12.4 Heisenberg uncertainty

12.4 Heisenberg uncertainty

We have seen that, describing the state of a free particle at a fixed time, one has $\delta$-function states corresponding to a well-defined position (in the position representation) or a well-defined momentum (in the momentum representation). But $Q$ and $P$ do not commute, and states with both well-defined position and well-defined momentum do not exist. An example of a state peaked at $q=0$ will be given by the Gaussian wavefunction

$$\psi (q)=e^{-\alpha \frac{q^2}{2}}$$

which becomes narrowly peaked for $\alpha$ large. By equation 11.6 the corresponding state in the momentum space representation is

$$\frac{1}{\sqrt{\alpha }}{e}^{-\frac{k^2}{2\alpha }}$$

which becomes uniformly spread out as $\alpha$ gets large. Similarly, as $\alpha$ goes to zero, one gets a state narrowly peaked at $k=0$ in momentum space, but uniformly spread out as a position space wavefunction.

For states with expectation value of $Q$ and P equal to zero, the width of the state in position space can be quantified by the expectation value of $Q^2$ and its width in momentum space by the expectation value of $P^2$ . One has the following theorem, which makes precise the limit on simultaneously localizability of a state in position and momentum space

Theorem (Heisenberg uncertainty)

Theorem (Heisenberg uncertainty).

$$\frac{⟨\psi |Q^2|\psi ⟩}{⟨\psi |\psi ⟩}\frac{⟨\psi |P^2|\psi ⟩}{⟨\psi |\psi ⟩}\ge \frac{1}{4}$$

Proof. For any real $\lambda$ one has

$$⟨(Q+i\lambda P)\psi |(Q+i\lambda P)\psi ⟩\ge 0$$

but, using self-adjointness of $Q$ and $P,$ as well as the relation $[Q,P]=i$ one has

$$\begin{array}{rl}⟨(Q+i\lambda P)\psi |(Q+i\lambda P)\psi ⟩& ={\lambda }^2⟨\psi |P^2\psi ⟩+i\lambda ⟨\psi |QP\psi ⟩-i\lambda ⟨\psi |PQ\psi ⟩+⟨\psi |Q^2\psi ⟩\\ & ={\lambda }^2⟨\psi |P^2\psi ⟩+\lambda (-⟨\psi |\psi ⟩)+⟨\psi |Q^2\psi ⟩\end{array}$$

This will be non-negative for all $\lambda$ if

$$⟨\psi |\psi {⟩}^2\le 4⟨\psi |P^2\psi ⟩⟨\psi |Q^2\psi ⟩$$

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