# 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 $$\hbar / 2$$.

This page titled 11.3.2: Uncertainty in Quantum Mechanics is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pieter Kok via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.