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

Question

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 1

There is a conceptually simple (but fiddly) way of relating this expansion to the usual Fourier expansion. TL;DR: Requiring $\phi$ to satisfy the Klein-Gordon equation divides the nonzero Fourier components into two classes, corresponding to particles and antiparticles.

For a general complex scalar field defined on spacetime, $$ \phi(x) = \int\frac{d^4 p}{(2\pi)^4} \hat{\phi}(p) e^{-ip \cdot x}. \label{fourier}\tag{1} $$ This is the ordinary four-dimensional Fourier expansion. Now, let us impose the Klein-Gordon equation $$ 0 = (\partial^2 + m^2) \phi = \int \frac{d^4 p}{(2\pi)^4} (-p^2 + m^2) \hat{\phi}(p) e^{-ip \cdot x}. $$ A general field satisfying this must consist of only modes where $p^2 = m^2$, i.e. on-shell modes. Thus, $$ \hat{\phi}(p) = 2\pi \delta(p^2 - m^2) f(p) $$ for some function $f$ (the factor $2\pi$ is convenient for comparing with the standard expansion). Writing $p = (p^0, \mathbf{p})$, $$ \delta(p^2 - m^2) = \delta\left((p^0)^2 - (\mathbf{p}^2 + m^2)\right). $$ The argument of the $\delta$-function has two zeros $p^0 = \pm \sqrt{\mathbf{p}^2 + m^2}$ (for fixed $\mathbf{p}$), so we use the general rule $$ \delta(f(x)) = \sum_{f(x_i) = 0} \frac{1}{|f'(x_i)|} \delta(x - x_i) $$ to find $$ \delta(p^2 - m^2) = \frac{1}{|2 p^0|} \left[ \delta\left(p^0 - \sqrt{\mathbf{p}^2 + m^2}\right) + \delta\left(p^0 + \sqrt{\mathbf{p}^2 + m^2}\right) \right]. $$ Putting this back into (\ref{fourier}) and performing the $p^0$ integral (abbreviating $\sqrt{\mathbf{p}^2 + m^2} = E_\mathbf{p}$): $$ \begin{align} \phi(x) &= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[ f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x} + f(-E_\mathbf{p}, \mathbf{p}) e^{-i(-E_\mathbf{p}, \mathbf{p}) \cdot x} \right] \\ &= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[ f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x} + f(-E_\mathbf{p}, \mathbf{p}) e^{i(E_\mathbf{p}, -\mathbf{p}) \cdot x} \right] \\ &= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[ f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x} + f(-E_\mathbf{p}, -\mathbf{p}) e^{i(E_\mathbf{p}, \mathbf{p}) \cdot x} \right] \\ \end{align} $$ where we have swapped $\mathbf{p} \mapsto -\mathbf{p}$ in the second term. We identify the usual (Heisenberg picture) expansion $$ \phi(x) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 E_\mathbf{p}}} \left[ a_\mathbf{p} e^{-ip \cdot x} + b^\dagger_\mathbf{p} e^{ip \cdot x} \right] $$ with $$ \begin{align} a_\mathbf{p} &= \frac{1}{\sqrt{2 E_\mathbf{p}}} f(E_\mathbf{p}, \mathbf{p}) \\ b_\mathbf{p}^\dagger &= \frac{1}{\sqrt{2 E_\mathbf{p}}} f(-E_\mathbf{p}, -\mathbf{p}) \end{align} $$ ($f(E_\mathbf{p}, \mathbf{p})$ and $f(-E_\mathbf{p}, -\mathbf{p})$ are the relativistically normalized annihilation and creation operators). If $\phi$ is real, we know from Fourier analysis that $\hat{\phi}(p) = \hat{\phi}(-p)^\dagger$, which immediately translates to $a_\mathbf{p} = b_\mathbf{p}$.

Up until now, the field is entirely classical ($a_\mathbf{p}$ and $b_\mathbf{p}$ are simply complex numbers, and $\dagger$ is complex conjugation). Thus even the classical field has two types of excitations: Positive-frequency solutions (with coefficients $a_\mathbf{p}$) and negative-frequency solutions (with coefficients $b_\mathbf{p}^\dagger$). After quantizing, they correspond to particles and antiparticles.

Answer 2

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.

Answer 3

Firstly and personally, I don't like the first few chapters of Peskin & Schroeder. I think it's a better book for Lecturers than it is for students learning the subject for the first time. (Otoh I think it gets better later on.)

I think it's more instructive to follow Srednicki for example in this instance. Also Srednicki uses $\mathrm{diag}(-1,1,1,1)$ metric convention, but you can in some instances go between metric conventions by inserting appropriate $-1$'s and $i$'s.

First we take a classical scalar field with the intention of Canonical Quantization à la Dirac.

We write down the Lagrangian for a free complex valued scalar field:

$$\mathscr{L}=-\frac{1}{2}\partial^\mu \phi^* \partial_\mu \phi-\frac{1}{2}m^2|\phi|^2+\Omega_0$$

Since at this moment $\phi$ is is just complex valued function, i.e. $\phi(x)$ is a number, we have $\phi^\dagger=\phi^*$, so you could equally well write the Lagrangian with daggers.

We note the equation of motion for $\phi$ is:

$$(-\Box+m^2)\phi=0$$

Now some functions satisfying this pde are $$\exp(i\mathbf{k}\cdot\mathbf{x}\pm i\omega t)$$

where $\mathbf{k}$ is an arbitrary real wave vector, and $\omega$ is: $$\omega=+\sqrt{\mathbf{k}^2+m^2}$$ One can imagine expanding a solution $\phi$ in terms of these plane waves, as you might do when solving other pdes. So we write (even in the case that $\phi$ is real valued) $\phi$ as:

$$\phi(x,t)=\int \frac{d^3k}{f(k)}\left[ a(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}- i\omega t}+b(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}+ i\omega t}\right]$$

with coefficients (not yet operators) $a(\mathbf{k})$ and $b(\mathbf{k})$, since there is no reason yet to assume they should be related. $f(k=|\mathbf{k}|)$ is inserted for the later reason of making the integration measure lorentz invariant and will be proportional to $\omega$.

If $\phi$ was real, $\phi^*=\phi$, then we would have: \begin{align} \phi^*(x,t)&=\int \frac{d^3k}{f(k)}\left[ a^*(\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{x}+ i\omega t}+b^*(\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{x}- i\omega t}\right]\\ &=\int \frac{d^3k}{f(k)}\left[ a^*(-\mathbf{k})e^{+i\mathbf{k}\cdot\mathbf{x}+ i\omega t}+b^*(-\mathbf{k})e^{+i\mathbf{k}\cdot\mathbf{x}- i\omega t}\right]\\ \phi(x,t)&=\int \frac{d^3k}{f(k)}\left[ a(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}- i\omega t}+b(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}+ i\omega t}\right] \end{align}

We simply changed $\mathbf{k}\to -\mathbf{k}$ in line $2$. Comparing we find that $a(\mathbf{k})=b^*(-\mathbf{k})$, or $a^*(-\mathbf{k})=b(\mathbf{k})$. Subbing in this for $\phi$, \begin{align} \phi(x,t)&=\int \frac{d^3k}{f(k)}\left[ a(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}- i\omega t}+a^*(-\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}+ i\omega t}\right]\\ &=\int \frac{d^3k}{f(k)}\left[ a(\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}- i\omega t}+a^*(\mathbf{k})e^{-i\mathbf{k}\cdot\mathbf{x}+ i\omega t}\right]\\ &=\int \frac{d^3k}{f(k)}\left[ a(\mathbf{k})e^{ik_\mu x^\mu}+a^*(\mathbf{k})e^{-ik_\mu x^\mu}\right]\\ \end{align}

To reconcile with P&S you can change $\mathbf{k}\to \mathbf{p}$ with a $\hbar\equiv 1$ in natural units. You can also then calculate $\pi$ by the conjugate momentum and differentiating. You can canonically quantise either $\phi$ and $\pi$ afterward or choose to quantise $a$'s. Change $a^*$ to $a^\dagger$ etc.

Anyway, some differences between real and complex scalar field will be that in the real field you get neutral particles, and the complex field you get two charged particles, of opposite charges. This comes from the symmetry of the phase choice of the field $\phi$ in the lagrangian, which you don't have for the real scalar field. You also have to do some re-interpreting of the $b$ operator in the complex case to avoid having negative energy states, the Dirac sea problem essentially. So $b$ should end up creating anti-particles. You'll see something of this sort with the quantisation of the Dirac field anyway.

Answer 4

The fundamental reason why we need two different species of creation/annihilation operators for complex scalar field is relativity!

A relativistic theory, such as quantum field theory, must be causal. This implies that the amplitude for a particle to being created at $x$ and annihilated at $y$, with $x-y$ space like, must vanish. The particle cannot propagate outside the light cone. Suppose that $$ \phi(x)=\int\widetilde{d^3p}a_p e^{-i p x}\tag1, $$ where $\widetilde{d^3p}$ denotes the invariant measure.Then the amplitude for a particle being created at $\vec x$, propagating and being annihilated at $\vec y$, instantaneously is $$\langle0|\phi(\vec y,t)\phi^\dagger(\vec x,t)|0\rangle=0.$$ Since $\phi(\vec y,t)|0\rangle=0$, then the above condition implies into $$\langle0|[\phi(\vec y,t),\phi^\dagger(\vec x,t)]|0\rangle=0.\tag2$$ On the other hand, plugging (1) into (2) and assuming $[a_p,a_q^\dagger]\propto\delta(\vec p-\vec q)$ we obtain $$\langle0|[\phi(\vec y,t),\phi^\dagger(\vec x,t)]|0\rangle\neq0.$$

In order to obtain that $[\phi(x),\phi^\dagger (y)]$ vanishes outside the light cone but not inside, we must allow for negative energy/frequency plane waves in the field expansion. The last step is to interpret these negative energy plane waves as corresponding to (positive energy) antiparticles with momenta opposite to the corresponding particles. We need therefore two species of creation/annihilation operators, $a_p, a_p^\dagger$ for particles and $b_p,b_p^\dagger$ for antiparticles. One can check that $$\phi(x)=\int\widetilde{d^3p}\left(a_p e^{-i p x}+b_p^\dagger e^{ipx}\right),$$ satisfies $$[\phi(x),\phi(y)^\dagger]=i\Delta(x,y),$$ where $\Delta(x,y)=0$ for space like intervals and $\Delta(x,y)\neq 0$ otherwise.

Answer 5

If $\phi$ only has one independent oscillator in the Fourier decomposition its not the most general solution of the Euler-Lagrange equations (e.o.m's). The field $\phi$ is, before quantisation, just a complex number and should thus have 2 independent degrees of freedom, not one as you write above.