juanrga Posted November 23, 2012 Share Posted November 23, 2012 (edited) Any textbook gives the interpretation of the density matrix in a SINGLE continuous basis [math]|\alpha\rangle[/math]: The diagonal elements [math]\rho(\alpha,\alpha) = \langle\alpha| \hat{\rho} |\alpha\rangle[/math] give the populations. The off-diagonal elements [math]\rho(\alpha,\alpha') = \langle\alpha| \hat{\rho} |\alpha'\rangle[/math] give the coherences. But what is the physical interpretation (if any) of the density matrix [math]\rho(\alpha,\beta) = \langle\alpha| \hat{\rho} |\beta\rangle[/math] for a DOUBLE continuous basis [math]|\alpha\rangle[/math], [math]|\beta\rangle[/math]? I already know that when the double basis are momentum [math]|p\rangle[/math] and position [math]|x\rangle[/math], then [math]\rho(p,x)[/math] (the well-known Wigner function) is interpreted as a pseudo-probability. I may confess that I have never completely understood the concept of pseudo-probability [*], but I would like to know if this physical interpretation as pseudo-probability can be extended to arbitrary continuous basis [math]|\alpha\rangle[/math], [math]|\beta\rangle[/math] for non-commuting operators [math]\hat{\alpha}[/math], [math]\hat{\beta}[/math] and if a probability interpretation holds for commuting operators. I.e. can [math]\rho(\alpha,\beta)[/math] be interpreted as a pseudo-probability for arbitrary non-commuting operators beyond x and p? Can [math]\rho(\alpha,\beta)[/math] be interpreted as a probability for arbitrary commuting operators? [*] Specially because [math]\rho(p,x)[/math] is bounded and cannot be 'spike'. Edited November 23, 2012 by juanrga Link to comment Share on other sites More sharing options...

## Recommended Posts

## Create an account or sign in to comment

You need to be a member in order to leave a comment

## Create an account

Sign up for a new account in our community. It's easy!

Register a new account## Sign in

Already have an account? Sign in here.

Sign In Now