Tuesday, 30 October 2018

commutator - Quantizing a complex Klein-Gordon Field: Why are there two types of excitations?


In most references I've seen (see, for example, Peskin and Schroeder problem 2.2, or section 2.5 here), one constructs the field operator $\hat{\phi}$ for the complex Klein-Gordon field as follows:


First, you take the Lagrangian density for the classical Klein-Gordon field


$$ \mathcal{L}=\partial_\mu \phi^\dagger\partial^\mu\phi-m^2\phi^\dagger\phi \tag{1} $$ and find the momentum conjugate to the field $\phi$ via


$$ \pi=\frac{\partial\mathcal L}{\partial \dot\phi}=\dot\phi^\dagger.\tag{2} $$ Then, one imposes the usual canonical commutation relations on $\hat\phi$ and $\hat\pi$:


$$ [\hat\phi(x),\hat\pi(y)]=i\delta^3(x-y).\tag{3} $$ So, one needs to find operators $\hat{\phi}$ and $\hat\pi$ such that they obey the above commutation relations, and such that $\hat\pi=\dot\phi^\dagger$. The textbooks then go on to show that defining


$$ \hat{\phi}(x)=\int\frac{d^3\vec{p}}{(2\pi)^3}\frac{1}{\sqrt{2p_0}}[a_p^\dagger e^{-i p_\mu x^\mu}+b_pe^{i p_\mu x^\mu}]\tag{4} $$ $$ \hat{\pi}(x)=i\int\frac{d^3\vec{p}}{(2\pi)^3}\sqrt{\frac{p_0}{2}}[a_p e^{i p_\mu x^\mu}-b_p^\dagger e^{-i p_\mu x^\mu}]\tag{5} $$ where $a$ and $b$ are bosonic annihilation operators, satisfies these properties.


My question is: Why do we need two different particle operators to define $\hat\phi$ and $\hat\pi$? It seems to me that one could simply define


$$ \hat{\phi}(x)=\int\frac{d^3\vec{p}}{(2\pi)^3}\frac{1}{\sqrt{2p_0}}a_p e^{-i p_\mu x^\mu}\tag{6} $$ $$ \hat{\pi}(x)=i\int\frac{d^3\vec{p}}{(2\pi)^3}\sqrt{\frac{p_0}{2}}a_p^\dagger e^{i p_\mu x^\mu}\tag{7} $$ with $\hat{a}_p$ a single bosonic annihilation operator. Then clearly $\hat{\pi}=\dot{\hat{\phi}}^\dagger$, and also


$$ \begin{array}{rcl} [\hat\phi(x),\hat\pi(y)]&=&i\int\frac{d^3p}{(2\pi)^3}\frac{d^3q}{(2\pi)^3}\sqrt{\frac{q_0}{4p_0}}e^{i(q_\mu y^\mu-p_\mu x^\mu)}[a_p,a_q^\dagger]\\ &=&i\int\frac{d^3p}{(2\pi)^3}\frac{d^3q}{(2\pi)^3}\sqrt{\frac{q_0}{4p_0}}e^{i(q_\mu y^\mu-p_\mu x^\mu)}(2\pi)^3\delta^3(p-q)\\ &=&i\int\frac{d^3p}{(2\pi)^3}\frac{1}{2}e^{ip_\mu (y^\mu-x^\mu)}\\ &=&\frac{i}{2}\delta^3(y-x)\\ \end{array}\tag{8} $$



which is, up to some details about normalizing the $\hat{a}_p$, correct. We would then have a Klein-Gordon field with just one kind of excitation, the $\hat{a}_p$ excitation. Why do all textbooks claim we need two separate bosonic excitations, $\hat{a}_p$ and $\hat{b}_p$?



Answer



The point is that the quantization procedure is usually only valid for real-valued physical observables. All versions of treat the classical observables as real functions on phase space (things get more complicated for fermions, which I will ignore for this issue), and associate quantum observables to those. For instance, the harmonic oscillator annihilation operator $a = x + \mathrm{i}p$ is not really an object one is allowed to look at in classical Hamiltonian mechanics - complex valued functions do not occur, or rather, they are no different from just a pair of real-valued functions that represent the real and imaginary part.


Therefore, to quantize a complex scalar field $\phi$, we must write it as $\phi = \phi_1(x) + \mathrm{i}\phi_2(x)$, and quantize both of the real scalar field separately. This yield the usual mode expansion of the complex scalar field with two different sets of creation/annihilation operators. For a real field, we can treat $a_p$ and $a^\dagger_p$ as operators because can obtain them from the Fourier transform of the fields $\phi(x)$ and $\pi(x)$, which are real-valued and hence operators after quantization. Both the Fourier transform and the computation of $a_p$ and $a_p^\dagger$ must be thought of as being carried out after quantization to be consistent with the derivation of the commutation relations of $a_p,a_p^\dagger$ from the CCR of $\phi$ and $\pi$.


Additionally, note that your attempt is inconsistent with the quantization of the real scalar field in another way: When we impose $\phi = \phi^\dagger$ on your scalar field, we also get $a = a^\dagger$ because $\dot{\phi} = \dot{\phi}^\dagger = \pi$ in that case, which contradicts their non-zero commutation relation. So your version of the quantization of the complex scalar field does not reduce to the quantization of the real scalar field, and is hence an entirely different quantization prescription.


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