The uncertainty relation

Intuition

Can a state be sharp for two observables at once? If they commute, yes. If they do not, their spreads are tied together: the product of the two uncertainties can be no smaller than half the size of the average of their commutator. It follows from nothing more than the Cauchy–Schwarz inequality of the first chapter, and for position and momentum it becomes Heisenberg’s famous limit.

A note played for a very short time has no definite pitch: the shorter the burst, the wider the spread of frequencies in it. Two quantities that do not commute are like duration and pitch — squeezing one widens the other.

For the real states $\cos\tfrac{\theta}{2}\ket{0}+\sin\tfrac{\theta}{2}\ket{1}$ , the spreads of $Z$ and $X$ against $\theta$ . Where one is zero the other is 1: the state cannot be sharp for both, because $X$ and $Z$ do not commute.

The Robertson uncertainty relation

For any two observables and any state, the product of their uncertainties is bounded below by half the modulus of the expectation value of their commutator. The bound depends on the state, and it is zero when the observables commute.

\Delta A\,\Delta B\ \ge\ \tfrac12\bigl|\langle[\hat{A},\hat{B}]\rangle\bigr|

What it says and what it does not

For $X$ and $Z$ , $[X,Z]=-2iY$ with $Y = (\begin{array}{cc} 0 & - \end{array}$ , so .

\Delta A\,\Delta B\ \ge\ \tfrac12\bigl|\langle[\hat{A},\hat{B}]\rangle\bigr|

Build two vectors from the state: the deviations of $\hat{A}$ and of $\hat{B}$ acting on it. Their lengths are the two uncertainties. Cauchy–Schwarz bounds their overlap by the product of the lengths, and the imaginary part of the overlap turns out to be half the average commutator.

Proof steps

\ket{f}=(\hat{A}-\langle\hat{A}\rangle)\ket{\psi},\quad \ket{g}=(\hat{B}-\langle\hat{B}\rangle)\ket{\psi}

Two deviation vectors; as in the last lesson, their lengths are $\Delta A$ and $\Delta B$ .

(\Delta A)^{2}(\Delta B)^{2}\ge|\braket{f|g}|^{2}

The Cauchy–Schwarz inequality for $\ket{f}$ and $\ket{g}$ .

|\braket{f|g}|^{2}\ge\Bigl(\frac{\braket{f|g}-\braket{g|f}}{2i}\Bigr)^{2}

The modulus squared of a number is at least the square of its imaginary part, and $\braket{g|f}=\braket{f|g}^{*}$ .

\braket{f|g}-\braket{g|f}=\langle[\hat{A},\hat{B}]\rangle

Multiplying out, the constant shifts cancel between the two orders and only the commutator survives.

\Delta A\,\Delta B\ge\tfrac12\bigl|\langle[\hat{A},\hat{B}]\rangle\bigr|

Put the pieces together and take square roots.

Applications

Practice

The Bound Uses the Commutator

The product of two uncertainties is at least half the size of the average commutator. For $X$ and $Z$ the commutator is $-2iY$ .

\Delta X\,\Delta Z\ge\tfrac12\bigl|\langle[X,Z]\rangle\bigr|=|\langle Y\rangle|

Try it

In a state with $\langle Y\rangle=0.6$ , where $Y=\begin{pmatrix}0&-i\\ i&0\end{pmatrix}$ , what is the smallest $\Delta X\,\Delta Z$ can be?