3
$\begingroup$

Suppose we have an 'initial' Hamiltonian $H_{i}$ whose eigenvalues are all non-degenerate, which we order as follows:

$ E^{0}_{i} < E^{1}_{i} < \dots < E^{N-2}_{i} < E^{N-1}_{i}$

Suppose also that we have a 'final' Hamiltonian $H_{f}$, again with non-degenerate eigenvalues which we can order as before:

$ E^{0}_{f} < E^{1}_{f} < \dots < E^{N-2}_{f} < E^{N-1}_{f}$

Now, if we have some time-dependent Hamiltonian $H(t)$, such that $H(t=0)=H_{i}$ and $H(t=T)=H_{f}$, then the quantum adiabatic theorem, as defined in the 'Proof' section of its Wikipedia page, states that if the state of the system at $t=0$ is an eigenstate of $H_{i}$, so that $|\psi(t=0)\rangle = |E_{i}^{j}\rangle$, then for sufficiently large $T$, $|\psi(t=T)\rangle$ will be an eigenstate of $H_{f}$.

One important application of the quantum adiabatic theorem is in the field of adiabatic quantum computing, where one encodes the solution to a problem in some problem Hamiltonian $H_{p}$, and evolves the system Hamiltonian from some easy-to-prepare initial Hamiltonian $H_{0}$ to $H_{p}$. The main idea is that if one uses some $H_{p}$ with a non-degenerate ground state that encodes the solution to your problem (e.g., to factor a number $N$, use a Hamiltonian $H_{p}=(N-\hat{x}\hat{y})^{2}$, where $\hat{x}$ and $\hat{y}$ are operators whose eigenvalues are natural numbers), and starts the system off in the ground state of $H_{0}$ (presumed non-degenerate), then the final state vector will be the ground state of $H_{p}$. In the case of the example I gave, this solves the problem because the ground state of $H_{p}$ has an energy of 0, and so calculating $\langle \hat{x} \rangle$ and $\langle \hat{y} \rangle$ gives us the natural numbers $x$ and $y$ such that $N = xy$, assuming that $\hat{x}$ and $\hat{y}$ commute with the Hamiltonian.

My question is this: how do we know that the system at time $T$ will be in the ground state of $H_{p}$ if it was prepared in the ground state of $H_{0}$, and not simply some arbitrary eigenstate of $H_{p}$? Put another way, does adiabatic quantum evolution preserve the ordering of eigenvalues? If the answer is no, what is special about the ground state that means that its time-evolved vector is always the ground state of the resultant Hamiltonian?

$\endgroup$
2
  • 3
    $\begingroup$ As far as I know, in order for the adiabatic theorem to work, there has to be a gap between the ground state and the rest of the spectrum at all times. The last part ensures that there cannot be an eigenvalue crossing, hence the ordering of the eigenvalues must be preserved. $\endgroup$
    – Martin
    Sep 7, 2015 at 15:52
  • $\begingroup$ There are however other versions of the adiabatic theorem which don't really need a gap condition at all, but I guess the proof must be very different... $\endgroup$
    – Martin
    Sep 7, 2015 at 15:56

1 Answer 1

3
$\begingroup$

The answer is given in the comment by Martin. The adiabatic theorem makes a stronger assertion than the one proposed in the question. It asserts that, if the conditions of the theorem hold, then the final state is not just any eigenstate of $H_f$, but the particular one whose energy-level quantum number is the same as that of the starting state. The conditions of the theorem include that $$ \tau \gg \frac{\hbar}{\Delta E} $$ where $\Delta E$ is energy gap between the starting state and nearest other state during the change and $\tau$ is the timescale of the change. In order that $\tau$ should be finite while obeying this condition, it follows that $\Delta E$ must not be zero. This means the starting state must not be degenerate, and there must be no level-crossing of its energy eigenvalue with that of another state during the evolution.

It is characteristic of ground states that they tend to be non-degenerate. For such a case $\Delta E$ is the smallest value of the energy gap between ground and first excited state during the adiabatic evolution.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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