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Spin-1/2 in a magnetic field

  1. Sep 30, 2015 #1
    A spin-1/2 is placed in a magnetic field with both x and z-components so that its
    Hamiltonian is [itex]H=-b_x \sigma^x-b_z\sigma^z[/itex], where [itex]\sigma^x[/itex] and [itex]\sigma^z[/itex] are the Pauli matrices. The real constants [itex]b_x[/itex] and [itex]b_z[/itex] have units of energy, and account for both the magnetic field components and coupling constants between the spin and the magnetic field.

    Consider that the spin-component along the z-axis of the spin-1/2 is known to be [itex]+\hbar /2[/itex] at [itex]t = 0[/itex].
    What is the probability that the spin component along the z-axis at time [itex]t ≥ 0[/itex] will be measured to be [itex]-\hbar /2[/itex]?

    Time-evolution of the initial state yields [itex]\vert \psi(t)\rangle=e^{-i \hat H t/ \hbar}\vert \uparrow_z \rangle[/itex], hence the probability of measuring [itex]-\hbar /2[/itex] is [itex]\vert \langle \downarrow_z \vert \psi(t) \rangle \vert ^2=0[/itex].

    Is this correct or am I missing something important here?
    Last edited: Sep 30, 2015
  2. jcsd
  3. Oct 1, 2015 #2


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    ##\vert \uparrow_z \rangle## and ##\vert \downarrow_z \rangle## are not the eigenstates of the problem's Hamiltonian. You may probably want to know how ##\vert \uparrow_z \rangle## and ##\vert \downarrow_z \rangle## expand into the basis made of the eigenstates of the Hamiltonian.
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