# Annihilation operators

• A
• etotheipi

#### etotheipi

I don't really know what I'm doing, I'd appreciate some nudges in the right direction. We defined ##\mathcal{S}## as the space of complex solutions to the Klein-Gordon equation, and for any ##\alpha, \beta \in \mathcal{S}## that ##(\alpha, \beta) =-\int_{\Sigma_0} d^3 x \sqrt{h} n_a j^a(\alpha, \beta)## where ##j(\alpha, \beta) = -i(\bar{\alpha} d\beta - \beta d\bar{\alpha})##. The exercise is to show that ##a_i' = \sum_j (\bar{A}_{ij} a_j - \bar{B}_{ij} a_j^{\dagger} )##, where the ##a_i## and ##a_i'## are the annihilation operators on ##\mathcal{S}_p## and ##\mathcal{S}_p'## respectively. So write ##a_i' = (\psi_i' , \Phi)## then\begin{align*}
a_i' &= \int_{\Sigma_0} d^3 x \sqrt{h} n^a j_a(\psi_i', \Phi) \\

&= -i\int_{\Sigma_0} d^3 x \sqrt{h} n^a \big{(} \sum_j \left[ \bar{B}_{ij} \psi_j + \bar{A}_{ij} \bar{\psi}_{j} \right] (d\Phi)_a \\

\end{align*}using the Bogoliubov transformation ##\bar{\psi}_i' = \sum_j \left[ \bar{B}_{ij} \psi_j + \bar{A}_{ij} \bar{\psi}_j \right]##. Is that right, and if so what's the next step toward the result? Thanks and sorry if I'm missing something obvious, I'm not really familiar with any of this subject.

Isn't this more quantum field theory than relativity? Should the thread go in the quantum forum?

The theory is covered in pages 389 to 418 of Wald GR as a preliminary to Hawking radiation, so I figured this sub-forum would make a good home for it. But if you think it would be better suited somewhere else, I don't mind. The theory is covered in pages 389 to 418 of Wald GR as a preliminary to Hawking radiation
Yes, that's a topic that is sort of in between GR and QFT. I think the particular questions you're asking might get better responses in the QM forum, so I'll move this thread there.

• etotheipi
It's obviously "QFT in curved spacetime". So it's both (general) relativistic and quantum.

The theory is covered in pages 389 to 418 of Wald GR
Indeed, there is nothing inherently quantum about Bogoliubov transformation, because it can be viewed as a formalism related to a change of basis in the expansion of a classical field (in flat or curved spacetime).

I don't really know what I'm doing, I'd appreciate some nudges in the right direction. We defined ##\mathcal{S}## as the space of complex solutions to the Klein-Gordon equation, and for any ##\alpha, \beta \in \mathcal{S}## that ##(\alpha, \beta) =-\int_{\Sigma_0} d^3 x \sqrt{h} n_a j^a(\alpha, \beta)## where ##j(\alpha, \beta) = -i(\bar{\alpha} d\beta - \beta d\bar{\alpha})##. The exercise is to show that ##a_i' = \sum_j (\bar{A}_{ij} a_j - \bar{B}_{ij} a_j^{\dagger} )##, where the ##a_i## and ##a_i'## are the annihilation operators on ##\mathcal{S}_p## and ##\mathcal{S}_p'## respectively. So write ##a_i' = (\psi_i' , \Phi)## then\begin{align*}
a_i' &= \int_{\Sigma_0} d^3 x \sqrt{h} n^a j_a(\psi_i', \Phi) \\

&= -i\int_{\Sigma_0} d^3 x \sqrt{h} n^a \big{(} \sum_j \left[ \bar{B}_{ij} \psi_j + \bar{A}_{ij} \bar{\psi}_{j} \right] (d\Phi)_a \\

\end{align*}using the Bogoliubov transformation ##\bar{\psi}_i' = \sum_j \left[ \bar{B}_{ij} \psi_j + \bar{A}_{ij} \bar{\psi}_j \right]##. Is that right, and if so what's the next step toward the result? Thanks and sorry if I'm missing something obvious, I'm not really familiar with any of this subject.
If I understand correctly, the bar means complex conjugate?
Then if that's right I would proceed as follow:
Let ##\{\psi_n\}## be a complete orthonormal set of solutions of the KG equation, complete in the sense that any solution can be written as a linear combination of ##\psi_i## and ##\bar{\psi}_i##.
Let ##\{\psi'_n\}## be another complete orthonormal set.
Then by definition we can write the solutions ##\{\psi'_n\}## in terms of ##\{\psi_n\}##, which will introduce the coefficients ##A_{ij}## and ##B_{ij}##.

Next, you can start with an arbitrary solution ##\phi##, which can be expressed in terms of ##\{\psi_n\}## with coefficients ##a_i## or in terms of ##\{\psi'_n\}## with coefficients ##a_i'##, substituting the relations between ##\{\psi_n\}## and ##\{\psi'_n\}## and using that both sets are complete and orthonormal, you should be able to find a relation between the ##a## coefficients in terms of the ##A,B## coefficients.

Maybe I'm wrong, but if this idea works, notice that (once you have the sets ##\{\psi_n\}## and ##\{\psi'_n\}##) you don't even need to use the KG equation neither the expression for the inner product.

• JD_PM and etotheipi
I highly recommend Winitzki's lecture notes on this topic.

• Demystifier, vanhees71 and etotheipi
Ah okay, I think I see how to do it in light of @Gaussian97's post. I think that, from the definition, the inner product is conjugate linear in the first argument and linear in the second, i.e. ##(u \psi_i, v\psi_j) = \bar{u}v(\psi_i, \psi_j)##. We also have the orthogonality relations ##(\psi_i, \bar{\psi}_j) = 0##, then ##(\psi_i, \psi_j) = \delta_{ij}##, and finally ##(\bar{\psi}_i, \bar{\psi}_j) = -\delta_{ij}##. Now express $$\Phi = \sum_i (a_i \psi_i + a_i^{\dagger} \bar{\psi}_i)= \sum_i (a_i' \psi_i' + a_i'^{\dagger} \bar{\psi}_i')$$then using the orthogonality relations we can write ##a_i' = (\psi_i', \Phi)##. The rest is just using the orthogonality and the (conjugate)-linearity; the ##(\psi_i, \bar{\psi}_j)## terms all get killed i.e.\begin{align*}
a_i' &= \left( \sum_j (A_{ij} \psi_j + B_{ij} \bar{\psi}_j), \sum_k (a_k \psi_k + a_k^{\dagger} \bar{\psi}_k) \right) \\
&= \left( \sum_j A_{ij} \psi_j, \sum_k a_k \psi_k \right) + \left( \sum_j B_{ij} \bar{\psi}_j, \sum_k a_k^{\dagger} \bar{\psi}_k \right) \\

&= \sum_k \sum_j \bar{A}_{ij} a_k (\psi_j, \psi_k) + \sum_k \sum_j \bar{B}_{ij} a_k^{\dagger}(\bar{\psi}_j, \bar{\psi}_k) \\

&= \sum_j (\bar{A}_{ij} a_j - \bar{B}_{ij} a_j^{\dagger})

\end{align*}having used the transformation ##\psi_i' = \sum_j (A_{ij} \psi_j + B_{ij} \bar{\psi}_j)## in the first line. How does that look? Last edited by a moderator:
• JD_PM and Gaussian97
It seems ok, you got exactly what you were supposed to obtain, right?
Just a little technical detail, be careful with using the same index ##i## for the free index and for the dummy index in the ##\Phi## expansion since can easily lead to confusions.

• vanhees71 and etotheipi
Just a little technical detail, be careful with using the same index ##i## for the free index and for the dummy index in the ##\Phi## expansion since can easily lead to confusions.
Oh yeah that was careless, I didn't even realize. I think I have fixed it now!

• vanhees71