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Lets consider a bosonic physical system in variables $t, x$ and $y(x)$ ($x$ dependent) with a classical Lagrangian $L$. To first order in fluctuations $x \to x+\xi_1$ and $y \to y+\xi_2$ the fluctuated action reads

$$S_{fl}=\int dt dx \sqrt{g} ~{\vec \xi}^T~{\tilde O}~{\vec \xi}$$

Where the differential operator ${\tilde O}$ is given by

$${\tilde O}=\begin{pmatrix} \nabla^2+M_1(x) & -\frac{1}{\sqrt{g}}\partial_x +M_{12} \\ \frac{1}{\sqrt{g}}\partial_x +M_{12} & \nabla^2+M_2(x) \end{pmatrix}$$

Following the standard procedure, one has to compute the functional determinant of ${\tilde O}$ to obtain the effective action for the fluctuations. Now, since the system is coupled (off-diagonal elements are present), the computation of $\det[{\tilde O}]$ will be rather hard, if not impossible:

$$\det[{\tilde O}]=\det\begin{pmatrix} \nabla^2+M_1(x) & -\frac{1}{\sqrt{g}}\partial_x +M_{12} \\ \frac{1}{\sqrt{g}}\partial_x +M_{12} & \nabla^2+M_2(x) \end{pmatrix}\\=\det\left(\left(\nabla^2+M_1(x)\right)\left(\nabla^2+M_2(x) \right)-\left(-\frac{1}{\sqrt{g}}\partial_x +M_{12}\right)\left(\frac{1}{\sqrt{g}}\partial_x +M_{12}\right)\right)$$

However, formally, before attempting to compute the determinant, one can multiply out all the terms in the fluctuation Lagrangian $L_{fl}={\vec \xi}^T~{\tilde O}~{\vec \xi}$, partially integrate once in the off-diagonal terms and rearrange the fields $\xi_1 \xi_2$ such that if one puts all the terms back into matrix notation, one finds:

$$S_{fl}=\int dt dx \sqrt{g} ~{\vec \xi}^T~{\tilde O}~'~{\vec \xi}$$

where now

$${\tilde O}~'=\begin{pmatrix} \nabla^2+M_1(x) & 0 \\ \frac{2}{\sqrt{g}}\partial_x +2M_{12} & \nabla^2+M_2(x) \end{pmatrix}$$

If one proceeds to compute the determinant now:

$$\det[{\tilde O}~']=\det\left(\left(\nabla^2+M_1(x)\right)\left(\nabla^2+M_2(x) \right)\right)\\=\det\left(\nabla^2+M_1(x)\right)\det\left(\nabla^2+M_2(x) \right)$$

one finds a much simpler problem to solve. Now my question - is this "simplification" actually allowed, or did I miss some subtlety which forbids such a rearrangement of bosonic fields at the Lagrangian level?

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What is to be achieved in your calculation? It seems to me the last part is incorrect as $\tilde{O}'$ must be brought to symmetric form before using the Gaussian integration formula. You can try your procedure for a simple two-dimensional Gaussian integral and check whether it produces the correct answer. –  Isidore Seville Jan 9 at 21:02

1 Answer 1

I asked my professor and in a discussion we came up with the following.

The process of establishing the effective action for a fluctuation Lagrangian to consist of the functional determinant of the initial differential operator involved, relies on the equality:

$$\det(A)=e^{Tr(\log(A))}$$

for a matrix $A$, which is only true for diagonalizable matrices. A generic matrix of the type

$$B=\begin{pmatrix} b_{11} & 0 \\ b_{21} & b_{22} \end{pmatrix}$$

cannot be diagonalized and therefore the procedure does not hold for these kind of matrices.

One must always choose a diagonalizable matrix structure for the differential operators to be able to apply the QFT machinery.

EDIT

Due to questions in the comments I decided to mention one further idea which forbids matrices $B$ as given above. Remember the basics of QFT on a discretized lattice. Usually the following relation is found for real scalar fields (use them as an example) by explicit evaluation of Gaussian integrals:

$$\int D\varphi ~e^{-\frac{1}{2}\varphi_i A_{ij}\varphi_j}=\frac{1}{\sqrt{\det A}}$$

Now, remember why we involve the notation $\det A$ and not some other symbol in the result. It stems from the fact that we recognize by inspection that the result is exactly a determinant of a symmetric matrix $A$. Since the Gaussian integrals are evaluated separately, in components, it is even unavoidable that if we use

$$B=\begin{pmatrix} b_{11} & 0 \\ b_{21} & b_{22} \end{pmatrix}$$

as our matrix, the result will come out to be $1/\sqrt{\det B'}$ with

$$B'=\begin{pmatrix} b_{11} & \frac{1}{2}b_{21} \\ \frac{1}{2}b_{21} & b_{22} \end{pmatrix}$$

Therefore, if we want to skip the part of explicitly evaluating Gaussian integrals and proceed directly to evaluate a determinant, we have to be careful to involve a matrix as it actually emerges from the formalism, namely symmetric.

Hope this different argument helps.

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Why $\det(A)=e^{Tr(\log(A))}$ doesn't work (at least to finite dimensional) matrices? as far as I can remember, this all holds as longs the whole operation is valid. –  user23873 May 9 '13 at 11:14
    
$A$ must be diagonalizable for the relation to hold. As a necessary condition. –  Kagaratsch May 9 '13 at 11:31
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Could you provide a reference, or a counter example for it? As far as I remembered, if you had, for example, $||A||<1$, using the spectral norm, this worked. I'll look again on my linear algebra books, but I believe that you if $\log A$ exists, everything is ok. –  user23873 May 9 '13 at 22:08
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@user23873: honest question, but how do you define $\ln A$ if $A$ isn't diagonalisable? Things like $\ln(1 + A)$ do indeed make sense if $\| A \| < 1.$ –  Vibert May 14 '13 at 16:54
    
@vibert Formally, if we define matrix logarithm $\ln{}A$ as the solution to matrix equation $\exp(B)=A$, then $\ln{}A$ exists if and only if $A$ is invertible. To see this, we first bring $A$ to the Jordan canonical form $J$ and then find $\ln{}J$ by using the power series. –  Isidore Seville Jan 9 at 21:00

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