Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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?

share|cite|improve this question
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 '14 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:


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.


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.

share|cite|improve this answer
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. – Hydro Guy 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
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. – Hydro Guy May 9 '13 at 22:08
@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 '14 at 21:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.