Canonical commutation relation

The canonical commutation relations used in quantum mechanics (QM) are:

${\displaystyle {\begin{aligned}\left[X_{i},X_{j}\right]&=0\quad i,j\in \{1,\ldots ,3\}\\\left[P_{i},P_{j}\right]&=0\\\left[X_{i},P_{j}\right]&=\mathrm {i} \hslash \delta _{i,j}\end{aligned}}}$

Designate here

• the ( Hermitian ) position operators (note: in general the operators in QM are marked with a "hat", this is omitted here for reasons of readability, so it applies, for example, to the position operator .) ${\displaystyle X}$ ${\displaystyle {\hat {x}}=x}$
• which are the (Hermitian) momentum operators from QM ${\displaystyle P}$
• the brackets around the operators, e.g. B. , the commutator ${\displaystyle [X_{i},X_{j}]}$
• ${\displaystyle \mathrm {i} }$the imaginary unit
• ${\displaystyle \hbar }$the reduced Planck quantum of action .

The position and momentum operators of different directions and “interchange” in pairs with one another, i. H. their commutator is zero. In practice, this means that these measurands (also called observables in QM ) can be measured simultaneously with any precision. ${\displaystyle i}$ ${\displaystyle j}$

If the commutator does not disappear, ie if it is not equal to zero, the associated operators do not “interchange”. The operators for position and momentum therefore represent an example of non-interchangeable operators . They describe quantities in the same quantum system that can not be measured at the same time as precisely as required, so their simultaneous measurement is subject to a certain degree of uncertainty . This leads directly to the famous uncertainty principle of Werner Heisenberg .

Derivation and justification

Since the product (i.e. the successive execution) of two linear operators is generally not commutative (i.e. the order of the successive execution cannot simply be reversed), the commutator (or commutation relation) of two linear operators and is defined as follows: ${\displaystyle A}$ ${\displaystyle B}$

${\displaystyle [A,B]=AB-BA}$

If we simply insert the operators for position and momentum into the above equation and let them act on a wave function, it follows: ${\displaystyle \psi (x)}$

${\displaystyle {\begin{aligned}\left[x,p_{x}\right]\psi (x)&=\left(x{\Big (}-\mathrm {i} \hbar {\frac {\partial }{\partial x}}{\Big )}\right)\psi (x)-\left({\Big (}-\mathrm {i} \hbar {\frac {\partial }{\partial x}}{\Big )}x\right)\psi (x)\\&=\left(-\mathrm {i} \hbar x{\frac {\partial }{\partial x}}+\mathrm {i} \hbar {\Big (}1+x{\frac {\partial }{\partial x}}{\Big )}\right)\psi (x)\\&=\mathrm {i} \hbar \psi (x)\end{aligned}}}$

The above calculation for the room components and lead to the same result. It is interesting that z. B. ${\displaystyle y,p_{y}}$ ${\displaystyle z,p_{z}}$

${\displaystyle {\begin{aligned}\left[x,p_{y}\right]\psi (x)&=\left(x{\Big (}-\mathrm {i} \hbar {\frac {\partial }{\partial y}}{\Big )}\right)\psi (x)-\left({\Big (}-\mathrm {i} \hbar {\frac {\partial }{\partial y}}{\Big )}x\right)\psi (x)\\&={\Big (}-\mathrm {i} \hbar x{\frac {\partial }{\partial y}}{\Big )}\psi (x)+{\Big (}\mathrm {i} \hbar {\frac {\partial }{\partial y}}{\Big )}{\Big (}x\psi (x){\Big )}\\&={\Big (}-\mathrm {i} \hbar x{\frac {\partial }{\partial y}}+\mathrm {i} \hbar x{\frac {\partial }{\partial y}}{\Big )}\psi (x)\\&=0\end{aligned}}}$

reversed. The proof that place and momentum components interchange with each other is simple. Overall, the above-mentioned canonical commutation relations then result.

literature

• Wolfgang Nolting : Basic course in theoretical physics. Volume 5/1: Quantum Mechanics Basics. Springer Verlag, 2009.
• G. Blatter: Quantum Mechanics I. Script for the lecture, ETH-Zurich, 2005.