Correlations between QM operators

[RedX]

In thermodynamics, two variables A and B are uncorrelated when:

<AB>=<A><B>

where <> are the expectation values in thermodynamics (for example calculated using Boltzmann distributions).

What are the conditions in quantum mechanics for two operators to be uncorrelated, i.e.,

&lt;AB&gt;=&lt;A&gt;&lt;B&gt;

where <> are now the expectation values in a particular quantum state?

Is it possible for two operators to be uncorrelated for every single state in quantum mechanics? What about thermodynamics: is it possible for two variables to be uncorrelated at every single temperature?

 

[fzero]

You can approach the QM question by inserting a complete set of eigenstates of A into the expectation value:

\langle \psi | A B | \psi \rangle = \sum_a \langle \psi | A|a\rangle\langle a| B | \psi \rangle .

You'll find that this can only equal \langle A\rangle \langle B\rangle if |a\rangle is also an eigenbasis for B. There is a condition that A and B must satisfy for this to be possible.

 

[arkajad]

You can calculate correlation for random variables that are jointly distributed. In quantum theory this happens only when A and B are represented by commuting operators. For noncommuting operators you can try to mimic some properties of the correlation but not all.

 

[A. Neumaier]

Is it possible for two operators to be uncorrelated for every single state in quantum mechanics?

This is the case if and only if the operators are multiples of the identity operator.

 

[RedX]

The reason I asked this is because combining statistical mechanics and quantum mechanics, the expectation value of an operator A for a Boltzmann distribution is:

\Sigma_k e^{-\beta E_k}&lt;E_k|A|E_k&gt; <br /> =\Sigma_k &lt;E_k|e^{-\beta H}A|E_k&gt;<br /> =Tr[e^{-\beta H}A]

Now if beta goes to zero (i.e., temperature gets really hot) then the expectation value becomes just the trace of the operator. In particular if you have the product of operators A and B, then the expectation value is Tr[AB].

But physically I thought that Tr[AB]=Tr[A]Tr even though this is not mathematically true. Because at high temperatures I thought things become uncorrelated (for example a ferromagnet at the Curie temperature), since randomess rules at high temperature: so you should have <AB>=<A><B>.

But obviously it is not true that Tr[AB]=Tr[A]Tr. So I thought something quantum mechanical must be happening to keep things from being uncorrelated.

Maybe my physical picture of what's going on is bad.

 

[A. Neumaier]

But physically I thought that Tr[AB]=Tr[A]Tr even though this is not mathematically true. Because at high temperatures I thought things become uncorrelated (for example a ferromagnet at the Curie temperature), since randomness rules at high temperature: so you should have <AB>=<A><B>.

But obviously it is not true that Tr[AB]=Tr[A]Tr. So I thought something quantum mechanical must be happening to keep things from being uncorrelated.

No. Even classically, the thermodynamic variables of a thermal state of an ideal gas is not uncorrelated.

You can understand this even with simple probability theory. If x is a Gaussian random vector in n variables and A is a nonsingular, nondiagonal matrix then the components of z=Ax are correlated. Thus uncorrelatedness is always a property of _particular_ observables, and does not generalize to arbitrary ones.

in particular, while the observables for single molecules in an ideal gas are uncorrelated, this no longer holds for the macroscopic observables, which combine all the microobservables in a similar 9but more complicated) fashion as in my simple example.