The regular order product

In quantum field theory, the regular order product (result comes an order cough British: normal ordered product) is the product that I sorted so that an extinction operator comes to the right side of the generation operator in the product of the operator of the place. I am called the regular product (I put result and come British: normal product). Add N to the top to express that is the regular order product; or both sides: A scale surrounding で is used. The regular order product has the property that vacuum expectation always becomes zero. When I do it when I quantize physical quantity such as the hamiltonian of the classic ground simply, the vacuum expectation may display it, but can take quantity to have a meaning physically by thinking about the regular order product. In addition, the time order product and the regular order product that I line up in order of an ingredient at time and changed are tied by a theorem of Wick of the operator level.

Table of contents

Definition

I assume a_α an extinction operator to support a generation operator of BOSE particle or the fermion in a_α ^†. About the monomial expression consisting of the products from a_α ^† and a_α, I call the product that I sorted so that an extinction operator comes to the right side of the generation operator with the regular order product then. But I shall not change the order about generation operators, extinction operators when I put it to sort it. In addition, of the order between operators of fermion, about replacing it, shall change the mark depending on the number of times, and, about replacing it except it, shall not change the mark. For example, when I assume a_α ^†, a_α the generation extinction operator of the BOSE particle

${\displaystyle N[a_{\alpha }a_{\beta }^{\dagger }]=a_{\beta }^{\dagger }a_{\alpha}}$ 

${\displaystyle N[a_{\alpha }^{\dagger }a_{\beta }a_{\gamma }^{\dagger }]=a_{\alpha }^{\dagger }a_{\gamma }^{\dagger }a_{\beta}}$ 

となる.

On the other hand, when I assume it a generation extinction operator of the fermion

${\displaystyle N[a_{\alpha }a_{\beta }^{\dagger }]=-a_{\beta }^{\dagger }a_{\alpha}}$ 

${\displaystyle N[a_{\alpha }^{\dagger }a_{\beta }a_{\gamma }^{\dagger }]=-a_{\alpha }^{\dagger }a_{\gamma }^{\dagger }a_{\beta}}$ 

である.

The regular order product defined about this monomial expression is expanded in the linear shape sum and the product of the generation extinction operator in form to keep the linear nature and distributive law.

Example

The neutral BOSE particle of spin 0 is described in true scalar field φ(x). φ(x) three dimensions are integral calculus in the momentum space then

${\displaystyle \phi (x)=\int {\frac {d^{3}\mathbf {k} }{(2\pi) ^{3}2k_{0}}}k_{0}(a(k)e^{ikx}+a^{\dagger }(k)e^{-ikx})}$ 

I can express で. But,

${\displaystyle k_{0}={\sqrt {\mathbf {k} ^{2}+m^{2}}},\,kx=k_{0}t-\mathbf {k} \cdot \mathbf {x}}$ 

である. Then,

${\displaystyle \phi (x)=\phi ^{(+)}(x)+\phi ^{(-)}(x)}$ 

${\displaystyle \phi ^{(+)}(x)=\int {\frac {d^{3}\mathbf {k} }{(2\pi) ^{3}2k_{0}}}k_{0}a(k)e^{ikx},\,\phi ^{(-)}(x)=\int {\frac {d^{3}\mathbf {k} }{(2\pi) ^{3}2k_{0}}}k_{0}a^{\dagger }(k)e^{-ikx}}$ 

When I divide it into φ⁽⁺⁾(x) only as for と extinction operator and φ^(-)(x) only as for the generation operator,

${\displaystyle N[\phi ^{(+)}(x)\phi ^{(-)}(y)]=N[\phi ^{(-)}(y)\phi ^{(+)}(x)]=\phi ^{(-)}(x)\phi ^{(+)}(y)}$ 

${\displaystyle N[\phi (x)\phi (y)]=N[(\phi ^{(+)}(x)+\phi ^{(-)}(x))(\phi ^{(+)}(y)+\phi ^{(-)}(y))]=\phi ^{(+)}(x)\phi ^{(+)}(y)+\phi ^{(-)}(x)\phi ^{(+)}(y)+\phi ^{(-)}(y)\phi ^{(+)}(x)+\phi ^{(-)}(x)\phi ^{(-)}(y)}$ 

But, it is managed.

Vacuum expectation

An extinction operator is vacuum state | When a generation operator serves to become zero when I act on 0> together in 〈 0|, it becomes zero and wants to do it, and, as for the regular order product of operator O comprised of a generation extinction operator, as for vacuum expectation 〈 0|N[O]|0 〉, it is zero by all means unless there are 恒等演算子 and the fixed number in doubling it.

Removal of the quantity of emission by the regular order product

The physical quantity in the classical theory of the place in simplicity plus associate; when quantize it, is infinite, and the vacuum expectation may include quantity to display. In this case I can take out meaningful quantity by thinking about the regular order product. For example, in an above-mentioned true scalar field quantity of classic in simplicity plus associate; hamiltonian and the momentum that quantized

${\displaystyle H={\frac {1}{2}}\int {\frac {d^{3}\mathbf {k} }{(2\pi) ^{3}2k_{0}}}k_{0}(a^{\dagger }(k)a(k)+a(k)a(k)^{\dagger})}$ 

${\displaystyle \mathbf {P} ={\frac {1}{2}}\int {\frac {d^{3}\mathbf {k} }{(2\pi) ^{3}2k_{0}}}\mathbf {k} (a^{\dagger }(k)a(k)+a(k)a(k)^{\dagger})}$ 

I display となるが, the vacuum expectation of Clause 2 of the integral calculus. Here,${\displaystyle H}$ と${\displaystyle \mathbf {P}}$ を took regular order beforehand${\displaystyle N[H]}$ と${\displaystyle N[\mathbf {P}]}$ If that is the case,

${\displaystyle H={\frac {1}{2}}\int {\frac {d^{3}\mathbf {k} }{(2\pi) ^{3}2k_{0}}}k_{0}(N[a^{\dagger }(k)a(k)+a(k)a(k)^{\dagger }])=\int {\frac {d^{3}\mathbf {k} }{(2\pi) ^{3}2k_{0}}}k_{0}a^{\dagger }(k)a(k)}$ 

${\displaystyle \mathbf {P} ={\frac {1}{2}}\int {\frac {d^{3}\mathbf {k} }{(2\pi) ^{3}2k_{0}}}\mathbf {k} (N[a^{\dagger }(k)a(k)+a(k)a(k)^{\dagger }])=\int {\frac {d^{3}\mathbf {k} }{(2\pi) ^{3}2k_{0}}}\mathbf {k} a^{\dagger }(k)a(k)}$ 

I can remove the neighbor, quantity of emission of the vacuum expectation.

References

• Michael E. Peskin and Daniel V. Schroeder, An Introduction To Quantum Field Theory, Addison-Wesley, Reading, 1995.

Allied item

• The time order product

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