How can we show that the annihilation operators satisfy the given equation?

[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 \\

&\quad\quad\quad\quad- \Phi \sum_j \left[ \bar{B}_{ij} (d\psi_j)_a + \psi_j (d\bar{B}_{ij})_a + \bar{A}_{ij} (d\bar{\psi}_j)_a + \bar{\psi}_j (d\bar{A}_{ij})_a \right] \big{)}
\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.

 

[PeterDonis]

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

 

[etotheipi]

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.

 

[PeterDonis]

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.

 

[vanhees71]

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

 

[Demystifier]

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).

 

[Gaussian97]

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 \\

&\quad\quad\quad\quad- \Phi \sum_j \left[ \bar{B}_{ij} (d\psi_j)_a + \psi_j (d\bar{B}_{ij})_a + \bar{A}_{ij} (d\bar{\psi}_j)_a + \bar{\psi}_j (d\bar{A}_{ij})_a \right] \big{)}
\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.

 

[haushofer]

I highly recommend Winitzki's lecture notes on this topic.

 

[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?

 

[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.

 

[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!