Textbook question - number of states

[trinitron]

Hi all,

I've been going through Becker and Schwarz this summer and I'm a bit stuck at the moment. I'm not sure if this is better suited for the homework forum, so feel free to move it.

For those that have the book, this is page 52, eq 2.143.

Given the number operator,

$$N = \sum_{i=1}^{D-2}\sum_{n=1}^\infty \alpha^i_{-n}\alpha^i_{n}$$

one can calculate the number of physical states of mass n-1 by calculating the coefficient of w^n-1 in the expansion of trace(w^N). That much I understand; however, there are a couple of things which seem unclear in the calculation of trace(w^N).

The book writes,

$$tr w^N = \prod_{n=1}^\infty\prod_{i=1}^{D-2} tr w^{\alpha_{-n}^i \alpha_{n}^i} = \prod_{n=1}^\infty (1-w^n)^{-24}$$

Thinking about the trace as a sum of eigenvalues, pulling the products out of the trace in the first equality seems to amount to changing the order of a sum and product (sum of eigenvalues and product over {i,n}). More directly, tr(AB)!=tr(A)tr(B). So what makes this equality valid?

I'm also having trouble with the second equality. It seems that the claim is $$tr w^{\alpha_{-n}^i\alpha_n^i} = w^n + w^{2n} + w^{3n} + \dots$$.

However, to me it seems that $$tr w^{\alpha_{-n}^i\alpha_n^i} = \infty w^n + \infty w^{2n} + \infty w^{3n} + \dots$$

Because if I want a state $$|\phi>$$ s.t. $$\alpha_{-n}^i\alpha_n^i |\phi> = m\times n|\phi>$$, I can simply take $$|0,k>$$ and act on it with $$m$$ copies of $$\alpha_{-n}^i$$ and an arbitrary number of $$\alpha_{-k}^j$$ for k!=n.

Any help would be greatly appreciated.

 

[wilsont23274]

Thanks in advance!</code>The first equality is valid because the trace of a product of two operators is equal to the product of the traces of the individual operators. This is just a result of the cyclic property of the trace. The second equality is not quite correct. The trace of w^{\alpha_{-n}^i \alpha_n^i} is equal to w^n + w^{2n} + w^{3n} + \dots + w^{mn}, where m is the maximum eigenvalue of the operator. Note that this is not necessarily infinity.

 

[charng]

I would first commend the person for their diligent study and for seeking help when faced with difficulties. I would also suggest that they reach out to their professor or a teaching assistant for further assistance with the material.

In terms of the specific questions raised, the first equality can be justified by using the fact that the trace of a product of operators is equal to the product of the traces, as long as the operators all commute with each other. In this case, the operators in the product all commute, so we can pull them out of the trace.

For the second equality, it is important to note that the trace of an operator is defined as the sum of its eigenvalues. In the case of w^{\alpha_{-n}^i\alpha_n^i}, the eigenvalues are all equal to 1 (since w is a complex number raised to a power). Therefore, the trace is just the sum of infinitely many 1's, which is equal to infinity. This is why we can write it as (1-w^n)^{-24}.

I hope this helps clarify the reasoning behind the equations and addresses any confusion. It is always important to thoroughly understand the material in textbooks, so do not hesitate to seek help and clarification when needed.