What are the explicit expressions for the creation and annihilation operators $\hat{a_\vec p}$ and $\hat{a}^{\dagger}_\vec p$ for bosons? I can't find them anywhere, as every source seems to introduce them when quantizing the fields $$\phi(\vec x) = \int \frac{d³p}{(2\pi)³} \frac{1}{\sqrt{2\omega_{\vec p}}} \left( \hat{a_\vec p} e^{i\vec p \cdot \vec x} + \hat{a}^{\dagger}_\vec p e^{-i\vec p \cdot \vec x}\right) $$ $$\pi(\vec x) = \int \frac{d³p}{(2\pi)³} (-i) \sqrt{\frac{\omega_{\vec p}}{2}} \left( \hat{a_\vec p} e^{i\vec p \cdot \vec x} - \hat{a}^{\dagger}_\vec p e^{-i\vec p \cdot \vec x}\right) $$

Without giving an explicit expression for them. I would like to know, because e.g. calculating the Hamiltonian for the Klein-Gordon field requires me to know what effect switching the sign of the momentum has, i.e. what $\hat{a}_{-\vec{p}}$ and $\hat{a}^\dagger_{-\vec{p}}$ are.

  • 3
    $\begingroup$ Are those equations by themselves not a definition of $a_p$? You can invert the Fourier transform if you want it even more explicit. $\endgroup$
    – knzhou
    Jan 22, 2020 at 18:50
  • 2
    $\begingroup$ The interpretation of $a_{k}^\dagger$ for any $k$ is that it creates a particle of momentum $k$. So you automatically know what $a_{-p}^\dagger$ does, it creates a particle of momentum $-p$. $\endgroup$
    – knzhou
    Jan 22, 2020 at 18:51
  • $\begingroup$ You have probably seen this: en.wikipedia.org/wiki/… but it is worth reading, esp. as regards the commutation relations, versus fermions (anti-commutation), which I think could be seen as part of their definition. $\endgroup$
    – user226006
    Jan 22, 2020 at 22:03

1 Answer 1


Note that you can also write $\phi$ and $\pi$ as $$ \phi(\vec{x}) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\vec{p}}}} \left(a_\vec{p} + a_{-\vec{p}}^\dagger \right) e^{i\vec{p} \cdot \vec{x}} $$ $$ \pi(\vec{x}) = \int \frac{d^3 p}{(2\pi)^3} (-i) \sqrt{\frac{\omega_{\vec{p}}}{2}} \left(a_\vec{p} - a_{-\vec{p}}^\dagger \right) e^{i\vec{p} \cdot \vec{x}} $$ (equations (2.27) and (2.28) in Peskin and Schroeder). From here, you can take the Fourier transform to get $$ a_\vec{p} + a_{-\vec{p}}^\dagger = \sqrt{2\omega_{\vec{p}}} \int d^3 x\, \phi(\vec{x}) e^{-i\vec{p} \cdot \vec{x}} $$ $$ a_\vec{p} - a_{-\vec{p}}^\dagger = i\sqrt{\frac{2}{\omega_{\vec{p}}}} \int d^3 x\, \pi(\vec{x}) e^{-i\vec{p} \cdot \vec{x}} $$ and adding these together gives $$ a_\vec{p} = \int d^3 x \left(\sqrt{\frac{\omega_\vec{p}}{2}} \phi(\vec{x}) + \frac{i}{\sqrt{2\omega_\vec{p}}} \pi(\vec{x}) \right) e^{-i\vec{p} \cdot \vec{x}} $$ and then, by Hermitian conjugation, $$ a_\vec{p}^\dagger = \int d^3 x \left(\sqrt{\frac{\omega_\vec{p}}{2}} \phi(\vec{x}) - \frac{i}{\sqrt{2\omega_\vec{p}}} \pi(\vec{x}) \right) e^{i\vec{p} \cdot \vec{x}}. $$

Using these equations, you can explicitly verify the commutation relation $[a_\vec{p}, a_\vec{q}^\dagger] = (2\pi)^3 \delta^{(3)}(\vec{p} - \vec{q})$. In practice, it seems like you rarely need the explicit expressions for the ladder operators; remembering the commutation relation is usually enough.


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