# Queries of proof of adiabatic theorem in QM

I have a few questions regarding the proof of the adiabatic theorem in the book "Introduction to Quantum Mechanics" by Griffiths:

The assumptions are that if the Hamiltonian changes with time then the eigenfunctions and eigenvalues themselves are time-dependent: $$H(t)\psi_{n}(t) = E_{n} \psi_{n}(t).$$

1. Firstly, how would we know that the spectrum remains discrete so as to write it in this way?

He then states that the eigenfunctions form an orthonormal set which is complete, this I understand comes from the postulates of QM. But then he states that the general solution to the time-dependent Schrodinger equation can be expresses as a linear combination of them: $$\Psi(t) = \sum_{n} c_{n}(t) \psi_{n}(t)e^{i \theta_{n}(t)} ~~~\text{where }~~\theta_{n}(t) := - \frac{1}{\hbar}\int_{0}^{t}E_{n}(t')dt'$$

1. How does he know that this is the form of the phase factor (he uses this later in the proof)?

1. One doesn't need to assume that the spectrum is discrete. As the page "Adiabatic theorem" that you linked to rightfully says, what is needed is that there is an energy gap around the energy value $E_n$ that we want to keep track of, i.e. that there exists a positive $\epsilon\gt 0$ such that $E_n$ is the only eigenvalue (in the spectrum) among the numbers from the interval $(E_n-\epsilon,E_n+\epsilon)$. The remaining states away from the gap may belong to a discrete spectrum or a mixed spectrum (it cannot be a purely continuous spectrum because the $E_n$ makes it at least partly discrete). If the spectrum is mixed, the notation is consistent with that because $n$ may be assumed to be a continuous number/label, or a multi-index that stores more information that just one number.

2. Griffiths may make the statement about the phase because it is a correct statement. If I remember well, Griffiths even actually proves that this is the correct phase. The OP's opinion that Griffiths should calculate the phase before he proves it is a logically flawed opinion about derivations in mathematics and physics in general. That's simply not how it works. Mathematicians and physicists often have to "guess" some correct result by some ingenious or heuristic tricks or by trying many viable candidate answers. If they are able to prove that it is the correct (and in many cases, the only correct) answer, then they have everything that is actually needed.

The proof is easy. In Schrödinger's equation, the time derivative of $\Psi$ picks the usual terms from the time-independent Hamiltonian case but also a term where the phase $\theta$ is differentiated, and that differentiation turns the indefinite integral of $E_n(t')$ to $E_n$ itself, and that's what is needed for the Schrödinger equation to hold.

On top of the proof, one could want some insight into the thinking of the first people who guessed the right answer. Well, different people could have thought differently but there would always be some heuristic quasi-rational component to that. It may be insightful to explain "how I discovered something" but it is in no way a "necessary" part of a discussion in a textbook.

• Thanks for your answer. For 2: I'm not sure where exactly 'he proves that this is the correct phase' as you stated? But I think it is just a generalization of the usual phase factor for the time evolution operator on the energy eigenstate, hence he is assuming that the Hamiltonian is time independent and that the $H$'s at different times commute, hence $\mathcal{U}(t,0) = \text{exp} \bigg[ - (\frac{i}{\hbar} \int_{0}^{t}dt'H(t'))\bigg]$ which when applied to a energy eigenstate yields the phase factor $e^{i \theta_n(t)}$ I was referring to. – Alex Apr 30 '17 at 15:05
• For 1: So the initial spectrum could be mixed but would it be possible that the spectrum after time $t$ to be purely continuous? – Alex Apr 30 '17 at 15:12
• Alex, the formula is valid even in the general case when the Hamiltonian doesn't commute at various times. Similar formulae with operators in the exponent would also have to include the time-ordering operator but this formula has the energy eigenvalue in the exponent. The phase just collects by how much the wave function is changing at every moment. The simple product $E_n t$ from the time-independent case gets generalized to $\int dt\,E_n$ in the time-dependent case. It's really straightforward. – Luboš Motl May 1 '17 at 4:42
• Yes, it is in principle possible for a time-dependent Hamiltonian to have a mixed spectrum at one moment, and continuous at another one. In the adiabatic evolution, the discrete state may diseappear into or emerge from a continuum part of the spectrum. Just consider some dip $V=-C\exp(-x^2)$ which allows a bound state and make the depth of the dip smaller and smaller. The negative eigenvalue will approach zero before the discrete eigenstate disappears as $E=0$. – Luboš Motl May 1 '17 at 4:44
• @LubošMotl Okay thanks for that explanation. If you have any time please take a look at my proposed working for my other post. – user101311 May 4 '17 at 17:53

This is just a partial answer. He writes the time dependent coefficient as $c_n(t)e^{i\theta_n(t)}$ instead of just $c_n(t)$ because he already knows the result and knows writing it this way will make the proceeding calculations easier (albeit marginally). You can try the proof starting with $\Psi(t)=\sum_nc_n(t)\psi(t)$ and you should still get the same result as Griffiths.

• I understand that there is a phase factor involved, that is the case even for a time independent Hamiltonian. I want to know how he gets the specific form of $\theta_{n}(t)$ in the theorem statement, before even starting the proof. – Alex Apr 29 '17 at 13:37