Commutability and precision of measurement
We have seen previously that operators may not always commute (i.e., \([A, B] \ne 0\)). An example of such operator pair is position \(\hat{x}\) and momentum \(\hat{p}_x\):
\[{\hat{p}_x\hat{x}\psi(x) = \hat{p}_x\left(x\psi(x)\right) = \left(\frac{\hbar}{i}\frac{d}{dx}\right)\left(x\psi(x)\right) = \frac{\hbar x}{i}\frac{d\psi(x)}{dx} + \frac{\hbar}{i}\psi(x)}\]
\[{\hat{x}\hat{p}_x\psi(x) = x\left(\frac{\hbar}{i}\frac{d\psi(x)}{dx}\right)}\]
\[{\Rightarrow \left[\hat{p}_x,\hat{x}\right]\psi(x) = \left(\hat{p}_x\hat{x} - \hat{x}\hat{p}_x\right)\psi(x) = \frac{\hbar}{i}\psi(x)}\]
\[{\Rightarrow \left[\hat{p}_x,\hat{x}\right] = \frac{\hbar}{i}}\]
In contrast, the kinetic energy operator and the momentum operators commute:
\[{\left[\hat{T},\hat{p}_x\right] = \left[\frac{\hat{p}_x^2}{2m},\hat{p}_x\right] = \frac{p_x^3}{2m} - \frac{p_x^3}{2m} = 0}\]
We had the uncertainty principle for the position and momentum operators:
\[\Delta x\Delta p_x \ge \frac{\hbar}{2}\]
In general, it turns out that for operators \(\hat{A}\) and \(\hat{B}\) that do not commute, the uncertainty principle applies in the following form:
\[{\Delta A\Delta B \ge \frac{1}{2}\left|\left<\left[\hat{A},\hat{B}\right]\right>\right|}\]
Example obtain the position/momentum uncertainty principle
Solution Denote \(\hat{A} = \hat{x}\) and \(\hat{B} = \hat{p}_x\).
\[\frac{1}{2}\left|\left<\left[\hat{A},\hat{B}\right]\right>\right| = \frac{1}{2}\left|\left<\left[\hat{x},\hat{p}_x\right]\right>\right| = \frac{1}{2}\left|\left<\frac{\hbar}{i}\right>\right|
= \frac{1}{2}\left|\left<\psi\left|\frac{\hbar}{i}\right|\psi\right>\right| = \frac{1}{2}\left|\frac{\hbar}{i}\underbrace{\left<\psi\left|\psi\right.\right>}_{=1}\right| = \frac{\hbar}{2}\]
\[\Rightarrow \Delta x\Delta p_x \ge \frac{\hbar}{2}\]
Example Show that if all eigenfunctions of operators \(\hat{A}\) and \(\hat{B}\) are identical, \(\hat{A}\) and \(\hat{B}\) commute with each other.
Solution Denote the eigenvalues of \(\hat{A}\) and \(\hat{B}\) by \(a_i\) and \(b_i\) and the common eigenfunctions by \(\psi_i\). For both operators we have then:
\[\hat{A}\psi_i = a_i\psi_i\textnormal{ and }\hat{B}\psi_i = b_i\psi_i\]
By using these two equations and expressing the general wavefunction \(\psi\) as a linear combination of the eigenfunctions, the commutator can be evaluated as:
\[\hat{A}\hat{B}\psi = \hat{A}\left(\hat{B}\psi\right) = \hat{A}\overbrace{\left(\hat{B}\sum\limits_{i=1}^{\infty}c_i\psi_i\right)}^{\textnormal{complete basis}}
= \hat{A}\overbrace{\left(\sum\limits_{i=1}^{\infty}c_i\hat{B}\psi_i\right)}^{\hat{B}\textnormal{ linear}} = \hat{A}\overbrace{\left(\sum\limits_{i=1}^{\infty}c_ib_i\psi_i\right)}^{\textnormal{eigenfunction of }\hat{B}}\]
\[= \overbrace{\sum\limits_{i=1}^{\infty}c_ib_i\hat{A}\psi_i}^{\hat{A}\textnormal{ linear}} = \overbrace{\sum\limits_{i=1}^{\infty}c_ib_ia_i\psi_i}^{\textnormal{eigenfunction of }\hat{A}} = \overbrace{\sum\limits_{i=1}^{\infty}c_ia_ib_i\psi_i}^{a_i\textnormal{ and }b_i\textnormal{ are constants}} = \sum\limits_{i=1}^{\infty}c_ia_i\hat{B}\psi_i\]
\[= \hat{B}\sum\limits_{i=1}^{\infty}c_ia_i\psi_i = \hat{B}\sum\limits_{i=1}^{\infty}c_i\hat{A}\psi_i = \hat{B}\hat{A}\sum\limits_{i=1}^{\infty}c_i\psi_i = \hat{B}\hat{A}\psi\]
\[\Rightarrow \left[\hat{A},\hat{B}\right] = 0\]
Note that the commutation relation must apply to all well-behaved functions and not just for some given subset of functions!
