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It's folklore dating back to von Neumann and Wigner that time-dependent Hamiltonian systems tend not to have level crossings of their energy eigenvalues. However, we can of course consider smoothly varying Hamiltonians which have been engineered to have level crossings. These don't even have to be complicated: for instance, any Hamiltonian of the form $$ H = -\sum_{(i,j)} \sigma^{(z)}_i \sigma^{(z)}_j + \epsilon P t $$ on a 1D spin chain, for $0 < \epsilon \ll 1$ and any Hermitian operator $P$, is an example: the all-up state and the all-down are ground-states for $t = 0$, though for $t \ne 0$ such symmetry is typically broken, so that for $|t| \ll 1$ we expect to have eigenstates close to the all-up and all-down states but with distinct eigenvalues.

After some investigation, I've come to suspect the following:

Conjecture. If $H$ has a level crossing between energy levels $E_0, E_1$ at some time $T$, and $\Pi$ is the projection onto the span of the $E_0$- and $E_1$-eigenstates then there are well-defined (continuously varying) eigenvectors through the level crossing only if $\Pi [H(T), \dot H(T)] = 0$ — that is, if the change in the Hamiltonian at the level crossing is only a change in the values of the two crossing eigenvalues, for a common pair of eigenvectors. Specifically: if this equality does not hold, then any (unitary) time-dependent change-of-basis operator from the standard basis to the energy eigenbasis of $H$ at time $t$ which is continuous for some neighborhood $t \in (T,T+\epsilon]$ will oscillate infinitely rapidly as $t \to T$.

  • Is this true generally? (If not, can you point to a counterexample?)

  • Is this a known result, and is there a reference that I can refer to where this question is treated clearly, and more-or-less formally for bounded operators (e.g. on finite-dimensional systems)?

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1 Answer 1

No, the eigenbasis is not unstable as the Hamiltonian approaches the time of the level crossing, and this can be seen by considering an adiabatic approximation which decouples the two crossing energy levels from the rest$\def\ket#1{|#1\rangle}\def\bra#1{\langle#1|}.$

How one could be fooled into thinking the eigenbasis is unstable

First, consider why one might think that the eigenvectors are inherently unstable. Let $\ket{E_0}, \ket{E_1}$ be the $E_0$-eigenvector and $E_1$-eigenvector respectively, and let $R: \mathbb C^2 \to \mathcal H$ be an operator mapping $\ket0 \mapsto \ket{E_0}$ and $\ket{1} \mapsto \ket{E_1}$ respectively. The rate of change of $R$, one might imagine, relates directly to how quickly the eigenstates change: $$ \ket{\dot E_j} = \tfrac{\mathrm d}{\mathrm dt} \Bigl[ R \ket{j} \Bigr] = \dot R \ket{j} .$$ Note that because $R^\dagger R = 1$ by construction, we have $$ 0 = \tfrac{\mathrm d}{\mathrm d t} \Bigl[ R^\dagger R \Bigr] = \dot R^\dagger R + R^\dagger \dot R ,$$ which implies that $R^\dagger \dot R$ is anti-hermitian. To get at the rate of change of $R$, consider the fact that $$ H = R D R^\dagger $$ for $D = \mathrm{diag}(E_0,E_1)$, so that $$ \dot H = \dot R D R^\dagger + R \dot D R^\dagger + R D \dot R^\dagger ,$$ which implies that for distinct $j,k \in \{0,1\}$, $$\begin{aligned}[b] \bra{E_j} \dot H \ket{E_k} &= \bra{E_j} \dot R D \ket{k} + \bra{j} \dot D \ket{k} + \bra{j} D \dot R^\dagger \ket{E_k} \\&=E_k \bra{j}R^\dagger \dot R \ket{k} + E_j \bra{j} \dot R^\dagger R \ket{k} \\&=(E_k - E_j) \bra{j} R^\dagger \dot R \ket{k} \\&=(E_k - E_j) \bra{E_j} \dot R R^\dagger \ket{E_k}. \end{aligned}$$ This would seem to imply that if $E_j - E_k$ vanishes, then the operator norm of $\dot R R^\dagger$ (and thus of $\dot R$ itself) will increase without bound unless the cross-terms of $\dot H$ in the $\ket{E_j}$-basis also vanish.

An observation which points the way forward

The key question when considering the cross-terms for $\dot H$ in the $\ket{E_j}$-basis is: how does one determine that basis to begin with, to evaluate the cross-terms? Without being able to solve for the eigenstates for times approaching the crossing, we are left only with the time of the crossing itself — and crucially, the eigenspace there is degenerate, which means that just because we have one eigenbasis in mind, does not make it the physically sensible choice.

My original conjecture (in a previous edit of the question) didn't involve the projector $\Pi$ onto $\mathrm{span}\{\ket{E_0},\ket{E_1}\}$. But it occurred to me later that it's irrelevant whether or not $\dot H$ fails to commute with $H$ if this is because some of the other eigenstates of $H$ are not eigenstates of $\dot H$. What really matters is whether $\dot H$ fails, for the two crossing eigenvalues alone (in a manner of speaking), to commute with $H$. So what we really care about is just the subspace spanned by $\ket{E_0}$ and $\ket{E_1}$, leading to the modification of the conjecture using the projector $\Pi$. But at the time of the crossing, $\Pi H = E_0 \Pi$ by that very fact: in the subspace, it is proportional to the identity, which commutes with everything. Thus we will have $$0 = [H, \Pi \dot H] = H \Pi \dot H - \Pi \dot H H = \Pi [H, \dot H].$$ That is, the conditions of the conjecture will always hold, which should indicate that the worry is over nothing.

Adiabatic restriction: a sketch

If the other eigenvalues of $H$ are bounded away from $E_0$ and $E_1$ by a constant near to the crossing, we don't really mind the extent to which $\ket{\dot E_j}$ for $j \in \{0,1\}$ overlaps the other eigenvectors of $H$: by the analysis above we expect them to do so by a finite amount. We're only really concerned with the magnitude of $\bra{E_j} \dot H \ket{E_k}$. So we may completely restrict our attention to the effective coupling of $\ket{E_0}$ and $\ket{E_1}$ by $\dot H$, which is to say that we may in effect replace $\dot H$ with $\Delta = \Pi \dot H \Pi$.

Having done so, we now have in effect a Hermitian operator $\Delta$ on a two dimensional subspace, which obviously has two eigenvectors spanning the space. These are the two common eigenvectors $\ket{\delta_0}, \ket{\delta_1}$ of $\dot H$ and $H$ at the level crossing, and the cross-terms of $\dot H$ between these two vectors will be zero near to the level crossing.

The two vectors $\ket{\delta_0},\ket{\delta_1}$ may not be eigenvectors of $\dot H$, but they do allow us to see that $R^\dagger \dot R$ may have bounded operator norm in a neighborhood of the time $T$ of the level crossing, when restricted to $\mathrm{span} \{ \ket{E_0}, \ket{E_1} \}$, if $\ket{E_0} = \ket{\delta_0}$ and $\ket{E_1} = \ket{\delta_1}$ at time $T$. For $t$ nearby to $T$, an adiabatic argument would indicate that $\ket{E_0}$ and $\ket{E_1}$ will mostly not interact with the other energy eigenstates if the evolution is slow enough, so that we expect $\ket{\delta_0}, \ket{\delta_1}$ to nearly be eigenvectors of $H$ for $t \approx T$. There is a question of how quickly $\ket{E_j}$ converges to $\ket{\delta_j}$, which determines how quickly the cross-terms $\bra{E_j} \dot H \ket{E_k}$ vanish; however, large cross-terms would correspond to large eigenvalues of $\Delta \propto \Pi H \Pi + \text{const.}$, which should cause $R^\dagger \dot R$ to quickly converge to an operator diagonal in the $\ket{\delta_j}$-basis.

The coefficients of $R^\dagger \dot R$ for the other energy levels in the energy eigenbasis should be bounded either because of eigenvalue gaps between them, or for similar reasons if they exhibit level crossings of their own.

So no: there should be no instability of the eigenbasis.

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