Is there any approximation to the two particle density matrix

[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!

 

[Rach3]

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

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

 

[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

 

[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":
https://www.physicsforums.com/showthread.php?t=124904

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

 

[samalkhaiat]

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

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

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

 

[Perturbation]

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

 

[CarlB]

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.

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

 

[samalkhaiat]

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, I know one or two things about the density matrix, so you don't need to point a "reference" to me

A physically acceptable (unique, hermitian & non-negative) density matrix must satisfy;

1) $$<\hat{A}> = Tr(\rho\hat{A})$$

2) the quantum Liouville equation;

$$i\partial_{t} \rho (x,x';t) = [H,\rho (x,x';t)]$$

So, Carl, be kind and show us how YOU prove that the talked about expession satisfies (1) & (2) ?

I am aware that some people who work on many-particle systems, non-relativistic field theory, DO , occasionally, call;

$$<\psi^{\dagger}(\vec{x},t)\psi(\vec{x'},t)>_{0}$$
(in Heisenberg picture)

or,

$$<\psi^{\dagger}(\vec{x})\psi(\vec{x'})>_{t}$$
(in Schrodinger picture)

a one-body density matrix!

Well, they are wrong, because these expressions are C-numbers. As such they do not satisfy (1) & (2).

Indeed, from (1) we see that these expressions are equal to;

$$Tr(\rho_{0}\psi^{\dagger}(x,t)\psi(x',t))=Tr(\rho_{t}\psi^{\dagger}(x)\psi(x'))$$

where the density matrix;

$$\rho(t) = exp(-iHt) \rho(0) exp(iHt)$$

does satisfy the quantum Liouville equation (2).

Don't ask me why, but even the number-density operator

$$n(\vec{x})=\psi^{\dagger}(\vec{x})\psi(\vec{x})$$

sometimes is called (by similar people) a density matrix!
I believe, it is a simple case of misuse of language, because these people do a fine work and make use of

$$<n(\vec{x})> = Tr \Left(\rho\psi^{\dagger}\psi\Right)$$

in their work!

regards

sam