Using Minkowski signature $(+,-,-,-)$, for the Lagrangian density

$${\cal L}=\partial_{\mu}\phi\partial^{\mu}\phi^{\dagger}-m^2\phi \phi^{\dagger}$$

of the complex scalar field, we have the field

$$\phi(x)=\int{\frac{d^3 \vec{p}}{2(2\pi)^3\omega_\vec{p}}}(a(\vec{p})e^{-ipx}+b^{\dagger}(\vec{p})e^{+ipx}).$$

I'm trying to now find an equation for the $a(\vec{p})$ and $b(\vec{p})$ (with the final goal of finding an expression for $[a(\vec{p}),b^{\dagger}(\vec{q})]$ using $[\phi(x),\Pi(y)]$).

What I should note is that we are considering all of this in the Schrodinger picture (t=0) so I suppose the very first thing to do is change all the $x$'s to $\vec{x}$ right?

The strategy I'm struggling to implement and failing at many points along the way:

  1. Find the momentum $\Pi^{\phi}(x)=\frac{\partial L}{\partial \dot{\phi}}=\dot{\phi^{\dagger}}$.

  2. Add some combination of $\phi(x)$ and $\Pi(x)$ to get rid of one of the creation/annihilation operators.

  3. Do an inverse Fourier transform to find $a(\vec{p})$ in terms of $\phi(x)$, for example.

None of the major text books seem to actually carry this through, and instead write something like "and it's easy to show…". However I don't find it too easy, especially part 3. as I'm not a Fourier transform expert.

Could anyone either direct me somewhere where the above is computed explicitly (in more than 2/3 lines), or help me understand each of the 3 steps above?

(I realise it is easy to find a reference where this is done for the real scalar field, in which case we have $a(\vec{p})$ and $a^{\dagger}(\vec{p})$. Even still, I find it hard to follow parts.)

  • 2
    $\begingroup$ So this seems like a very broad question to me, it might be useful to break it into smaller parts. For example, what do you have for part 1 (computing the momentum)? You should find that the momentum conjugate to $\phi^\dagger$ is $\dot{\phi}$ and that the momentum conjugate to $\phi$ is $\dot{\phi}^\dagger$, is that what you find? $\endgroup$ – Andrew Apr 14 '14 at 17:41
  • 1
    $\begingroup$ Yes but you can easily get that just from looking at the Lagrangian. Perhaps I'll add it in then $\endgroup$ – Phibert Apr 14 '14 at 17:43
  • 2
    $\begingroup$ Good :) Just trying to figure out exactly where you are stuck. OK so you can get the momentum, and you know that $[\phi,\Pi] = i \delta$ on equal time slices. Can you express $\Pi$ as a mode expansion of $a,b$ and their daggers? If so what do you get when you plug the mode expansions of $\phi$ and $\Pi$ into $[\phi,\Pi]=i\delta$? Again it's useful for giving an answer to have a clearer idea of exactly what you tried and where things aren't working. $\endgroup$ – Andrew Apr 14 '14 at 17:51

The fields satisfy the wave equation. We can therefore write \begin{equation} \begin{split} \phi(x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ a({\bf p}) e^{i p \cdot x} + b^\dagger({\bf p} ) e^{- i p \cdot x} \right] \\ \phi^\dagger (x) = \int \frac{ d^3 p}{ (2\pi)^3} \frac{1}{2 \omega_{\bf p} } \left[ b({\bf p}) e^{i p \cdot x} + a^\dagger({\bf p} ) e^{- i p \cdot x} \right] \\ \end{split} \end{equation} where $\omega_{\bf p} = \sqrt{ {\bf p}^2 + m^2 } $. Inverting this, we find (VERIFY THIS) \begin{equation} \begin{split} a^\dagger ( {\bf p} ) &= - i \int d^3 x e^{i p \cdot x} \overleftrightarrow{\partial_0} \phi^\dagger(x) \\ b^\dagger ( {\bf p} ) &= - i \int d^3 x e^{i p \cdot x} \overleftrightarrow{\partial_0} \phi(x) \\ a ( {\bf p} ) &= i \int d^3 x e^{-i p \cdot x} \overleftrightarrow{\partial_0} \phi(x) \\ b ( {\bf p} ) &= i \int d^3 x e^{-i p \cdot x} \overleftrightarrow{\partial_0} \phi^\dagger(x) \\ \end{split} \end{equation} where $A \overleftrightarrow{\partial_0} B = A \partial_0 B - (\partial_0 A ) B$.

The conjugate momenta can be determined from the Lagrangian as $$ \pi = \partial_0 \phi^\dagger,~~ \pi^\dagger = \partial_0 \phi $$ The commutation relations in terms of the fields are $$ [ \phi(t, {\bf x}), \pi(t, {\bf y}) ] = i \delta^3 ( {\bf x} - {\bf y} ) $$ Using this information, you should be able to compute the brackets of the mode coefficients.

PS - I should add that I'm using the $(-+++)$ signature for the metric.

  • 5
    $\begingroup$ Nice answer; most likely the OP will also need the well-known identity: \begin{equation} \delta^n(k) = \int \frac{\mathrm{d}^n x }{(2 \pi)^n} \; e^{i k \cdot x} \end{equation} $\endgroup$ – Hunter Apr 14 '14 at 18:05
  • 2
    $\begingroup$ The conceptually easy but 'algebraically hard' way that I did it in my QFT class was simply plug in the Fourier expansion of the field in the commutation relation. Using all field/momentum commutation relations will give you the right relation between all creationg/annihilation operators, no inverting necessary. $\endgroup$ – Danu Apr 14 '14 at 18:12
  • $\begingroup$ @Danu This is what I originally started off doing. A page and a half later and I realised that I was just going to get a some commutation relations of the creation and annihilation operators equal to $i\delta^3(\vec{x}-\vec{y})$? $\endgroup$ – Phibert Apr 14 '14 at 23:41
  • $\begingroup$ @Prahar Good answer. However where you say "verify this", this was the main problem I was having. Where have you got those four equations from? Did you use the method I suggested as number 2) in my question? Is there some general formula for investing a Fourier transform? $\endgroup$ – Phibert Apr 14 '14 at 23:43
  • 1
    $\begingroup$ For example, in the case of Minkowski space, it is standard to choose $\phi_k^\pm = e^{ \mp i k \cdot x }$. The creation and annihilation operators are then defined as $a^\dagger_k = (\phi^+_k, \phi)$ and $a_k = (\phi_k^-, \phi)$. Working out all these explicitly, one obtains the formula that I have described. $\endgroup$ – Prahar Mitra Apr 15 '14 at 1:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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