# Why does this integral go to zero?

1. Apr 28, 2010

### vertices

Why does this integral go to zero? Where alpha+ is a creation operator?

$$\int \frac{d^3k}{w_k} k^j \left(\alpha^+ (\underline{k}) \alpha^+ (-\underline{k})e^{2iw_k x^0}\right)$$

2. Apr 29, 2010

### ansgar

odd integrand

3. Apr 29, 2010

### vertices

Sorry - stupid question, why is it an odd integrand? If it is an odd integrand, alpha(+k) and alpha(-k) are either odd and even (or even and odd) or both odd. An odd function has the property that f(-x)=-f(x) and an even function has the property that f(-x)=f(x) - why must the creation operators have these properties? As I understand it, they are just coefficients of the general expansion of the field, so they should be totally arbitrary..

4. Apr 29, 2010

### ansgar

as far as I see, there is a k in front of the parenthesis..

5. Apr 29, 2010

### vertices

yes but that k is just a constant. We are integrating wrt vector k, I think.

EDIT: In anycase, if the k was a variable in the integration, why would this make the integral zero? (sorry if this is a stupid question)..

6. Apr 29, 2010

### ansgar

why isn't it the j-th component? j = 1,2,3

d^3k = dk1 dk2 dk3

so the integral is a 3-vector relation.

where did you find this expression?

7. Apr 29, 2010

### vertices

Oh okay, that makes sense.

The original question itself is this:

find

$$\int d^3.\underline{x}T^{0j}(x)$$

in terms of creation and annihilation operators.

Basically, when I expand it out (very long), I get two integrals like the one in the OP - which should go to zero according to the solutions I've been given... I just dont understand why.

8. Apr 29, 2010

### ansgar

well the integral you posted first is odd, so it is zero

and from

$$\int d^3.\underline{x}T^{0j}(x)$$

you that it is a j0 tensor which is what you have posted in your post. i.e it SHOULD depend on the j-th coordinate

Last edited: Apr 29, 2010
9. Apr 29, 2010

### vertices

Yes it should depend on the jth coordinate.

Sorry I'm sure this is a really stupid question, but would I be right in saying that the alpha functions must both be even (or one even one odd, or both odd), for the integral to be zero (as evenxevenxodd=odd (the standalone k is odd).

Why must this be the case? The alpha functions are Fourier coefficients of a general field, why can we impose such a constraint to these functions?

Last edited: Apr 29, 2010
10. Apr 29, 2010

### ansgar

you see that the integral is zero since it is odd due to the k^j factor

Also you have a(-x)a(x) which is even under x -> -x

so it doesen't matter if a(x) is odd or even under x-> -x, since they come in this combination.