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The canonical commutation relations for a complex scalar field are of the form

$$[\phi(t,\vec{x}),\pi(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$$ $$[\phi^{*}(t,\vec{x}),\pi^{*}(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$$

How can these commutation relations be obtained from the commutation relations for two free real scalar fields?

nightmarish
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  • Comment to the post (v3): Why did you change the CCRs/ move the star? Did you originally get this from some reference? – Qmechanic Sep 18 '16 at 22:20
  • Yes, I got them from David Tong's lecture notes in http://www.damtp.cam.ac.uk/user/tong/qft.html. The commutation relations are in equation (2.71) in page 34 of his notes. – nightmarish Sep 18 '16 at 22:47
  • But what made you write it differently in v1? Was it just a typo? – Qmechanic Sep 18 '16 at 23:10
  • Yes. My original commutation relations were $[\phi(t,\vec{x}),\pi^{}(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$ $[\phi^{}(t,\vec{x}),\pi(t,\vec{y})]=i\delta^{(3)}(\vec{x}-\vec{y})$. But I don't think they're correct. – nightmarish Sep 18 '16 at 23:42

1 Answers1

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I) The complex scalar field comes from two real/Hermitian scalar fields with equal-time CCRs

$$[\hat{\phi}^j(t,\vec{x}),\hat{\phi}^k(t,\vec{y})]~=~0, $$ $$[\hat{\phi}^j(t,\vec{x}),\hat{\pi}_k(t,\vec{y})]~=~i\hbar{\bf 1}~\delta^j_k~ \delta^{3}(\vec{x}-\vec{y}), $$ $$[\hat{\pi}_j(t,\vec{x}),\hat{\pi}_k(t,\vec{y})]~=~0, \qquad j,k~\in~\{1,2\}, \tag{A}$$

and the definitions

$$ \hat{\phi}~=~\frac{1}{\sqrt{2}}(\hat{\phi}^1+i\hat{\phi}^2),\tag{B}$$ $$ \hat{\pi}~=~\frac{1}{\sqrt{2}}(\hat{\pi}_1\color{red}{-}i\hat{\pi}_2),\tag{C}$$

cf. e.g. this Phys.SE post. This leads to OP's mentioned CCRs.

II) If the $\color{red}{\text{minus sign}}$ in eq. (C) seems strange, consider the following classical argument. The Lagrangian density is $$ {\cal L}~=~|\dot{\phi}|^2 - |\nabla \phi |^2 - {\cal V} ~=~\frac{1}{2}(\dot{\phi}^1)^2+\frac{1}{2}(\dot{\phi}^2)^2-\frac{1}{2}(\nabla \phi^1)^2-\frac{1}{2}(\nabla \phi^2)^2 - {\cal V} .\tag{D} $$ Therefore the momenta read

$$ \pi_j~=~\frac{\partial {\cal L}}{\partial \dot{\phi}^j}~=~\dot{\phi}^j, \qquad j~\in~\{1,2\}, \tag{E}$$

$$ \pi~=~\frac{\partial {\cal L}}{\partial \dot{\phi}} ~=~\frac{1}{\sqrt{2}}\left(\frac{\partial {\cal L}}{\partial \dot{\phi}^1}\color{red}{-}i \frac{\partial {\cal L}}{\partial \dot{\phi}^2} \right) ~=~\dot{\phi}^{\ast} ~=~\frac{1}{\sqrt{2}}(\pi_1\color{red}{-}i\pi_2).\tag{F}$$

III) For reference, let us mention that the Hamiltonian Lagrangian density reads

$$ {\cal L}_H~=~\pi \dot{\phi}+\pi^{\ast} \dot{\phi}^{\ast} -{\cal H} ~=~\pi_1 \dot{\phi}^1+\pi_2 \dot{\phi}^2 -{\cal H}, \tag{G}$$

where the Hamiltonian density is

$$ {\cal H}~=~|\pi|^2 + |\nabla \phi |^2 + {\cal V}~=~\frac{1}{2}(\pi_1)^2+\frac{1}{2}(\pi_2)^2+\frac{1}{2}(\nabla \phi^1)^2+\frac{1}{2}(\nabla \phi^2)^2 + {\cal V} .\tag{H}$$

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