Wigner Weyl Transforms

1. Dec 27, 2012

aim1732

For Wigner transforming the function of operators x and p : (xp+px)/2 we need to evaluate something like:

g(x,p) = ∫dy <x - y/2 | (xp+px)/2 | x+y/2> e(ipy/h)
where h is h/2π.

Now I am not sure how to evaluate <x - y/2 | (xp+px)/2 | x+y/2> . I mean what I did was think of |x+y/2> as a delta function whose eigenvalue is x+y/2 and the basis to use is (from the bra) x-y/2.But that gives

∫(xp+px)/2 * δ(-y) e(ipy/h)
which comes out to be a constant where I took x=x and p=(h/i)∂/∂x.
I was expecting g(x,p)=xp

Actually I realise its quite a stupid doubt, rather a problem of me not understanding notations.I would be grateful if somebody gets me out of this mess.

2. Dec 27, 2012

vanhees71

First of all let's calculate the matrix elements. For the first expression you have
$$\langle x_1|\hat{x} \hat{p}|x_2 \rangle=x_1 \langle x_1|\hat{p} x_2 \rangle = -\mathrm{i} x_1 \partial_{x_1} \langle x_1 \rangle x_2 = -\mathrm{i} x_1 \partial_{x_1} \delta(x_1-x_2).$$
The other term is
$$\langle x_1 |\hat{p} \hat{x} |x_2 \rangle = x_2 \langle x_1|\hat{p}| x_2 \rangle = -\mathrm{i} x_2 \partial_{x_1} \delta(x_1-x_2).$$
Now you set
$$x=\frac{x_1+x_2}{2}, \quad y=x_2-x_1$$
which means
$$x_1=x-y/2, \quad x_2=x+y/2.$$
Then we have
$$\partial_{x_1}=\frac{\partial x}{\partial x_1} \partial_x+\frac{\partial y}{\partial x_1} \partial_y=\frac{1}{2}\partial_x-\partial_y$$.
From this we get (in your convention for the Fourier transform, which differs from what I'm used to, but anyway):
$$g(x,p)=\frac{1}{2} \int \mathrm{d} y \exp(\mathrm{i} p y) [+\mathrm{i} (x-y/2) \partial_y \delta(y)+\mathrm{i} (x+y/2) \partial_y \delta(y)]=\mathrm{i} x \int \mathrm{d} y \exp(\mathrm{i} p y) \partial_y \delta(y) =x p,$$
as you expected.

The difficulty is that one has to be careful with the expression of the matrix elements with help of the operators in the position representation and the resulting distributions. That's why we had to evaluate the matrix elements first in the original arguments $x_1$ and $x_2$ and then transform into the "macroscopic position" $x$ and the "relative postion" $y$ variables afterwards.

Last edited: Dec 27, 2012
3. Jan 2, 2013

aim1732

Thanks a lot for the reply.I realized I should have seen something like this.My original mistake was in assuming that the operators themselves have an inherent basis when in fact they do not.
However my original (revised) attempt involved working with the key idea that the operators 'get' their basis from the left hand side term of the matrix element, as in <u|xp|v> will be written as [u*(h/i)∂u] δ(u-v).That worked out gives the result-xp but when I showed that to my professor he told me that it was not right.Specifically I was not justified in writing the matrix element like that because <u|p|v> was p δ(u-v) but terms involving products with other operators would not be necessarily so.He then did it the same way a you did.
Am I wrong?And why exactly?

4. Jan 2, 2013

vanhees71

I don't see what's the difference between my derivation and yours. I only used other names for the eigenvalues, or do I miss something?

5. Jan 2, 2013

aim1732

The point of difference between the two is whether we can write

<u|xp|v> = [u*(h/i)∂u] δ(u-v)

6. Jan 2, 2013

vanhees71

If $|u \rangle$ and $|v \rangle$ are position eigenvectors, it's correct. I used this myself in my derivation. I've only called these vectors $|x_1 \rangle$ and $|x_2 \rangle$.

7. Jan 2, 2013

aim1732

And the position eigenvectors figure only w.r.t the delta functions used so that you can write any two operators multiplied with each other and appearing in the <u|OPERATOR1*OPERATOR2|v> as OPERATOR1*OPERATOR2* [Eigenvector of the type |u> and |v> with basis u and eigenvalue v] where the basis of the operators are themselves u?