{eq_minusx}\tag{5.17}\\ | +y \rangle =\mathstrut \amp \sqrt{\frac{1}{2}} \,| +z \rangle + i\sqrt{\frac{1}{2}} \,| -z \rangle\label{eq_plusy}\tag{5.18}\\ | -y \rangle =\mathstrut \amp \sqrt{\frac{1}{2}} \,| +z \rangle - i\sqrt{\frac{1}{2}} \,| -z \rangle \text{.}\label{eq_minusy}\tag{5.19} \end{align}

Don't forget the factors of “\(i\)” in the equations for \(| +y \rangle\) and \(| -y \rangle\text{;}\) they are important.

Using Equations (5.16) and (5.17), show that we can write the spin-up \(S_z\) state \(| +z \rangle\) as a linear superposition of the states \(| +x \rangle\) and \(| -x \rangle\text{.}\)

Solution.

If we add equations (5.16) and (5.17) we obtain

\begin{align*} | +x \rangle + | -x \rangle =\mathstrut \amp \left[ \sqrt{\frac{1}{2}}| +z \rangle + \sqrt{\frac{1}{2}}| -z \rangle \right] + \left[ \sqrt{\frac{1}{2}}| +z \rangle - \sqrt{\frac{1}{2}}| -z \rangle \right]\\ =\mathstrut \amp 2 \cdot \sqrt{\frac{1}{2}} | +z \rangle \text{.} \end{align*}

Solving for \(| +z \rangle\) we get

\begin{equation} | +z \rangle = \sqrt{\frac{1}{2}}\ | +x \rangle + \sqrt{\frac{1}{2}}\ | -x \rangle\text{.}\tag{5.20} \end{equation}

Similarly, if we subtract equations (5.16) and (5.17) and solve for \(| -z \rangle\) we obtain

\begin{equation} | -z \rangle = \sqrt{\frac{1}{2}} | +x \rangle - \sqrt{\frac{1}{2}} | -x \rangle\text{.}\tag{5.21} \end{equation}

In a similar manner we can use equations (5.16) \(-\)(5.19) to determine any of the basis spin vectors in terms of the others. The complete set of transformations among the basis spin states appears in Table 5.9.

Table 5.9. Transformation of the basis spin vectors
\(| +z \rangle\) \(=\) \(\sqrt{\frac{1}{2}} | +x \rangle + \sqrt{\frac{1}{2}} | -x \rangle\) \(=\) \(\sqrt{\frac{1}{2}} | +y \rangle + \sqrt{\frac{1}{2}} | -y \rangle\)
\(| -z \rangle\) \(=\) \(\sqrt{\frac{1}{2}} | +x \rangle - \sqrt{\frac{1}{2}} | -x \rangle\) \(=\) \(-i\sqrt{\frac{1}{2}} | +y \rangle + i\sqrt{\frac{1}{2}} | -y \rangle\)
\(| +x \rangle\) \(=\) \(\sqrt{\frac{1}{2}} | +z \rangle + \sqrt{\frac{1}{2}} | -z \rangle\)
\(| -x \rangle\) \(=\) \(\sqrt{\frac{1}{2}} | +z \rangle - \sqrt{\frac{1}{2}} | -z \rangle\)
\(| +y \rangle\) \(=\) \(\sqrt{\frac{1}{2}} | +z \rangle + i\sqrt{\frac{1}{2}} | -z \rangle\)
\(| -y \rangle\) \(=\) \(\sqrt{\frac{1}{2}} | +z \rangle - i\sqrt{\frac{1}{2}} | -z \rangle\)

The previous example demonstrates exactly what we discovered in the experiment of Figure 5.6. If we have a beam of electrons in the state \(| +z \rangle\) and measure the \(x\)-component of spin \(S_x\text{,}\) we find that we obtain \(S_x = +\hbar /2\) (spin-up along \(x\)) and \(S_x = -\hbar /2\) (spin-down along \(x\)) with equal probabilities \(|\frac{1}{\sqrt{2}}|^2 = \frac{1}{2}\text{.}\)

Consider an electron in the state given by

\begin{equation} |\psi\rangle = \sqrt{\frac{2}{3}}| +z \rangle + \sqrt{\frac{1}{3}}| -z \rangle\text{.}\label{eq_ex3}\tag{5.22} \end{equation}

(a) Calculate the probability of obtaining the value \(+\hbar/2\) and \(-\hbar/2\) when the \(z\)-component of the spin angular momentum is measured. (b) Calculate the probability that an electron in the state \(|\psi\rangle\) will be measured to have an \(x\)-component of spin of \(-\hbar/2\text{.}\) (c) Calculate the probability that an electron in the state \(|\psi\rangle\) will be measured to have a \(y\)-component of spin of \(-\hbar/2\text{.}\)

Solution.

(a) This is straightforward because \(|\psi\rangle\) is written as a superposition of the \(| \pm z \rangle\)-states. The probability of the spin being found as “spin up” is the square of the coefficient of the base state \(| +z \rangle\text{.}\) That is,

\begin{equation} \mbox{Prob} \bigl(\mbox{spin measured to be spin-up} \bigr) = P(+z) = \left|\sqrt{\frac{2}{3}}\right|^2 = \frac{2}{3}\text{.}\tag{5.23} \end{equation}

Likewise, the probability of the spin being found as “spin down” is

\begin{equation} \mbox{Prob} \bigl(\mbox{spin measured to be spin-down} \bigr) = P(-z) = \left|\sqrt{\frac{1}{3}}\right|^2 = \frac{1}{3}\text{.}\tag{5.24} \end{equation}

(b) For the \(x\)-component of spin, we need to rewrite \(| \psi \rangle\) as a linear superposition of the \(S_x\) basis states \(| +x \rangle\) and \(| -x \rangle\text{.}\) This can be accomplished by using the results of Example 5.8 and Table 5.9 to write the states \(| +z \rangle\) and \(| -z \rangle\) in terms of the basis states \(| +x \rangle\) and \(| -x \rangle\text{.}\)

\begin{align*} | \psi \rangle =\mathstrut \amp \sqrt{\frac{2}{3}} \left[ \sqrt{\frac{1}{2}} \left[ | +x \rangle + | -x \rangle \right] \right] + \sqrt{\frac{1}{3}} \left[ \sqrt{\frac{1}{2}} \left[ | +x \rangle - | -x \rangle \right] \right]\\ =\mathstrut \amp \sqrt{\frac{1}{2}} \left[ \sqrt{\frac{2}{3}} + \sqrt{\frac{1}{3}} \right] | +x \rangle + \sqrt{\frac{1}{2}} \left[ \sqrt{\frac{2}{3}} - \sqrt{\frac{1}{3}} \right] | -x \rangle\\ =\mathstrut \amp \left( \sqrt{\frac{1}{3}} + \sqrt{\frac{1}{6}}\right)| +x \rangle + \left( \sqrt{\frac{1}{3}} - \sqrt{\frac{1}{6}}\right)| -x \rangle \text{.} \end{align*}

Therefore, the probability that the electron is in the spin-down state of \(S_x\) is equal to the square of the coefficient of the \(| -x \rangle\) basis state

\begin{equation} \mbox{Prob} \bigl(\mbox{spin-down in \(S_x\)} \bigr) = \left|\sqrt{\frac{1}{3}} - \sqrt{\frac{1}{6}}\right|^2 \approx 0.029\text{.}\tag{5.25} \end{equation}

(c) The calculation here is very similar to that for part (b) except that we should write the states \(| \pm z \rangle\) in terms of the basis states \(| \pm y \rangle\text{.}\) Since we are interested in measuring a value of the \(y\)-component of spin, \(S_y\text{,}\) we need to express our quantum state as a superposition of the \(S_y\) basis states. We will leave this to you to finish, but you should obtain an answer of 0.5 for the probability.