1.5 Quantum Uncertainty

The most famous appearance of uncertainty in quantum mechanics is the Heisenberg uncertainty principle between position and momentum.

\Delta x\Delta p\geq\frac{\hbar}{2}.

(1.48)

This means that, in contrast to classical mechanics, there are no possible states a particle can be in where it has both definite momentum and position. This has dramatic consequences, such as the loss of a well-defined trajectory for quantum particles.

It also implies that particles confined to finite spaces must have nonzero kinetic energy.

Mathematically, the quantum uncertainty of any observable is defined as the standard deviation of the probability distribution prior to measurement. If we have a particle in an eigenstate of a given observable and we measure that observable, we know which eigenvalue we will get back and so there is no uncertainty. For any superposition state, there is nonzero uncertainty. This represents the fact that the outcome of a measurement is truly random, quantum uncertainty does not represent our lack of knowledge of a definite state.

The uncertainty in position is

\Delta x=\sqrt{\langle\hat{x}^{2}\rangle-\langle\hat{x}\rangle^{2}}.

(1.49)

One way to remember which way round the two terms in the square root are is by noting that uncertainty must be real. The first term, which is an average of squares i.e. things that are $\geq 0$ , must therefore be greater than the second term, which is an average of things that can be positive or negative.

Does this formula generalise to all operators? Suppose $\hat{O}$ is an operator with a discrete set of measurement outcomes $O_{i}$ with probabilities $P(O_{i})$ . Then the standard deviation is defined as

$\displaystyle\Delta O$	$\displaystyle=\sqrt{\sum_{i}(O_{i}-\langle\hat{O}\rangle)^{2}P(O_{i})}$	(1.50)
	$\displaystyle=\sqrt{\sum_{i}O_{i}^{2}P(O_{i})-2\langle\hat{O}\rangle\sum_{i}O_% {i}P(O_{i})+\langle\hat{O}\rangle^{2}\sum_{i}P(O_{i})}$	(1.51)
	$\displaystyle=\sqrt{\langle\hat{O}^{2}\rangle-2\langle\hat{O}\rangle^{2}+% \langle\hat{O}\rangle^{2}}$	(1.52)
	$\displaystyle=\sqrt{\langle\hat{O}^{2}\rangle-\langle\hat{O}\rangle^{2}},$	(1.53)

so the relation holds.

Definition 1.2.

For an observable operator $\hat{O}$ , the uncertainty is given by

\Delta O=\sqrt{\langle\hat{O}^{2}\rangle-\langle\hat{O}\rangle^{2}}.

(1.54)

Example 1.1.

Suppose we are going to measure the momentum of a particle, which has probability $0.5$ to have momentum $+p_{0}$ , and probability $0.5$ to have momentum $-p_{0}$ . What is the uncertainty in the measurement?

Using equation 1.54, we have

\Delta p=\sqrt{\langle\hat{p}^{2}\rangle-\langle\hat{p}\rangle^{2}}.

(1.55)

We need to calculate both terms inside the square root.

Since the two possible measurement outcomes are equidistant from zero with equal probability, we expect the average value of measurement to be zero. We can verify this by calculating:

\langle\hat{p}\rangle=p_{0}\times 0.5-p_{0}\times 0.5=0.

(1.56)

On the other hand, the average value of squared momentum is

\langle\hat{p}^{2}\rangle=p_{0}^{2}\times 0.5+(-p_{0})^{2}\times 0.5=p_{0}^{2}.

(1.57)

Then the uncertainty is

\Delta p=\sqrt{p_{0}^{2}-0}=p_{0}.

(1.58)