4 replaced http://physics.stackexchange.com/ with https://physics.stackexchange.com/ edited Apr 13 '17 at 12:39 I) It is true that operator ordering procedures are idempotent operations $$\tag{1} T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots).$$ But it is not true that the outermost ordering cancels the effect of the innermost ordering $$\tag{2} T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \quad (\longleftarrow \text{Both Wrong!})$$ In fact, the opposite is true $$\tag{3} T(N(\ldots))~=~N(\ldots)\quad\text{and}\quad N(T(\ldots))~=~T(\ldots),$$ as a special case of a nested$${}^1$$ Wick's Theorem, cf. Section II below. Example. If for two operators $$a$$ and $$b$$, we have the relation $$\tag{4} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$ where the contraction $$\langle ab\rangle$$ is a $$c$$-number, then $$T(N(ab))~\stackrel{(4)}{=}~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~\stackrel{\text{linearity}}{=}~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$ $$\tag{5} ~\stackrel{(1)}{=}~T(ab)- \langle ab\rangle {\bf 1} ~\stackrel{(4)}{=}~N(ab)~\neq~ T(ab) .$$ II) More generally, if we want to bring a nested expression of the form $$\tag{6} T\left(N(\ldots) \ldots N(\ldots)\right)$$ on normal ordered form, there is a nested$${}^1$$ Wick's Theorem, which states that we should only include contractions between pairs of operators who belong to different normal order symbols. Example. In OP's case (5), this means for the lhs. that $$\tag{7} T(N(ab))~=~N(ab),$$ since $$a$$ and $$b$$ belong to the same normal order symbol $$N(ab)$$, while the rhs. is $$\tag{8} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1}.$$ [Note that $$N(a)=a$$ and $$N(b)=b$$.] See also e.g. thisthis and thisthis Phys.SE posts. -- $${}^1$$ A nested Wick's Theorem (between radial order and normal order) is briefly stated on p. 39 in J. Polchinski, String Theory, Vol. 1. Beware that radial order is often only implicitly written in CFT texts. I) It is true that operator ordering procedures are idempotent operations $$\tag{1} T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots).$$ But it is not true that the outermost ordering cancels the effect of the innermost ordering $$\tag{2} T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \quad (\longleftarrow \text{Both Wrong!})$$ In fact, the opposite is true $$\tag{3} T(N(\ldots))~=~N(\ldots)\quad\text{and}\quad N(T(\ldots))~=~T(\ldots),$$ as a special case of a nested$${}^1$$ Wick's Theorem, cf. Section II below. Example. If for two operators $$a$$ and $$b$$, we have the relation $$\tag{4} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$ where the contraction $$\langle ab\rangle$$ is a $$c$$-number, then $$T(N(ab))~\stackrel{(4)}{=}~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~\stackrel{\text{linearity}}{=}~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$ $$\tag{5} ~\stackrel{(1)}{=}~T(ab)- \langle ab\rangle {\bf 1} ~\stackrel{(4)}{=}~N(ab)~\neq~ T(ab) .$$ II) More generally, if we want to bring a nested expression of the form $$\tag{6} T\left(N(\ldots) \ldots N(\ldots)\right)$$ on normal ordered form, there is a nested$${}^1$$ Wick's Theorem, which states that we should only include contractions between pairs of operators who belong to different normal order symbols. Example. In OP's case (5), this means for the lhs. that $$\tag{7} T(N(ab))~=~N(ab),$$ since $$a$$ and $$b$$ belong to the same normal order symbol $$N(ab)$$, while the rhs. is $$\tag{8} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1}.$$ [Note that $$N(a)=a$$ and $$N(b)=b$$.] See also e.g. this and this Phys.SE posts. -- $${}^1$$ A nested Wick's Theorem (between radial order and normal order) is briefly stated on p. 39 in J. Polchinski, String Theory, Vol. 1. Beware that radial order is often only implicitly written in CFT texts. I) It is true that operator ordering procedures are idempotent operations $$\tag{1} T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots).$$ But it is not true that the outermost ordering cancels the effect of the innermost ordering $$\tag{2} T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \quad (\longleftarrow \text{Both Wrong!})$$ In fact, the opposite is true $$\tag{3} T(N(\ldots))~=~N(\ldots)\quad\text{and}\quad N(T(\ldots))~=~T(\ldots),$$ as a special case of a nested$${}^1$$ Wick's Theorem, cf. Section II below. Example. If for two operators $$a$$ and $$b$$, we have the relation $$\tag{4} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$ where the contraction $$\langle ab\rangle$$ is a $$c$$-number, then $$T(N(ab))~\stackrel{(4)}{=}~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~\stackrel{\text{linearity}}{=}~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$ $$\tag{5} ~\stackrel{(1)}{=}~T(ab)- \langle ab\rangle {\bf 1} ~\stackrel{(4)}{=}~N(ab)~\neq~ T(ab) .$$ II) More generally, if we want to bring a nested expression of the form $$\tag{6} T\left(N(\ldots) \ldots N(\ldots)\right)$$ on normal ordered form, there is a nested$${}^1$$ Wick's Theorem, which states that we should only include contractions between pairs of operators who belong to different normal order symbols. Example. In OP's case (5), this means for the lhs. that $$\tag{7} T(N(ab))~=~N(ab),$$ since $$a$$ and $$b$$ belong to the same normal order symbol $$N(ab)$$, while the rhs. is $$\tag{8} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1}.$$ [Note that $$N(a)=a$$ and $$N(b)=b$$.] See also e.g. this and this Phys.SE posts. -- $${}^1$$ A nested Wick's Theorem (between radial order and normal order) is briefly stated on p. 39 in J. Polchinski, String Theory, Vol. 1. Beware that radial order is often only implicitly written in CFT texts. 3 Corrections to the answer (v2): 1. The word _linarity_ in the first eq. (5) should be _linearity._ 2. Second eq. (5) should be eq. (6). edited Feb 2 '17 at 23:58 Qmechanic♦ 114k1313 gold badges225225 silver badges13521352 bronze badges I) It is true that operator ordering procedures are idempotent operations $$\tag{1} T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots).$$ But it is not true that the outermost ordering cancels the effect of the innermost ordering $$\tag{2} T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \quad (\longleftarrow \text{Both Wrong!})$$ In fact, the opposite is true $$\tag{3} T(N(\ldots))~=~N(\ldots)\quad\text{and}\quad N(T(\ldots))~=~T(\ldots),$$ as a special case of a nested$${}^1$$ Wick's Theorem, cf. Section II below. Example. If for two operators $$a$$ and $$b$$, we have the relation $$\tag{4} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$ where the contraction $$\langle ab\rangle$$ is a $$c$$-number, then $$T(N(ab))~\stackrel{(4)}{=}~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~\stackrel{\text{linarity}}{=}~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$$$T(N(ab))~\stackrel{(4)}{=}~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~\stackrel{\text{linearity}}{=}~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$ $$\tag{5} ~\stackrel{(1)}{=}~T(ab)- \langle ab\rangle {\bf 1} ~\stackrel{(4)}{=}~N(ab)~\neq~ T(ab) .$$ II) More generally, if we want to bring a nested expression of the form $$\tag{5} T\left(N(\ldots) \ldots N(\ldots)\right)$$$$\tag{6} T\left(N(\ldots) \ldots N(\ldots)\right)$$ on normal ordered form, there is a nested$${}^1$$ Wick's Theorem, which states that we should only include contractions between pairs of operators who belong to different normal order symbols. Example. In OP's case (5), this means for the lhs. that $$\tag{7} T(N(ab))~=~N(ab),$$ since $$a$$ and $$b$$ belong to the same normal order symbol $$N(ab)$$, while the rhs. is $$\tag{8} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1}.$$ [Note that $$N(a)=a$$ and $$N(b)=b$$.] See also e.g. this and this Phys.SE posts. -- $${}^1$$ A nested Wick's Theorem (between radial order and normal order) is briefly stated on p. 39 in J. Polchinski, String Theory, Vol. 1. Beware that radial order is often only implicitly written in CFT texts. I) It is true that operator ordering procedures are idempotent operations $$\tag{1} T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots).$$ But it is not true that the outermost ordering cancels the effect of the innermost ordering $$\tag{2} T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \quad (\longleftarrow \text{Both Wrong!})$$ In fact, the opposite is true $$\tag{3} T(N(\ldots))~=~N(\ldots)\quad\text{and}\quad N(T(\ldots))~=~T(\ldots),$$ as a special case of a nested$${}^1$$ Wick's Theorem, cf. Section II below. Example. If for two operators $$a$$ and $$b$$, we have the relation $$\tag{4} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$ where the contraction $$\langle ab\rangle$$ is a $$c$$-number, then $$T(N(ab))~\stackrel{(4)}{=}~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~\stackrel{\text{linarity}}{=}~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$ $$\tag{5} ~\stackrel{(1)}{=}~T(ab)- \langle ab\rangle {\bf 1} ~\stackrel{(4)}{=}~N(ab)~\neq~ T(ab) .$$ II) More generally, if we want to bring a nested expression of the form $$\tag{5} T\left(N(\ldots) \ldots N(\ldots)\right)$$ on normal ordered form, there is a nested$${}^1$$ Wick's Theorem, which states that we should only include contractions between pairs of operators who belong to different normal order symbols. Example. In OP's case (5), this means for the lhs. that $$\tag{7} T(N(ab))~=~N(ab),$$ since $$a$$ and $$b$$ belong to the same normal order symbol $$N(ab)$$, while the rhs. is $$\tag{8} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1}.$$ [Note that $$N(a)=a$$ and $$N(b)=b$$.] See also e.g. this and this Phys.SE posts. -- $${}^1$$ A nested Wick's Theorem (between radial order and normal order) is briefly stated on p. 39 in J. Polchinski, String Theory, Vol. 1. Beware that radial order is often only implicitly written in CFT texts. I) It is true that operator ordering procedures are idempotent operations $$\tag{1} T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots).$$ But it is not true that the outermost ordering cancels the effect of the innermost ordering $$\tag{2} T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \quad (\longleftarrow \text{Both Wrong!})$$ In fact, the opposite is true $$\tag{3} T(N(\ldots))~=~N(\ldots)\quad\text{and}\quad N(T(\ldots))~=~T(\ldots),$$ as a special case of a nested$${}^1$$ Wick's Theorem, cf. Section II below. Example. If for two operators $$a$$ and $$b$$, we have the relation $$\tag{4} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$ where the contraction $$\langle ab\rangle$$ is a $$c$$-number, then $$T(N(ab))~\stackrel{(4)}{=}~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~\stackrel{\text{linearity}}{=}~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$ $$\tag{5} ~\stackrel{(1)}{=}~T(ab)- \langle ab\rangle {\bf 1} ~\stackrel{(4)}{=}~N(ab)~\neq~ T(ab) .$$ II) More generally, if we want to bring a nested expression of the form $$\tag{6} T\left(N(\ldots) \ldots N(\ldots)\right)$$ on normal ordered form, there is a nested$${}^1$$ Wick's Theorem, which states that we should only include contractions between pairs of operators who belong to different normal order symbols. Example. In OP's case (5), this means for the lhs. that $$\tag{7} T(N(ab))~=~N(ab),$$ since $$a$$ and $$b$$ belong to the same normal order symbol $$N(ab)$$, while the rhs. is $$\tag{8} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1}.$$ [Note that $$N(a)=a$$ and $$N(b)=b$$.] See also e.g. this and this Phys.SE posts. -- $${}^1$$ A nested Wick's Theorem (between radial order and normal order) is briefly stated on p. 39 in J. Polchinski, String Theory, Vol. 1. Beware that radial order is often only implicitly written in CFT texts. 2 Added explanation edited Jan 31 '15 at 16:15 Qmechanic♦ 114k1313 gold badges225225 silver badges13521352 bronze badges I) It is true that operator ordering procedures are idempotent operations $$T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots).$$$$\tag{1} T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots).$$ But it is not true that the outermost ordering cancels the effect of the innermost ordering $$T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \qquad (\longleftarrow \text{Both Wrong!})$$$$\tag{2} T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \quad (\longleftarrow \text{Both Wrong!})$$ EIn fact, the opposite is true $$\tag{3} T(N(\ldots))~=~N(\ldots)\quad\text{and}\quad N(T(\ldots))~=~T(\ldots),$$ as a special case of a nested$${}^1$$ Wick's Theorem, cf.g Section II below. if Example. If for 2two operators $$a$$ and $$b$$, we have the relation $$T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$ where $$\tag{4} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$ where the contraction $$\langle ab\rangle$$ is a $$c$$-number, then $$T(N(ab))~=~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~=~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$$$T(N(ab))~\stackrel{(4)}{=}~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~\stackrel{\text{linarity}}{=}~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$ $$~=~T(ab)- \langle ab\rangle {\bf 1}~\neq~ T(ab) .$$$$\tag{5} ~\stackrel{(1)}{=}~T(ab)- \langle ab\rangle {\bf 1} ~\stackrel{(4)}{=}~N(ab)~\neq~ T(ab) .$$ II) More generally, if we want to bring a nested expression of the form $$\tag{5} T\left(N(\ldots) \ldots N(\ldots)\right)$$ on normal ordered form, there is a nested$${}^1$$ Wick's Theorem, which states that we should only include contractions between pairs of operators who belong to different normal order symbols. Example. In OP's case (5), this means for the lhs. that $$\tag{7} T(N(ab))~=~N(ab),$$ since $$a$$ and $$b$$ belong to the same normal order symbol $$N(ab)$$, while the rhs. is $$\tag{8} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1}.$$ [Note that $$N(a)=a$$ and $$N(b)=b$$.] See also e.g. this and this Phys.SE posts. -- $${}^1$$ A nested Wick's Theorem (between radial order and normal order) is briefly stated on p. 39 in J. Polchinski, String Theory, Vol. 1. Beware that radial order is often only implicitly written in CFT texts. It is true that operator ordering procedures are idempotent operations $$T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots).$$ But it is not true that the outermost ordering cancels the effect of the innermost ordering $$T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \qquad (\longleftarrow \text{Both Wrong!})$$ E.g. if for 2 operators $$a$$ and $$b$$, we have the relation $$T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$ where the contraction $$\langle ab\rangle$$ is a $$c$$-number, then $$T(N(ab))~=~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~=~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$ $$~=~T(ab)- \langle ab\rangle {\bf 1}~\neq~ T(ab) .$$ See also e.g. this and this Phys.SE posts. I) It is true that operator ordering procedures are idempotent operations $$\tag{1} T(T(\ldots))~=~T(\ldots)\quad\text{and}\quad N(N(\ldots))~=~N(\ldots).$$ But it is not true that the outermost ordering cancels the effect of the innermost ordering $$\tag{2} T(N(\ldots))~=~T(\ldots)\quad\text{and}\quad N(T(\ldots))~=~N(\ldots). \quad (\longleftarrow \text{Both Wrong!})$$ In fact, the opposite is true $$\tag{3} T(N(\ldots))~=~N(\ldots)\quad\text{and}\quad N(T(\ldots))~=~T(\ldots),$$ as a special case of a nested$${}^1$$ Wick's Theorem, cf. Section II below. Example. If for two operators $$a$$ and $$b$$, we have the relation $$\tag{4} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1},$$ where the contraction $$\langle ab\rangle$$ is a $$c$$-number, then $$T(N(ab))~\stackrel{(4)}{=}~T\left(T(ab)- \langle ab\rangle {\bf 1}\right) ~\stackrel{\text{linarity}}{=}~T(T(ab))- T\left(\langle ab\rangle {\bf 1})\right)$$ $$\tag{5} ~\stackrel{(1)}{=}~T(ab)- \langle ab\rangle {\bf 1} ~\stackrel{(4)}{=}~N(ab)~\neq~ T(ab) .$$ II) More generally, if we want to bring a nested expression of the form $$\tag{5} T\left(N(\ldots) \ldots N(\ldots)\right)$$ on normal ordered form, there is a nested$${}^1$$ Wick's Theorem, which states that we should only include contractions between pairs of operators who belong to different normal order symbols. Example. In OP's case (5), this means for the lhs. that $$\tag{7} T(N(ab))~=~N(ab),$$ since $$a$$ and $$b$$ belong to the same normal order symbol $$N(ab)$$, while the rhs. is $$\tag{8} T(ab) ~=~ N(ab) + \langle ab\rangle {\bf 1}.$$ [Note that $$N(a)=a$$ and $$N(b)=b$$.] See also e.g. this and this Phys.SE posts. -- $${}^1$$ A nested Wick's Theorem (between radial order and normal order) is briefly stated on p. 39 in J. Polchinski, String Theory, Vol. 1. Beware that radial order is often only implicitly written in CFT texts. 1 answered Jan 30 '15 at 11:20 Qmechanic♦ 114k1313 gold badges225225 silver badges13521352 bronze badges