I am looking at a spin 1/2 particle in a magnetic field. This has Hamiltonian $$H=-\mu s\cdot B_0$$ For simplicity, assume $B_0=B_0\hat z$ so $H=-\mu B_0$. I then apply a perturbative magnetic field such that $$V'=-\mu B_1 s_x$$ First I wanted to compute $E^{(1)}$ $$E^{(1)}_n=\langle\psi_n^{(0)}|-\mu B_1s_x|\psi_n^{(0)}\rangle=\mp \mu B_1 \hbar/2$$ Now I am looking to find the first order correction to the ground state wavefunction. I know that this is given as $$\psi^{(1)}_n=\sum_{n\neq n'} \psi^{(0)}_{n'}\frac{\langle\psi_{n'}^{(0)}|-\mu B_1s_x|\psi_{n}^{(0)}\rangle}{E_n^{(0)}-E_{n'}^{(0)}}$$ I am confused as to how to treat the summation. The only term I would get is if $n=n'$, but that would be degerate. So I am thinking that this first order correction is 0. Is this correct?

  • $\begingroup$ The term $n=n'$ is not even included in the sum $\endgroup$ – Antonio Ragagnin Apr 14 '14 at 7:19

Spin1/2 particle

Ususally, in this kind of Hamiltonian, people uses $s=s_z$, where

$$s=s_z=\left[ \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right].$$

Then, your unperturbed hamiltonian $H_0$ is: $$H_0=-\mu s\cdot B_0 = -\mu \left[ \begin{array}{cc} 1 & 0 \\ 0 & -1\end{array} \right]B_{0,z}. $$

Then the eigen vectors of energy are:

$$|\psi^0_+\rangle=\left[ \begin{array}{c} 1 \\ 0\end{array} \right],$$ $$|\psi^0_-\rangle=\left[ \begin{array}{c} 0\\ 1 \end{array} \right].$$

Perturbation solution

Then you want to compute $|\psi_+\rangle$ and $|\psi_-\rangle$ for the perturbed Hamiltonian $H=H_0-\mu B_1 s_x$, where $$s_x=\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0\end{array} \right].$$

As you said, you have to compute the following quantities (note I use $+,-$ instead of $n=0,1$. Which became:

$$\psi^{(1)}_+=\sum_{n\neq +} \psi^{(0)}_{n'}\frac{\langle\psi_{n'}^{(0)}|-\mu B_1s_x|\psi_{+}^{(0)}\rangle}{E_+^{(0)}-E_{n'}^{(0)}}=\psi^{(0)}_{-}\frac{\langle\psi_{-}^{(0)}|-\mu B_1s_x|\psi_{+}^{(0)}\rangle}{E_+^{(0)}-E_{-}^{(0)}}$$

$$\psi^{(1)}_-=\sum_{n\neq -} \psi^{(0)}_{n'}\frac{\langle\psi_{n'}^{(0)}|-\mu B_1s_x|\psi_{-}^{(0)}\rangle}{E_-^{(0)}-E_{n'}^{(0)}}=\psi^{(0)}_{+}\frac{\langle\psi_{+}^{(0)}|-\mu B_1s_x|\psi_{-}^{(0)}\rangle}{E_-^{(0)}-E_{+}^{(0)}}$$

Put here vectors and matrices we just found and let me know if you get zero.

  • $\begingroup$ I ended up with $\psi_{up}^{(1)}=(0,B_1/2B_0)$ $\endgroup$ – yankeefan11 Apr 14 '14 at 18:27
  • $\begingroup$ And $\psi_{down}^{(1)}=(B_1/2B_0 , 0)$ $\endgroup$ – yankeefan11 Apr 14 '14 at 18:30
  • $\begingroup$ There is a wrong sign, in fact the $up$ is multipled by $1/\Delta E$ and the $down$ by $-1/\Delta E$. Besides this, I put wrong signes in my reply too (corrected now). $\endgroup$ – Antonio Ragagnin Apr 14 '14 at 18:47
  • $\begingroup$ Do my values seem correct? $\endgroup$ – yankeefan11 Apr 14 '14 at 21:35
  • $\begingroup$ Nope, I'd say $\psi^{(1)}_{down}=(-B1/2B0,0)$, while $\psi^{(1)}_{up}$ is correct. $\endgroup$ – Antonio Ragagnin Apr 15 '14 at 8:37

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