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.


  • $\begingroup$ 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

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)

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