Given the Hamiltonian: $$%H = \omega \left(|0\rangle \langle1| + |1\rangle \langle0| \right) = \begin{bmatrix} 0 & \omega \\ \omega & 0 \end{bmatrix}$$, I want to find the final state $\rho(t_f)$of the given density operator: $$\rho(0) =|0\rangle \langle0| = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $$

To do so I started by stating: \begin{equation} \rho(t_f) = U\rho(0)U^\dagger\\ U=e^{-i\frac{t_f}{\hbar}H} \approx 1-i\frac{t_f}{\hbar}H\; \; \Rightarrow \;\;U^\dagger \approx 1+i\frac{t_f}{\hbar}H \end{equation} Although once I compute $\rho(t_f)$ using the above formula I obtain a non normalized state: $$Tr(\rho(t_f))\neq 1 \; \; \forall \omega \neq 0$$ Of course this problem could be solved if out of nowhere I multiplied my $\rho(t_f)$ with a normalization constant N: $$N = \frac{1}{Tr(\rho(t_f))}$$

My question is: is there something wrong with my thought process or calculations? Or do I really just have to introduce a new normalization constant? I would not mind an explanation in the option that the latter was the case(even if just as a reference).

I worked with it for a bit, and this is what I got:

As suggested, I fully expand the U operator: $$\%mathbf{U}=e^{-i\frac{t_f}{\hbar}\mathbf{H}} = \sum_n^\infty \left(\frac{c^n}{n!}\mathbf{H}^n \right)$$ Where for simplification I defined $c =i\frac{t_f}{\hbar}$.
By introducing a new operator denoted as $\mathbf{H}'$ ($\mathbf{H'} = \frac{1}{\omega}\mathbf{H}$), I notice the property:

$$\mathbf{H}^n=\left\{\begin{matrix}\omega^{n} \mathbf{I},& if \;\; n = even \\ \omega^n \mathbf{H}',& \; \; \; \; if \;\; n = odd \end{matrix}\right.$$ Hence, the problem to solve becomes: $$\mathbf{\rho}(t_f) = -\left(\sum_n^\infty \frac{c^n}{n!}\mathbf{H}^n \right)\rho(0) \left(\sum_n^\infty \frac{c^n}{n!}\mathbf{H}^n \right)$$

$$=%-\left( \sum_n^\infty\frac{c^{2n}}{2n!}\omega^{2n}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega^{2n+1}\mathbf{H'}\right)\rho(0)\left( \sum_n^\infty\frac{c^{2n}}{2n!}\omega^{2n}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega^{2n+1}\mathbf{H'}\right)$$

$$=-\left[ \sum_n^\infty \omega^{2n} \left( \frac{c^{2n}}{2n!}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega\mathbf{H'}\right )\right]\rho(0)\left[ \sum_n^\infty \omega^{2n} \left( \frac{c^{2n}}{2n!}\mathbf{I} + \frac{c^{2n+1}}{(2n+1)!}\omega\mathbf{H'}\right )\right]$$

$$=-\left( \sum_n^\infty \omega^{2n} \begin{bmatrix} \frac{c^{2n}}{2n!} & \omega\frac{c^{2n+1}}{(2n+1)!}\\ \omega\frac{c^{2n+1}}{(2n+1)!}&\frac{c^{2n}}{2n!} \end{bmatrix} \right)\left( \sum_n^\infty \omega^{2n} \begin{bmatrix} \frac{c^{2n}}{2n!} & \omega\frac{c^{2n+1}}{(2n+1)!}\\ 0&0 \end{bmatrix} \right)$$

$$%= \sum_n^\infty \omega^{4n}\begin{pmatrix}\frac{c^{4n}}{\left(2n\right)!\left(2n\right)!}&\omega\frac{c^{4n+1}}{\left(2n\right)!\left(2n+1\right)!}\\ \omega \frac{c^{4n+1}}{\left(2n\right)!\left(2n+1\right)!}& \omega^2\frac{c^{4n+2}}{\left(2n+1\right)!\left(2n+1\right)!}\end{pmatrix} = \mathbf{\rho}(t_f)$$

  • $\begingroup$ Note that the normalization of the density matrix at any instance of time $t$ is fixed by $\mathrm{Tr}(\rho(t)) = \mathrm{Tr}(\rho(0))$, which follows from the properties of the trace and from $U\,U^\dagger = 1$. $\endgroup$ – Jakob Jan 21 at 10:34
  • $\begingroup$ Yes and that is why this confuses me so much, my result does not seem to really care about the unitarity of U, which made me wonder if I was missing something. I did and redid the calculations, but I always get back the same $\rho(t_f)$ $\endgroup$ – Oti Dioti Jan 21 at 10:37
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    $\begingroup$ Why have you 'linearized' $U$, actually? $\endgroup$ – Jakob Jan 21 at 10:38
  • $\begingroup$ I would nott really know how to apply it to $\rho(0)$ otherwise. Is there something wrong there? $\endgroup$ – Oti Dioti Jan 21 at 10:39
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    $\begingroup$ You can use that the full power series, where the $n$-th power of an operator is just the operator multiplied by itself $n$ times. Then you can try to calculate $H^{(n)}$; actually, there is a 'pattern'- you will see it if you calculate $H^2, H^3 \ldots$. Edit: No need for feeling dumb. :) $\endgroup$ – Jakob Jan 21 at 11:07

Additionally to Norbert Schuch's answer and comments, where he points out your problem of normalization, I want to add a brief note about the exact calculations that can be performed in this example.

First, for convenience, I want to take the factor $\omega$ out of the definition of the Hamiltonian. Now we have to notice (you also stated it in a comment) that $$ H^{n}=\begin{cases} |0\rangle\langle 0| + |1\rangle\langle 1| &\quad \text{for } n \text{ even} \\ H &\quad \text{for } n \text{ odd} \end{cases} \quad .$$

To proceed, the exponential of an operator is defined by $$e^{c\,H} \equiv \sum\limits_n \frac{c^n}{n!}\, H^n \quad,$$ for $c\in\mathbb{C}$. You can use this relation for your operators $U$ and $U^\dagger$. To make us of the elaborated properties of the Hamiltonian, you have to split the series into even and odd terms. You'll find a very simple expression for these operators. From this, you can calculate $U\, \rho(0) \,U^\dagger$ and thus find a form of $\rho(t)$.

Edit: Of course, if you calculate the exact $\rho(t)$, then $\mathrm{Tr}\rho(t) = 1$.

Edit 2: I think it would be easier to first consider only the expansion of $U$, which reads ($\hbar=1$): $$U(t) = \underbrace{\sum\limits_{n=0}^{\infty} \frac{(-i\,\omega\,t)^{2n}}{(2n)!} H^{2n}}_{\text{even}} + \underbrace{\sum\limits_{n=0}^{\infty} \frac{(-i\,\omega\,t)^{2n+1}}{(2n+1)!} H^{2n+1}}_{\text{odd}} \quad.$$ If you now use the properties of $H$ for the even and odd series, you will find $$ U(t) = \sum\limits_{n=0}^{\infty} \frac{(-i\,\omega\,t)^{2n}}{(2n)!}\, \left(|0\rangle\langle 0| +|1\rangle\langle 1| \right) + \sum\limits_{n=0}^{\infty} \frac{(-i\,\omega\,t)^{2n+1}}{(2n+1)!}\, H \quad.$$ Now you can simplify this expression with the help of the sine and cosine function, i.e. their respective series expansions. From there it is easy to obtain $U^{\dagger}(t)$ and also straightforward to calculate $\rho(t)$. Still, if you have questions, let me know.

  • $\begingroup$ I edited the post. I added some calculations, but I am still confused on whether I am missing some fundamental steps that could simplify my expression even more $\endgroup$ – Oti Dioti Jan 21 at 13:15
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    $\begingroup$ Thank you, I got it (something like I*cos(x) -isin(x)*H'). You really helped me a lot today. I appreciate it $\endgroup$ – Oti Dioti Jan 21 at 16:22
  • $\begingroup$ Hey Jakob, dont want to bother you too much given all the help you gave me last time. But I just posted a new question regarding a similar way to approach the same topic we previously covered: physics.stackexchange.com/questions/610527/…. I hoped you may be inclined to follow up and help me to get an understanding also of this. $\endgroup$ – Oti Dioti Jan 27 at 11:21

If you expand $U$ to linear order in $t$, your density matrix will also only have trace one to linear order $t$, so $\mathrm{tr}(\rho(t))=1+O(t^2)$. As long as you get this, you did everything fine. Of course, your results will only be correct as long as the terms of order $t^2$ and higher will be small compared to the rest.

  • $\begingroup$ Well, that would make sense I guess. Is there maybe a way of proceeding that does not require a linear expansion? $\endgroup$ – Oti Dioti Jan 21 at 10:49
  • $\begingroup$ For the H at hand, it is easy to compute the matrix exponential. Just try to write down the Taylor series! $\endgroup$ – Norbert Schuch Jan 21 at 10:57

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