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Along the lines of Birrell and Davies, which contrary to Mukhanov-Winitzki (which is actually my level) gives a quite general but short account on Bogolubov transformations, I tried to follow the formula and obtained a little different result. Before thinking too much about it I rather ask here, but first a summary of the notation and logic:

  1. Suppose there are two "complete" sets of solutions $\{u_i\}, \{\bar{u}_i\}$ to some eigenvalue problem that enable us to expand a general solution to the operator equation with periodic boundary conditions (see my question Scalar product for scalar field mode functions ) as $\phi(x) = \sum \left[ a_i u_i(x) + a^\dagger_i u^*_i(x)\right]$ and $\phi(x) = \sum \left[ \bar{a}_i \bar{u}_i(x) + \bar{a}^\dagger_i \bar{u}^*_i(x)\right]$ and there is further some scalar product and both sets are orthonormal, i.e. $(u_i,u_j)=\delta_{ij},(u_i^*,u_j^*)=-\delta_{ij}, (u_i,u_j^*)=0$ and similarly for the other set.

  2. It is clear that we must have $\bar{u}_j = \sum_i (\alpha_{ji}u_i + \beta_{ji}u_i^*)$ for some coefficients.

  3. It is also clear that these coefficients are given by $\alpha_{ij} = (\bar{u}_i,u_j), \beta_{ij} = -(\bar{u}_i,u_j^*) $

  4. It then follows from the properties of the scalar product that the expansion of $u$ in terms of $\bar{u}$ has related coefficients, $u_i = \sum_j (\alpha_{ji}^*\bar{u}_j - \beta_{ji}\bar{u}_j^*)$

  5. From equating both $\phi$'s and doing the and the other replacements it is then clear that $a_i = \sum_j(\alpha_{ji}\bar{a}_j+\beta_{ji}^*\bar{a}^\dagger_j)$ and $\bar{a}_j = \sum_i(\alpha_{ji}^*a_i-\beta_{ji}^*a^\dagger_i)$

  6. From demanding that both $a$ and $\bar{a}$ have the right commutation relations we find a condition for the Bogolubov coefficients.

There are two ways to do this (starting with the commutation relation of $a$ or $\bar{a}$) and what I find is:

  1. $\delta_{ik}=\sum_n ( \alpha_{kn} \alpha_{in}^* -\beta_{in}\beta^*_{kn})$ and
  2. $\delta_{ik}=\sum_n ( \alpha_{ni} \alpha_{nk}^* -\beta_{nk}\beta^*_{ni})$

This suggests that $\alpha_{ij} = \alpha_{ji}^*$. Question: is this true? How to see it? (Maybe it is very easy, but I am too tired now)

Then, what Birrell-Davies says, is instead

$\delta_{ik}=\sum_n ( \alpha_{in} \alpha_{kn}^* -\beta_{in}\beta^*_{kn})$

Now, this would only be true (modulo calculation mistakes I made) if $\sum_n \alpha_{in} \alpha_{kn}^*$ is real! Question: is this true? How to see it?

Thank you!

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First question

No, it is almost never true that $\alpha_{ij}=\alpha^*_{ji}$. This statement would mean that the matrix $\alpha$ is Hermitian. You don't have any evidence for that; at most, you have some suggestion that it could be naturally unitary. But even this is not true if the creation and annihilation operators mix, and this mixing is the whole point of Bogoliubov transformations. In that general case, the right unitarity-like identities are obeyed by $\alpha,\beta$. Both matrices appear in the equations and you actually wrote what these conditions are.

Note that $i,j$ in $\alpha_{ij}$ are indices of a completely different type, corresponding to two completely different, independent, and arbitrary bases, so exchanging the values of these two indices means to mix apples with oranges.

Second question

In the same way, it is not true that $\sum_n\alpha_{in}\alpha^*_{kn}$ is real. It is not clear at all why you decided it should be. You may easily see that this conjecture is indefensible because you may change the phase of $\alpha_{in}$ in a way that depends on $i$, i.e. is different for $i\to k$, and then the phase of the sum changes, too. But it's still an equally plausible set of coefficients $\alpha_{ij}$.

The formula you attribute to Birrell-Davies is actually identical to the formula you identify as "1" a few lines earlier or "2" on the following line. The only difference is that it uses a different convention: all entries of $\alpha^*$ (but not $\beta$) or vice versa are complex conjugated in one convention relatively to the other.

It is also pretty clear that the incoherence in the notation was introduced by yourself. On the line "2. It is clear we must have... for some coefficients" you arbitrarily labeled the coefficients $\alpha$ and $\beta$ and then you assumed they're the same ones as $\alpha,\beta$ used by others. That clearly doesn't have to be the case and it isn't the case here, either. A more conventional way to write your point "2" would probably have $\beta^*$ instead of $\beta$ in the second term.

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Thank you, this clarified things a lot. –  Andreas Finke Oct 9 '12 at 17:39
    
I concluded that the sum must be real because when I equate Birrell and Davies line with my 1. version the beta-part drops out. Then I can move out the complex conjugate of the remaining alpha-sum and I remain with the statement that it is real. The part that confused me here is that Birrell and Davies itself have some inconsistent notation on one page then: they introduce what you call uncommen notation themselves. I just checked what I typed here. I copied all expressions exactly as on page 46 of a 1984 edition I have here. –  Andreas Finke Oct 9 '12 at 17:43
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