Hi all, I have a question about observables and the absolute value

I am considering an expectation value of a density operator of some system, and it appears as though taking the absolute value prematurely gives the wrong results

Why is this? I always thought the absolute value may be considered distributive

It also seems as if taking it prematurely makes the system non-integrable

@1010011010 I'm not sure what you mean by "distributive", but $\langle \lvert A\rvert\rangle \neq \lvert \langle A\rangle \rvert$.

In other words, you need to be clear whether you want to take the absolute value of the expectation value or the expectation value of the absolute value; those are not the same thing.

Certainly, though suppose my density operator is given by two factors, certainly one would expect the absolute value of the 2 state matrix element to be the absolute value of the individual factors, no?

In technical terms $\lvert <\hat\rho> \rvert = \lvert A\rvert \cdot \lvert B\rvert$, where $A$ and $B$ are those two factors

It appears as though in my system these two factors "communicate" for the lack of a better wording, or so the math suggests

I don't understand how the "two factors" are supposed to be related to $\rho$

Oh, sorry, ok, so basically the quantum operator is calculated between two states, which I'm calling a matrix element of the state space, so the matrix element could be $<\Psi|\hat\rho|\Phi>$ for any two states psi and phi

I know what a matrix element and a density operator are, I do not know what "my density operator is given by two factors" means.

Ok suppose the expectation value is given by a product of two functions, $f(x)g(x)$ both of which are allowed to be complex

Though it is not necessary that they are

What's $x$, and why would an expectation value be complex rather than real?

If I calculate the absolute value of this matrix element, why can't I just say it is equal to $|f(x)|$ times $|g(x)|$

Where $x$ is an otherwise indeterminate system parameter on which $\rho$ depends

$\lvert f(x) g(x)\rvert$ = $\lvert f(x) \rvert \lvert g(x)\rvert$ is rather trivially true - you'll have to describe more precisely what you're doing here and how your result differs from that.

I calculated my density operator of the 1d bose gas by the conventional method and by using that identity I too thought was trivial (using the absolute value in that way) and by keeping all the phases

If I keep the phases and use a geometrical argument, parts of the expression become Dirac deltas after the thermodynamic transition

This imposes constraints that don't seem to be there when I use that rule

Oh, there's a limit involved! It is not true that $\lim_{x\to a} \lvert f(x)\rvert = \lvert \lim_{x\to a} f(x)\rvert$

What is the condition for it to be equal?\

Is it correct for a root of some function? Or for a simple rational product?

@1010011010 If $\lim_{x\to a}f(x)$ exists, then $g(\lim_{x\to a} f(x)) = \lim_{x\to a} g(f(x))$ for any continuous function $g$.

But the fact that you are talking about $\delta$-functions there makes me suspect these limits do not rigorously exist

Good point. In this case they are the result of a limit though

I have made a plan and will start with the derivation of the two state matrix element to see if the absolute value is already applied, that would explain a lot

Thanks