Agmon’s Estimate for the Schrödinger Equation

One of the most important equations in mathematical physics is the time-independent Schrödinger equation,

$\left(-\frac{\hbar^2}{2m}\Delta + V(x)\right) \psi = E \psi, \quad x \in \mathbb{R}^n$

(see this post for an introduction to the math behind the Schrodinger equation), which governs the evolution of the quantum wavefunction $\psi$ associated to a quantum particle with total energy $E$ in a potential $V(x).$

One of the interesting notions regarding solutions to this equation (which one can recognize as an eigenvalue problem for the operator $H = -\frac{\hbar^2}{2m} \Delta + V$ ) is the physical interpretation of the square magnitude $|\psi|^2$ as a probability distribution for the associated quantum particle. In particular, one feature that distinguishes this equation from classical mechanics is the existence of a classically forbidden region $R = \{x \in \mathbb{R}^n: V(x) > E\}.$

In classical mechanics, a particle with total energy $E = V(x) + \frac12 mv^2$ will never be found in this region, for it would require the kinetic energy of the particle to be negative, which is impossible. However, solutions to the Schrödinger equation exhibit an explicitly quantum phenomenon known as tunneling, which reflects that there is a positive probability that one finds a quantum particle in a region which, if it were purely classical, would be impossible to reach.

It would make sense though that even if the probability of finding a particle in the classically forbidden region is nonzero, it would have to be quite small. An explicit calculation of a particle in a finite well (which is often done in a first quantum mechanics course) shows that the magnitude of the wavefunction $|\psi|$ decays exponentially in the classically forbidden region. However, despite remaining a nice heuristic argument, this argument does not answer many of the questions that one might ask about classically forbidden regions – namely, how does the geometry of the potential $V$ control the rate of exponential decay? Even if one takes the simplest realistic potential of quantum mechanics – the quantum harmonic oscillator – one cannot definitely explain how the potential determines the decay rate of the solution.

Agmon’s Estimate

Agmon’s idea was to incorporate the geometric structure of the potential $V$ by creating a new geometrical object known as the Agmon metric.

Definition 1: For two vectors $\xi,\eta \in \mathbb{R}^n,$ we define the inner product $\langle \xi, \eta\rangle_A := (V(x)-E)_+ \langle \xi, \eta \rangle,$ where $f_+ = \max(f,0).$

Notice that this inner product induces the norm $\|f\|_A = \sqrt{\langle f,f\rangle_A}.$ Similarly to how one may define the Euclidean distance between two points as the length of the shortest path between two points, one may now define the Agmon metric:

Definition 2: For $x,y \in \mathbb{R}^n,$ define the Agmon metric

$\rho(x,y) = \inf_\gamma \int_0^1 \sqrt{(V(x)-E)_+}\|\gamma'(t)\|_A dt,$

where the infimum is taken over all absolutely continuous paths from $x$ to $y.$

One may easily verify that this is indeed a metric. There are two other important properties satisfied by $\rho:$

Lemma 1: $\rho: \mathbb{R}^n \times \mathbb{R}^n \to [0,\infty)$ is locally Lipschitz continuous (thus differentiable a.e. by Rademacher’s theorem), and satisfies $|\nabla_y \rho(x,y)|^2 \leq (V(y)-E)_+$ whenever $\rho$ is differentiable.

Proof: The first statement follows almost immediately from the fact that $\rho$ is a metric. On the other hand, when $\rho$ is differentiable, by the triangle inequality one has

$\limsup_{h \to 0} \frac{|\rho(x,y+h)-\rho(x,y)|}{\|h\|} \leq \limsup_{h \to 0} \frac{\rho(y,y+h)}{\|h\|} \leq \limsup_{h \to 0} \int_0^1 (V(y+th)-E)_+ dt \leq (V(y)-E)_+.$

Now that we have defined the metric, we may now state Agmon’s main result:

Theorem (Agmon): Let $H=-\Delta+V(x)$ be a closed bounded below operator with real spectrum, where $V$ is real and continuous. Suppose that $(E-V(x))_+$ is a compactly supported function, and let $\psi \in L^2(\mathbb{R}^n)$ be an eigenfunction of $H.$ Then, for any $\epsilon>0,$ one has

$\int e^{2(1-\epsilon)\rho(x,0)}|\psi(x)|^2 dx = O_\epsilon(1).$

Remark: The theorem essentially states that $\psi = O(e^{-(1-\epsilon)\rho(x,0)})$ as $\|x\| \to \infty,$ i.e. it gives a rate of exponential decay of $\psi$ depending on the shape of the potential $V.$ For example, if $V(x) = \|x\|^2,$ then $\rho(x,0) \sim \|x\|,$ so the theorem gives $\psi = O(e^{-(1-o(1))\|x\|}).$

Before jumping into the proof of the theorem, which is somewhat complicated, here’s one possible way for deriving the intuition and the key ideas for how the theorem can be proven.

Proof Idea: Our goal is to bound the integral of $|\psi|^2$ multiplied by some exponentially growing function. The problem naturally comes from the fact that the integral is over all of $\mathbb{R}^n,$ whereas we can easily handle the integral on any compact set, and in particular, the admissible region $A = \{x: V-E<0\}.$ Thus, we only need to worry about the tail $A^c.$

We should certainly use in our proof at some point the fact that $\psi$ is an eigenfunction of our Schrödinger operator $H,$ i.e. $H\psi = E\psi.$ Since we are interested in bounding $\|\psi e^\rho\|_2$ above, one potential step would involve showing that some version of $H-E$ is bounded below on $L^2,$ i.e. $\|\psi e^\rho\|_2 \leq C \|(H-E)(\psi e^\rho)\|_2.$ Of course, so far we don’t even know that $\psi e^\rho \in L^2,$ but we can fix that by throttling the exponential at some point.

Now, if it were somehow possible to move the tail of $\|H(\psi e^\rho)\|_2$ , then by preceding remark we would be done. Let $\eta$ be the cutoff function for the tail (that is, a smooth function that equals 1 on $A^c$ and 0 on some compact set). Why would one expect $\|H(\eta \psi e^\rho)\|$ to be compactly supported? Well, in since $\eta \psi e^\rho \equiv \psi e^\rho$ on the tail and $H$ is a local operator (as a sum of a multiplication and a differential operator), then $((H-E)e^{-\rho})(\psi e^\rho) = (H-E)\psi=0$ in a neighborhood of the tail. But since $H-E$ is bounded below by the previous step, the total mass of the function cannot dissipate – in particular, it has to move onto the set where $\eta<1,$ which is by construction a compact set.

Notice that we had to multiply our operator $H-E$ by $e^{-\rho}$ in order for this computation to make sense – this is precisely what motivates the conjugated version of the operator that appears in the lemmas.

Proof

Based on the outline above, we first want to show that the transformed version of $H-E$ is bounded below. This gives us the following lemma:

Lemma: Let $f \equiv (1-\epsilon) \rho_\epsilon.$ Let $\phi \in H^1(\R^n) \cap D(V)$ be supported in the set $\{x: V(x)-E)>\delta\}$ for some $\delta>0.$ Then, there exists $\delta_1>0$ s.t.

$\text{Re} \langle \phi, e^f (H-E) e^{-f} \phi \rangle \geq \delta_1 \|\phi\|^2,$

where $e^f (H-E) e^{-f}$ is the transformed Hamiltonian.

Agmon’s Estimate

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