Quantization of Klein-Gordon Field

Canonical Quantization of real-valued Scalar Field

Here we only consider the real-valued Scalar Field. The result for charged one (complex-valued) is easily to be obtained. The Hamiltonian of Scalar Field system is:

$$H=\frac 1 2 \int \td^3 \bm{x} \ \big(\pi^2+\nabla\phi\cdot \nabla\phi+m^2\phi^2 \big)$$

To quantized this system, we need let the canonical coordinate and momentum be operators on Hilbert space: $\pi(\bm{x})\rightarrow \hat \pi(\bm{x}) \ ; \ \phi(\bm{x})\rightarrow \hat \phi(\bm{x})$ . For simplicity, we will omit the ^ of the operator symbol. Because in the future we will discuss the Quantum Field Theory only.

The operator should satisfy the commutation relation:

$$[\pi(\bm{x}),\pi(\bm{x}')]=[\phi(\bm{x}),\phi(\bm{x}')]=0\ ; \ [\phi(\bm{x}),\pi(\bm{x}')]=\ti \delta^3(\bm{x}-\bm{x}')$$

They are operators in Schrodinger Picture, which are independent of time. Equivalently, we can say they are operators in Heisenberg Picture with the same time. That is why these relations are so-called as equal-time commutation relation

One can rewrite them into the plane wave expansion. Like what we did in Classical Theory. In Schrodinger Picture:

$$\begin{aligned} \phi(\bm{x})&=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \Big\{ e^{\ti \bm{k}\cdot \bm{x}} a(\bm{k})+e^{-\ti \bm{k}\cdot \bm{x}} a(\bm{k})^{\dagger} \Big\} \\ \pi(\bm{x})&=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac {\ti \omega_{|\bm{k}|}} {\sqrt{2\omega_{|\bm{k}|}}} \Big\{ -e^{\ti \bm{k}\cdot \bm{x}} a(\bm{k})+e^{-\ti \bm{k}\cdot \bm{x}} a(\bm{k})^{\dagger} \Big\} \\ \end{aligned}$$

Where we let the coefficient $a^*$ to be operator $a^\dagger$ so that $\phi$ will be a Hermitian operator, corresponding to the real-valued field requirement, and $\omega_{|\bm{k}|}=\sqrt{|\bm{k}|^2+m^2}$

We can write the inverse transform, which will represent $a(\bm{k})$ by $\phi,\pi$ ::

$$\begin{aligned} a(\bm{k})&=\int \frac {\td^3\bm{x}} {(2\pi)^{3/2}} \Big\{\sqrt{\frac {\omega_{|\bm{k}|} } 2} e^{-\ti\bm{k}\cdot \bm{x}}\phi(\bm{x})+\ti \sqrt{\frac 1 {2\omega_{|\bm{k}|}}} e^{-\ti \bm{k}\cdot \bm{x}} \pi(\bm{x}) \Big\} \\ a(\bm{k})^\dagger&=\int \frac {\td^3\bm{x}} {(2\pi)^{3/2}} \Big\{\sqrt{\frac {\omega_{|\bm{k}|} } 2} e^{\ti\bm{k}\cdot \bm{x}}\phi(\bm{x})-\ti \sqrt{\frac 1 {2\omega_{|\bm{k}|}}} e^{\ti \bm{k}\cdot \bm{x}} \pi(\bm{x}) \Big\} \\ \end{aligned}$$

And their commutator:

$$[a(\bm{k}),a(\bm{k}')]=[a(\bm{k})^\dagger,a(\bm{k}')^\dagger]=0 \ ; \ [a(\bm{k}),a(\bm{k}')^\dagger]=\delta^3(\bm{k}-\bm{k}')$$

Proof

Using the 3-d delta function: $\delta^3(\bm{x}-\bm{y})=\int \frac {\td^3 \bm{k}} {(2\pi)^3} e^{\ti \bm{k}\cdot(\bm{x}-\bm{y})}$ , we have: $$\begin{aligned} \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \Big(a(\bm{k})+a(-\bm{k})^\dagger\Big)&=\int \frac {\td^3\bm{x}} {(2\pi)^{3/2}} e^{-\ti \bm{k}\cdot \bm{x}} \phi(\bm{x})\\ \frac {-\ti\omega_{|\bm{k}|}} {\sqrt{2\omega_{|\bm{k}|}}} \Big(a(\bm{k})-a(-\bm{k})^\dagger\Big)&=\int \frac {\td^3\bm{x}} {(2\pi)^{3/2}} e^{-\ti \bm{k}\cdot \bm{x}} \pi(\bm{x}) \end{aligned}$$ That is: $$\begin{aligned} a(\bm{k})&=\int \frac {\td^3\bm{x}} {(2\pi)^{3/2}} \Big\{\sqrt{\frac {\omega_{|\bm{k}|} } 2} e^{-\ti\bm{k}\cdot \bm{x}}\phi(\bm{x})+\ti \sqrt{\frac 1 {2\omega_{|\bm{k}|}}} e^{-\ti \bm{k}\cdot \bm{x}} \pi(\bm{x}) \Big\} \\ a(\bm{k})^\dagger&=\int \frac {\td^3\bm{x}} {(2\pi)^{3/2}} \Big\{\sqrt{\frac {\omega_{|\bm{k}|} } 2} e^{\ti\bm{k}\cdot \bm{x}}\phi(\bm{x})-\ti \sqrt{\frac 1 {2\omega_{|\bm{k}|}}} e^{\ti \bm{k}\cdot \bm{x}} \pi(\bm{x}) \Big\} \\ \end{aligned}$$ We can compute some commutator: $$\begin{aligned}[] [a(\bm{k}),a(\bm{k}')] &= \int \frac {\td^3\bm{x}\td^3\bm{x}'} {(2\pi)^3} \Big\{\ti\sqrt{\frac {\omega} {4\omega'}} e^{-\ti(\bm{k}\cdot\bm{x}+\bm{k}'\cdot \bm{x}')} [\phi(\bm{x}),\pi(\bm{x}')]+\ti\sqrt{\frac {\omega'} {4\omega}}e^{-\ti(\bm{k}\cdot\bm{x}+\bm{k}'\cdot \bm{x}')} [\pi(\bm{x}),\phi(\bm{x}')] \Big\} \\ &=\int \frac {\td^3\bm{x}\td^3\bm{x}'} {(2\pi)^3} \Big\{-\sqrt{\frac {\omega} {4\omega'}} e^{-\ti(\bm{k}\cdot\bm{x}+\bm{k}'\cdot \bm{x}')} \delta^3(\bm{x}-\bm{x}')+\sqrt{\frac {\omega'} {4\omega}}e^{-\ti(\bm{k}\cdot\bm{x}+\bm{k}'\cdot \bm{x}')} \delta^3(\bm{x}-\bm{x}') \Big\} \\ &=\int \frac {\td^3 \bm{x}} {(2\pi)^3} \Big\{-\sqrt{\frac {\omega} {4\omega'}} e^{-\ti(\bm{k}+\bm{k}')\cdot \bm{x}}+ \sqrt{\frac {\omega'} {4\omega}} e^{-\ti(\bm{k}+\bm{k}')\cdot \bm{x}} \Big\} \\ &= -\sqrt{\frac {\omega} {4\omega'}}+ \sqrt{\frac {\omega'} {4\omega}} \\ &=0 \end{aligned}$$ Similarly, one can prove $[a(\bm{k})^\dagger,a(\bm{k}')^\dagger]=0$ Then we compute: $$\begin{aligned}[] [a(\bm{k}),a(\bm{k}')^\dagger] &= \int \frac {\td^3\bm{x}\td^3\bm{x}'} {(2\pi)^3} \Big\{-\ti\sqrt{\frac {\omega} {4\omega'}} e^{-\ti(\bm{k}\cdot\bm{x}-\bm{k}'\cdot \bm{x}')} [\phi(\bm{x}),\pi(\bm{x}')]+\ti\sqrt{\frac {\omega'} {4\omega}}e^{\ti(\bm{k}\cdot\bm{x}-\bm{k}'\cdot \bm{x}')} [\pi(\bm{x}),\phi(\bm{x}')] \Big\} \\ &=\int \frac {\td^3\bm{x}\td^3\bm{x}'} {(2\pi)^3} \Big\{\sqrt{\frac {\omega} {4\omega'}} e^{-\ti(\bm{k}\cdot\bm{x}-\bm{k}'\cdot \bm{x}')} \delta^3(\bm{x}-\bm{x}')+\sqrt{\frac {\omega'} {4\omega}}e^{\ti(\bm{k}\cdot\bm{x}-\bm{k}'\cdot \bm{x}')} \delta^3(\bm{x}-\bm{x}') \Big\} \\ &=\int \frac {\td^3 \bm{x}} {(2\pi)^3} \Big\{\sqrt{\frac {\omega} {4\omega'}} e^{-\ti(\bm{k}-\bm{k}')\cdot \bm{x}}+ \sqrt{\frac {\omega'} {4\omega}} e^{\ti(\bm{k}-\bm{k}')\cdot \bm{x}} \Big\} \\ &=2 \sqrt{\frac {\omega_{|\bm{k}|}} {4\omega_{|\bm{k}'|}}}\delta^3(\bm{k}-\bm{k}')\\ &=\delta^3(\bm{k}-\bm{k}') \end{aligned}$$

With this result and the hold-order discussion in Classical theory, we can write the Hamiltonian as:

$$\hat H =\frac 1 2 \int \td^3\bm{k} \ \omega_{|\bm{k}|} \big(a(\bm{k})^\dagger a(\bm{k})+a(\bm{k})a(\bm{k})^\dagger\big)=E_0+\int \td^3 \bm{k} \ \omega_{|\bm{k}|} a(\bm{k})^\dagger a(\bm{k})$$

Where $E_0=\frac 1 2 \int\td^3\bm{k} \ \omega_{|\bm{k}|}\delta(0)$ .. This is a $\infty$ quantity. But it will not bother us if we admit that the absolute value of energy is non-measurable. But the relative value is work.

But in General Relativistic theory, the energy value will influence the space-time structure. This argument will not work at that situation.

With this argument, we can simply write the Hamiltonian as:

$$\hat H= \int \td^3 \bm{k} \ \omega_{|\bm{k}|} a(\bm{k})^\dagger a(\bm{k})$$

One can also write these operators in Heisenberg Picture :

$$\begin{aligned} \phi(x)&=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \Big\{ e^{-\ti k_\mu x^\mu} a(\bm{k})+e^{\ti k_\mu x^\mu} a(\bm{k})^{\dagger} \Big\} \\ \pi(x)&=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac {\ti \omega_{|\bm{k}|}} {\sqrt{2\omega_{|\bm{k}|}}} \Big\{ -e^{-\ti k_\mu x^\mu} a(\bm{k})+e^{\ti k_\mu x^\mu} a(\bm{k})^{\dagger} \Big\} \\ \end{aligned}$$

Proof

With the commutator of $a,a^\dagger$ and $\hat H$ :: $$\begin{aligned}[] [a(\bm{k}),\hat H]&=\omega_{|\bm{k}|} a(\bm{k}) \\ [a(\bm{k})^\dagger,\hat H]&=-\omega_{|\bm{k}|} a(\bm{k})^\dagger \end{aligned}$$ Noting that $\hat H$ : is independent of time, we have: $$\begin{aligned} a_H(\bm{k}) &=e^{\ti t\hat H} a(\bm{k}) e^{-\ti t \hat H}=e^{-\ti \omega_{|\bm{k}|} t} a(\bm{k}) \\ a_H(\bm{k})^\dagger &= e^{\ti \omega_{|\bm{k}|} t} a(\bm{k})^\dagger \end{aligned}$$ That is, in Heisenberg Picture, the operators which is dependent of time: $$\begin{aligned} \phi(x)&=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \Big\{ e^{-\ti k_\mu x^\mu} a(\bm{k})+e^{\ti k_\mu x^\mu} a(\bm{k})^{\dagger} \Big\} \\ \pi(x)&=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac {\ti \omega_{|\bm{k}|}} {\sqrt{2\omega_{|\bm{k}|}}} \Big\{ -e^{-\ti k_\mu x^\mu} a(\bm{k})+e^{\ti k_\mu x^\mu} a(\bm{k})^{\dagger} \Big\} \\ \end{aligned}$$

Hilbert Space of Scalar Field

The Hamiltonian of Real-valued Scalar Field is the same as the Hamiltonian of a set of non-interaction harmonic oscillators. Then we can construct the Hilbert space by the ground state and creation/annihilation operators. This is also called as Fock space.

[Definition] : vacuum of the Klein-Gordon Field is the ground state of Scalar Field. Usually denote it as $\ket{0}$ , and it is zero if annihilation operator acts on it:

$$a(\bm{k}) \ket{0}=0 \ ; \ \forall \bm{k}$$

Assume that it is normalized as $\bra{0}0\rangle=1$ . Then the energy of vacuum is zero: $\bra{0}\hat H\ket{0}=0$ ..

[Definition] : 1-particle state is proportional to a creation operator acting on vacuum. Usually denote it as $\ket{\bm{k}}$ :

$$\ket{\bm{k}} = \sqrt{2\omega_{|\bm{k}|}} a(\bm{k})^\dagger \ket{0}$$

And the orthonormal relation is:

$$\bra{\bm{k}}\bm{k}'\rangle = 2\omega_{|\bm{k}|} \delta^3(\bm{k}-\bm{k}')$$

One can prove that under the Lorentz Transformation that leads $\Lambda_{\bm{k}\rightarrow \bm{0}} k=(m,0,0,0)$ ,, one has

$$\bra{\Lambda_{\bm{k}\rightarrow \bm{0}} \bm{k}} \Lambda_{\bm{k}\rightarrow \bm{0}} \bm{k}'\rangle = \bra{\bm{k}}\bm{k}'\rangle$$

in this degree we say this quantity is invariant under Lorentz transformation.

And the complete relation in 1-particle subspace:

$$1_{\text{one-particle}}=\int \frac {\td^3\bm{k}} {2\omega_{|\bm{k}|}} \ket{\bm{k}}\bra{\bm{k}}$$

Where $\int \td^3\bm{p} (2\omega_{|\bm{p}|})^{-1}=\int_{p^0\gt 0} \td^4p \ \delta(p^\mu p_\mu-m^2)$ is covariant measure.

Proof

Consider $$\begin{aligned} \bra{\bm{k}}\bm{p}\rangle &= 2\omega_{|\bm{k}|} \delta^3(\bm{k}-\bm{p}) \\ &=2\sqrt{m^2+|\bm{k}|^2}\delta(k^1-p^1)\delta(k^2-p^2)\delta(k^3-p^3) \end{aligned}$$ Consider Lorentz transform acting on $\bm{k},\bm{p}$ : $$\begin{aligned} \bra{\Lambda\bm{k}}\Lambda \bm{p}\rangle &= 2\sqrt{m^2+|\Lambda \bm{k}|^2}\prod_i\delta(\Lambda^i_{\indent \mu}(k^\mu - p^\mu)) \\ &=2\frac {|\Lambda^0_{\indent \mu}k^\mu|} {|\text{det}J|} \prod_i \delta(k^i-p^i) \end{aligned}$$ Where $\Lambda \bm{k}$ means the spatial components of $\Lambda(\omega_{|\bm{k}|},\bm{k})$ . And: $$\begin{aligned} \det J &= \det\Big[ \frac {\partial} {\partial k^j}\Big(\Lambda^i_{\indent 0}(\sqrt{m^2+|\bm{k}|^2})-p_0)+\Lambda^i_{\indent l} (k^l-p^l)\Big)\Big] \\ &=\det\Big[\Lambda^i_{\indent 0} \frac {k^j} {\sqrt{m^2+|\bm{k}|^2}}+\Lambda^i_{\indent j} \Big] \\ &=\det \begin{bmatrix}\Lambda^1_{\indent 1}+\Lambda^1_{\indent 0} k^1/k^0 & \Lambda^1_{\indent 2}+\Lambda^1_{\indent 0} k^2/k^0 & \Lambda^1_{\indent 3}+\Lambda^1_{\indent 0} k^3/k^0 \\ \Lambda^2_{\indent 1}+\Lambda^2_{\indent 0} k^1/k^0 & \Lambda^2_{\indent 2}+\Lambda^2_{\indent 0} k^2/k^0 & \Lambda^2_{\indent 3}+\Lambda^2_{\indent 0} k^3/k^0 \\ \Lambda^3_{\indent 1}+\Lambda^3_{\indent 0} k^1/k^0 & \Lambda^3_{\indent 2}+\Lambda^3_{\indent 0} k^2/k^0 & \Lambda^3_{\indent 3}+\Lambda^3_{\indent 0} k^3/k^0 \end{bmatrix} \\ &=\det \Lambda'_0+\frac {k^1} {k^0} \det \Lambda'_1-\frac {k^2} {k^0}\det \Lambda'_2 +\frac {k^3} {k^0}\det \Lambda'_3 \\ &=\frac 1 {k^0}\sum_{\mu=0}^3 (-1)^{\mu}k_\mu \det \Lambda'_\mu \end{aligned}$$ Where $\det \Lambda'_\mu$ are determinants of minors of $\Lambda$ with cancelling $0$ --row-- $\mu$ --column. That is: $$\det \Lambda =\sum_{\mu=0}^3 (-1)^\mu \Lambda^0_{\indent \mu} \det \Lambda'_\mu $$ For $k_\mu= m \Lambda^0_{\indent \mu}$ , we have that: $\Lambda^0_{\indent \mu}k^\mu = m \Lambda^0_{\indent \mu} \Lambda^{0\mu}=m\Lambda^0_{\indent \mu} \Lambda_0^{\indent \mu}=m\delta_0^0=m$ , which means that in this case: $\Lambda k=(m,0,0,0)$ . And $\det J=\frac m {k^0}\det \Lambda=m/k^0$ That means under this special Lorentz Transformation: $\Lambda_{\bm{k}\rightarrow \bm{0}} (\sqrt{m^2+\bm{k}^2},k^1,k^2,k^3) =(m,0,0,0)$ : $$\langle\Lambda_{\bm{k}\rightarrow \bm{0}} \bm{k}|\Lambda_{\bm{k}\rightarrow \bm{0}}\bm{p}\rangle =2k^0 \prod_i \delta(k^i-p^i)=\bra{\bm{k}}\bm{p}\rangle $$ q.e.d. Consider the integral: $$\begin{aligned} \int \td^4 p \ \delta(p^\mu p_\mu-m^2) f(p)&=\int \td^3 \bm{p} \int \td p^0 \ \delta\Big((p^0)^2-m^2-|\bm{p}|^2\Big)f(p^0,\bm{p}) \\ &=\int \td^3 \bm{p} \frac {f(p^0=\sqrt{m^2+|\bm{p}|^2},\bm{p})} {2\sqrt{m^2+|\bm{p}|^2}} + \int \td^3 \bm{p} \frac {f(p^0=-\sqrt{m^2+|\bm{p}|^2},\bm{p})} {2\sqrt{m^2+|\bm{p}|^2}} \\ &=\int \frac {\td^3 \bm{p}} {2\omega_{|\bm{p}|}} \tilde{f}(\bm{p}) \end{aligned}$$ Where: $$\tilde{f}(\bm{p})=f(p^0=\omega_{|\bm{p}|},\bm{p})+f(p^0=-\omega_{|\bm{p}|},\bm{p})$$ If $f$ is independent of $p^0$ , it is simply $\tilde{f}=2f$ .. In this case, we can write down the equality of integral measure: $$\int \frac {\td^3 \bm{p}} {2\omega_{|\bm{p}|}}=\frac 1 2\int \td^4 p \ \delta(p^\mu p_\mu -m^2)=\int_{p^0\gt 0} \td^4 p \ \delta(p^\mu p_\mu -m^2)$$ The right-hand-side of the equality is obvious invariant under (Orthochronous) Lorentz Transformation.

In Fock space, the Lorentz transformation for coordinates induces the Unitary transformation for operators (In finite-dimensional space Lorentz group has no unitary representation because it is not tight. But Fock space is infinite-dimensional.).

[Definition] : Consider the coordinate transformation(special Lorentz Transformation) $x\rightarrow x'=\Lambda x$ , the corresponding operator on Hilbert space is $U(\Lambda)$ , then we have:

$$U(\Lambda)\ket{\bm{k}}=\ket{\Lambda \bm{k}}$$

The vacuum is unique: $U(\Lambda)\ket{0}=\ket{0}$ . And the creation operator:

$$U(\Lambda) a(\bm{k})^\dagger U(\Lambda)^{-1} = \sqrt {\frac {\omega_{|\Lambda \bm{k}|}} {\omega_{|\bm{k}|}}} a(\Lambda \bm{k})^\dagger$$

Same for annihilation operator. And one can check that the field operator will be:

$$\phi'(x'=\Lambda x)\equiv U(\Lambda)\phi(x)U(\Lambda)^{-1} = \phi(x'=\Lambda x)$$

Proof

For the field operator: $$\phi(x)=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac 1 {\sqrt{2\omega_{|\bm{k}|}}} \Big\{ e^{-\ti k_\mu x^\mu} a(\bm{k})+e^{\ti k_\mu x^\mu} a(\bm{k})^{\dagger} \Big\}$$ We have: $$\begin{aligned} U(\Lambda)\phi(x)U(\Lambda)^{-1}&=\int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}} \frac {\sqrt{2\omega_{|\Lambda \bm{k}|}}} {2\omega_{|\bm{k}|}} \Big\{ e^{-\ti k_\mu x^\mu} a(\Lambda\bm{k})+e^{\ti k_\mu x^\mu} a(\Lambda\bm{k})^{\dagger} \Big\} \\ &=\int \frac {\td^3 \Lambda\bm{k}} {(2\pi)^{3/2}} \frac {\sqrt{2\omega_{|\Lambda \bm{k}|}}} {2\omega_{|\Lambda\bm{k}|}} \Big\{ e^{-\ti k_\mu x^\mu} a(\Lambda\bm{k})+e^{\ti k_\mu x^\mu} a(\Lambda\bm{k})^{\dagger} \Big\} \\ &=\phi(x'=\Lambda x) \end{aligned}$$ Where we used $k_\mu x^\mu = (\Lambda k)_\mu (\Lambda x)^\mu$ , and the covariant of the integral measure: $$\int \frac {\td^3 \bm{k}} {2\omega_{|\bm{k}|}}$$ together with that $\Lambda$ is the special Lorentz Transformation.

For general cases, if $\Lambda$ is a general Lorentz Transformation, this conclusion works for operators in Schrodinger equation for they are independent of time, or equivalently, the zero-component of $k$ :

$$\phi'(\bm{x}'=\Lambda \bm{x})\equiv U(\Lambda)\phi(\bm{x})U(\Lambda)^{-1} = \phi(\bm{x}'=\Lambda \bm{x})$$

Some discussion about this and the classical form

This seems to violate the transformation regulation in Classical theory which is: $$\psi(x)\rightarrow \psi'(x'=\Lambda_{\text{spacetime}}(\omega)x) = \big(\Lambda_{\text{Field}}(\omega)\psi\big)(x=\Lambda_{\text{spacetime}}(\omega)^{-1}x') $$ But actually not. Here we use the field operator $\phi$ to create or annihilate at some spacetime point. But in classical theory we discussed that the coordinate transform with a `fixed` field. Like the passive transformation & active transformation, they are just mutually inverse. We will see this at spinor and vector field, in which the inverse relationship will be much more clear. For the $\Lambda_{\text{Field}}$ in those cases are inversion too.

Causality of Scalar Field

Our quantization will not violent to the causality demanded by special Relativistic Theory. That is to say:

[Theorem] : In Heisenberg Picture, for any two point $x,y$ with spacelike interval: $\|x-y\|^2\lt 0$ . Then the field operators are commutable:

$$[\phi(x),\phi(y)]=0 \ ; \ \|x-y\|^2\lt 0$$

Proof

With the quantization that $[a(\bm{k}),a(\bm{k}')^\dagger]=\delta^3(\bm{k}-\bm{k}')$ . we have: $$\begin{aligned}[] [\phi(x),\phi(y)]&= \int \frac {\td^3 \bm{k}_1\td^3 \bm{k}_2} {(2\pi)^{3} \sqrt{4\omega_1\omega_2}}\Big\{[a(\bm{k}_1),a(\bm{k}_2)^\dagger]e^{-\ti k_1\cdot x+\ti k_2\cdot y}+[a(\bm{k}_1)^\dagger,a(\bm{k}_2)]e^{\ti k_1\cdot x-\ti k_2\cdot y} \Big\} \\ &=\int \frac {\td^3 \bm{k}} {(2\pi)^3 2\omega_{|\bm{k}|}}\big(e^{-\ti k\cdot (x-y)}-e^{\ti k\cdot (x-y)} \big) \\ & \equiv \ti \Delta(x-y) \end{aligned}$$ One can rewrite the integral as a covariant form: $$\ti \Delta(x-y)=\int \frac {\td^4 k} {(2\pi)^3} \delta(k^\mu k_\mu-m^2)\Theta(k^0)\big(e^{-\ti k\cdot (x-y)}-e^{\ti k\cdot (x-y)}\big)$$ For $x-y$ is a spacelike vector, we can always find a Lorentz Transformation so that $\ti\Delta(x-y)=f(\bm{x}-\bm{y})$ . Then: $$f(\bm{x})=\int \frac {\td^4 k} {(2\pi)^3} \delta(k^\mu k_\mu-m^2)\Theta(k^0)\big(e^{-\ti \bm{k}\cdot \bm{x}}-e^{\ti \bm{k}\cdot \bm{x}}\big)=0$$ which can be easily find by transform the parameter $\bm{k}\rightarrow -\bm{k}$ . q.e.d.

Complex-valued Scalar Field

We write down the conclusion about complex-valued Scalar Field here. The derivation is similar to the real-valued case. But we need two kinds of creation operator corresponding to the internal degrees of freedom of Complex-valued Scalar Field:

The field operator in Heisenberg Picture:

$$\phi(x) = \int \frac {\td^3 \bm{k}} {(2\pi)^{3/2}2\omega_{|\bm{k}|}} \Big\{e^{-\ti k\cdot x} a_+(\bm{k}) + e^{\ti k\cdot x} a_-(\bm{k})^\dagger \Big\}$$

Where $k\cdot x=k_\mu x^\mu$ . And the Hamiltonian:

$$\hat H=\int \td^3 \bm{k} \ \omega_{|\bm{k}|} \big(a_+(\bm{k})^\dagger a_+(\bm{k})+a_-(\bm{k})^\dagger a_-(\bm{k}) \big)$$

And the commutation relationship:

$$\begin{aligned}[] [a_+(\bm{k}),a_+(\bm{k}')] &= [a_+(\bm{k})^\dagger,a_+(\bm{k}')^\dagger]=0 \\ [a_-(\bm{k}),a_-(\bm{k}')] &= [a_-(\bm{k})^\dagger,a_-(\bm{k}')^\dagger]=0 \\ [a_+(\bm{k}),a_-(\bm{k}')] &= [a_+(\bm{k})^\dagger,a_-(\bm{k}')^\dagger]=0\\ [a_+(\bm{k}),a_+(\bm{k}')^\dagger]&=[a_-(\bm{k}),a_-(\bm{k}')^\dagger]=\delta^3(\bm{k}-\bm{k}') \end{aligned}$$

The conserved charge:

$$Q=\int\td^3 \bm{k} \ \big(a_+(\bm{k})^\dagger a_+(\bm{k})-a_-(\bm{k})^\dagger a_-(\bm{k}) \big) = \sum_{\bm{k}} \big(N_{\bm{k}}^+-N_{\bm{k}}^-\big)$$

Propagator of Scalar Field

[Definition] : Feynman Propagator or two-point time-ordered Green's Function of Scalar Field is defined as :

$$\ti \Delta_F(x-y):=\bra{0}\mathcal{T}\{\phi(x)\phi(y)^\dagger\}\ket{0} $$

Here the time-ordered product (for bosonic operators) is:

$$\mathcal{T}\{A(x)B(y)\}=\Theta(x^0-y^0)A(x)B(y)+\Theta(y^0-x^0)B(y)A(x)$$

One can compute the propagator:

$$\ti \Delta_F(x-y)=\int \frac {\td^4 p} {(2\pi)^4} e^{-\ti p\cdot (x-y)} \frac {\ti} {p_\mu p^\mu-m^2+\ti \epsilon}$$

Where $\epsilon\rightarrow 0^+$ .

Proof

We compute these two terms in time-ordered product: $$\begin{aligned} \bra{0}\phi(x)\phi(y)^\dagger\ket{0}&= \int \frac {\td^3 \bm{k}_1 \td^3 \bm{k}_2} {(2\pi)^3 \sqrt{4\omega_1\omega_2}}e^{-\ti k_2\cdot x+\ti k_1\cdot y} \bra{0}a(\bm{k}_2)a(\bm{k}_1)^\dagger \ket{0} \\ &=\int \frac {\td^3 \bm{k}} {(2\pi)^3 2\omega_{|\bm{k}|}} e^{-\ti k\cdot (x-y)} \\ \bra{0}\phi(y)^\dagger\phi(x)\ket{0}&=\int \frac {\td^3 \bm{k}_1 \td^3 \bm{k}_2} {(2\pi)^3 \sqrt{4\omega_1\omega_2}}e^{-\ti k_2\cdot y+\ti k_1\cdot x} \bra{0}a(\bm{k}_2)a(\bm{k}_1)^\dagger \ket{0} \\ &=\int \frac {\td^3 \bm{k}} {(2\pi)^3 2\omega_{|\bm{k}|}} e^{\ti k\cdot (x-y)} \\ \end{aligned}$$ Then we have: $$\ti \Delta_F(x-y)=\int \frac {\td^3 \bm{p}} {(2\pi)^3 2\omega_{|\bm{p}|}} \Big\{\Theta(x^0-y^0)e^{-\ti p\cdot (x-y)} + \Theta(y^0-x^0)e^{\ti p\cdot (x-y)} \Big\}$$ Then we consider the integral: $$\begin{aligned} \int \frac {\td^4 p} {(2\pi)^4} e^{-\ti p\cdot x} \frac {\ti} {p_\mu p^\mu-m^2+\ti \epsilon}&=\int \frac {\td^3\bm{p}} {(2\pi)^4} \int \td p^0 \frac {\ti e^{-\ti p\cdot x}} {(p^0-\omega_{|\bm{p}|}+\ti \epsilon)(p^0+\omega_{|\bm{p}|}-\ti \epsilon)} \end{aligned}$$ If $x^0\lt 0$ , the Jordon's Lemma allow us to use the upper half plane. Then the pole is $-\omega_{|\bm{p}|}+\ti \epsilon$ , we have: $$\text{integral} =2\pi \ti \int \frac {\td^3 \bm{p}} {(2\pi)^4} \frac {\ti e^{\ti \omega_{|\bm{p}|}x^0+\ti \bm{p}\cdot \bm{x}}} {-2\omega_{|\bm{p}|}}=\int \frac {\td^3 \bm{p}} {(2\pi)^3 2\omega_{|\bm{p}|}} e^{\ti \omega_{|\bm{p}|}x^0 - \ti \bm{p}\cdot \bm{x}} = \int \frac {\td^3 \bm{p}} {(2\pi)^3 2\omega_{|\bm{p}|}} e^{\ti p\cdot x } $$ And if $x^0\gt 0$ , we can only use the lower half plane. Then the pole is $\omega_{|\bm{p}|}-\ti \epsilon$ , we have: $$\text{integral} =-2\pi \ti \int \frac {\td^3 \bm{p}} {(2\pi)^4} \frac {\ti e^{-\ti \omega_{|\bm{p}|}x^0+\ti \bm{p}\cdot \bm{x}}} {2\omega_{|\bm{p}|}}=\int \frac {\td^3 \bm{p}} {(2\pi)^3 2\omega_{|\bm{p}|}} e^{-\ti \omega_{|\bm{p}|}x^0 + \ti \bm{p}\cdot \bm{x}} = \int \frac {\td^3 \bm{p}} {(2\pi)^3 2\omega_{|\bm{p}|}} e^{-\ti p\cdot x } $$ That is : $$\int \frac {\td^4 p} {(2\pi)^4} e^{-\ti p\cdot x} \frac {\ti} {p_\mu p^\mu-m^2+\ti \epsilon}=\int \frac {\td^3 \bm{p}} {(2\pi)^3 2\omega_{|\bm{p}|}} \Big\{\Theta(x^0-y^0)e^{-\ti p\cdot (x-y)} + \Theta(y^0-x^0)e^{\ti p\cdot (x-y)} \Big\}$$ Or equivalently: $$\ti \Delta_F(x-y) = \int \frac {\td^4 p} {(2\pi)^4} e^{-\ti p\cdot (x-y)} \frac {\ti} {p_\mu p^\mu-m^2+\ti \epsilon}$$

Discrete Symmetry of Scalar Field

Spatial Inversion

We hope the spatial inversion induces a unitary operator so as to:

$$U_P \phi(x^0,\bm{x})U_P^{-1}= e^{\ti \theta_P} \phi(x^0,-\bm{x})$$

Where $e^{\ti\theta_P}$ is a phase-factor. And $U_P\ket{0}=\ket{0}$ .

For creation operators:

$$U_P a_+(\bm{p})U_P^{-1}=e^{\ti \theta_P} a_+(-\bm{p}) \ ; \ U_P a_-(\bm{p})^\dagger U_P^{-1}=e^{\ti \theta_P} a_-(-\bm{p})^\dagger$$

For complex-valued Klein-Gordon Field, $\theta_P$ has no physical meaning. But for real-valued Klein-Gordon Field, $e^{\ti \theta_P}=\pm 1$ ,, denoting the particle’s internal parity.

We can also represent $U_P$ by creation/annihilation operators:

$$U_P=\exp\Bigg\{\ti \frac {\pi} 2 \int \td^3 \bm{k} \Big(a_+(\bm{k})^\dagger a_+(-\bm{k})+a_-(\bm{k})^\dagger a_-(-\bm{k}) - e^{\ti \theta}\big(a_+(\bm{k})^\dagger a_+(\bm{k})+a_-(\bm{k})^\dagger a_-(\bm{k})\big) \Big) \Bigg\}$$

Charge Conjugation

We hope the charge conjugation induces a unitary operator so as to:

$$U_C \phi(x)U_C^{-1}=e^{\ti \theta_C}\phi(x)^\dagger$$

Where $e^{\ti\theta_C}$ is a phase-factor, and also is unmeasurable for complex-valued field. For real-valued Klein-Gordon Field $e^{\ti \theta_C}=\pm 1$ .

And for creation operators:

$$U_C a_+(\bm{p})U_C^{-1}=e^{\ti \theta_C} a_-(\bm{p}) \ ; \ U_C a_-(\bm{p})^\dagger U_C^{-1}=e^{\ti \theta_C} a_+(\bm{p})^\dagger$$

Time Reversal

Time reversal induces an anti-unitary operator. With conjugation operator $K$ we can write it as $U_T K$ , where $U_T$ is a unitary operator:

$$(U_T K) \phi(x^0,\bm{x})(U_T K)^{-1} = e^{\ti \theta_T} \phi(-x^0,\bm{x})$$

And for creation operators:

$$(U_T K) a_+(\bm{k})(U_T K)^{-1} = e^{\ti \theta_T} a_+(-\bm{k}) \ ; \ (U_T K) a_-(\bm{k})(U_T K)^{-1} = e^{-\ti \theta_T} a_-(-\bm{k})$$