The Hamiltonian of a free real scalar field theory $$\mathcal{L} = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2 \phi^2$$ after quantization is \begin{equation} H = \int \frac{\mathrm{d}^3p}{(2\pi)^3} \: E_\mathbf{p} a_\mathbf{p}^\dagger a_\mathbf{p} \end{equation} with $E_\mathbf{p} = \sqrt{\mathbf{p}^2 + m^2}$ and \begin{equation} a_\mathbf{p} = \int \mathrm{d}^3 x \: \bigg( \sqrt{\frac{E_\mathbf{p}}{2}} \phi(x) + \frac{i}{\sqrt{2E_\mathbf{p}}} \pi(x) \bigg) e^{ip \cdot x} \end{equation} where $\pi(x)$ is the canonical momentum. Obviously, the charge \begin{equation} Q = \int \frac{\mathrm{d}^3 p}{(2\pi)^3} \: a_\mathbf{p}^\dagger a_\mathbf{p}, \end{equation} which corresponds to the total number of particles, is conserved, as $[H, Q] = 0$. In complex scalar field theory the analogous conserved charge is the number of particles minus the number of antiparticles. There we get the conserved charge (classically) from the symmetry of the action under phase rotation \begin{equation} \psi \rightarrow e^{i\varphi} \psi \end{equation} via Noether's theorem. This conserved charge carries over to the quantum theory. Now as in real scalar field theory we can't rotate any phase, because the field is real, the conservation law for $Q$ cannot come from the same symmetry as in the complex case. Therefore I was wondering, which symmetry corresponds to the conservation law for $Q$ in the real scalar field theory?
2 Answers
There might be a more formal answer, but at least morally the reason number is conserved for a free scalar field is a combination of the Hamiltonian being quadratic and energy conservation. The fact that the Hamiltonian is quadratic (or equivalently that the equations of motion are linear, or equivalently that there are no interactions) means that there are no transitions between modes. So the energy in each mode is constant. The particle number in a given mode is proportional to the energy of that mode. So, mode by mode, particle number in that mode is conserved because energy in that mode is conserved. Then you just sum over modes and particle number is conserved.
If you add interactions, it's easy to violate particle number conservation. For example, in a $\lambda \phi^3$ theory (where you add a $\phi^3$ term), you can have processes where two $\phi$ particles annihilate into one $\phi$ particle.
Also, if you violate time translation invariance, you will violate particle number conservation. For example, if you add a time dependent mass to the scalar field, you can have particle production. See, eg, Section 4 of these notes https://www.astro.umd.edu/~ricotti/NEWWEB/teaching/ASTR688s08/Ryan_CosmologyPaper.pdf. Eq 29 describes an oscillator with a time-dependent mass, and Eq 38 shows the result of a calculation for the number of particles created.
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Notice that the particle number $$N~=~\sum_ka^{\ast}_k a_k\tag{1}\label{eq:1}$$ is formulated in terms of complex Fourier modes $a_k$ & $a^{\ast}_k$ (as opposed to the real field $\phi$). So to have an honest discussion/application of Noether's theorem, we should also formulate the action for a real field theory in terms of complex modes. The main point is that the complex modes still have complex phase transformations even if the real$^1$ field has none.
In detail, the Hamiltonian action for a real field takes the form $$\begin{align} S_H~=~&\int\!\mathrm{d}t L_H, \cr L_H~=~&\frac{i}{2}\sum_k(a^{\ast}_k\dot{a}_k-a_k\dot{a}^{\ast}_k)-H(a,a^{\ast}). \end{align} \tag{2}\label{eq:2}$$ The non-zero fundamental Poisson bracket is$^2$ $$ \{a_k,a^{\ast}_{k^{\prime}}\}_{PB}~=~-i\delta_{k,k^{\prime}}.\tag{3}$$ The pertinent infinitesimal transformation $$\begin{align} \delta a_k~=~& \epsilon\{a_k,N\}_{PB} ~\stackrel{(1)+(3)}{=}~-i\epsilon a_k, \cr \delta a^{\ast}_k~=~& \epsilon\{a^{\ast}_k,N\}_{PB} ~\stackrel{(1)+(3)}{=}~+i\epsilon a^{\ast}_k, \end{align}\tag{4}\label{eq:4}$$ is generated by the particle number \eqref{eq:1}, cf. e.g. this related Phys.SE post. So if the complex phase transformation \eqref{eq:4} is a quasisymmetry of the action \eqref{eq:2}, then the Noether current \eqref{eq:1} is conserved $$ \{N,H\}_{PB}~=~0. \tag{5}$$
$^1$ By the way, 1 complex field is equivalent to 2 real fields.
$^2$ The corresponding non-zero fundamental commutator is $$ [\hat{a}_k,\hat{a}^{\dagger}_{k^{\prime}}]~=~\hbar\delta_{k,k^{\prime}}\hat{\bf 1}.\tag{6}$$
