1
$\begingroup$

I came across this,

\begin{post} To each state there corresponds a unique state operator.The average value of a dynamical variable \textbf{R}, represented by the operator $R$ ,in the virtual ensemble of events that may result from a preparation procedure for the state, represented by the operator $\rho$, is $$\langle \textbf{R}\rangle=\dfrac{Tr(\rho {R})}{Tr\rho}$$

\end{post}

$\rho$ is called density matrix or statistical operator.

What's meant by this and why does this give average?

$\endgroup$
  • 3
    $\begingroup$ this follows directly from it's definition. en.wikipedia.org/wiki/Density_matrix $\hat\rho = \sum_i p_i |\psi_i \rangle \langle \psi_i|$ and then $\langle A \rangle = \sum_i p_i \langle \psi_i | \hat{A} | \psi_i \rangle = \sum_{mn} \langle u_m | \hat\rho | u_n \rangle \langle u_n | \hat{A} | u_m \rangle = \sum_{mn} \rho_{mn} A_{nm} = \operatorname{tr}(\rho A).$ both copied from Wikipedia. I can recommend the book by shankar for a good aclimation to QM. $\endgroup$ – pindakaas May 10 '15 at 21:33
3
$\begingroup$

I believe the language in the text you quote is quite confusing and unusual (but this may only be my perception due to the books I use).

Consider an ensemble of systems, it is prepared in a way, that a system in the ensemble is in state $\left| n \right>$ with probability $p_n$.

A density matrix is an operator of the form: $$ \rho = \sum_n \left|n\right> p_n \left<n\right|. $$

Where $\left| n \right>$ is a complete (not necessarily orthogonal) set of states.

If we now consider an observable $R$, and consider the expression (for simplicity let $\left| n \right>$ be orthogonal, but it also works out otherwise): $$ \text{Tr}(\rho R) = \sum_{m, n} \left<m\middle|n \right> p_n \left<n\middle| R \middle| m \right> = \sum_n p_n \left<n \middle| R \middle|n\right> = \sum_n p_n \left< R \right>_n = \left< R \right>$$ Where we evaluate the trace in the same basis set as chosen to specify the statistical operator. That is, the considered expression evaluates to what we expect of an average (the quantum mechanical average of $R$ in the states averaged over the probability of the states)!

$\endgroup$
  • $\begingroup$ I do not know the book. But in what way is this answer not abstract enough (the relevant parts are just using mathematics and definitions)? And even with a book on an abstract level it is important to have some intuition about the involved concepts. $\endgroup$ – Sebastian Riese May 11 '15 at 8:21
  • $\begingroup$ I just glanced in the book, and as far as I understands it, he simply defines the density matrix the way: $\text{Tr}(\rho R) = \langle R \rangle$ (equation (2.1)) by definition (or rather by axiom). So you can either choose to accept his axioms, or the derivation and definition in standard QM. That is, an average is expressed that way, because the theory resulting from the axioms is consistent with the experiments. $\endgroup$ – Sebastian Riese May 11 '15 at 12:12
  • $\begingroup$ That I gave! (in terms of standard QM). $\endgroup$ – Sebastian Riese May 11 '15 at 14:40
2
$\begingroup$

Observables are given by self adjoint operators, what we seek is the expectation values for these observables. So we need a probability measure that assigns to each eigenvalue it's probability of occurrence. This reduces the problem to finding the measures on separable Hilbert space $\mathcal{H}$.

Spectral theorem: If $A$ be a self adjoint operator then it can be decomposed as, $$A=\sum_i \lambda_i P_i$$ where $\lambda_i$s are eigenvalues and $P_i$ are the corresponding projection operators.

Gleason's theorem: The only possible measures on complex separable Hilbert spaces $\mathcal{H}$ of dimension greater than 2 are measures of the form $$\mu(\lambda_i)=Tr(\rho E_i).$$ where $\rho$ is a positive semidefinite self adjoint operator of unit trace and $E$ is the spectral measure of the self adjoint operator $A$.

Now the expectation value of the observable will be given by, $$\langle A\rangle= \sum_i\lambda_i \mu(\lambda_i)=Tr(\rho A)$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.