Adiabatic theorem states that if the Hamiltonian of the system $H(t)$ is slowly changed, and if the initial state is in the $n$th eigenstate of $H(0)$, then the final state will remain in the $n$th eigenstate of $H(t)$ assuming that the Hamiltonian does not have degeneracy and the energy is discrete. I wonder if there is some generalization of this theorem to cases when the Hamiltonian is degenerate throughout the entire evolution time and the dimension of each degenerate space remains unchanged? A similar result might be that the state will remain in the $n$th degenerate subspace permitted the change of the Hamiltonian is slow?

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    $\begingroup$ Are you asking what happens if you have a degeneracy somewhere along the time evolution ($H(t)$ for some particular $t$), or if you have a degeneracy throughout the entire time? The latter situation will be hard to construct without a symmetry. And in which case the degenerate states might be separated by transforming under different representations of that symmetry - so for example if the symmetry is parity, parity even states will remain even the entire time. Perhaps if you had an example it would be easier to see what version of this problem you were most interested in. $\endgroup$
    – AXensen
    May 26, 2023 at 9:03
  • $\begingroup$ @AXensen Thank you for your comment. I want to know the simple case, that is if the Hamiltonian is degenerate throughout the whole time and the dimension of the degenerate space remains unchanged all the time. $\endgroup$
    – narip
    May 26, 2023 at 9:08

1 Answer 1


Yes there is. Naturally these degeneracies should be motivated by symmetry considerations or else they would typically destroyed by perturbations. In particular, to have non trivial irreducible representations of the symmetry group, you’ll need a non-abelian group of symmetry. For things to be interesting, the representation of the group in the Hilbert space must vary in time so that you cannot simply keep the same basis.

Let us consider the following example. For a non abelian group, I chose $SO(3)$ for its physical relevance. To minimize the dimension of the Hilbert space and make the example as simple as possible, take 2 spin 1/2 particles: $\vec S_1,\vec S_2$ of total spin $\vec S= \vec S_1+\vec S_2$. Consider the Hamiltonians of the form: $$ H = -J\vec S_1\cdot (R\vec S_2) $$ with $J$ a non zero real number and $R$ a rotation matrix. These Hamiltonians are now invariant by the rotations generated by $\vec S_R= \vec S_1+R\vec S_2$. It is unitarily equivalent to the one generated by $\vec S$, so there is the singlet eigenspace $H^R_0$, $S_R=0$, and the triplet eigenspace $H^R_1$, $\vec S_R=1$. Note that I have chosen these Hamiltonians because, the rotation group representations preserving the Hamiltonians are different.

Now, you can vary $R$ (still a rotation matrix) and $J$ (still non zero) slowly. In the slow limit, you recover the analogue of the adiabatic theorem in the degenerate case.

Instead of reasoning in terms of kets, it is best to reason in terms of eigenspaces. You’ll have one dynamical phase per eigenspace given by the same formula. However, the geometric phases per basis vector becomes a geometric unitary transformation per eigenspace.

Schrödinger’s equation is: $$ \frac{d}{dt}|\psi\rangle = -iH|\psi\rangle $$ For actual computations, you can define a rotating basis. For the non-degenerate adiabatic case, you have a $U(1)^{\otimes n}$ gauge invariance ($n$ is the dimension of the Hilbert space) due to the freedom in the choice of the phases of the basis vectors. In the degenerate case, you now have the gauge group $\bigotimes_{i=1}^mU(n_i)$ with $m$ the number of eigenspaces and and $(n_1, …, n_m)$ their respective dimensions (by assumption $m$ and $(n_i)$ are constant). This is why the geometric phase is upgraded to a unitary transformation. In fact, if you are familiar with non-abelian gauge theories like Yang-Mills, the mathematical formalism is the same. In my example, the gauge group would be $U(1)\otimes U(3)$ the first factor corresponding to the singlet and the second one to the triplet.

You can then choose up to the gauge transformation a co-moving basis indexed by $|i,j\rangle$ with $i=1…m$ and $j=1…n_i$. In my example, consider the basis $|0,R\rangle, |1,-1,R\rangle, |1,0,R\rangle, |1,1,R\rangle$ a comoving basis with the first one spanning the singlet and the last three spanning the triplet. Expanding the kets in this basis: $$ |\psi\rangle=\psi_0 |0,R\rangle+\psi_{1,-1}|1,-1,R\rangle+\psi_{1,0}|1,0,R\rangle+\psi_{1,1}|1,1,R\rangle $$ the Schrödinger equation becomes: \begin{align} \dot \psi_0 + \psi_0 \langle 0,R|\frac{d}{dt} |0,R\rangle \color{red}{+\psi_{1,-1} \langle 0,R|\frac{d}{dt} |1,-1,R\rangle+\psi_{1,0} \langle 0,R|\frac{d}{dt} |1,0,R\rangle+\psi_{1,1} \langle 0,R|\frac{d}{dt} |1,1,R\rangle} &= -iE_0 \psi_0\\ \dot \psi_{1,-1} \color{red}{+{\psi_0 \langle 1,-1,R|\frac{d}{dt} |0,R\rangle}}+\psi_{1,-1} \langle 1,-1,R|\frac{d}{dt} |1,-1,R\rangle+\psi_{1,0} \langle 1,-1,R|\frac{d}{dt} |1,0,R\rangle+\psi_{1,1} \langle 1,-1,R|\frac{d}{dt} |1,1,R\rangle &= -iE_1 \psi_{1,-1}\\ \dot \psi_{1,0} \color{red}{+{\psi_0 \langle 1,0,R|\frac{d}{dt} |0,R\rangle}}+\psi_{1,-1} \langle 1,0,R|\frac{d}{dt} |1,-1,R\rangle+\psi_{1,0} \langle 1,0,R|\frac{d}{dt} |1,0,R\rangle+\psi_{1,1} \langle 1,0,R|\frac{d}{dt} |1,1,R\rangle &= -iE_1 \psi_{1,0}\\ \dot \psi_{1,1} \color{red}{+{\psi_0 \langle 1,1,R|\frac{d}{dt} |0,R\rangle}}+\psi_{1,-1} \langle 1,1,R|\frac{d}{dt} |1,1,R\rangle+\psi_{1,0} \langle 1,-1,R|\frac{d}{dt} |1,0,R\rangle+\psi_{1,1} \langle 1,1,R|\frac{d}{dt} |1,1,R\rangle &= -iE_1 \psi_{1,1} \end{align} The red terms are neglected according to the adiabatic approximation (neglecting crossing between eigenspaces).

The first component is calculated using the usual formulas: $$ \begin{align} \psi_0(t) &= e^{-i\phi_0(t)}U_0(t)\psi_0(0)\\ \phi_0 &= \int_0^t E_0(t’)dt’ \\ U_0(t) &= \exp\left(-\int \langle 0,R|d|0,R\rangle\right) \end{align} $$

The triplet has a similar dynamical phase, but the geometric part is given by a time ordered integral:

$$ \begin{align} \psi_1(t) &= e^{-i\phi_1(t)}U_1(t)\psi_1(0)\\ \phi_1 &= \int_0^t E_1(t’)dt’ \\ U_1(t) &= T\exp\left(-\int \langle 1,R|d|1,R\rangle\right) \end{align} $$

All this discussion generalises quite well with more eigenspaces of higher dimensions. Indeed, indexing by $i=1...m$ the eigenspace of respective degeneracy $n_i$. Let $(|i,j\rangle)_{i\leq m,j\leq n_i}$ be a moving eigenbasis, $H$ will be by construction block diagonal, with eigenvalues $E_i$ so that: $$ H|i,j\rangle = E_i|i,j\rangle $$ Let $|\psi\rangle$ be a ket decomposed in the basis as: $$ |\psi\rangle = \sum \psi_{ij}|i,j\rangle $$ then in the adiabatic limit: $$ \begin{align} \psi_{ij} &= e^{-i\phi_i}[U_i]_{jj'}\psi_{ij'} \\ \phi_i &= \int_0^t E_i(t’)dt’ \\ [U_i]_{jj'} &= \left[T\exp\left(-\int \langle i|\frac{d}{dt}|i\rangle dt\right)\right]_{jj'} \end{align} $$ where I used the shorthand for the $n_i\times n_i$ matrix: $$ [\langle i|\frac{d}{dt}|i\rangle]_{kk'} = \langle i,k|\frac{d}{dt}|i,k'\rangle $$

The same caveats apply as in the non-degenerate case. The $U_i$ depend only on the path taken in parameter space, which is why you interpret them as purely geometrical. However, they do depend generally on the choice of the moving basis, i.e. on the gauge. To get a gauge invariant quantity, you'll need to take the trace of $U_i$ over a closed loop, a Wilson loop.

Hope this helps.


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