as many times as we choose, and in any order, the average value of the result will be zero. Or, if we apply the same operators to ${\displaystyle u}$, and finally take the average value, it will be the same as the value obtained by writing the sign of average separately as ${\displaystyle u}$, and on ${\displaystyle \epsilon }$, ${\displaystyle A_{1}}$, ${\displaystyle A_{2}}$, etc., in all the operators. If ${\displaystyle u}$ is independent of the momenta, formulae similar to the preceding, but having ${\displaystyle \epsilon _{q}}$ in place of ${\displaystyle \epsilon }$, may be derived from equation (179).