1
$\begingroup$

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.

$\endgroup$
3
  • 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
    Commented 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
    Commented 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
    Commented Jan 22, 2020 at 22:03

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.