# First order wave function in adiabatic approximation

If the Hamiltonian is slowly varying in time and suppose the initial state is the n-th eigenstate of the initial Hamiltonian H(0), the adiabatic theorem says that the state will still on the n-th eigenstate of the Hamiltonian H(t) during the evolution, accumulating a dynamic phase and geometric phase.

The above theorem can be shown by considering only the zero-th order wave function, however, what does the first order wave function looks like?

i.e. when we consider the the transition between the different energy levels, just like in the perturbation method in the time independent perturbation theorem.

Consider a Hamiltonian $$\hat{H}\left(\mathbf{P}\right)$$, where $$\mathbf{P}(t)$$ is some time-varying parameter vector. This Hamiltonian has a set of time-dependent eigenstates $$|n\left(t\right)\rangle$$ with corresponding energies $$E_n\left(t\right)$$ which evolve via the Schrodinger's equation $$i\hbar \partial_t|\Psi\left(t\right)\rangle = \hat{H}\left(t\right)|\Psi\left(t\right)\rangle$$.

If $$\mathbf{P}\left(t\right)$$ is constant in time and the system starts out in one of the eigenstates $$|n\left(0\right)\rangle$$ at $$t = 0$$, the state accumulates a phase factor with time so that $$|\Psi\left(t\right)\rangle = e^{-iE_nt/\hbar}|n\left(0\right)\rangle$$.

For finite $$\partial_t\mathbf{P}\left(t\right)$$, the eigenstates and eigenvalues change in time. Still, let us set $$|\Psi\left(t\right)\rangle = c_n(t)|n\left(t\right)\rangle$$. Putting this into the Schrodinger's equation gives

$$$$i\hbar\partial_t \left[ c_n(t)|n\left(t\right)\rangle\right] = E_n\left(t\right)c_n(t)|n\left(t\right)\rangle\,.$$$$

Taking the time derivative of the left-hand side and then applying $$\langle n\left(t\right)|$$ to both sides, we get

$$$$\dot{c}_n\left(t\right) = -\frac{i}{\hbar}c_n\left(t\right)E_n\left(t\right) -c_n\left(t\right)\langle n\left(t\right)|\dot{n}\left(t\right)\rangle \,.$$$$

This differential equation can be solved using the separation of variables:

$$$$c_n\left(t\right) = \exp\left[-\frac{i}{\hbar}\int_0^t E_n\left(t'\right)dt' - \int_0^t\langle n\left(t'\right)|\dot{n}\left(t'\right)\rangle dt'\right]\,.$$$$

It is possible to use the chain-rule to rewrite the second integrand as $$\langle n\left(t'\right)|\dot{n}\left(t'\right)\rangle = \langle n\left(\mathbf{P}\right)|\nabla_\mathbf{P}n\left(\mathbf{P}\right)\rangle\cdot \dot{\mathbf{P}}\left(t'\right)$$, where $$\nabla_\mathbf{P}$$ is the gradient in $$\mathbf{P}$$ space. Since $$\dot{\mathbf{P}}\left(t'\right)dt' = d\mathbf{P}$$, we have

$$$$c_n\left(t\right) =e^{-\frac{i}{\hbar}\int_0^t E_n\left(t'\right)dt'}e^{i\int_{\mathbf{P}_i}^{\mathbf{P}_f}\mathcal{A}_n\cdot d\mathbf{P}}\,,$$$$

where $$\mathbf{P}_i = \mathbf{P}_i\left(0\right)$$, $$\mathbf{P}_f = \mathbf{P}\left(t\right)$$, and $$\mathcal{A}_n = i\langle n\left(\mathbf{P}\right)|\nabla_\mathbf{P}n\left(\mathbf{P}\right)\rangle$$.

So, here you have the dynamic and the geometric (Berry) phase that you mentioned.

Up to now, we assumed that the system remains in the same eigenstate by asserting that the Hamiltonian changes very slowly and the system is always in the $$n$$th eigenstate. Let us now relax this assumption and write $$|n\left(t\right)\rangle \rightarrow |n\left(t\right)\rangle + f\left(\dot{\mathbf{P}}\left(t\right)\right)|\delta n\left(t\right)\rangle$$, where $$f\left(\dot{\mathbf{P}}\left(t\right)\right)$$ is some function. For $$\dot{\mathbf{P}}\left(t\right)\rightarrow 0$$, the second term vanishes as we restore the adiabatic regime. This means that, to the leading order, $$f\left(\dot{\mathbf{P}}\left(t\right)\right)$$ is linear in $$\dot{\mathbf{P}}\left(t\right)$$. By keeping only this linear term and absorbing it into the correction, we write the corrected state $$|n\left(t\right)\rangle \rightarrow |n\left(t\right)\rangle + |\delta \left(t\right)\rangle$$, where $$|\delta \left(t\right)\rangle$$ is linear in $$\dot{\mathbf{P}}\left(t\right)$$.

Above, we showed that the phase of the state does not depend on the rate of change of $$\mathbf{P}$$. So, we keep $$c_n\left(t\right)$$ for the state with the correction. Putting $$c_n\left(t\right)\left[|n\left(t\right)\rangle + |\delta \left(t\right)\rangle\right]$$ into the Schrodinger's equation and dividing by $$c_n\left(t\right)$$ yields

\begin{align} i\hbar\left\{\left( -\frac{i}{\hbar}E_n\left(t\right)+i\mathcal{A}_n\left(t\right) \cdot \dot{\mathbf{P}}\left(t\right) \right) \left[|n\left(t\right)\rangle + |\delta\left(t\right)\rangle\right] + \dot{\mathbf{P}}\left(t\right)\cdot \nabla_\mathbf{P} \left[|n\left(t\right)\rangle + |\delta\left(t\right)\rangle\right]\right\} &= \hat{H}\left(t\right)\left[|n\left(t\right)\rangle + |\delta\left(t\right)\rangle\right]\,, \end{align}

where we have replaced the time derivative acting on $$|n\left(t\right)\rangle + |\delta\left(t\right)\rangle$$ by $$\dot{\mathbf{P}}\left(t\right)\cdot \nabla_\mathbf{P}$$.

There are a few simplifications we make here. First, the $$E_n\left(t\right)|n\left(t\right)\rangle$$ term on the left gets cancelled by $$\hat{H}\left(t\right)|n\left(t\right)\rangle$$ on the right. Also, $$\dot{\mathbf{P}}\left(t\right) |\delta\left(t\right)\rangle$$ and $$\dot{\mathbf{P}}\left(t\right) \cdot \nabla_\mathbf{P}|\delta\left(t\right)\rangle$$ are quadratic in $$\dot{\mathbf{P}}\left(t\right)$$, so we eliminate them. The remaining terms are

\begin{align} i \hbar\dot{\mathbf{P}}\left(t\right)\cdot \left[ i \mathcal{A}_n\left(t\right) + \nabla_\mathbf{P} \right] |n\left(t\right)\rangle &= \left[\hat{H}\left(t\right) - E_n \left(t\right)\right]|\delta\left(t\right)\rangle \,. \end{align}

Finally, we apply $$\langle m\left(t\right)|$$ to both sides of this expression to get

$$$$|\delta\left(t\right)\rangle = i\hbar \sum_{m\neq n}|m\left(t\right)\rangle\frac{\langle m\left(t\right) |\partial_t n\left(t\right)\rangle }{E_m\left(t\right)-E_n\left(t\right)}\,, \label{eqn:delta_3}$$$$

where we have recombined $$\dot{\mathbf{P}}\left(t\right)\cdot \nabla_\mathbf{P}$$ into $$\partial_t$$.

Using $$\langle m\left(t\right)|\partial_t \left[\hat{H}\left(t\right)|n\left(t\right)\rangle\right] = \langle m\left(t\right)|\partial_t\hat{H}\left(t\right) |n\left(t\right)\rangle + \langle m\left(t\right)|E_m\left(t\right)|\partial_tn\left(t\right)\rangle = E_n\left(t\right)\langle m\left(t\right)|\partial_t n\left(t\right)\rangle$$, we write the time-dependent state as

$$$$|\Psi\left(t\right)\rangle =c_n\left(t\right)\left[|n\left(t\right)\rangle - i\hbar \sum_{m\neq n}|m\left(t\right)\rangle\frac{\langle m\left(t\right) |\partial_t \hat{H}\left(t\right)|n\left(t\right)\rangle }{\left[E_m\left(t\right)-E_n\left(t\right)\right]^2}\right] \,.$$$$

• $c_n(t)$ is absorbed into $|n(t)>$ or dropped completely in your last formula? Is it justified in time dependent S.E. – an offer can't refuse Aug 2 at 11:45
• Does $\delta n(t)$ has $n(t)$ component? I only see you discuss $m \neq n$ case. I know we can absorb that part in the original $n(t)$, but that means the coefficient of the second term of last equation should also change. Maybe your logic is considering it as a second order $\dot{P}(t)$ effect. – an offer can't refuse Aug 2 at 13:12
• I reinstated the $c_n(t)$. In principle, if you are not looking for interference effects and only care about the expectation value, $c_n(t)$ doesn't play a role. For the correction, there is no $m = n$ case. This would cause the denominator to go to zero and the expression to diverge. The absence of $n$ is the standard result in perturbation theory (like the 1st order correction in time-independent systems) – IcyOtter Aug 2 at 13:39
• Diverge in denominator is not the excuse. – an offer can't refuse Aug 2 at 15:18
• Your final answer does not satisfy the initial condition $|\psi(0)\rangle = |n(0)\rangle$. – Hosein Aug 28 at 11:42