Polchinski Excercise Question - delta function

[selfAdjoint]

I am going again over Polchinski's excercises, trying to work them and using http://schwinger.harvard.edu/~headrick/polchinski.html when I get stuck. In problem 2.1, P. wants us to show that
$$\partial \bar{\partial} ln \vert z \vert^2 = 2 \pi \delta^2(z,\bar{z})$$

and Headrick, introducing a test function f(z) under the integral sign,
$$\int_R d^2z \partial \bar{\partial} ln \vert z \vert^2 f(z)$$

eventually gets
$$\partial \bar{\partial} ln \vert z \vert^2 = 2 \pi f(0)$$

Can anybody spell out for me how this arbitrary f(o) is the delta function?

 

[lethe]

Originally posted by selfAdjoint

Can anybody spell out for me how this arbitrary f(o) is the delta function?

the defining characteristic of the delta function is exactly this property you show above:

namely
$$\int\delta(x)f(x)=f(0)$$

any function(al) for which this is satisfied for an arbitrary f is the (unique) delta function

so f(0) isn't the delta function, it is the result of integration with a delta function

 

[selfAdjoint]

Thanks lethe. You notice the answer is a squared delta function, but Headrick's test function isn't (explicitly) squared. How does that work?

 

[lethe]

Originally posted by selfAdjoint
Thanks lethe. You notice the answer is a squared delta function, but Headrick's test function isn't (explicitly) squared. How does that work?

i think that 2 in the exponent just indicates that you integrate over z and z*. it's a two dimensional delta function

 

[lethe]

Originally posted by lethe
i think that 2 in the exponent just indicates that you integrate over z and z*. it's a two dimensional delta function

just like we write the 3 dimensional dirac delta

$$\delta^3(\mathbf{r})$$

 

[lethe]

Originally posted by selfAdjoint
Thanks lethe. You notice the answer is a squared delta function, but Headrick's test function isn't (explicitly) squared. How does that work?

in other words, just because the delta function is squared, it doesn't mean the function gets squared. for example, we have:

$$\int\delta^3(\mathbf{r})f(\mathbf{r})d^3\mathbf{r} = \int\delta(x)\delta(y)\delta(z)f(x,y,z)dx\ dy\ dz=f(0,0,0)$$

 

[selfAdjoint]

OK, I think. So in this two dimensional case, we know the delta gets squared because of the double integration of the double derivative. But it still just picks out f(0) from the test function. Have I got that right?

Afterthought: Would it be a general method to test any expression with a delta in it by putting it under an appropriate integral with a test function and seeing if it picks out the appropriate value of the function? I never had any course work in deltas so I'm kind of flying blind here.

 

[lethe]

Originally posted by selfAdjoint
OK, I think. So in this two dimensional case, we know the delta gets squared because of the double integration of the double derivative. But it still just picks out f(0) from the test function. Have I got that right?

yes. i am thinking of the function as a function of z and z*, except that the functional dependence of a holomorphic on z* is trivial, so that factor of the delta function integrates away trivially.

Afterthought: Would it be a general method to test any expression with a delta in it by putting it under an appropriate integral with a test function and seeing if it picks out the appropriate value of the function? I never had any course work in deltas so I'm kind of flying blind here.

yes: by definition, two expressions with delta functions in them are defined to be equal if, when you integrate them, you pick out the appropriate values.

 

[selfAdjoint]

Thanks a lot, Lethe. I am clear on that. This exercise turns out to be a key step in proving that the normal ordered product is a harmonic function, which is in turn key to developing the OPE for the $$X^{\mu}$$ physics.

 

[Haelfix]

Btw, it drives me nuts how physicists insist on using the above eqn (the first one from SA).

A bare operator valued distribution function ! Horror!

Put an integral sign around both sides, and then its well defined, and all mathematicians may rest easy!

 

[selfAdjoint]

Yeah, and the operator product expansion is an "operator" equation. Only meaningful up to expectation. Abus de langage. You just have to get used to it.