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The problem: find how the Hamiltonian $$ H= -\frac{\hbar\omega_0}{2}Z-\frac{\hbar \omega_1}{2}(\sigma_+e^{i\omega t}+\sigma_- e^{-i\omega t}) $$ where $\sigma_{\pm}=\frac{1}{ 2}(X\pm i Y)$ changes under a rotation $U=\exp(\frac{i\omega t}{2}Z)$.


As $H'=UHU^{-1}$, the first term remains unchanged as $[A, e^{tA}]=0$ for any operator $A$ and thus $U$ and $U^{-1}$ cancel out. As for the second term, we should find how $\sigma_ {\pm}$ transforms: in terms of $X$ and $Y$, by observing $$ X'=e^{i\omega t Z/2}Xe^{-i\omega tZ/2}=(\cos(\omega t/2)\mathbb{I}+i\sin(\omega t/2)Z)X(\cos(\omega t/2)\mathbb{I}-i\sin(\omega t/2)Z) \\ =X(\cos^2(\omega t/2)-\sin^2(\omega t/2))+2Y\sin(\omega t/2)\cos(\omega t/2) \\ =X\cos(\omega t)-Y\sin(\omega t)$$ and similarly $Y'=Y\cos(\omega t)+X\sin(\omega t)$. Therefore $$\sigma_{\pm}'=\frac{1}{2}(X\cos(\omega t)-Y\sin(\omega t)\pm i(Y\cos(\omega t)+X\sin(\omega t)))=\frac{1}{2}(Xe^{\pm i\omega t}\pm iYe^{\pm i\omega t})=\sigma_\pm e^{\pm i\omega t}. $$ According to my notes however, $$ H'=\frac{\hbar (\omega-\omega_0)}{2}Z-\frac{\hbar\omega_1}{2}X$$ which means I have made two mistakes: first, it should be $\sigma_\pm'=\sigma_\pm e^{\mp i\omega t}$, and second, there is a term $(\hbar\omega/2)Z$ I have not accounted for. I can live with having made a sign error somewhere in the calculation above (if someone can point it out, all the better); however I am not sure where this last term should come from. Can anyone help me out?

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  • $\begingroup$ Are you sure the $Z$ operator should be in the exponential of the unitary operator? You wrote the problem statement differently from how you solved the problem. $\endgroup$
    – prolyx
    Feb 8, 2021 at 1:59
  • $\begingroup$ Yes, sorry. $Z$ is in the exponential. And I think that the first "mistake" comes from the fact that there is a typo in the source and the operator is missing a minus sign... $\endgroup$ Feb 8, 2021 at 2:03
  • $\begingroup$ A couple of other writing mistakes: expression for $X'$ has wrong expansion for exponential in it (in the first one is wrong, second one looks right). The equation for $\sigma'_{\pm}$ is missing an $i$ beside the $Y$ somewhere. I haven't caught the math mistake yet, though. $\endgroup$
    – prolyx
    Feb 8, 2021 at 2:19
  • $\begingroup$ I worked through the math, and also got your result, sign error and all. So it's either the problem statement/answer do not match, or we're both getting tripped up on a similar mistake. I submitted an edit of a couple of typos I saw in your expessions, though. $\endgroup$
    – prolyx
    Feb 8, 2021 at 2:45
  • $\begingroup$ @JonathanJeffrey The correct operator should be $U=e^{-i\omega tZ/2}$. If that $-$ sign is missing, the rotation is induced in the opposite direction, thus leading to a result where the phase in the exponential is doubled, rather than the elimination of the time dependence. This should fix the first problem, and the second one should be fixed by the answer below. Thanks to all who replied. $\endgroup$ Feb 8, 2021 at 21:32

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Regarding the $\frac{\hbar\omega}{2} Z$ term: it is not as simple as $H' = U H U^\dagger$. We have to demand that the Schrödinger equation $$ \mathrm i \hbar\, \partial_t |\psi'(t)\rangle = H'(t) |\psi'(t)\rangle , $$ still holds, where $|\psi'(t)\rangle = U(t) |\psi(t)\rangle$ is the transformed state. We calculate the right hand side using the product rule: $$ \partial_t \bigl[ U(t) |\psi(t)\rangle \bigr] = U'(t) |\psi(t)\rangle + \frac{1}{\mathrm i \hbar} U(t) H(t) |\psi(t)\rangle . $$ Now compare both equations to find $H'$.

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