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Let $$H := \sigma_z \otimes \sum_{k=1}^{n}\mathrm{id} \otimes ...\otimes g_k \sigma_z \otimes...\otimes \mathrm{id}$$ be the Hamiltonian ($\sigma_z$ is the Pauli-matrix of couse)

and $$|\psi_0 \rangle:=(a|\!\uparrow\rangle+b |\!\downarrow\rangle ) \otimes_{k=1}^{n} (\alpha_k |\!\uparrow \rangle + \beta_k |\!\downarrow\rangle).$$

Then my script says that the time-evolution of this state is

\begin{align} |\psi(t)\rangle &=a|\!\uparrow\rangle \otimes_{k=1}^{n} (\alpha_k e^{i g_kt}|\!\uparrow \rangle + e^{-i g_k t} \beta_k |\!\downarrow\rangle) \\ &\quad +b |\!\downarrow\rangle\otimes_{k=1}^{n} (\alpha_k e^{-i g_kt}|\!\uparrow \rangle + e^{i g_k t} \beta_k |\!\downarrow\rangle). \end{align}

Does anybody have an explanation for this?

My concern about this equation arises from the fact that for $n=0$, we would have $|\psi(t)\rangle= |\psi(0)\rangle$ which should not be case, but rather $|\psi(t)\rangle= e^{-it}a|\!\uparrow\rangle+ e^{it} b|\!\downarrow\rangle.$

What do you think?

If anything is unclear, please let me know.

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    $\begingroup$ For starters, dimensional analysis should offer a valuable flag here. In your example $\hbar=1$ but time and energy are not dimensionless, with $H$ and the $g_k$ carrying dimensions of angular frequency. Thus $e^{ig_k t}$ is fine, but combinations like $e^{±it}$ should throw a big red flag of invalid dimensions. $\endgroup$ – Emilio Pisanty Feb 15 '16 at 17:48
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From a simpler perspective, take a look at your hamiltonian, $$ H := \sigma_z \otimes \sum_{k=1}^{n}\mathrm{id} \otimes ...\otimes g_k \sigma_z \otimes...\otimes \mathrm{id}, $$ when $n=0$. Here the sum is over a null set, so the convention is that the empty sum is zero, i.e. $H=0$. Under that hamiltonian, the time evolution $$|\psi(t)⟩\equiv |\psi(0)⟩$$ is perfectly reasonable.

In general, finding solutions of the Schrödinger equation - like all other ODEs - can be complicated, but verifying if a given function is a solution is in principle very simple. It is relatively simple (though it's a pain to typeset in LaTeX) to plug the solution you've been given into the Schrödinger equation and see that it's a solution.

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  • $\begingroup$ so you are saying it is correct? $\endgroup$ – The Homeworker Feb 15 '16 at 17:55
  • $\begingroup$ Yes, precisely. $\endgroup$ – Emilio Pisanty Feb 15 '16 at 18:03
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You seem to be making the assumption $e^{i(H_1 \otimes H_2)t} = e^{iH_1t}\otimes e^{iH_2t}.$ This is not true in general. Write down the spectral decompositions of the operators and then exponentiate them carefully. To be more specific $$e^{i (\sigma _z \otimes g\sigma _z) t} \neq e^{i \sigma _z t}\otimes e^{i g\sigma _zt}.$$ You can write the matrices explicitly and check it yourself.

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  • $\begingroup$ right, I was falsely assuming this. But do you see a way to calculate the time evolution of such a state explicitly? Somehow I believe also the time-evolution for the state given in the script is wrong, as for $n=0$ you would just stay in state $a|\uparrow>+b |\downarrow >$, according to the script formula $\endgroup$ – The Homeworker Feb 15 '16 at 17:19
  • $\begingroup$ For $n=0$, the $g_0 \sigma _z$ operstor will come in the very first position where already another $\sigma _z$ is sitting. So the Hamiltonian becomes proportional to identity meaning that there is no evolution. This explains the formula in your script. The formula agrees for higher n also $\endgroup$ – biryani Feb 15 '16 at 17:42
  • $\begingroup$ I don't feel is fair to assume that the formula works for $n=0$ given that the summation starts from 1. The above explanation seems rather coincidental. $\endgroup$ – biryani Feb 15 '16 at 17:45

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