# Is there any approximation to the two particle density matrix

1. Jul 28, 2006

### chuckschuldiner

Let phi(x) and phi_dagger(x) be field operators which satisfy the appropriate commutation relations.

Then is there any analytic approximation for the two particle density matrix given by

<phi_dagger(x)phi_dagger(x')phi(x')phi(x)>

Thanks!

2. Jul 28, 2006

### Rach3

$$\langle \phi^{\dagger} (x) \phi^{\dagger} (y) \phi(y) \phi (x) \rangle$$

I'm not thinking this morning - this is a density matrix?

3. Jul 28, 2006

### chuckschuldiner

Hi
The author calls it the two particle density matrix. The reference is Phys Rev B, Vol 71, 165104 (2005)

The page number is 165104-3

4. Jul 28, 2006

### CarlB

It most certainly is the field operator version of the density matrix. By the way, this fact has to do with why I called the state |0><0| the "vacuum" in the discussion on the "superposition and kets":

The best reference I've seen for explaining why this is the case is Julian Schwinger's "Quantum Kinematics and Dynamics", which is a small classic paperback book that is available cheaply at most good physics bookstores and also on Amazon.

Carl

5. Jul 30, 2006

### samalkhaiat

NO, It is a number, a single number. It is neither a density nor a matrix.

the phi's are operator-valued distributions, putting them inside < |...| > gives you a number.

regards

sam

Note from Hurkyl: I fixed the formatting tags so that this will display properly

Last edited by a moderator: Jul 30, 2006
6. Jul 31, 2006

### Perturbation

Depending on the theory, perturbation theory...? And yeah, it's an amplitude not a density matrix.

7. Jul 31, 2006

### CarlB

See, for example, equation (5) of:
http://arxiv.org/abs/nucl-th/9508008

No reason to argue over terminology. And $$\rho(x,x')$$ is a lot more complicated than just a number.

Carl

Last edited: Jul 31, 2006
8. Aug 3, 2006

### samalkhaiat

Last edited: Aug 4, 2006