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So, I am given a Hamiltonian of a system, represented in the $|e_{i}\rangle$ basis as: $$H=\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}$$ where, $|e_{1}\rangle = \begin{bmatrix}1\\0\\0\end{bmatrix}$, $|e_{2}\rangle= \begin{bmatrix}0\\1\\0\end{bmatrix}$, $|e_{3}\rangle = \begin{bmatrix}0\\0\\1\end{bmatrix}$. The eigenvalues and eigenvectors of H are $$|E_{1}=\bar{h}\omega\rangle=\frac{1}{\sqrt{2}}\begin{bmatrix}0\\1\\-1\end{bmatrix}, |E_{2}=\bar{h}\omega,1\rangle= \frac{1}{\sqrt{2}}\begin{bmatrix}0\\1\\1\end{bmatrix}, |E_{2}=\bar{h}\omega,2\rangle=\begin{bmatrix}1\\0\\0\end{bmatrix}$$ Let C be represented in the $|e_{i}\rangle$ basis as $$C=\begin{bmatrix}0&0&2\\0&1&0\\2&0&0\end{bmatrix}$$ At t=0, the system is in the state $|\alpha(t=0)\rangle=\frac{1}{\sqrt{2}}|e_{1}\rangle+\frac{1}{\sqrt{2}}|e_{2}\rangle$

The question asks to determine the representation of time evolution operator $U(t,t_{0}=0)$ in the $|e_{i}\rangle$ representation.

Attempt at a solution: Well, I know the usual form of the time evolution operator but I am confused as to how to approach this problem and represent the operator in the given basis.

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closed as off-topic by sammy gerbil, Norbert Schuch, Jon Custer, Gert, John Rennie Jan 9 '17 at 7:23

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  • $\begingroup$ what is the relevance of $C$ in your problem? $\endgroup$ – ZeroTheHero Jan 8 '17 at 1:50
  • $\begingroup$ well the first part of the question asked what are the possible values resulting from measurement of C and with what probabilities. that was simple enough $\endgroup$ – Blitz Jan 8 '17 at 2:12
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The time evolution operator will be the exponential of $H$. This is easy enough to work out since $H^2$ is diagonal (and in fact the unit matrix). Thus $$ U(t)=\exp(itH):=\sum_n \frac{(it)^n H^n}{n!} $$ can be summed easily and will satisfy the boundary condition at $t=0$. (I have used $\hbar=1$ throughout.) To be explicit, $U$ is the series \begin{align} U&=1_{3\times 3}+\frac{itH}{1!}+\frac{(i t H)^2}{2!}+\frac{(itH)^3}{3!}+\ldots\\ &=1_{3\times 3}\left(1+\frac{(it)^2}{2!}+\frac{(it)^4}{4!}\right)+ \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \\ \end{array} \right) \left(it+\frac{(it)^3}{3!} +\ldots\right)\, , \end{align} the expression for which can be broken into even and odd powers as indicated. By comparing with the series for sine and cosine the series can be summed easily to recover $$ U(t)=\left( \begin{array}{ccc} e^{i t} & 0 & 0 \\ 0 & \cos (t) & i \sin (t) \\ 0 & i \sin (t) & \cos (t) \\ \end{array} \right)\, . $$ It can be hard to resum the series unless the Hamiltonian is simple. A more general method is to write $$ H=T\cdot D\cdot T^{-1} \qquad \leftrightarrow \qquad D=T^{-1}\cdot H\cdot T $$ where $D$ is diagonal with the eigenvalues as diagonal entries, and $T$ is a matrix constructed using the eigenvectors of $H$, arranged in columns. This way $$ H^2= T\cdot D\cdot T^{-1} \cdot T\cdot D\cdot T^{-1} T\cdot D\cdot T^{-1} = T\cdot D^2\cdot T^{-1} $$ and by induction $$ H^n= T\cdot D^n\cdot T^{-1} $$ so that $$ \exp(i t H)= T\cdot \exp(i t D)\cdot T^{-1}\, . $$ In your specific case, $$ D=\hbox{diagonal}(1,-1,1)\, ,\qquad \exp(i t D)= \left( \begin{array}{ccc} e^{i t} & 0 & 0 \\ 0 & e^{-i t} & 0 \\ 0 & 0 & e^{i t} \\ \end{array} \right)\, , $$ so it only remains to multiply by $T$ and $T^{-1}$, but you already have the eigenvectors.

This approach will work in general.

The time-evolution operator depends on the basis because $H$ depends on the basis, but does not depend on the initial state as $H$ does not depend on the initial state either. Of course different initial states will evolve differently in time, but this is because the action if $U$ on different input states will produce different outputs for a given $U$.

(Nota: the final result is correct but they may be typos in the answer.)

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  • $\begingroup$ sorry, i am not fully clear on what you are implying. Can you please elaborate on the steps to a solution? $\endgroup$ – Blitz Jan 8 '17 at 2:13
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    $\begingroup$ sorry but I don't understand your comment. $\endgroup$ – ZeroTheHero Jan 8 '17 at 2:16
  • $\begingroup$ updated with additional details. I hope this will fully answer your question. $\endgroup$ – ZeroTheHero Jan 8 '17 at 3:22
  • $\begingroup$ Thanks loads! This does answer my question. So can I take this as a general method to solve questions like these? i.e Expand the time evolution operator in power series and try to separate out the even and odd parts and find a matrix representation of U? $\endgroup$ – Blitz Jan 8 '17 at 3:51
  • $\begingroup$ I actually just added additional information on the fully general method. $\endgroup$ – ZeroTheHero Jan 8 '17 at 4:02

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