0
$\begingroup$

The density matrix of the system is given by:

$$ [\rho_{S}(t)]_{mn} = [\rho_{S}(0)]_{mn} e^{-i\omega_{0}(m - n)t} e^{-i \delta(t)(m^2 - n^2) - \gamma(t)(m - n)^2}, $$

where $[...]_{mn}$ denotes the matrix elements of the relevant operator. The basis is defined below. The problem is to:

calculate that the probability that at $t = \tau$, the system is in the state $$ \mid \psi \rangle = \frac{1}{\sqrt{2}} \Big ( \mid e \; \rangle \; + \;\mid g \; \rangle \Big ), $$

where the kets are the eigentstaes of the $\sigma_{z}$ operator of the two level system.

We make the following two simplifications:

  1. Before each measurement, we apply the rotation $U_{S}(\tau) = e^{iH_{S}\tau}$, which removes the system evolution induced by the system itself. This has the effect of ignoring the first exponential in the first equation.

  2. "Let us neglect any measurement-induced disturbance to the environment (at least valid for weak system-environment coupling)." I think that this statement means that we can ignore the term involving the second exponential in the first equations, because it defines the indirect system-environment couplong.

With these two assumptions, how would one go about getting the answer: I get completely non-sense as that with probability 1, the density operator will be in the same initial state. The correct answer is:

$$ 1 - \frac{1}{2} \Bigg [1 - e^{-\gamma(\tau)} \Bigg ] = \frac{1}{2} \Bigg [1 - e^{-\gamma(\tau)} \Bigg ]. $$

Any suggestions:

Note: $\hbar = 1$ in the above discussion.

Please let me know if you'd also like the expressions for $\delta(t)$ and $\gamma(t)$, although it'll be really tedious to write them down and explain the context in detail.

Thanks!

$\endgroup$
  • $\begingroup$ Are you considering only a two level system here? Do $ m, n $ only take values 0,1? Or maybe -1,1 since the Pauli matrices are involved? $\endgroup$ – Mikael Fremling Jun 19 '16 at 21:38
  • $\begingroup$ Two level system. I think m and n take (eigen)values of -1 and 1 since sigma z is involved. $\endgroup$ – Junaid Aftab Jun 20 '16 at 7:45
0
$\begingroup$

This is just a quick stab, and it might show my ignorance more than anything else. Since you are working with a two, level spin system i'm actually giessting $m,n=\pm\frac{1}{2}$ . You can then explicitly write your density matrix as

$$ \rho\left(t\right)=\begin{pmatrix}\rho_{\frac{1}{2},\frac{1}{2}} & \rho_{-\frac{1}{2},\frac{1}{2}}e^{-i\omega t}e^{-\gamma\left(t\right)}\\ \rho_{\frac{1}{2},-\frac{1}{2}}e^{i\omega t}e^{-\gamma\left(t\right)} & \rho_{-\frac{1}{2},-\frac{1}{2}} \end{pmatrix} $$ Note that $\delta\left(t\right)$ is not present at all now.

Using the unitary transformation you remove the energy dependence and get

$$ \rho\left(t\right)=\begin{pmatrix}\rho_{\frac{1}{2},\frac{1}{2}} & \rho_{-\frac{1}{2},\frac{1}{2}}e^{-\gamma\left(t\right)}\\ \rho_{\frac{1}{2},-\frac{1}{2}}e^{-\gamma\left(t\right)} & \rho_{-\frac{1}{2},-\frac{1}{2}} \end{pmatrix} $$ where by symmetry $\rho_{1,-1}=\rho_{-1,1}$. I think it's hard to say anything at this point without knowing the initial values for $\rho_{m,n}$. You can work out the probability to find a state $\left|\psi\right>$ by consicering the expenctatino value of the projector $$ P_{\psi}=\left|\psi\right>\left<\psi\right| $$

and computing $$ \left\langle P\right\rangle =\mathrm{tr}\left(\rho\left(\tau\right)P_{\psi}\right) $$ assuming your state is $$ \left|\psi\right>=\frac{1}{\sqrt{2}}\left|\frac{1}{2}\right>+\left|-\frac{1}{2}\right> $$ the projector (as density matrix) is $$ P_{\psi}=\frac{1}{2}\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix} $$ Evaluating this expectation value becomes \begin{eqnarray*} \left\langle P\right\rangle & = & \mathrm{tr}\left(\rho\left(\tau\right)P_{\psi}\right)=\mathrm{tr}\left(\begin{pmatrix}\rho_{\frac{1}{2},\frac{1}{2}} & \rho_{-\frac{1}{2},\frac{1}{2}}e^{-\gamma\left(t\right)}\\ \rho_{\frac{1}{2},-\frac{1}{2}}e^{-\gamma\left(t\right)} & \rho_{-\frac{1}{2},-\frac{1}{2}} \end{pmatrix}\frac{1}{2}\begin{pmatrix}1 & 1\\ 1 & 1 \end{pmatrix}\right)\\ & = & \mathrm{tr}\left(\begin{pmatrix}\frac{1}{2}\rho_{\frac{1}{2},\frac{1}{2}}+\frac{1}{2}\rho_{-\frac{1}{2},\frac{1}{2}}e^{-\gamma\left(t\right)} & \frac{1}{2}\rho_{\frac{1}{2},\frac{1}{2}}+\frac{1}{2}\rho_{-\frac{1}{2},\frac{1}{2}}e^{-\gamma\left(t\right)}\\ \frac{1}{2}\rho_{\frac{1}{2},-\frac{1}{2}}e^{-\gamma\left(t\right)}+\frac{1}{2}\rho_{-\frac{1}{2},-\frac{1}{2}} & \frac{1}{2}\rho_{\frac{1}{2},-\frac{1}{2}}e^{-\gamma\left(t\right)}+\frac{1}{2}\rho_{-\frac{1}{2},-\frac{1}{2}} \end{pmatrix}\right)\\ & = & \frac{1}{2}\rho_{\frac{1}{2},\frac{1}{2}}+\frac{1}{2}\rho_{-\frac{1}{2},\frac{1}{2}}e^{-\gamma\left(t\right)}+\frac{1}{2}\rho_{\frac{1}{2},-\frac{1}{2}}e^{-\gamma\left(t\right)}+\frac{1}{2}\rho_{-\frac{1}{2},-\frac{1}{2}}\\ & = & \frac{1}{2}\left(\rho_{\frac{1}{2},\frac{1}{2}}+\rho_{-\frac{1}{2},-\frac{1}{2}}\right)+\frac{1}{2}\left(\rho_{-\frac{1}{2},\frac{1}{2}}+\rho_{\frac{1}{2},-\frac{1}{2}}\right)e^{-\gamma\left(t\right)} \end{eqnarray*} Here $\rho_{\frac{1}{2},\frac{1}{2}}+\rho_{-\frac{1}{2},-\frac{1}{2}}=1$ from normalization and also $\rho_{-\frac{1}{2},\frac{1}{2}}=\rho_{\frac{1}{2},-\frac{1}{2}}$ so we get

\begin{eqnarray*} \left\langle P\right\rangle & = & \frac{1}{2}+\rho_{-\frac{1}{2},\frac{1}{2}}e^{-\gamma\left(t\right)} \end{eqnarray*} which is pretty close to that we expect (given that we have used no assumptions)

$\endgroup$

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.