- #1
Thomas Brady
- 12
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I'm pretty new to quantum, so I'm pretty sure I'm missing something basic here. I've got a 4x4 Hamiltonian with eigenkets $$\psi_{U} = 1/(\sqrt 2) (\psi_{1up} \pm \psi_{2up})$$ and $$\psi_{D} = 1/(\sqrt 2) (\psi_{1down} \pm \psi_{2down})$$ The only difference between the two states is the spin as signified by the subscripts up (U) and down (D). The plus states have the eigenvalue ##E_0 - t## and the minus states have the eigenvalue ##E_0 + t##. Knowing this, how can I say what the probability is of an electron being in any of the ##\psi_{1up}##, ##\psi_{1down}##, ##\psi_{2up}##, and ##\psi_{2down}## states, without knowing the probability it is in ##\psi_{U}## or ##\psi_{D}##?
Just to be clear there are of course 4 eigenkets, but the difference between each of the plus and minus eigenkets is the spin
Just to be clear there are of course 4 eigenkets, but the difference between each of the plus and minus eigenkets is the spin