1
$\begingroup$

I am a mathematician studying the dynamics of the $N$-Body density matrix $\rho_{N}(x;y)$ for $n$ particles, defined by

$$\rho_{N, t}^{(n)} (x_1,..,n_n; y_1,...,y_n) = \begin{cases} \int \rho_{N,t}(x_1,...,x_n,x_{n+1},...,x_N; y_1,...,y_n,x_{n+1},...,x_N) dx_{n+1}...d_{x_N},\,\,\,\,\, 1 \leq n \le N\\ 0,\,\,\,\,\, o.w. \end{cases}$$

and the Wigner transform for one particle is:

$$w_N(x;v):= \frac{1}{(3 \pi)^{3 N}} \int e^{- i v. y} \rho_N \left( x + \frac{y}{2}, x- \frac{y}{2}\right) dy.$$

I cannot understand how to obtain that

$$\int w_N(x,v) dx dv =1.$$

$\endgroup$

1 Answer 1

1
$\begingroup$

I think your coefficients have typo's, but in any case, you first want to make sure that your reduced density matrix is properly normalized to unit trace. For one particle, that would mean that $$\int\rho^{(1)}_N(x,x)\,dx=1.$$ The Wigner transform can be then be defined as $$w_N^{(1)}(x,v)= \frac{1}{2\pi} \int e^{- i v y} \rho^{(1)}_N \left( x + \frac{y}{2}, x- \frac{y}{2}\right)\, dy.$$ Now check that $$\int w^{(1)}_N(x,v) \,dx dv =\int \delta(y) \rho^{(1)}_N \left( x + \frac{y}{2}, x- \frac{y}{2}\right)\, dxdy$$ $$\qquad=\int\rho^{(1)}_N(x,x)\,dx=1,$$ where I used that $\int e^{-ivy}\,dy=2\pi\delta(y)$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.