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 $\hbar=1$.


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 $\frac{1}{\sqrt 2}\begin{pmatrix} -i\\ 1 \end{pmatrix}$

  • $\begingroup$ 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 $e^{-iHt}:\;e^{-iHt}=e^{-iPDP^{-1}t}=Pe^{-iDt}P^{-1}.$ $\endgroup$ – Adrian Keister Jan 31 at 16:37
  • $\begingroup$ @AdrianKeister Thank you for your comment. I figured out that $$ H=\begin{pmatrix}-1&1\\ 1&1\end{pmatrix}\begin{pmatrix}i&0\\ 0&-i\end{pmatrix}\begin{pmatrix}-\frac{1}{2}&\frac{1}{2}\\ \frac{1}{2}&\frac{1}{2}\end{pmatrix} $$ $\endgroup$ – UnknownW Jan 31 at 17:20
  • $\begingroup$ Looks like you're on the right track! $\endgroup$ – Adrian Keister Jan 31 at 17:26
  • $\begingroup$ @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. $\endgroup$ – UnknownW Jan 31 at 18:15
  • $\begingroup$ 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. $\endgroup$ – Adrian Keister Jan 31 at 18:20

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