
#1
Oct1405, 05:31 PM

P: 20

We have a spin state described by a timedependent density matrix
[tex]\rho(t) = \frac{1}{2}\left(\mathbf{1}+\mathbf{r}(t)\cdot \mathbf{\sigma} \right)[/tex] Initial condition for the motion is [tex]\mathbf{r} = \mathbf{r}_0[/tex] at [tex]t = 0[/tex]. We are then asked to give a general expression for [tex]\rho(t)[/tex] in terms of the time evolution (TE) operator, and use that to find the timedependent vector [tex]\mathbf{r}(t)[/tex]. The density matrix expression in terms of the TE operator i got to be [tex]\rho(t) = \mathcal{U}(t)\rho_0 [\mathcal{U}(t)]^{\dagger}[/tex] where [tex]\rho_0 = \rho(t = 0)[/tex]. Now that I'm going to find the timedependent vector [tex]\mathbf{r}(t)[/tex] I'm having a bit more trouble. I've started with the equation [tex]\rho_0 = \frac{1}{2}\left(\mathbf{1}+\mathbf{r}_0\cdot \mathbf{\sigma} \right)[/tex] let the TE operator and it's adjoint operate on it from the left and right respectively. That has left me with the relation [tex]\mathcal{U}(t) \mathbf{r}_0 \cdot \mathbf{\sigma}[\mathcal{U}(t)]^{\dagger} = \mathbf{r}(t)\cdot \mathbf{\sigma}[/tex] Furthermore, the Hamiltonian for the system is [tex]H = \frac{1}{2}\hbar \omega_c \sigma_z[/tex] So, assuming that I'm on track so far, does anyone have any suggestions as to how I may proceed next? What I tried was to first use Euler's formula to write out the TE operator and its adjoint. Then I expanded the cosine and the sine part separately and found that the cosine part only contains even powers of the exponent, thus making all [tex]\sigma_z[/tex] become unity. For the sine part, which contains only odd powers of the exponent, all powers of [tex]\sigma_z[/tex] are equal to the matrix itself. The problem became when I put all this into the relation I need to solve, as it gave many parts containing [tex]\sigma_z[/tex], [tex]\mathbf{\sigma}[/tex] and/or [tex]\mathbf{r}_0[/tex] multiplied in different orders, and I'm not really sure how to handle that. So, what I need to know is if I'm on the right track, or maybe I'm ignoring something or perhaps there's an easier way to do this that I should look into. Suggestions are appreciated. 



#2
Oct1505, 02:27 AM

Emeritus
Sci Advisor
PF Gold
P: 6,238

Just a suggestion, I didn't work it out myself. 



#3
Oct1505, 05:59 PM

P: 20

Of course! Since [tex]\sigma_z[/tex] is diagonal, when it's the exponential, you just get a the same matrix with each diagonal element exponentiated!
Now I finally got an expression for the time dependent vector! It remains, however, to be seen if it's the correct expression. Anyway, thanks for helping me on my way, vanesch. I'll probably be back with more questions though. 


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