2.3. Uncertainty relation

Here, we derive the following general relation:

(2.61)\[\Delta A \cdot \Delta B \ge \frac{1}{2} \left|\langle\psi\middle\vert[\hat{A},\hat{B}]\middle\vert\psi\rangle\right|\]

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

(2.62)\[\Delta A = \sqrt{\langle \hat{A}^2\rangle - \langle \hat{A}\rangle^2} \qquad \Delta B = \sqrt{\langle \hat{B}^2\rangle - \langle \hat{B}\rangle^2}\]

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

(2.63)\[\hat{a} = \hat{A} - \langle A \rangle \qquad \hat{b} = \hat{B} - \langle B \rangle\]

With these, the rmsds can be written as

(2.64)\[(\Delta A)^2 = \langle\hat{a}^2\rangle \qquad (\Delta B)^2 = \langle\hat{b}^2\rangle\]

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:

(2.65)\[\left\langle(x\hat{a}+\mathrm{i}\hat{b})\psi\middle|(x\hat{a}+\mathrm{i}\hat{b})\psi\right\rangle\ge 0\]

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

(2.66)\[x^2 \langle\hat{a}\psi|\hat{a}\psi\rangle +\mathrm{i}x \langle\hat{a}\psi|\hat{b}\psi\rangle -\mathrm{i}x \langle\hat{b}\psi|\hat{a}\psi\rangle + \langle\hat{b}\psi|\hat{b}\psi\rangle \ge 0\]

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

(2.67)\[x^2 \langle\psi|\hat{a}^2|\psi\rangle + \mathrm{i}x \langle\psi|(\hat{a}\hat{b}-\hat{b}\hat{a})|\psi\rangle + \langle\psi|\hat{b}^2|\psi\rangle\]

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

(2.68)\[x^2 \langle\psi|\hat{a}^2|\psi\rangle + x \langle\psi|\mathrm{i}\hat{c}|\psi\rangle + \langle\psi|\hat{b}^2|\psi\rangle \ge 0\]

or, in a more abbreviated form,

(2.69)\[x^2 \langle\hat{a}^2\rangle + x \langle\mathrm{i}\hat{c}\rangle + \langle\hat{b}^2\rangle \ge 0\]

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

(2.70)\[\langle\hat{a}^2\rangle \left( x + \frac{\langle\mathrm{i}\hat{c}\rangle}{2\langle\hat{a}^2\rangle} \right)^2 + \langle\hat{b}^2\rangle - \frac{\langle\mathrm{i}\hat{c}\rangle^2}{4\langle\hat{a}^2\rangle} \ge 0\]

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

(2.71)\[\langle\hat{b}^2\rangle - \frac{\langle\mathrm{i}\hat{c}\rangle^2}{4\langle\hat{a}^2\rangle} \ge 0\]

Moving the second term to the right-hand side and multiplying by \(\langle\hat{a}^2\rangle\) gives

(2.72)\[\langle\hat{a}^2\rangle \cdot \langle\hat{b}^2\rangle \ge \frac{1}{4} \langle\mathrm{i}\hat{c}\rangle^2 = \frac{1}{4} \langle\hat{c}\rangle^2\]

Taking the square root and inserting all the definitions gives

(2.73)\[\Delta A \cdot \Delta B \ge \frac{1}{2} \left|\langle\psi\middle\vert[\hat{A},\hat{B}]\middle\vert\psi\rangle\right|\]

which is the general relation we set out to prove.