Normalization Constant in Time Evolution of Density Matrix

Question

Given the Hamiltonian:
$$%H = omega left(|0rangle langle1| + |1rangle langle0|  right) = begin{bmatrix}
0 & omega  
  omega & 0 
end{bmatrix}$$, I want to find the final state $rho(t_f)$of the given density operator:
$$rho(0) =|0rangle langle0| = begin{bmatrix}
1 & 0  
 0 & 0 
end{bmatrix} $$
To do so I started by stating:
begin{equation}
rho(t_f) = Urho(0)U^dagger
 U=e^{-ifrac{t_f}{hbar}H} approx 1-ifrac{t_f}{hbar}H; ; Rightarrow ;;U^dagger approx 1+ifrac{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:

P.S.
As suggested, I fully expand the U operator:
$$%mathbf{U}=e^{-ifrac{t_f}{hbar}mathbf{H}} = sum_n^infty left(frac{c^n}{n!}mathbf{H}^n right)$$
Where for simplification I defined $c =ifrac{t_f}{hbar}$.
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:
$$mathbf{rho}(t_f) = -left(sum_n^infty frac{c^n}{n!}mathbf{H}^n right)rho(0) left(sum_n^infty frac{c^n}{n!}mathbf{H}^n right)$$
$$=%-left( sum_n^inftyfrac{c^{2n}}{2n!}omega^{2n}mathbf{I} + frac{c^{2n+1}}{(2n+1)!}omega^{2n+1}mathbf{H'}right)rho(0)left( sum_n^inftyfrac{c^{2n}}{2n!}omega^{2n}mathbf{I} + frac{c^{2n+1}}{(2n+1)!}omega^{2n+1}mathbf{H'}right)$$
$$=-left[ sum_n^infty omega^{2n} left( frac{c^{2n}}{2n!}mathbf{I} + frac{c^{2n+1}}{(2n+1)!}omegamathbf{H'}right )right]rho(0)left[ sum_n^infty omega^{2n} left( frac{c^{2n}}{2n!}mathbf{I} + frac{c^{2n+1}}{(2n+1)!}omegamathbf{H'}right )right]$$
$$=-left( sum_n^infty omega^{2n} begin{bmatrix}
frac{c^{2n}}{2n!} & omegafrac{c^{2n+1}}{(2n+1)!} 
 omegafrac{c^{2n+1}}{(2n+1)!}&frac{c^{2n}}{2n!} 
end{bmatrix} right)left( sum_n^infty omega^{2n} begin{bmatrix}
frac{c^{2n}}{2n!} & omegafrac{c^{2n+1}}{(2n+1)!} 
 0&0 
end{bmatrix} right)$$
$$%= sum_n^infty omega^{4n}begin{pmatrix}frac{c^{4n}}{left(2nright)!left(2nright)!}&omegafrac{c^{4n+1}}{left(2nright)!left(2n+1right)!} omega frac{c^{4n+1}}{left(2nright)!left(2n+1right)!}& omega^2frac{c^{4n+2}}{left(2n+1right)!left(2n+1right)!}end{pmatrix} = mathbf{rho}(t_f)$$

Norbert Schuch · Answer

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.

Answered by Norbert Schuch on February 3, 2021

Jakob · Answer

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
$$ H^{n}=begin{cases} |0ranglelangle 0| +  |1ranglelangle 1| &quad text{for } n text{ even}  H &quad text{for } n text{ odd} end{cases}  quad .$$
To proceed, the exponential of an operator is defined by
$$e^{c,H} equiv sumlimits_n frac{c^n}{n!}, H^n  quad,$$
for $cinmathbb{C}$.
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 ($hbar=1$):
$$U(t) = underbrace{sumlimits_{n=0}^{infty} frac{(-i,omega,t)^{2n}}{(2n)!} H^{2n}}_{text{even}} + underbrace{sumlimits_{n=0}^{infty} frac{(-i,omega,t)^{2n+1}}{(2n+1)!} H^{2n+1}}_{text{odd}}  quad.$$
If you now use the properties of $H$ for the even and odd series, you will find
$$ U(t) = sumlimits_{n=0}^{infty} frac{(-i,omega,t)^{2n}}{(2n)!}, left(|0ranglelangle 0| +|1ranglelangle 1| right)
 + sumlimits_{n=0}^{infty} frac{(-i,omega,t)^{2n+1}}{(2n+1)!}, H   quad.$$
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.

Normalization Constant in Time Evolution of Density Matrix

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