## Bogolubov transformations

Hi,

So in a general curved spacetime we have no preferred choice of modes and the Bogolubov transformations allow us to convert between the fields expanded in the various complete sets of modes.

If we have one set of modes $$f_{i}$$ and another $$g_i$$ both normalized like normalized as $$(f_i,f_j)=\delta_{ij}$$ and $$(f^{*}_i,f^{*}_j)=-\delta_{ij}$$, then we can tranform between the modes as:

$$f_i=\sum_j \left(\alpha^{*}_{ji}g_j-\beta_{ji}g^{*}_j\right)$$, where the Bogolubov coeffs are given as $$\alpha_{ij}=(g_i,f_j)$$ and $$\beta_{ij}=-(g_i,f^{*}_j)$$

Thus it follows that

$$\delta_{ij}=(f_i,f_j)=\left(\sum_m \left(\alpha^{*}_{mi}g_m-\beta_{mi}g^{*}_m\right),\sum_n \left(\alpha^{*}_{nj}g_n-\beta_{nj}g^{*}_n\right) \right)$$

expanding this and using the normalization of the g modes $$(g_i,g_j)=\delta_{ij}$$ and $$(g^{*}_i,g^{*}_j)=-\delta_{ij}$$, others zero:

$$\delta_{ij}= \sum_m \left( \alpha^{*}_{mi}\alpha^{*}_{mj}-\beta_{mi}\beta_{mj} \right)$$

Giving the normalization of the coefficients. However this answer differs to the one quoted in say Birrell and Davies:

$$\delta_{ij}= \sum_m \left( \alpha_{im}\alpha^{*}_{jm}-\beta_{im}\beta^{*}_{jk} \right)$$

Just wondering if anyone can spot how to get this normalization?

thanks alot
 Recognitions: Science Advisor Aren't the modes f_i complex?

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 Quote by LAHLH Hi, So in a general curved spacetime we have no preferred choice of modes and the Bogolubov transformations allow us to convert between the fields expanded in the various complete sets of modes. If we have one set of modes $$f_{i}$$ and another $$g_i$$ both normalized like normalized as $$(f_i,f_j)=\delta_{ij}$$ and $$(f^{*}_i,f^{*}_j)=-\delta_{ij}$$, then we can tranform between the modes as: $$f_i=\sum_j \left(\alpha^{*}_{ji}g_j-\beta_{ji}g^{*}_j\right)$$, where the Bogolubov coeffs are given as $$\alpha_{ij}=(g_i,f_j)$$ and $$\beta_{ij}=-(g_i,f^{*}_j)$$
These equations are inconsistent, since

$$( g_i , f_j ) = \sum_k \alpha^{*}_{kj}(g_i,g_k) = \alpha^*_{ij}.$$

You should double-check the other term as well.

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## Bogolubov transformations

The inner product is linear on one side and antilinear on the other, so when you pull the coefficients out, one side should get complex conjugated.

 Quote by Bill_K The inner product is linear on one side and antilinear on the other, so when you pull the coefficients out, one side should get complex conjugated.
Oh yes, that makes sense from the definition of the inner product here.

So I should have really got:

$$\delta_{ij}= \sum_m \left( \alpha^{*}_{mi}\alpha_{mj}-\beta_{mi}\beta^{*}_{mj} \right)$$

comparing to the B&D result :

$$\delta_{ij}= \sum_m \left( \alpha_{im}\alpha^{*}_{jm}-\beta_{im}\beta^{*}_{jm} \right)$$

These would be equivalent if $$\alpha^{*}_{mi}=\alpha_{im}$$

But this doesn't seem to hold from the inner product and the definition of alpha:

$$\alpha_{im} \equiv (g_i,f_m) =-i\int\, \mathrm{d}x (g_i \nabla f^{*}_m-f^{*}_m\nabla g_i)$$

and
$$\alpha_{mi} \equiv (g_m,f_i) =-i\int\, \mathrm{d}x (g_m \nabla f^{*}_i-f^{*}_i\nabla g_m)$$

so

$$\alpha^{*}_{mi} =+i\int\, \mathrm{d}x (g^{*}_m \nabla f_i-f_i\nabla g^{*}_m)$$

 Quote by fzero These equations are inconsistent, since $$( g_i , f_j ) = \sum_k \alpha^{*}_{kj}(g_i,g_k) = \alpha^*_{ij}.$$ You should double-check the other term as well.

Taking into account the anti-linearity in the second arg of inner product as Bill_K noted, means the equations are consistent also.
 one also would require $$\beta_{im}=\pm \beta_{mi}$$. I started with the $$\delta_{ij}=(g_i,g_j)$$ and found that it led me to the Birrel and Davies result. So I think my earlier result must hold with these conditions on alpha and beta being used to show the two normalizations are in fact equivalent but I can't see how the alpha coeffs are Hermitian and beta symmetric/antisymmetric.