2.3. Uncertainty relation

Here, we derive the following general relation:

(2.59)$$\mathrm{\Delta }A\cdot \mathrm{\Delta }B\ge \frac{1}{2}|⟨\psi |[\hat{A},\hat{B}]|\psi ⟩|$$

This is called the Robertson-Schrödinger uncertainty relation and is valid for any Hermitian operators $\hat{A}$ and $\hat{B}$ and any normalized wavefunction $\psi$. On the left-hand side, we have a product of two root-mean-square deviations (rmsds), which are defined as

(2.60)$$\mathrm{\Delta }A=\sqrt{⟨{\hat{A}}^2⟩-⟨\hat{A}{⟩}^2}\mathrm{\Delta }B=\sqrt{⟨{\hat{B}}^2⟩-⟨\hat{B}{⟩}^2}$$

where the wavefunction $\psi$ is implicit (i.e. $⟨\hat{A}⟩=⟨\psi |\hat{A}|\psi ⟩$ etc.). Let’s define operators that represent the derivations from the expectation values:

(2.61)$$\hat{a}=\hat{A}-⟨A⟩\hat{b}=\hat{B}-⟨B⟩$$

With these, the rmsds can be written as

(2.62)$$(\mathrm{\Delta }A{)}^2=⟨{\hat{a}}^2⟩(\mathrm{\Delta }B{)}^2=⟨{\hat{b}}^2⟩$$

To start the proof, we construct the operator $x\hat{a}+\mathrm{i}\hat{b}$ with arbitrary real-valued $x$, apply it to a general wavefunction $\psi$, and calculate the normalization integral of the result. This is non-negative:

(2.63)$$⟨(x\hat{a}+\mathrm{i}\hat{b})\psi |(x\hat{a}+\mathrm{i}\hat{b})\psi ⟩\ge 0$$

Multiplying out (and properly taking the complex-conjugates) gives

(2.64)$$x^2⟨\hat{a}\psi |\hat{a}\psi ⟩+\mathrm{i}x⟨\hat{a}\psi |\hat{b}\psi ⟩-\mathrm{i}x⟨\hat{b}\psi |\hat{a}\psi ⟩+⟨\hat{b}\psi |\hat{b}\psi ⟩\ge 0$$

Since $\hat{a}$ and $\hat{b}$ are Hermitian operators, this can be rewritten as

(2.65)$$x^2⟨\psi |{\hat{a}}^2|\psi ⟩+\mathrm{i}x⟨\psi |(\hat{a}\hat{b}-\hat{b}\hat{a})|\psi ⟩+⟨\psi |{\hat{b}}^2|\psi ⟩$$

The operator in the middle term is a commutator and we can abbreviate it as $\hat{c}=[\hat{a},\hat{b}]$, so that we get

(2.66)$$x^2⟨\psi |{\hat{a}}^2|\psi ⟩+x⟨\psi |\mathrm{i}\hat{c}|\psi ⟩+⟨\psi |{\hat{b}}^2|\psi ⟩\ge 0$$

or, in a more abbreviated form,

(2.67)$$x^2⟨{\hat{a}}^2⟩+x⟨\mathrm{i}\hat{c}⟩+⟨{\hat{b}}^2⟩\ge 0$$

The left-hand side is a quadratic expression in $x$. We can rewrite it by completing the square:

(2.68)$$⟨{\hat{a}}^2⟩{(x+\frac{⟨\mathrm{i}\hat{c}⟩}{2⟨{\hat{a}}^2⟩})}^2+⟨{\hat{b}}^2⟩-\frac{⟨\mathrm{i}\hat{c}{⟩}^2}{4⟨{\hat{a}}^2⟩}\ge 0$$

Since the first of the three terms contains only squares, it is never negative, and it is zero only for $x=-⟨\mathrm{i}\hat{c}⟩/2⟨{\hat{a}}^2⟩$. The inequality must be satisfied for this $x$ as well, so we can write

(2.69)$$⟨{\hat{b}}^2⟩-\frac{⟨\mathrm{i}\hat{c}{⟩}^2}{4⟨{\hat{a}}^2⟩}\ge 0$$

Moving the second term to the right-hand side and multiplying by $⟨{\hat{a}}^2⟩$ gives

(2.70)$$⟨{\hat{a}}^2⟩\cdot ⟨{\hat{b}}^2⟩\ge \frac{1}{4}⟨\mathrm{i}\hat{c}{⟩}^2=\frac{1}{4}⟨\hat{c}{⟩}^2$$

Taking the square root and inserting all the definitions gives

(2.71)$$\mathrm{\Delta }A\cdot \mathrm{\Delta }B\ge \frac{1}{2}|⟨\psi |[\hat{A},\hat{B}]|\psi ⟩|$$

which is the general relation we set out to prove.