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Why is it the case that for a two level system, say a particle which is a spin $1/2$ system (hence can either be spin up or spin down), in the absence of any external perturbation by a magnetic field or electric field, the Hamiltonian can be considered by $\hat{H} = \hbar \frac{\omega_0}{2}\hat{\sigma}_z$?

Also I don't really see how it is possible to have a Hamiltonian with no external influence to account for it?

Thanks for any assistance.

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  • $\begingroup$ Your instincts are good (specifically, you should be suspicious of such a Hamiltonian with no external influences). But what makes you think this is the correct Hamiltonian for your system? Are you sure you can assume no external field? $\endgroup$ – Mike Jul 19 '17 at 11:19
  • $\begingroup$ @Mike I'm not sure, but I was told by a Prof that this is the general Hamiltonian for a two level system in the absence of any external radiation or influence. Hence I am trying to understand what the source of this Hamiltonian is or if this whole assumption is incorrect, quite confusing. $\endgroup$ – user101311 Jul 19 '17 at 11:23
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The effect of themagnetic field on the system - irrespective of how it is generated - is to split the energy level, i.e. raise the spin-up state and lower the spin-down state. The Hamiltonian you have does exactly this. The stength of the field is "hidden" in the parameter $\omega_0$, and is written this way so the units conveniently simplify: $\hbar \omega_0$ has units of energy and $\sigma_z$ is dimensionless.

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  • $\begingroup$ Okay thanks but how do we know there will always be a magnetic field? I am assuming no electromagnetic radiation. $\endgroup$ – user101311 Jul 19 '17 at 11:13
  • $\begingroup$ What field though? I am assuming no field at all. Are you saying that there is always a magnetic field for any two level spin system? $\endgroup$ – user101311 Jul 19 '17 at 11:40
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    $\begingroup$ No I'm saying the above Hamiltonian models the effect of a static magnetic field on a spin-1/2 system. The strength of $B$ is related to $\omega_0$ by the Larmor relation: $\omega_0=e B/m$ (depending on the units) see hyperphysics.phy-astr.gsu.edu/hbase/Nuclear/larmor.html. Radiation produces a time-dependent field, so you Hamiltonian does not include radiation as per your assumption. $\endgroup$ – ZeroTheHero Jul 19 '17 at 11:46
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    $\begingroup$ magnetic and static; something that splits the enegy levels of the spins, and magnetic static is the easiest. Anything else is equivalent to this in a 2-level system. $\endgroup$ – ZeroTheHero Jul 19 '17 at 12:01
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    $\begingroup$ In vacuum there would be no energy difference between the spin up and spin down states.. The Hamiltonian $\sim \sigma_z$ always has an energy difference, so it would be appropriate to describe a system with no energy difference between the two states.. $\endgroup$ – ZeroTheHero Jul 19 '17 at 12:47

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