Talk:Normal order

Normal Ordered Function
If F(x) = (x^2-1)/(x-1) then f(x) = x+1 would be a normal function of F(x) on the real numbers for given any epsilon neighborhood around c in the real numbers including c = 0; f(c) equals F(c). —The preceding unsigned comment was added by 72.71.209.132 (talk • contribs) 16:25, 2 June 2006.


 * I've never seen the normal ordering concept applied to functions like that. Do you have more examples? --HappyCamper 15:57, 4 September 2006 (UTC)


 * This is not normal ordening, this is just removing a singular point. Greets, David [[image:Da Vit in Chu Nôm.jpg|23px|Da Vit]] 13:28, 4 December 2006 (UTC)

Error in example 4 of 2.1.1
Example 4 of the single-boson section 2.1.1 seems incorrect. If the normal ordering is removed from the left hand side of the equation in that example what remains is an operator identity, i.e. it is true that
 * $$ \exp (\lambda \hat{a}^\dagger \hat{a}) \,= \sum^\infty_{n=0} \frac{(e^\lambda -1)^n}{n!} \hat{a}^{\dagger n} \hat{a}^n$$

I guess the example which is meant is
 * $$ \exp (\lambda \hat{a}^\dagger \hat{a}) \,= : \exp((e^\lambda -1)\hat{a}^\dagger \hat{a}) : $$

which follows from the above. KoreanApple (talk) 03:40, 5 September 2008 (UTC)


 * I have corrected the above example. The corrected equation follows directly from expanding the exponential and normal ordering each term. (I decided it was best not to include the above operator identity as it is not related to normal-ordering and is probably just confusing in this context.) KoreanApple (talk) 06:48, 10 September 2008 (UTC)

Your corrected example is incorrect and I've removed it - normal-ordering is not linear, so you can't normal-order the terms in the power series individually.Ted.tem.parker (talk) 07:21, 15 July 2017 (UTC)