- #1

JD_PM

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- Homework Statement
- Let [itex]\varphi (x)[/itex] be a real scalar field (operator) satisfying the Klein-Gordon equation [itex](\Box + m^{2}) \varphi (x) = 0[/itex] (where ##\Box:= \partial^{\mu}\partial_{\mu}##). Assuming that the commutator [itex]\big[ \varphi (x) , \varphi (y) \big][/itex] is a Lorentz invariant c-number (i.e., function), show that [tex]\big[\varphi (x) , \varphi (0) \big] = C \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x} ,[/tex] with [itex]C[/itex] being an arbitrary real constant.

- Relevant Equations
- [tex]\big[\varphi (x) , \varphi (0) \big] = C \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x} ,[/tex]

This exercise was proposed by samalkhaiat here (#9). I am going to work using natural units.

OK I think I got it (studying pages 46 & 47 from Mandl & Shaw was really useful) . However I took the lengthy approach. If there is a quicker method please let me know

We first Fourier-expand ##\varphi(x)## in a complete set of solutions of the KG equation (##(\Box + m^{2}) \varphi (x) = 0##); ##\varphi(x)=\varphi^+(x)+\varphi^-(x)##, where

$$\varphi^+(x)=\sum_{\vec k} \Big(\frac{1}{2V \omega_{\vec k}} \Big)^{1/2} a(\vec k)e^{-ik \cdot x}$$

$$\varphi^-(x)=\sum_{\vec k} \Big(\frac{1}{2V \omega_{\vec k}} \Big)^{1/2} a^{\dagger}(\vec k)e^{ik \cdot x}$$

Where ##k## is the wave four-vector of a particle of mass ##m##, momentum ##\vec k## and energy ##E=\omega_{\vec k}=+ \sqrt{m^2+(\vec k)^2}:=k_0##

Now let's evaluate the given commutator ##\big[ \varphi (x) , \varphi (y) \big]## (using the rule ##\big[ A+B , C+D \big]=\big[ A , C \big]+\big[ A , D \big]+\big[ B , C \big]+\big[ B , D \big]## and ##\big[ \varphi^+ (x) , \varphi^+ (y) \big]=\big[ \varphi^- (x) , \varphi^- (y) \big]=0##) we get

$$\big[ \varphi (x) , \varphi (y) \big]=\big[ \varphi^+ (x) , \varphi^- (y) \big]+\big[ \varphi^- (x) , \varphi^+ (y) \big]$$

Then

$$\big[ \varphi^+ (x) , \varphi^- (y) \big]=\frac{1}{2V}\sum_{\vec k \vec k'}\frac{1}{(\omega_{\vec k}\omega_{\vec k'})^{1/2}}\big[ a (\vec k) , a^{\dagger} (\vec k') \big] e^{-ik \cdot x} e^{ik' \cdot y}$$

Using ##\big[ a (\vec k) , a^{\dagger} (\vec k') \big]=\delta_{\vec k \vec k'}## and taking the lim ##V \rightarrow \infty## (i.e. ##\frac{1}{V} \sum_{\vec k} \rightarrow \frac{1}{(2\pi)^3}##) we get

$$\big[ \varphi^+ (x) , \varphi^- (y) \big]=\frac{1}{2(2\pi)^3}\int d^3 \vec k \frac{1}{\omega_{\vec k}}e^{-ik\cdot(x - y)} \tag{1}$$

We now define

$$\Delta^+(x) := \frac{-i}{2(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}e^{-ik\cdot x}=\frac{1}{2i(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}e^{-ik\cdot x} \tag{2}$$

Plugging ##(2)## into ##(1)## we get

$$\big[ \varphi^+ (x) , \varphi^- (y) \big]=i\Delta^+(x-y) \tag{3}$$

Analogously we obtain

$$\big[ \varphi^- (x) , \varphi^+ (y) \big]=\frac{-1}{2(2\pi)^3}\int d^3 \vec k \frac{1}{\omega_{\vec k}}e^{ik\cdot(x-y)} \tag{4}$$

$$\Delta^-(x) := \frac{i}{2(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}e^{ik\cdot x}=\frac{-1}{2i(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}e^{ik\cdot x} \tag{5}$$

Plugging ##(5)## into ##(4)## we get

$$\big[ \varphi^- (x) , \varphi^+ (y) \big]=i\Delta^-(x-y) \tag{6}$$

Thus we can finally define

$$\Delta(x) := \Delta^+(x)+\Delta^-(x)=\frac{-1}{(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}\Big(\frac{e^{ik\cdot x}-e^{-ik\cdot x}}{2i}\Big)=\frac{-1}{(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}\sin (k \cdot x) \tag{7}$$

And get the desired commutation relation

$$\big[\varphi(x), \varphi(y) \big]=i\Delta(x-y) \tag{8}$$

In this particular exercise we're asked to show ##\big[\varphi(x), \varphi(0) \big]=i\Delta(x)=\frac{1}{(2 \pi)^3} \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x}##

Nice! So we only have to show that ##(7)## is equivalent to ##\frac{-i}{(2 \pi)^3} \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x}##. To get it done, we first play a bit with ##\delta (k^{2} - m^{2})##; using the definition of the 4-inner product ##k^2=k_0^2-(\vec k)^2## and ##m^2+(\vec k)^2=(k_0)^2 \Rightarrow m^2=(k_0)^2-(\vec k)^2=\omega_{\vec k}^2-(\vec k)^2## we get

$$\delta (k^{2} - m^{2})=\delta (k_0^{2} - (\vec k)^2 - \omega_{\vec k}^2+(\vec k)^2)=\delta(k_0^2-\omega_{\vec k}^2)=\frac{1}{2 \omega_{\vec k}}\Big(\delta(k_0+\omega_{\vec k})+\delta(k_0-\omega_{\vec k})\Big)$$

Plugging the above into ##\frac{-i}{(2 \pi)^3} \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x}## and integrating the ##k_0## variable we indeed get ##(7)## (I may type the explicit integration below; I am trying to shorten #1).

This means we can legitimately write

$$\big[\varphi(x), \varphi(0) \big]=i \Delta(x)=\frac{1}{(2 \pi)^3} \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x}$$

Where ##C=\frac{1}{(2 \pi)^3} \in \Bbb R##, as stated by samalkhaiat

OK I think I got it (studying pages 46 & 47 from Mandl & Shaw was really useful) . However I took the lengthy approach. If there is a quicker method please let me know

We first Fourier-expand ##\varphi(x)## in a complete set of solutions of the KG equation (##(\Box + m^{2}) \varphi (x) = 0##); ##\varphi(x)=\varphi^+(x)+\varphi^-(x)##, where

$$\varphi^+(x)=\sum_{\vec k} \Big(\frac{1}{2V \omega_{\vec k}} \Big)^{1/2} a(\vec k)e^{-ik \cdot x}$$

$$\varphi^-(x)=\sum_{\vec k} \Big(\frac{1}{2V \omega_{\vec k}} \Big)^{1/2} a^{\dagger}(\vec k)e^{ik \cdot x}$$

Where ##k## is the wave four-vector of a particle of mass ##m##, momentum ##\vec k## and energy ##E=\omega_{\vec k}=+ \sqrt{m^2+(\vec k)^2}:=k_0##

Now let's evaluate the given commutator ##\big[ \varphi (x) , \varphi (y) \big]## (using the rule ##\big[ A+B , C+D \big]=\big[ A , C \big]+\big[ A , D \big]+\big[ B , C \big]+\big[ B , D \big]## and ##\big[ \varphi^+ (x) , \varphi^+ (y) \big]=\big[ \varphi^- (x) , \varphi^- (y) \big]=0##) we get

$$\big[ \varphi (x) , \varphi (y) \big]=\big[ \varphi^+ (x) , \varphi^- (y) \big]+\big[ \varphi^- (x) , \varphi^+ (y) \big]$$

Then

$$\big[ \varphi^+ (x) , \varphi^- (y) \big]=\frac{1}{2V}\sum_{\vec k \vec k'}\frac{1}{(\omega_{\vec k}\omega_{\vec k'})^{1/2}}\big[ a (\vec k) , a^{\dagger} (\vec k') \big] e^{-ik \cdot x} e^{ik' \cdot y}$$

Using ##\big[ a (\vec k) , a^{\dagger} (\vec k') \big]=\delta_{\vec k \vec k'}## and taking the lim ##V \rightarrow \infty## (i.e. ##\frac{1}{V} \sum_{\vec k} \rightarrow \frac{1}{(2\pi)^3}##) we get

$$\big[ \varphi^+ (x) , \varphi^- (y) \big]=\frac{1}{2(2\pi)^3}\int d^3 \vec k \frac{1}{\omega_{\vec k}}e^{-ik\cdot(x - y)} \tag{1}$$

We now define

$$\Delta^+(x) := \frac{-i}{2(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}e^{-ik\cdot x}=\frac{1}{2i(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}e^{-ik\cdot x} \tag{2}$$

Plugging ##(2)## into ##(1)## we get

$$\big[ \varphi^+ (x) , \varphi^- (y) \big]=i\Delta^+(x-y) \tag{3}$$

Analogously we obtain

$$\big[ \varphi^- (x) , \varphi^+ (y) \big]=\frac{-1}{2(2\pi)^3}\int d^3 \vec k \frac{1}{\omega_{\vec k}}e^{ik\cdot(x-y)} \tag{4}$$

$$\Delta^-(x) := \frac{i}{2(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}e^{ik\cdot x}=\frac{-1}{2i(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}e^{ik\cdot x} \tag{5}$$

Plugging ##(5)## into ##(4)## we get

$$\big[ \varphi^- (x) , \varphi^+ (y) \big]=i\Delta^-(x-y) \tag{6}$$

Thus we can finally define

$$\Delta(x) := \Delta^+(x)+\Delta^-(x)=\frac{-1}{(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}\Big(\frac{e^{ik\cdot x}-e^{-ik\cdot x}}{2i}\Big)=\frac{-1}{(2 \pi)^3} \int \frac{d^3 \vec k}{\omega_{\vec k}}\sin (k \cdot x) \tag{7}$$

And get the desired commutation relation

$$\big[\varphi(x), \varphi(y) \big]=i\Delta(x-y) \tag{8}$$

In this particular exercise we're asked to show ##\big[\varphi(x), \varphi(0) \big]=i\Delta(x)=\frac{1}{(2 \pi)^3} \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x}##

Nice! So we only have to show that ##(7)## is equivalent to ##\frac{-i}{(2 \pi)^3} \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x}##. To get it done, we first play a bit with ##\delta (k^{2} - m^{2})##; using the definition of the 4-inner product ##k^2=k_0^2-(\vec k)^2## and ##m^2+(\vec k)^2=(k_0)^2 \Rightarrow m^2=(k_0)^2-(\vec k)^2=\omega_{\vec k}^2-(\vec k)^2## we get

$$\delta (k^{2} - m^{2})=\delta (k_0^{2} - (\vec k)^2 - \omega_{\vec k}^2+(\vec k)^2)=\delta(k_0^2-\omega_{\vec k}^2)=\frac{1}{2 \omega_{\vec k}}\Big(\delta(k_0+\omega_{\vec k})+\delta(k_0-\omega_{\vec k})\Big)$$

Plugging the above into ##\frac{-i}{(2 \pi)^3} \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x}## and integrating the ##k_0## variable we indeed get ##(7)## (I may type the explicit integration below; I am trying to shorten #1).

This means we can legitimately write

$$\big[\varphi(x), \varphi(0) \big]=i \Delta(x)=\frac{1}{(2 \pi)^3} \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x}$$

Where ##C=\frac{1}{(2 \pi)^3} \in \Bbb R##, as stated by samalkhaiat