# Unitary transformation of the Hamiltonian with spin-orbital coupling

I am reading this Paper recently. The author says that: for this Hamiltonian: $$H(t) = \frac{p^2}{2m} + \frac{m\omega^2}{2}x^2 + \alpha p_x \sigma_y$$

If we make a unitary transformation $\mathcal{A_{\alpha}} = e^{-imx\alpha \sigma_y/\hbar}$, The Hamiltonian will be transformed to $$H_0 = \frac{p^2}{2m} + \frac{m\omega^2}{2}x^2$$ And after that, we can solve the Schrodinger Equation and the evolution of the states can be calculated.

I cannot figure out how to do the transform. Is it just $\mathcal{A_{\alpha}}H(t)\mathcal{A_{\alpha}}^{\dagger}$? (I failed while trying to calculate this). Does anybody knows how to do the transformation?

For the right A take the CC of the exponent and for the Pauli spin matrix take adjoint. Hope it works.

• Thanks for you attention SAKhan! what do you mean by "CC of the exponent"? – TimQian Dec 8 '14 at 7:14
• I mean take the complex conjugate. – SAKhan Dec 8 '14 at 7:36

First we must see that a unitary transform is much like a change of basis in linear algebra:

$$\text{If: }i\hbar\dot\psi=H\psi, \text{ and } \psi=\mathcal{A_{\alpha}^\dagger}\phi,$$

then

$$i\hbar (\mathcal{A_{\alpha}^\dagger}\dot\phi+\dot {\mathcal{A_{\alpha}^\dagger}}\phi)=H\phi \to i\hbar\dot\phi=(\mathcal{A_{\alpha}}H\mathcal{A_{\alpha}^\dagger}-i\hbar\mathcal{A_{\alpha}}\dot{\mathcal{A_{\alpha}^\dagger}} )\phi.$$

We are lucky that $\dot{\mathcal{A_{\alpha}^\dagger}}=0$ in this case.

Now, to see the effects of this transform in this Hamiltonian we do

$$x \to \mathcal{A_{\alpha}} x \mathcal{A_{\alpha}}^\dagger=x\\ p \to \mathcal{A_{\alpha}} p \mathcal{A_{\alpha}}^\dagger=p-m \alpha\sigma_y\\ p^2 \to \mathcal{A_{\alpha}} p^2 \mathcal{A_{\alpha}}^\dagger=\mathcal{A_{\alpha}} p \mathcal{A_{\alpha}}^\dagger \mathcal{A_{\alpha}} p \mathcal{A_{\alpha}}^\dagger = p^2 -2m\alpha\sigma_y p+m^2\alpha^2$$

so

$$H_1=\frac{1}{2m}p^2+\frac{m}{2}(x^2\omega^2-\alpha^2).$$

Adding a constant to the Hamiltonian contributes only to the global phase, so we can discard this and obtain the desired result

$$H_0=\frac{1}{2m}p^2+\frac{m}{2}x^2\omega^2.$$

• A constant addition to the hamiltonian doesn't matter physically. – Mikewins Jun 6 '17 at 18:54
• you are prefectly correct @MichaelWiner. Thanks for reminding me, I'll update the answer. – ivbc Jun 6 '17 at 19:35