2.3. Uncertainty relation
Here, we derive the following general relation:
(2.59)\[\Delta A \cdot \Delta B
\ge
\frac{1}{2}
\left|\langle\psi\middle\vert[\hat{A},\hat{B}]\middle\vert\psi\rangle\right|\]
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
(2.60)\[\Delta A = \sqrt{\langle \hat{A}^2\rangle - \langle \hat{A}\rangle^2}
\qquad
\Delta B = \sqrt{\langle \hat{B}^2\rangle - \langle \hat{B}\rangle^2}\]
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:
(2.61)\[\hat{a} = \hat{A} - \langle A \rangle
\qquad
\hat{b} = \hat{B} - \langle B \rangle\]
With these, the rmsds can be written as
(2.62)\[(\Delta A)^2 = \langle\hat{a}^2\rangle
\qquad
(\Delta B)^2 = \langle\hat{b}^2\rangle\]
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:
(2.63)\[\left\langle(x\hat{a}+\mathrm{i}\hat{b})\psi\middle|(x\hat{a}+\mathrm{i}\hat{b})\psi\right\rangle\ge 0\]
Multiplying out (and properly taking the complex-conjugates) gives
(2.64)\[x^2
\langle\hat{a}\psi|\hat{a}\psi\rangle
+\mathrm{i}x
\langle\hat{a}\psi|\hat{b}\psi\rangle
-\mathrm{i}x
\langle\hat{b}\psi|\hat{a}\psi\rangle
+
\langle\hat{b}\psi|\hat{b}\psi\rangle
\ge 0\]
Since \(\hat{a}\) and \(\hat{b}\) are Hermitian operators, this can be rewritten as
(2.65)\[x^2
\langle\psi|\hat{a}^2|\psi\rangle
+
\mathrm{i}x
\langle\psi|(\hat{a}\hat{b}-\hat{b}\hat{a})|\psi\rangle
+
\langle\psi|\hat{b}^2|\psi\rangle\]
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
(2.66)\[x^2
\langle\psi|\hat{a}^2|\psi\rangle
+
x
\langle\psi|\mathrm{i}\hat{c}|\psi\rangle
+
\langle\psi|\hat{b}^2|\psi\rangle
\ge
0\]
or, in a more abbreviated form,
(2.67)\[x^2
\langle\hat{a}^2\rangle
+
x
\langle\mathrm{i}\hat{c}\rangle
+
\langle\hat{b}^2\rangle
\ge
0\]
The left-hand side is a quadratic expression in \(x\). We can rewrite it by completing the square:
(2.68)\[\langle\hat{a}^2\rangle
\left(
x
+
\frac{\langle\mathrm{i}\hat{c}\rangle}{2\langle\hat{a}^2\rangle}
\right)^2
+
\langle\hat{b}^2\rangle
-
\frac{\langle\mathrm{i}\hat{c}\rangle^2}{4\langle\hat{a}^2\rangle}
\ge
0\]
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
(2.69)\[\langle\hat{b}^2\rangle
-
\frac{\langle\mathrm{i}\hat{c}\rangle^2}{4\langle\hat{a}^2\rangle}
\ge
0\]
Moving the second term to the right-hand side and multiplying by \(\langle\hat{a}^2\rangle\) gives
(2.70)\[\langle\hat{a}^2\rangle
\cdot
\langle\hat{b}^2\rangle
\ge
\frac{1}{4}
\langle\mathrm{i}\hat{c}\rangle^2
=
\frac{1}{4}
\langle\hat{c}\rangle^2\]
Taking the square root and inserting all the definitions gives
(2.71)\[\Delta A \cdot \Delta B
\ge
\frac{1}{2}
\left|\langle\psi\middle\vert[\hat{A},\hat{B}]\middle\vert\psi\rangle\right|\]
which is the general relation we set out to prove.