I'm looking for a Lorentz covariant expression of Noether charges and I found this article: https://arxiv.org/abs/hep-th/0701268, section II-A in particular.

Consider specifically eq. (20-21), they claim: $$Q_\mu=\frac{1}{2}(\phi, P_\mu \rhd \phi),$$ $Q$ is the conserved charge, "$\rhd$" is just an "acting on" symbol and the inner product is defined by ($\varepsilon$ is the sign function): $$ (\phi_1, \phi_2)=\int d^4p \, \delta(p^2-m^2) \varepsilon(p_0)\tilde{\phi}_1^*(-p)\tilde{\phi}_2(p). $$ $\tilde{\phi}$ is the fourier transform of $\phi$.

Hence the Noether Charge is \begin{equation} \label{1} \tag{1} Q_\mu = \frac{1}{2}\int d^4p \, \delta(p^2-m^2) \varepsilon(p_0)p_\mu\tilde{\phi}^*(-p)\tilde{\phi}(p). \end{equation}

Now I'm struggling to get the well known quantised expression in QFT: $$ Q_\mu= \int d^3p\,\, p_\mu \,\,a^\dagger (\vec{p}) \,a (\vec{p}), \,\,\,\,[a^\dagger (\vec{p}), \,a (\vec{q})]=-\delta^3 (\vec{p}-\vec{q}) $$ from (1), by plugging in usual scalar Klein-Gordon field with creation and annihilation operators.

If I'm not wrong (1) in coordinate space looks like $$ \int d^4x \,\,d^4y \,\, \phi (x) \Delta (x-y)\left( -i\frac{\partial \phi(y)}{\partial y^\mu}\right)=Q_\mu, \tag{2} $$ where $\Delta$ is the usual commutator function $$\Delta (x-y)=\int d^4p \,\, \varepsilon(p_0) \delta (p^2-m^2)e^{-ip\cdot (x-y)}.$$ Just by substituting in (2) $$\phi(x)= \int \frac{d^3p}{\sqrt{2\omega}_\vec{p}}(a(\vec{p}) e^{-ip\cdot x}+a^\dagger(\vec{p}) e^{ip\cdot x} )$$ I don't seem to be getting the right answer.

Maybe I'm doing some calculation wrong or misinterpreted the article. Any help would be greatly appreciated!


For instance writing $\phi(x)=\int d^4p \,\, a(p) \delta(p^2-m^2) e^{-ip \cdot x}$ then $\tilde{\phi}(p)=a(p)\delta(p^2-m^2)$ and (1) becomes $$ Q_\mu= \frac{1}{2} \int d^4p \,\, \varepsilon(p_0) \, p_\mu a(-p)a(p)\,\,\delta(p^2-m^2)\delta(p^2-m^2)\delta(p^2-m^2). $$ Is this right? How to work out the three deltas? I could use the identity $\delta(x)f(x)=\delta(x)f(0)$ with $f=\delta$ twice to get$$ \delta(p^2-m^2)\delta(p^2-m^2)\delta(p^2-m^2)=\delta(p^2-m^2)\delta(0)\delta(0)=\delta(p^2-m^2)\cdot \mathcal{S}, $$ where $\mathcal{S}$ is an (infinite) surface contribution which I currently fail to see how cancels out. what am I missing?


1 Answer 1


The standard convention is to take $\tilde \phi(p)\equiv \delta(p^2-m^2)a(\vec p)$. It should be clear that this expression cannot be correct in this context, precisely for the reason OP found: the scalar product $(\phi_1,\phi_2)$ would be undefined, inasmuch as it would contain several $\delta(0)=\infty$ factors. Therefore, in this context, the notation $\tilde \phi(p)$ must have a different meaning than the standard one; a meaning that does not include on-shell Dirac deltas, so as to avoid the factors of $\delta(0)=\infty$. The $\delta(p^2-m^2)$ in the definition of $(\phi_1,\phi_2)$ already puts $p$ on-shell, so it would be redundant to include another delta in $\tilde\phi(p)$.

It appears that, in this context, $\tilde\phi(\omega_{\vec p},\vec p)$ is just a synonym for $a(\vec p)$: $$ \tilde\phi(p)\delta(p^0-\omega_{\vec p})\equiv a(\vec p)\delta(p^0-\omega_{\vec p}) $$

With this, and using (cf. this PSE post) $$ \delta(p^2-m^2) \Theta(p_0)\,\mathrm dp^0=\frac{1}{2\omega(\vec p)}\delta(p^0-\omega_{\vec p})\,\mathrm dp^0 $$ you get \begin{equation} \begin{aligned} Q^\mu &= \frac{1}{2}\int \delta(p^2-m^2) \varepsilon(p^0)p^\mu\tilde{\phi}^*(-p)\tilde{\phi}(p)\,\mathrm dp\\ &=\int \frac{1}{2\omega(\vec p)} p^\mu\tilde{\phi}^*(-\omega_{\vec p},-\vec p)\tilde{\phi}(\omega_{\vec p},\vec p)\,\mathrm d\vec p\\ &=\int p^\mu a^*(\vec p)a(\vec p)\,\widetilde{\mathrm dp} \end{aligned} \end{equation} where $$ \widetilde{\mathrm dp}\equiv \frac{1}{2\omega(\vec p)}\mathrm d\vec p $$ is the Lorentz-invariant measure.

Note that $Q^\mu$ agrees with the standard expression for the generator of space-time translations, up to the conventional Fourier factor $(2\pi)^3$, which is usually included in $\widetilde{\mathrm dp}$.

  • $\begingroup$ A silly note for anyone interested: maybe you misinterpreted my notation, $\varepsilon$ is the sign function. So $\varepsilon(p_0)=\Theta(p_0)-\Theta(-p_0)$. Then all you wrote is correct except for an extra term that, after all, is unimportant when normal ordering the operators. $\endgroup$
    – user78618
    Commented Aug 2, 2017 at 15:34
  • 1
    $\begingroup$ @rhetoricalphysicist I'm glad I could help :-) [BTW, yes, I thought that $\varepsilon$ was the step function, so my result was off by a factor of $\frac12$; I've fixed it, I hope it is now OK.] $\endgroup$ Commented Aug 2, 2017 at 16:04

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