Working out ##\big[\varphi (x) , \varphi (0) \big]## commutator

[JD_PM]

Homework Statement

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

Relevant Equations

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

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

 

[samalkhaiat]

Expansion in terms of creation and annihilation operators is a textbook material, so you can look it up in almost all textbooks. The exercise does not tell you anything about $a$, $a^{\dagger}$ or $[ a , a^{\dagger}]$. The exercise asks you to show that the requirement that $[\varphi (x) , \varphi (y)]$ be an invariant c-number fixes (apart from a constant) the value of the commutator. So there is no $a, a^{\dagger}$ or $[a , a^{\dagger}]$ in the exercise.

 

[JD_PM]

I just used the creation and annihilation commutator to show equations $(1)$ and $(4)$, which lead to equation $(8)$. I see it was unnecessary to show $(8)$ as in this exercise we're assuming it.

However; don't you accept the equivalence trick I used?

 

[samalkhaiat]

I just used the creation and annihilation commutator to show equations $(1)$ and $(4)$, which lead to equation $(8)$. I see it was unnecessary to show $(8)$ as in this exercise we're assuming it.

However; don't you accept the equivalence trick I used?

Again, what you have done is the usual textbook method which take the algebra of the operators $(a, a^{\dagger})$ as a postulate. In the exercise, the operators $(a,a^{\dagger})$ and their algebra are not given to you. So, you cannot use them at all. In other words, you can do the exercise without knowing anything about the operators $(a,a^{\dagger})$. Hint: re-read the post where the exercise was originally posted. That post contains almost all the information needed to work out the exercise.

The purpose of the exercise was to show you the power of symmetry and to teach you something about Lorentz invariance.

 

[vanhees71]

It may depend on the textbook, but one way is to use the equal-time (anti-)commutation relations as a postulate ("canonical quantization"). For free fields you can then introduce the mode decomposition with energy-momentum modes to build the annihilation and creation operators. Their (anti-)commutators follow then from the equal-time (anti-)commutators uniquely. The next step is then the calculation of the various two-point correlation functions, one of which is the commutator of field operators with arbitrary arguments, as is discussed here.

Of course the setup of the problem tells you to use the Lorentz/Poincare symmetry properties only and not the commutation relations for annihilation and creation operators.

 

[JD_PM]

Hi samalkhaiat

What I see so far is that if we assume

$$\big[ \varphi (x) , \varphi (y) \big] = i\Delta(x-y)$$

And using the following properties you provided

$$\ (\Lambda k) \cdot (\Lambda x) = k \cdot x$$

$$\ p \cdot (\Lambda x) = (\Lambda^{-1}p) \cdot x$$

$$\ \left(\Lambda^{-1}k \right)^{2} = k^{2}$$

We can show that $\Delta(x-y)$ is Lorentz invariant

$$\Lambda \Delta(x-y)=\frac{-i}{(2 \pi)^3} \int d^{4}k \ \epsilon(k_{0}) \delta ((\Lambda^{-1} k)^{2} - m^{2}) \ \Big(e^{- i (\Lambda k) \cdot (\Lambda x-\Lambda y)}-e^{ i (\Lambda k) \cdot (\Lambda x-\Lambda y)}\Big)=\Delta(x-y)$$

Where $\Lambda$ is orthogonal and symmetric so $\Lambda^{-1} = \Lambda^T = \Lambda$

In our case $\big[ \varphi (x) , \varphi (0) \big] = i\Delta(x)$ so

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

show that $$\big[\varphi (x) , \varphi (0) \big] = C \int d^{4}k \ \epsilon(k_{0}) \delta (k^{2} - m^{2}) \ e^{- i k \cdot x} ,$$ with $C$ being an arbitrary real constant.

The issue I have is that I do not see how to show the above equation without using $\big[ a (\vec k) , a^{\dagger} (\vec k') \big]=\delta_{\vec k \vec k'}$. Might you please give me another hint/give me an analog solved example?

Thank you very much for your time

 

[samalkhaiat]

The problem says that the commutator is a c-number function. So, you may define the function $$[\varphi (x) , \varphi (y)] = F(x,y),$$ and try to discover its properties:

1) Translation invariance implies that $F(x + a , y + a) = F(x,y)$, meaning that $F$ can only be a function of $(x - y)$, i.e., $[\varphi (x) , \varphi (y)] = F (x - y),$ or $$[\varphi (x) , \varphi (0)] = F(x) .$$

2) Lorentz invariance of the commutator implies that $$F(\Lambda x) = F (x).$$

3) Since $[\varphi (x) , \varphi (0)] = - [\varphi (0) , \varphi (x)]$, $F(x)$ must be an odd function, i.e., $$F(x) = - F(-x) .$$

4) Since, $\varphi^{\dagger}(x) = \varphi (x)$, we conclude $$F^{\ast}(x) = - F (x).$$ So, by considering (3), we write $$F^{\ast}(x) = F(-x) = - F(x) . \ \ \ \ \ \ (3 + 4)$$

5) Since $\varphi (x)$ satisfy the Klein-Gordon equation, we conclude that $$(\partial^{2} + m^{2})[\varphi (x) , \varphi (0)] = 0 = (\partial^{2} + m^{2})F(x) .$$

Thus, In one sentence, $F(x)$ is an odd invariant solution of the Klein-Gordon equation.

Now, (5) allows you to write $$F(x) = \int d^{4}p \ e^{-ip \cdot x} \ \delta (p^{2} - m^{2}) \ G(p) , \ \ \ \ \ \ (i)$$ where $G(p)$ is defined only on the two hyperboloids $p_{0} = \pm \sqrt{\mathbf{p}^{2} + m^{2}}$. This means that $p$ is a time-like vector.

Next, show that (2) implies $$G(\Lambda p) = G(p).$$ Now, since $p^{2}$ and $\epsilon (p_{0})$ are the only two invariants which can be formed from the time-like vector $p$, we may write $$G(p) = G_{1}(p^{2}) + \epsilon (p_{0}) \ G_{2}(p^{2}) . \ \ \ \ \ (ii)$$ Now, if you put (ii) in (i) and use the property $f(x) \delta (x-a) = f(a)\delta (x-a)$, you get $$F(x) = \int d^{4}p \ \delta (p^{2} - m^{2}) \left( C_{1} + \epsilon(p_{0}) C_{2}\right) \ e^{- i p \cdot x} ,$$ where $C_{n} = G_{n}(m^{2}), \ n = 1,2$ are constants.

Now, from the oddness of $F(x)$(property 3 above) you can easily show that $$C_{1} = - C_{1}, \ \Rightarrow \ \ C_{1} = 0.$$

Finally, use (3+4) to show that $C_{2} = C^{\ast}_{2}$, meaning that $C_{2}$ is a real constant. Thus, you end up with $$[\varphi (x) , \varphi (0)] = C_{2} \int d^{4}k \ e^{-ikx} \ \epsilon (k_{0}) \ \delta (k^{2} - m^{2}) .$$

 

[vanhees71]

Finally $C_2$ is determined from the equal-time canonical commutation relation
$$[\Pi(0,\vec{x}),\varphi(0)]=[\dot{\varphi}(0,\vec{x}),\varphi(0)]=-\mathrm{i} \delta^{(3)}(\vec{x}).$$