# Time evolution of a finite dim. quantum system

We will study the time-evolution of a finite dimensional quantum system. To this end, let us consider a quantum mechanical system with the Hilbert space $$\mathbb{C}^2$$. We denote by $$\left . \left | 0 \right \rangle\right .$$ and $$\left . \left | 1 \right \rangle\right .$$ the standard basis elements $$(1,0)^T$$ and $$(0,1)^T$$. Let the Hamiltonian of the system in this basis be given by $$H=\begin{pmatrix} 0 &-i \\ -i &0 \end{pmatrix}$$ and assume that for $$t=0$$ the state of the system is just given by $$\psi(t=0)=\left . \left | 0 \right \rangle\right .$$. In the following, we also assume natural units in which $$\hbar=1$$.

Problems:

a) Determine the eigenvalues $$\lambda_i$$ and the normalized eigenvectors $$f_i$$ of $$H$$.

b) Compute the time evolution operator $$U(t)=e^{-iHt}$$ of the system, according to $$f(H)=\sum_{i=1}^{2}f(\lambda_i)\left . \left | e_i \right \rangle\right . \left \langle \left . e_i \right | \right .$$ for $$f$$ and analytic function for all $$\lambda_i$$. Compute the time evolved state $$\psi(t)=U(t)\psi(t=0)$$.

I do not understand what to do for problem b). I need help for this one. For a), I found out that the eigenvalues $$\lambda_i$$ are $$\pm 1$$, and the normalized eigenvectors $$f_i$$ are $$\frac{1}{\sqrt 2}\begin{pmatrix} i\\ 1 \end{pmatrix}$$ and $$\frac{1}{\sqrt 2}\begin{pmatrix} -i\\ 1 \end{pmatrix}$$

• You need to diagonalize $H$. If you can write $H=PDP^{-1},$ with $D$ diagonal and $P$ orthogonal, then it's much more straight-forward to take the matrix exponential $e^{-iHt}:\;e^{-iHt}=e^{-iPDP^{-1}t}=Pe^{-iDt}P^{-1}.$ – Adrian Keister Jan 31 at 16:37
• @AdrianKeister Thank you for your comment. I figured out that $$H=\begin{pmatrix}-1&1\\ 1&1\end{pmatrix}\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}\begin{pmatrix}-\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}\end{pmatrix}$$ – UnknownW Jan 31 at 17:20
• Looks like you're on the right track! – Adrian Keister Jan 31 at 17:26
• @AdrianKeister That's good to know. I'm still confused to what exactly I am doing. What connection does it have with the $f(H)$? I apologize if the questions seem easy as I am kind of new to this subject. – UnknownW Jan 31 at 18:15
• The formula given for $f(H)$ corresponds to my $e^{-iHt}=Pe^{-iDt}P^{-1},$ because you can essentially act on the eigenvalues of the matrix instead of on the matrix itself, once you've diagonalized it. The entries on the diagonal matrix $D$ are the eigenvalues. – Adrian Keister Jan 31 at 18:20