Vertex Operators

Posted by DickT on October 04, 2003 at 11:01:50:

In Reply to: Re: Actions, Amplitudes, and the geometry of it all posted by DickT on October 02, 2003 at 15:48:10:

Mike,

Here is the second installment.

When we left off last time we had a two dimensional "blob" representing a string interaction, and a collection of punctured discs on the blob which had been properly transformed (using math derived from the freedom of the action) from tubes representing incoming and outgoing strings. Now we are going to put in the centers of those discs. And we'll finally get to the path integrals for that.

But first - there's always a but first - the operator state correspondence. product correspondence. You know that quantum theory is not really about particles, or fields either. It's about states and the operators which act on them. And in these punctured discs we can show (conformal complex variables math) that there is a one to one correspondence (an isomorphism) between the quantum states and the quantum operators. To each state there is an operator that produces that state, and each operator only produces one state. This is a remarkable property of stringy physics that isn't true in other realms of the quantum continent.

Now taking advantage of the state-operator correspondence we are going to put points, called vertices, into the centers of those discs. Recall that the discs are the true representatives of the tubes representing the incoming and outgoing strings, and the vertices will represent - will carry - the quantum amplitudes of the string in question, derived by addition and subtraction operators, as in QFT, but now with unique creation of states. And these vertex operators, these states, will be functions of the spacetime coordinates, and those functions, wait for it, will be computed by path integrals!

Next time, path integrals and insertions.

Regards,
Dick

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