11.3.2: Uncertainty in Quantum Mechanics

In order to bring this into quantum mechanics, we already know how to calculate the average \(\langle a\rangle\), which we call the “expectation value”. If the state of the system is \(|\psi\rangle\) and the operator corresponding to the observable \(a\) is \(\hat{A}\), then

\[\langle a\rangle=\langle\psi|\hat{A}| \psi\rangle\tag{11.20}\]

Similarly, now that we recognize that we can interpret \(\hat{A}^{2}\) as just applying the operator \(\hat{A}\) twice, we can calculate \(\left\langle a^{2}\right\rangle\):

\[\left\langle a^{2}\right\rangle=\left\langle\psi\left|\hat{A}^{2}\right| \psi\right\rangle\tag{11.21}\]

For example, let’s consider the state \(|\psi\rangle=|+z\rangle\) and the observable spin-\(z\). We expect the uncertainty here to be zero, because we know exactly what we’ll get if we measure spin-\(z\). Let’s see if it works out that way:

\[\begin{aligned}
\left\langle s_{z}\right\rangle &=\left\langle\psi\left|\hat{S}_{z}\right| \psi\right\rangle \\
&=\frac{\hbar}{2}\left[\begin{array}{ll}
1 & 0
\end{array}\right]\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{l}
1 \\
0
\end{array}\right] \\
&=\frac{\hbar}{2}\left[\begin{array}{ll}
1 & 0
\end{array}\right]\left[\begin{array}{l}
1 \\
0
\end{array}\right] \\
&=\frac{\hbar}{2}
\end{aligned}\tag{11.22}\]

As expected, the expectation value for spin-\(z\) is \(+\hbar / 2\). For the other part:

\[\begin{aligned}
\left\langle s_{z}^{2}\right\rangle &=\left\langle+z\left|\hat{S}_{z} \hat{S}_{z}\right|-z\right\rangle \\
&=\frac{\hbar^{2}}{4}\left[\begin{array}{ll}
1 & 0
\end{array}\right]\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{l}
1 \\
0
\end{array}\right] \\
&=\frac{\hbar^{2}}{4}\left[\begin{array}{ll}
1 & 0
\end{array}\right]\left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]\left[\begin{array}{l}
1 \\
0
\end{array}\right] \\
&=\frac{\hbar^{2}}{4}\left[\begin{array}{ll}
1 & 0
\end{array}\right]\left[\begin{array}{l}
1 \\
0
\end{array}\right] \\
&=\frac{\hbar^{2}}{4}
\end{aligned}\tag{11.23}\]

If we take the difference \(\left\langle s_{z}{ }^{2}\right\rangle-\left\langle s_{z}\right\rangle^{2}\), we get \(\hbar^{2} / 4-\hbar^{2} / 4=0\), as expected.

What if we want to know the uncertainty on \(S_{x}\) for this state?

\[\begin{aligned}
\left\langle s_{x}\right\rangle &=\left\langle+z\left|\hat{S}_{x}\right|+z\right\rangle \\
&=\frac{\hbar}{2}\left[\begin{array}{ll}
1 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{l}
1 \\
0
\end{array}\right] \\
&=\frac{\hbar}{2}\left[\begin{array}{ll}
1 & 0
\end{array}\right]\left[\begin{array}{l}
0 \\
1
\end{array}\right] \\
&=0
\end{aligned}\tag{11.24}\]

If the system is in the state \(|+z\rangle\), we know that we have a 50% chance each for finding spin-\(x\) to be \(+\hbar / 2\) or \(-\hbar / 2\). Thus, it’s no surprise that the average value of spin-\(x\) is zero, even though zero isn’t a value we might measure. To figure out the variance:

\[\begin{aligned}
\left\langle s_{x}^{2}\right\rangle &=\left\langle+z\left|\hat{S}_{x} \hat{S}_{x}\right|+z\right\rangle \\
&=\frac{\hbar^{2}}{4}\left[\begin{array}{ll}
1 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{l}
1 \\
0
\end{array}\right] \\
&=\frac{\hbar^{2}}{4}\left[\begin{array}{ll}
1 & 0
\end{array}\right]\left[\begin{array}{ll}
0 & 1 \\
1 & 0
\end{array}\right]\left[\begin{array}{l}
0 \\
1
\end{array}\right] \\
&=\frac{\hbar^{2}}{4}\left[\begin{array}{ll}
1 & 0
\end{array}\right]\left[\begin{array}{l}
1 \\
0
\end{array}\right] \\
&=\frac{\hbar^{2}}{4}
\end{aligned}\tag{11.25}\]

Thus, in this case, the formal uncertainty \(\Delta s_{x}\) on the \(x\)-spin is \(\hbar / 2\).