Single particle Density Matrix meaning

[jajabinker]

Hey guys!
In an n-electron system,
The second order reduced DM is defined as
\Gamma (x_{1},x_{2}) = \frac{N(N-1)}{2}\int{\psi(x_{1},x_{2}...,x_{n})\psi^{*}(x_{1},x_{2}...,x_{n})}dx_{3}...dx_{n}

It can be intepreted as the probability of finding two electrons at x_{1} and x_{2} respectively for all possible configurations of the other electrons because of the probabilistic interpretation of |\psi\psi*

The single particle density matrix is defined as :
\gamma (x_{1},x'_{1}) = N\int{\psi(x_{1},x_{2}...,x_{n})\psi^{*}(x'_{1},x_{2}...,x_{n})}dx_{2}...dx_{n}

Where the prime variable appears only in the complex conjugate of the function.

Can this be interpreted in probabilistic terms? why is there a prime? What is the meaning of the product of a wave function with its complex conjugate and having a different variable.

I know that for the diagonal terms it reduces to \rho (r).

Any replies are much appreciated.

 

[cgk]

The main point of defining the 1-particicle reduced density matrix (or 1-RDM) is that its knowledge is sufficient to evaluate the expectation values of all 1-particle operators for the system. Consider the matrix as expanded in a basis set in 2nd quantization:
<br /> \gamma_{rs}:=\langle c^\dagger_r c_s\rangle<br />
where r/s are the basis labels and the c's are creation and destruction operators. Then for any 1-particle operator o
<br /> \hat o = \sum_{rs} o_{rs} c^\dagger_r c_s<br />
its expectation value \langle \hat o \rangle can obviously be evaluated with the RDM only, without knowing the actual wave function.

This does not say anything about the direct interpretation of the 1-RDM of course. But in simple kinds of wave functions (Slater determinants), the off-diagonal elements of the 1-RDM are closely related to covalent bonding. For example, the Wiberg bond order (pure covalent bond order) is defined as
B_{AB}=\sum_{r\in A,s\in B}|\gamma_{rs}|^2
where r are AO basis functions of the first atom A and s the AO basis functions of the second atom B (how those are defined is another matter, they are not technically observable, but there are sensible ways of defining them).
However, this interpretation is relying (indirectly) on the fact that for such determinant wave functions the 2-RDM can also be expressed in terms of the 1-RDM, not only the other way around---bond order is really a 2-particle property.
Nevertheless, there is one thing one can learn from that: Unlike the diagonal elements, the off-diagonal elements of the 1-RDM have no simple classical analogon and represent pure quantum effects.

 

[jajabinker]

Thanks for your answer but it did not answer my question, which was why do we need primed variables in the complex conjugate?
Anyways here is the answer from Mcweeny and Sutcliffe:

Consider the expectation value of an operator F:
\langle F\rangle=\int\psi ^{*} (x) F \psi(x)dx
If we want to write this in terms of the electron density, \rho=\psi ^{*} \psi
we need F to be able to commute wit \psi*, which is generally not true for an integral or differential operator.

So we give dummy variables to \psi ^{*} called x'. These no longer interact with the operator F and we can integrate safely at x=x'.

So this is just a trick used to be able to calculate stuff through the electorn density which is more easily available than the wave function.