# Normalization Constant in Time Evolution of Density Matrix

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: $$$$\rho(t_f) = U\rho(0)U^\dagger\\ U=e^{-i\frac{t_f}{\hbar}H} \approx 1-i\frac{t_f}{\hbar}H\; \; \Rightarrow \;\;U^\dagger \approx 1+i\frac{t_f}{\hbar}H$$$$ Although once I compute $$\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:

P.S.
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: $$\mathbf{\rho}(t_f) = -\left(\sum_n^\infty \frac{c^n}{n!}\mathbf{H}^n \right)\rho(0) \left(\sum_n^\infty \frac{c^n}{n!}\mathbf{H}^n \right)$$

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

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

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

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

• 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 $U\,U^\dagger = 1$. – Jakob Jan 21 at 10:34
• 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 $\rho(t_f)$ – Oti Dioti Jan 21 at 10:37
• Why have you 'linearized' $U$, actually? – Jakob Jan 21 at 10:38
• I would nott really know how to apply it to $\rho(0)$ otherwise. Is there something wrong there? – Oti Dioti Jan 21 at 10:39
• 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. :) – Jakob Jan 21 at 11:07

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.

• 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 – Oti Dioti Jan 21 at 13:15
• Thank you, I got it (something like I*cos(x) -isin(x)*H'). You really helped me a lot today. I appreciate it – Oti Dioti Jan 21 at 16:22
• 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. – 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.

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