6.2. Proof of adiabatic theorem

Expected prerequisites

Before the start of this lecture, you should be able to:

1. Write down the Schrödinger equation.
2. Distinguish between time-dependent and time-independent Schrödinger equation.
3. Solve first order ordinary differential equations.

Learning goals

After this lecture you will be able to:

1. Write down the wave function of a system under adiabatic changes.
2. Explain the origin of the dynamic and geometrical phases.
3. Write down the criteria required for the adiabatic theorem to hold.

6.2.1. Proof of the theorem

So far, we have just stated the theorem and shown a few examples. Now, we will formulate the theorem in a mathematical way and prove it.

1. Our starting point is the time-dependent Schrödinger equation: $$i \hbar \frac{\partial}{\partial t} \left|\Psi(t)\right> = H(t) \left|\Psi(t)\right>\,$$ with a time-dependent Hamiltonian $H(t)$. For notational simplicity, we will in the following often leave out the explicit bra-ket notation, and simply write $\Psi(t)$ instead of $\left|\Psi(t)\right>$.

   We now want to consider a Hamiltonian $H(t)$ that changes slowly in time. But what does slowly mean? Slow compared to what? In contrast to the WKB approximation, this is less obvious, but we will find a proper criterion in the course of the proof.

2. If we have a time-independent Hamiltonian $H$, the time-dependent Schrödinger equation is readily solved as, $$H \phi_n = E_n \phi_n,\quad \phi_n(t)= \phi_n(0) e^{-i E_n t / \hbar},$$ where we have used $\phi$ to denote the eigenstates of $H$.

3. To solve the time-dependent problem, we will now start with a definition: We define $\psi_n(t)$ to be those eigenfunctions that solve the equation $$H(t) \psi_n(t) = E_n(t) \psi_n(t)\,$$ where we now consider time $t$ as a parameter of the Hamiltonian that we can fix to some arbitrary value.

   Instantaneous eigenstates

   The states $\psi_n(t)$ do not solve the time-dependent Schrödinger equation! (despite the occurrence of $H(t)$ in the equation). In contrast, the $\psi_n(t)$, which we have just defined, solve a stationary Schrödinger equation (we fix $t$ to some value). As a result, they must constitute a complete, orthogonal set for any time $t$. That is: $\langle{\psi_n(t) | \psi_m(t)} \rangle = \delta_{nm}$.

   If we take two different times $t_0$ and $t_1$, we cannot say anything about the inner product $\langle {\psi_n(t_0) | \psi_m(t_1)} \rangle$ in general, since the $\psi_m$ could in principle evolve in any way such that the overlap between $\psi_n$ and $\psi_m$ is no longer 0.

4. Because $\{\psi_n(t)\}$ form a complete orthonormal set, we can express the full quantum state $\Psi(t)$ as a linear combination of the states $\psi_n(t)$:

   $$\Psi(t) = \sum_n c_n(t) \psi_n(t) e^{i \theta_n(t)}.$$

   where $\theta_n(t) = -\frac{1}{\hbar} \int_0^t E_n(t') dt'$. Here, we expand the wave function $\Psi(t)$ in a different basis for every value time $t$ (using the set of $\{\psi_n(t)\}$). This is allowed; for a given value of time $t$, the wave function $\Psi(t)$ can be expanded in any arbitrary basis set.

   We could have absorbed the exponential factor involving $\theta$ into the $c_n(t)$, but as we will see, it is convenient to include it explicitly. This term may look strange, but in fact, it is a straightforward generalisation of the phase factor that an eigenstate picks up, which would be $\exp (-i E_n t/ \hbar)$, when using a constant Hamiltonian.

5. Now, we have an expression for $\Psi(t)$ and we want to see what that gives us when we apply the time-dependent Schrödinger equation to it. We get $$i \hbar \sum_n \left[ \dot{c}_n \psi_n + c_n \dot{\psi}_n + c_n \psi_n i \dot{\theta}_n \right] e^{i \theta_n (t)} = \sum_n c_n(t) H(t) \psi_n(t) e^{i \theta_n(t)}.$$

6. Recall that we defined $\psi_n$ to satisfy $H \psi_n = E_n \psi_n$, and we recognize that

   $$i \dot{\theta}_n = -\frac{i}{\hbar} E_n(t)$$

   so the $\dot{\theta}$ derivative term on the left-hand side cancels with the right-hand side of the previous equation. What we are left with is

   $$i \hbar \sum_n \left( \dot{c}_n \psi_n + c_n \dot{\psi}_n \right) e^{i \theta_n} = 0 \\ \sum_n \dot{c}_n \psi_n e^{i \theta_n} = - \sum_n c_n \dot{\psi}_n e^{i \theta_n}$$

7. By projecting these expressions on the $\psi_m$ eigenstate (which amounts to applying $\langle\psi_m|$ from the left on both sides), we find that

   $$\dot{c}_m = - \sum_n c_n \langle{\psi_m| \dot\psi_n} \rangle e^{i (\theta_n - \theta_m)}$$ where we used the orthogonality of $\psi_n$ and $\psi_m$ to eliminate the sum on the left-hand side.

8. In order to make progress evaluating what $\langle{\psi_m| \dot\psi_n} \rangle$ might be, we take the derivative of the time-independent Schrödinger equation, which we used to define the $\psi_n$ in the first place. That is,

   $$\dot{H}\psi_n + H \dot{\psi}_n = \dot{E}_n \psi_n + E_n \dot{\psi}_n$$

9. Again projecting this onto the $\psi_m$ gives us:

   $$\begin{aligned} \langle{\psi_m | \dot{H} | \psi_n }\rangle + \langle{\psi_m | {H} | \dot \psi_n }\rangle = \dot{E}_n \langle{\psi_m| \psi_n} \rangle + E_n \langle{\psi_m| \dot\psi_n} \rangle \\ \langle{\psi_m | \dot{H} | \psi_n }\rangle + E_m\langle{\psi_m| \dot\psi_n} \rangle = \dot{E}_n \delta_{nm} + E_n \langle{\psi_m| \dot\psi_n} \rangle \\ \end{aligned}$$

   So for $n\neq m$, we find that

   $$\langle{\psi_m| \dot\psi_n} \rangle = \frac{\langle{\psi_m | \dot{H} | \psi_n }\rangle}{E_n - E_m}.$$

10. When we have a slow, gradual change being applied to the system, $\dot{H}$ is small. Now, we are finally ready to make the approximation that you must have been anticipating since first reading the title of these notes. We neglect the contributions when $n \neq m$ and just get:

    $$\dot{c}_m = -c_m \langle{\psi_m| \dot\psi_m} \rangle.$$

    Adiabatic approximation

    What is considered slow is determined by $E_n - E_m$. If the energy difference is large, then a gradual change could mean something that is much faster than we might expect. We also see that in cases of degeneracy (when $E_n = E_m$ despite $n \neq m$), there isn't any definition of slow that is slow enough. The adiabatic approximation breaks down here. Similarly if we change the system in such a way that two energy levels that were separated come together or switch places, the approximation will again break down.

11. Solving the equation for the coefficients gives

    $$c_m(t) = c_m(0) e^{i \gamma(t)}, \quad \text{ where } \gamma(t) = i \int_0^t \langle{\psi_m| \dot\psi_m} \rangle dt'$$

    (You may notice that the factor $i$ appears twice in the above expression, both in the expression $e^{i \gamma(t)}$ and in the definition of $\gamma(t)$. This is not a mistake, as we will see later.)

12. So, if we start in the $n^{\text{th}}$ eigenstate of the initial Hamiltonian, that means that $\Psi = \psi_n$. So $c_n(0) = 1$ and all $c_m(0) = 0$ for $n \neq m$. Then, the full wave function after the gradual change has taken place is

    $$\Psi(t) = e^{i \theta_n(t)} e^{i\gamma_n(t)} \psi_n(t).$$ The $n^{\text{th}}$ eigenstate stays in the $n^{\text{th}}$ eigenstate.

6.2.2. Adiabatic criterion: how slow is slow?

In order to obtain the result above, the terms $n\neq m$ have been neglected by considering their contribution as small and by arguing that there should be a large enough energy gap between the instantaneous eigenenergies. Let us now derive a more quantitative criterion.

1. To derive it, we will proceed in the spirit of perturbation theory: we start with the initial approximation and then insert it back to get a higher-order approximation. In our derivation for the adiabatic theorem, we found an intermediate result $$\begin{align} \dot{c}_{n} &= - \sum_k c_k \langle{\psi_n| \dot\psi_k} \rangle e^{i (\theta_k - \theta_n)}\\ &= \sum_k c_k \frac{\langle \psi_n | \dot H |\psi_k \rangle}{E_n(t) - E_k(t)} e^{-\frac{i}{\hbar} \int_{0}^t E_k(t') - E_n(t') dt'}. \end{align}$$
2. We then found the approximate solution $c_m(t) = 1$ and $c_{n\neq m}(t)=0$. Let us now put this approximate solution back into the still exact equation to find $$\dot c_{n \neq m} = \frac{\langle \psi_n | \dot H |\psi_m \rangle}{E_n(t) - E_m(t)} e^{-\frac{i}{\hbar} \int_{0}^t E_m(t') - E_n(t') dt'}\,.$$

3. Integrating this equation, we then arrive at $$\rightarrow c_{n \neq m}(T) = \int_{0}^T \frac{\langle \psi_n | \dot H |\psi_m\rangle}{E_n(t) - E_m(t)} e^{\frac{i}{\hbar} \int_{0}^t E_n(t') - E_m(t') dt'} dt.$$

   Of course we know that $c_{n \neq m}(T)$ should be zero - that was what we input into the equation to begin with! We will now be able to find our quantitative criterion for adiabaticity by finding out under which conditions $c_{n \neq m}(T)$ will be very small.

4. There are several time-dependent quantities in the previous expression. To find a solution, we will approximate some of them by using the following constant bounds:

   • The largest contribution from the rate of change in the Hamiltonian will come from the largest matrix element:

   $$\langle \psi_n (t) | \dot H |\psi_m(t) \rangle \approx \overline{\langle \psi_n | \dot H |\psi_n\rangle}$$

   • The smallest energy difference will contribute the most:

   $$E_n(t) - E_m(t) \approx \overline{E_n - E_m}$$

5. Replacing these time-dependent quantities with these extremal bounds, we arrive at a simpler problem that we can now solve analytically: $$\begin{align} c_{n\neq m}(T) &\approx \int _0^T \frac{\overline{\langle \psi_n | \dot H |\psi_m\rangle}}{\overline{E_n - E_m}} e^{\frac{i}{\hbar} \int_0^t \overline{E_n - E_m} dt'} dt\\ &=\frac{\overline{\langle \psi_n | \dot H |\psi_m\rangle}}{\overline{E_n - E_m}} \int_0^T e^{\frac{i}{\hbar} \overline{E_n - E_m} t} dt\\ &= \frac{\overline{\langle \psi_n | \dot H |\psi_m\rangle}}{\overline{E_n - E_m}} \frac{i\hbar }{\overline{E_n - E_m}}\left( e^{-i \overline{E_n - E_m} T /\hbar } - 1 \right)\\ &\approx \frac{\overline{\langle \psi_n | \dot H |\psi_m\rangle}}{\overline{E_n - E_m}} \frac{i\hbar }{\overline{E_n - E_m}} \end{align}$$ In the last line, we made use of the fact that the last term of the previous line is oscillating. Since we are only interested in upper bounds, we can approximate it with a constant 1.

6. To get our adiabaticity criterion, we now remember that we need to have $|c_{n\neq m}(T)| \ll 1$, so that $$\frac{\hbar \overline{\langle \psi_m | \dot H |\psi_n\rangle} }{\overline{E_n - E_m}^2} \ll 1$$ or in other words $$\frac{\overline{\langle \psi_m | \dot H |\psi_n\rangle} }{\overline{E_n - E_m}} \ll \frac{\overline{E_n - E_m}}{\hbar}\,.$$ This is a quantitative estimate for the adiabaticity criterion.

6.2.3. Summary

Take-home message

Under adiabatic evolution, the wave function of the system is given as,

$$\Psi(t) = e^{i \theta_n(t)} e^{i\gamma_n(t)} \psi_n(t).$$

This means that:

• The system remains in the instantaneous eigenstate.
• It acquires two phase shifts depending on the evolution. The dynamical phase $\theta_n(t)$, and the geometrical phase $\gamma_n(t)$. They are given as,

$$\theta_n(t) = -\frac{1}{\hbar} \int_0^t E_n(t') dt', \quad \gamma_n(t) = i \int_0^t \langle{\psi_n| \dot\psi_n} \rangle dt'.$$ In order for this description to hold, we require that the system evolves slowly enough. That is, the following condition must be satisfied at all times: $$\frac{\hbar \overline{\langle \psi_n | \dot H |\psi_m\rangle} }{\overline{E_n - E_m}^2} << 1, \quad n \neq m.$$ Here, the overline indicates the largest matrix element and the smallest energy difference.