0
$\begingroup$

I am trying to figure out the Berry phase in case of 2D Harmonic Oscillator under adiabatic cyclic condition of Hamiltonian. I know that in 1 D since Hamiltonian has only 1 time dependent parameter (spring constant $k_{x}$), the loop integral in Berry phase $$\gamma_{n}(T) = i\oint \langle\psi_{n}|\nabla_{R}\psi_{n}\rangle.d\textbf{R}$$

vanishes. I also know that if eigenfunctions are real, then also Berry phase vanishes independent of number of parameters in Hamiltonian. Since for 2D case also eigenfunction is some gaussean times Hermite polynomial, which is real, can I say that Berry phase will vanish. Can you refer me to some paper in this topic. I want someone to guide me through the step by step calculation of Berry phase even if it is 0.

Can someone also refer me to some article or any textbook where Berry phase for many different systems has been calculated and shown explicitly. In any quantum mechanics book I find only as introduction to Berry phase with no problem solving questions.

$\endgroup$

Before answering, please see our policy on resource recommendation questions. Please write substantial answers that detail the style, content, and prerequisites of the book, paper or other resource. Explain the nature of the resource so that readers can decide which one is best suited for them rather than relying on the opinions of others. Answers containing only a reference to a book or paper will be removed!

  • $\begingroup$ For detailed exposition with few examples you can look into "Geometric Phases in Classical and Quantum Mechanics" by Chruciski and Jamiokowski. $\endgroup$ – Sunyam Oct 24 '17 at 4:27
1
$\begingroup$

For 2D oscillator the wave function is the product of $\Psi(x,y)=\Psi_{1D}(x)\Psi_{1D}(y)$ so in this case one can integrate independetly and you get also zero for Berry phase. It may be easy proven that in any dimensions if the wave function is real, Berry phase has zero value.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.