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 $\hat{b}_p$?

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    $\begingroup$ In short, the textbook is cheating. Simply finding a pair of operators that satisfy the commutation relation isn't sufficient, you also have to use up all the degrees of freedom in the field. Otherwise what you wind up with will only be an incomplete description of the field's behaviour, e.g., you'll leave out some of the particle types. :-) (Now, how exactly you know when you've used up all the degrees of freedom I have no idea.) $\endgroup$ Commented Nov 27, 2016 at 0:09
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    $\begingroup$ In short, OP's ansatz (6) is lacking the negative frequency/energy modes. $\endgroup$
    – Qmechanic
    Commented Aug 29, 2018 at 11:03
  • 1
    $\begingroup$ Although ACuriousMind's answer is interesting, the real reason is given by Elias Riedel Gårding. Indeed the problem already arises before quantization. In the draft of the book by Srednicki (3.21) p.40, one finds the expression of $a$ function of $\varphi$ and $\pi$. (He previously uses $\varphi^*=\varphi$ to identify $b=a^*$, but if one does not, one could find a different explicit formula for $b$) $\endgroup$
    – Noix07
    Commented Jan 9, 2020 at 15:57

5 Answers 5


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.


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.

  • $\begingroup$ Why, then, do we get a single creation operator when doing the same procedure to the Schrodinger equation to get a Schrodinger field? There, we have a complex valued wavefunction (thought of as a classical field), but a single creation operator suffices to determine the field operator. $\endgroup$ Commented Nov 26, 2016 at 16:33
  • $\begingroup$ @JahanClaes I'm not sure what a Schrödinger field is but I'm pretty sure whatever you start with is not a classical Hamiltonian system to begin with so your "quantization" is entirely ad hoc and not an instance of canonical quantization. $\endgroup$
    – ACuriousMind
    Commented Nov 26, 2016 at 16:36
  • $\begingroup$ A Schrodinger field is a classical field that obeys the Schrodinger equation. It has Hamiltonian $\mathcal{H}=\pi(1-\frac{d}{dt}-i\nabla^2)\psi$, which I guess should be viewed as two coupled real fields? $\endgroup$ Commented Nov 27, 2016 at 15:21
  • $\begingroup$ @JahanClaes That's not the correct Hamiltonian, Hamiltonians do not contain time-derivatives. In any case, the correct canonical momentum of a Schrödinger field is just $\pi = \mathrm{i}\psi^\dagger$ so $\pi$ and $\psi$ are not independent and you have a constrained Hamiltonian system and the formal quantization would require applying the Dirac-Bergmann recipe, during which some variables get eliminated. That Wiki seems to take the shortcut here that you can just simply take $\psi$ and $\psi^\dagger$ as your variables is probably the end result, but not a priori justified. $\endgroup$
    – ACuriousMind
    Commented Nov 27, 2016 at 15:44
  • $\begingroup$ How could the Hamiltonian not contain a time derivative? The canonical momentum, as you pointed out, does not have a time derivative, so you can't eliminate the derivative of $\psi$ in favor of $\pi$. Do you mean you just move the time derivate onto the $\pi$ via integration by parts? $\endgroup$ Commented Nov 27, 2016 at 17:23

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:


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.

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    $\begingroup$ This is nice, but it doesn't explain why $\hat a$ should be a bosonic annihilation operator when you quantize the theory. You still need to follow the canonical quantization procedure to get commutation relations, no? $\endgroup$ Commented Nov 27, 2016 at 1:57
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    $\begingroup$ @JahanClaes So that's actually another thing Srednicki shows. You either choose to canonically quantise $\phi$ and $\pi$ like you would for position and momentum, except with the Dirac delta function. Or you can just start with the Hamiltonian you would've got in terms of the $a$'s, and canonically quantise the $a$'s. But it has to be Bosonic for spin zero, and in general integer spin. Otherwise you hit all sorts of problems like the Hamiltonian being constant and things not vanishing outside the lightcone. $\endgroup$
    – snulty
    Commented Nov 27, 2016 at 11:13
  • $\begingroup$ @JahanClaes You can also read the sections for yourself online. On the page I linked, the author makes a draft copy available online, which would be closer to the First Printed edition. In that respect it would also have similar typos, I don't think they're too major in the first few chapters. They're listed on the web page anyway. $\endgroup$
    – snulty
    Commented Nov 27, 2016 at 16:15

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.


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.

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    $\begingroup$ But when we do the same to the Schrodinger equation to get a Schrodinger field, only one creation/annihilation operator is needed, even though the Schrodinger field is complex valued. $\endgroup$ Commented Nov 26, 2016 at 16:34
  • $\begingroup$ The Schrödinger equation is different. Putting $\hbar$ $\endgroup$
    – Nid
    Commented Nov 26, 2016 at 21:04
  • $\begingroup$ For Christ sake. I made a mess. Why can't I delete my comment? The Schrödinger equation doesn't allow for oscillators. The identity $$ i \dot{\Psi} = E \Psi $$ just put them to a constant. For example, have a look at the free particle on en.wikipedia.org/wiki/Schr%C3%B6dinger_equation (you also have to add the time dependent part there). $\endgroup$
    – Nid
    Commented Nov 26, 2016 at 21:09

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