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One wants to calculate the quantity, $\det'(\frac{\partial}{\partial t} - i [\alpha, ])$ where the prime on the "det" means that one wants to do a product over only non-zero eigenvalues of the operator $\frac{\partial}{\partial t} - i [\alpha, ]$. This operator is acting on the adjoint representation of a Lie algebra. ($\alpha$ itself is in the adjoint representation of the Lie algebera)

Now one claims that one can find an eigenbasis basis of matrix functions for $\alpha$ such that whose eigenvalues are $\{ \lambda _i \}_{i=1} ^ {i = n}$ and whose $t$ dependence is $\exp(\frac{i2\pi n t}{\beta})$

  • Can someone write down the exact equation into which the above translates?

I would vaguely guess that if $X_i$ is such an eigenvector then because of the adjoint nature of the representation it means that $[\alpha, X_i] = \lambda _ i X_i$ But I don't seem to see exactly where to fix the exponential dependence.

Now one claims that in this basis the determinant is equal to the following expression,

$$\prod _{n \neq 0} \prod _ {i,j} [ \frac{i2\pi n}{\beta} - i (\lambda _ i - \lambda _j)]$$

and the above can apparently be simplified to give,

$$\det'(\frac{\partial}{\partial t} - i [\alpha, ]) = \left ( \prod _{m \neq 0} \frac{i2\pi m}{\beta} \right )\prod _ {i,j} \frac{2}{\beta (\lambda _i - \lambda _j)}\sin (\frac{\beta (\lambda _ i - \lambda _j)}{2})$$

  • It would be great if someone can help explain the above simplification.
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1 Answer 1

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$\def\a{\alpha} \def\at{\widetilde\a} \def\b{\beta} \def\d{\delta} \def\j{\psi} \def\J{\Psi} \def\JT{\widetilde\J} \def\l{\lambda} \def\L{\Lambda} \def\p{\pi} \def\w{\omega} \def\det{\mathrm{det'}\,} \def\dt{\frac{\partial}{\partial t}} \def\diag{\mathrm{diag}} \def\su{\mathfrak{su}}$ A simpler determinant

Let's first have a look at a related problem, to determine $$\det\left(\dt - i \a\right)$$ where $\a$ is some scalar. (Recall that the analog of commutation is multiplication.) Some of the difficulties of the full problem exist already in this simpler one.

We are looking for the eigenvectors of the operator $\dt - i \a$, that is, for solutions to the equation $$\left(\dt - i \a\right)\j = \L \j.$$ The eigenvectors are $\j(t) = e^{i\w t}$, with eigenvalues $\L = i(\w-\a)$. Imposing periodic boundary conditions, $\j(\b) = \j(0)$, we find $\w_m = 2m\p/\b$, where $m\in\mathbb{Z}$. Thus, the eigenvalues are quantized, $\L_m = i\left(\frac{2 m\p}{\b}-\a\right).$

We assume that $\a\ne 2m\p/\b$ for any $m\in\mathbb{Z}$, so $\L_m \ne 0$. In particular, $\L_0 = -i\a \ne 0$. Then the product $\det$ ranges over all $m$, $$\begin{eqnarray*} \det\left(\dt - i \a\right) &=& \prod_{m=-\infty}^\infty \L_m \\ &=& \L_0 \prod_{m\ne 0} \L_m \\ &=& \L_0 \prod_{m\ne 0} i\left(\frac{2 m\p}{\b}-\a\right) \\ &=& \L_0 \prod_{m\ne 0} \frac{2m\p i}{\b} \left(1-\frac{\a\b}{2\p m}\right) \\ &=& \L_0 \left(\prod_{m=1}^\infty \left(\frac{2m\p}{\b}\right)^2 \right) \prod_{m=1}^\infty \left(1-\left(\frac{\a\b}{2\p m}\right)^2\right) \\ &=& \L_0 \frac{2}{\a\b}\sin\frac{\a\b}{2} \prod_{m=1}^\infty \left(\frac{2m\p}{\b}\right)^2 \\ &=& -\frac{2i}{\b}\sin\frac{\a\b}{2} \prod_{m=1}^\infty \left(\frac{2m\p}{\b}\right)^2. \end{eqnarray*}$$ Above we have used the infinite product representation for the sine function, $\sin x = x\prod_{m=1}^\infty \left(1-\frac{x^2}{m^2\pi^2}\right)$.

The determinant in general

We assume the algebra is a finite dimensional semisimple Lie algebra over $\mathbb{R}$ or $\mathbb{C}$.

Consider the equation $$\begin{equation*} \dt\J - i [\a,\J] = \L \J.\tag{1} \end{equation*}$$ The objects $\a$ and $\J$ are now $n\times n$ matrices in the adjoint representation of the Lie algebra where $n$ is the number of elements in the algebra. We are free to apply a similarity transformation to (1), and instead solve $$\begin{equation*} \dt\JT - i [\at,\JT] = \L \JT,\tag{2} \end{equation*}$$ where $\widetilde A = U^{-1}AU$, and $U$ is independent of time. In fact, we can diagonalize $\alpha$ with this transformation, $\at = \diag(\l_1,\ldots,\l_n)$, so $\at$ is in the Cartan subalgebra. (See the example for $\su(3)$ below.)

Let's choose as our basis for matrices $e_{ab}$, where $a,b=1,\ldots,n$, such that the components are $$(e_{ab})_{ij} = \d_{ai} \d_{bj},$$ where $\d$ is the Kronecker delta. Then $$[\at,e_{ab}] = (\l_a - \l_b) e_{ab},$$ so the $e_{ab}$s are eigenvectors of $\at$. (This is straightforward to prove using the fact that $\at = \sum_a \l_a e_{aa}$.)

Thus, the eigenvectors of (2) are $\JT = e_{ab}e^{i\w t}$, since $$\begin{eqnarray*} \dt\JT - i [\at,\JT] &=& \left(i\w - i(\l_a-\l_b)\right)\JT. \end{eqnarray*}$$ Imposing the boundary condition $\JT(\beta) = \JT(0)$ we find $$\L_m^{ab} = i\left(\frac{2m\pi}{\beta} - (\l_a-\l_b)\right).$$ For every $a,b$ there is an infinite tower of eigenvalues corresponding to the index $m$.

Notice that $\L\ne 0$ if and only if $\l_a-\l_b \ne 2m\p/\b$. For $m\ne0$ and a generic $\b$, $\l_a-\l_b \ne 2m\p/\b$. For $m=0$, $\L \ne 0$ if and only if $\l_a - \l_b \ne 0$.

Therefore, $$\begin{eqnarray*} \det\left(\dt - i [\a,\,]\right) &=& \left(\prod_{\l_a\ne\l_b} \L_0^{ab} \right) \left(\prod_{m\ne0\atop a,b} \L_m^{ab}\right) \\ &=& \left(\prod_{\l_a\ne\l_b} \L_0^{ab}\right) \left(\prod_{m\ne0\atop a,b} i\left(\frac{2m\pi}{\beta} - (\l_a-\l_b)\right) \right) \\ &=& \left(\prod_{\l_a\ne\l_b} \L_0^{ab}\right) \left(\prod_{a,b} \frac{2}{(\l_a-\l_b)\b}\sin\frac{(\l_a-\l_b)\b}{2} \prod_{m=1}^\infty \left(\frac{2m\p}{\b}\right)^2\right) \\ &=& \left(\prod_{\l_a\ne\l_b} \L_0^{ab}\right) \left(\prod_{\l_a\ne\l_b} \frac{2}{(\l_a-\l_b)\b}\sin\frac{(\l_a-\l_b)\b}{2}\right) \left(\prod_{m=1\atop a,b}^\infty \left(\frac{2m\p}{\b}\right)^2\right) \\ &=& \left(\prod_{\l_a\ne\l_b} -\frac{2i}{\b}\sin\frac{(\l_a-\l_b)\b}{2} \right) \left(\prod_{m=1\atop a,b}^\infty \left(\frac{2m\p}{\b}\right)^2\right). \end{eqnarray*}$$ Convince yourself that the correct way to interpret $\prod_{\l_a=\l_b} \frac{2}{(\l_a-\l_b)\b}\sin\frac{(\l_a-\l_b)\b}{2}$ is $\prod_{\l_a=\l_b}1 = 1$. Intuitively, $\lim_{x\to 0}\frac{\sin x}{x} = 1$.

Notice the similarity between this result and the previous one. The formula for $\det\left(\dt-i[\a,\,]\right)$ given in the question statement is missing some factors.

For readers unfamiliar with such calculations, which appear in physics quite regularly, and disturbed by the product $\prod_{m=1\atop a,b}^\infty \left(\frac{2m\p}{\b}\right)^2$, it may relieve you (a little) to know that this is just $\det\left(\dt\right)$ and typically we are interested in objects of the form $$\frac{\det\left(\dt-i[\a,\,]\right)}{\det\left(\dt\right)}.$$

Example: Eigenvalues of $\at$ for $\su(3)$

We use the structure constants typical to physics, $f_{abc}$, to construct the adjoint representation $(t_a)_{bc} = -i f_{abc}$.

The Cartan subalgebra is two dimensional. Simultaneously diagonalize $t_3$ and $t_8$ so $\at = x \widetilde t_3 + y \widetilde t_8$, where $x$ and $y$ are some constants. To be explicit, $$t_3 = \textstyle\left( \begin{array}{cccccccc} 0 & -i & 0 & 0 & 0 & 0 & 0 & 0 \\ i & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -\frac{i}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{i}{2} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & \frac{i}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & -\frac{i}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right)$$ $$t_8 = \textstyle\left( \begin{array}{cccccccc} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & -\frac{i \sqrt{3}}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & \frac{i \sqrt{3}}{2} & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -\frac{i \sqrt{3}}{2} & 0 \\ 0 & 0 & 0 & 0 & 0 & \frac{i \sqrt{3}}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{array} \right)$$ Up to permutation, $$\at = \diag\left( x, -x, \frac{1}{2}(x+\sqrt{3} y), \frac{1}{2}(x-\sqrt{3} y), -\frac{1}{2}(x+\sqrt{3} y), -\frac{1}{2}(x-\sqrt{3} y), 0,0\right).$$ The diagonal elements above are the eigenvalues $\l_a$.

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Thanks for the reply! Apologies for my belated response! Let me study your answer in details and then I will get back to you again! –  user6818 Jul 18 '12 at 19:36
    
@user6818: You're welcome. Glad to help. Let me know if you have any questions. –  user26872 Jul 18 '12 at 22:45
    
Thanks a lot for this extraordinarily detailed answer! So is there any general reference which you used for this? Is there a general idea behind these which I can use to learn? I am reminded of another such similar calculation which I was stuck wit - on page 32 of this paper, arxiv.org/pdf/0704.3740v1.pdf It seems to be a similar calculation that has been done to go to equation B.2 from B.1 and the simplifications of B.2 which follow on page 33-36. May be you can lead me to a reference which helps understand what has been done! (..or should I put that up as a separate question?..) –  user6818 Jul 19 '12 at 18:05
    
@user6818: Thank you for the interesting question! I don't remember seeing this particular calculation done in any text so I just worked it out and tried to make the answer self-contained. In terms of machinery, we need to be able to find the eigenvalues of differential equations, recognize product formulas for trigonometric functions, and know some facts about the representations of Lie algebras. –  user26872 Jul 20 '12 at 0:33
    
@user6818: Functional determinants are usually seen in a course on quantum field theory and/or advanced statistical mechanics, so if you have any texts on hand look there first. Some references: (1) Peskin and Schroeder's An introduction to quantum field theory, (2) Ramond's Field theory: A modern primer, (3) Cahn's Semi-simple Lie algebras and their representations (available here), (4) Wikipedia also has some examples. –  user26872 Jul 20 '12 at 0:33
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