Convergence to Planewaves for Solutions to Schrödinger's equation III

In the last installment we studied the problem $i \partial_t \phi(t,x) + \partial^2_{xx}\phi(t,x) = 0$ where the unknown function $\phi: \mathbb{R}\times\mathbb{R}\to \mathbb{C}$, subject to the initial condition $$\phi(0,x) = \Xi(x) e^{i k x} \label{initdata}$$ where $\Xi(x)$ is the Heaviside function that vanishes on the negative real line and equals 1 on the positive real line.

There we approached the problem using the Laplace transform; we saw that the final states (for fixed $x$ as $t\to \infty$) of the wave function is given by the poles of the (analytic continuation of the) Laplace transform $p\mapsto \hat{\phi}(p,x)$, and that the rate of convergence to the steady state is typically $t^{-1/2}$ and determined by the branch cut along the negative real axis. That the rate is $t^{-1/2}$ is due to the amplitude of $\hat{\phi}(p,x)$ growing approximately like $|p|^{-1/2}$ with its phase oscillating also at frequency $\sqrt{|p|}$ as $|p| \to 0$.

The singular behavior of the amplitude as $|p| \to 0$ is known as a resonance at 0, and is an unstable feature of the free Schrödinger equation. In this post we will discuss some aspects in which this resonance manifest, and how it can be destroyed by lower-order perturbations of the equation.

Schrödinger operators, resolvents, eigenvalues

Consider now the Schrödinger operator $$H = - \partial^2_{xx} + V(x)$$ where $V(x)$ is a real-valued function. For convenience we will assume $V$ is continuous, and bounded with compact support.

We are interested in answering the question: for which values $z\in \mathbb{C}$ is the operator $(H-z)$ invertible?1 To see why this question is important, we can go back to our discussion of the Laplace transform. After taking the Laplace transform, the Schrödinger equation $i \partial_t \phi = H\phi$ with initial data $\phi(0,x) = \phi_0(x)$ becomes the parametrized family of ordinary differential equations $ip \hat{\phi}(p,x)= H\hat{\phi}(p,x) + i \phi_0(x).$ The solution is the formally $\hat{\phi}(p,x) = -i (H - ip)^{-1} \phi_0(x)$ and we see the connection with the invertibility of $(H-z)$. The poles of the Laplace transform, which gives the stationary states to which the solution converges, correspond to singularities of $(H-ip)^{-1} \phi_0$, and one easily imagines this being correlated to where $(H-z)$ fails to be invertible.

A note on terminology: the set of $z\in \mathbb{C}$ where $(H-z)$ is invertible is called the "resolvent set", and the operator $(H-z)^{-1}$ defined on this set is called the "resolvent". To properly talk about invertibility normally we need to talk about some functional analysis. But in the case of the 1 dimensional equation we can look at it from the point of ordinary differential equations and do everything by hand.

The Wronskian

When $V$ is continuous, by standard theory of ordinary differential equations, we know that there exists a two parameter family of $C^2$ solutions to the linear equation $$(H-z) \psi = 0.$$ Linearity implies that the family is spanned by two linearly independent solutions; the linear independence of $\psi_1, \psi_2$ can be checked by checking their Wronskian: $$W(\psi_1, \psi_2) = \psi_1' \psi_2 - \psi_2' \psi_1.$$ Obviously when the two functions are linearly dependent, the Wronskian is identically vanishing. One can check2 by explicitly differentiating that if $\psi_1, \psi_2$ both solve $(H-z)\psi =0$, then the Wronskian $W(\psi_1, \psi_2)$ is a constant function. And therefore if at any point two solutions have non-vanishing Wronskian, they must be linearly independent.

Now, for generic $z$, there is a way to pick out two special solutions of $(H-z)\psi = 0$. Observe that by assumption that $V$ has compact support, so for $|x|$ sufficiently large, the equation satisfied by $\psi$ is $- \partial^2_{xx} \psi = z\psi$ and we can formally write the solutions as the linear combination $\psi(x;z) = C_+ \exp( i \sqrt{z} x) + C_- \exp(-i \sqrt{z} x).$ Note that this function depends on the square-root function $z\mapsto\sqrt{z}$ and again we have to make a choice of a branch cut. Outside of this (fixed) branch cut, this function (for fixed values of $C_\pm$) is holomorphic in $z$.

For convenience, we will take our branch cut along the positive real axis, such that on its domain $\sqrt{z}$ always have positive imaginary part. Solutions with generic $C_\pm$ necessarily grows exponentially as $|x|\to \infty$. We define the two special solutions $\psi_\pm$ to $(H-z)\psi = 0$ by setting $\psi_+(x;z) = \exp( i \sqrt{z} x)$ for all large positive $x$, and $\psi_-(x;z) = \exp(-i \sqrt{z} x)$ for all large negative $x$. The solution $\psi_+$ is the unique solution to $(H-z)\psi = 0$ that is bounded on the right half line, and $\psi_-$ is the unique solution to $(H-z)\psi = 0$ that is bounded on the left half line.

We note that writing $z = \lambda + i \eta$ into its real and imaginary parts, for $\lambda > 0$ we have that $\lim_{\eta \to 0^+} \psi_+(x; \lambda + i \eta) = \exp(i \sqrt{\lambda} x)$ where the square root is taken in the usual sense of positive real numbers. Similarly $\lim_{\eta \to 0^-} \psi_+(x; \lambda + i \eta) = \exp(-i \sqrt{\lambda} x).$

It is easy to check that $W_H = W(\psi_+, \psi_-)$ is a holomorphic function of $z$ on $\mathbb{C} \setminus \mathbb{R}_+$.

The points where $W_H = 0$ have special significance: when $W_H = 0$ this means that $\psi_+$ and $\psi_-$ are in fact linearly dependent. This means that there exists a non-trivial solution to $(H-z)\psi = 0$ that decays exponentially both as $z\to\pm\infty$. For any reasonable choice of function space as a domain, this means that $(H-z)$ cannot be injective, and hence cannot be invertible! Another way to say the same thing is then that such a function $\psi$ satisfies $H\psi = z\psi$ and hence is an eigenfunction of the operator $H$ with eigenvalue $z$.

The Green's function

Where $W_H \neq 0$, we can construct the Green's function of the operator $(H-z)$. Since $\psi_+$ and $\psi_-$ are linearly independent, for every $y\in \mathbb{R}$, there exists a unique pair $(a(y), b(y))$ such that $a(y) \psi_+(y) = b(y) \psi_-(y), \qquad a(y) \psi_+'(y) = b(y) \psi_-'(y) + 1.$ Note that the corresponding linear system is solvable as the Wronskian is non-zero. Consider the function $$\label{defgreen} G(x,y;z) = \begin{cases} a(y) \psi_+(x) & x > y \newline b(y) \psi_-(x) & x < y \end{cases}$$ For any fixed $y\in \mathbb{R}$ and $z\in \mathbb{C}\setminus \mathbb{R}_+$, this function is decaying exponentially as $x\to \pm \infty$. By construction it is continuous and Lipschitz continuous in $x$.

Additionally, for all $x$ sufficiently large and positive, there exists a unique pair of constants $c_{-+}, c_{--}$ (with $c_{--} \neq 0$) such that $$\label{decomplefteigen} \psi_-(x) = c_{-+} \exp( i \sqrt{z} x) + c_{--} \exp(-i \sqrt{z} x)$$ we see that for all $y$ sufficiently large the linear system for $a(y), b(y)$ is solved by $$\label{expby} b(y) = \frac{1}{2 i \sqrt{z} c_{--}} e^{i \sqrt{z} y}$$ and $$\label{expay} a(y) = \frac{c_{-+}}{2i\sqrt{z}c_{--}} e^{i \sqrt{z} y} + \frac{1}{2i\sqrt{z}} e^{-i\sqrt{z} y} .$$ Now, as for fixed $x$ and with $y\to +\infty$, we have that $G(x,y;z)$ becomes $b(y) \psi_-(x)$ for all sufficiently large $y$, we have that $G(x,y;z)$ decays exponentially in $y$ for fixed $x$. A similar argument shows that $G(x,y;z)$ also decays exponentially as $y\to -\infty$ with $x$ held fixed. In particular, this means that for any $\xi$ bounded and locally integrable the integral $\int G(x,y;z) \xi(y) ~\mathrm{d}y$ converges for every $x\in\mathbb{R}$ and $z\in \mathbb{C}\setminus \mathbb{R}_+$.

Now let us compute the action of $(H - z)$ on this function in the sense of distributions. Given any test function $\eta(x)$, we can evaluate $\iint G(x,y;z) \xi(y) (H-z)\eta(x) ~\mathrm{d}y ~\mathrm{d} x = ( \iint_{x> y} + \iint_{x < y} ) G(x,y;z) \xi(y) (H-z) \eta(x) ~\mathrm{d}y ~\mathrm{d}x.$ The exponential decay of the Green's function allows us to use Fubini with impunity, and as $G(x,y;z)$ is $C^2$ in the regions $\{x > y\}$ and $\{ x < y \}$ we can integrate by parts $\iint_{x < y} G(x,y;z) \xi(y) (H-z) \eta(x) ~\mathrm{d}y ~\mathrm{d}x = \iint_{x < y} (H-z)G(x,y;z) \xi(y) \eta(x) ~\mathrm{d}y ~\mathrm{d}x + \int G(y,y;z) \xi(y) \eta'(y) ~\mathrm{d}y - \int \lim_{u\to y^-} \partial_xG(u,y;z) \xi(y) \eta(y) ~\mathrm{d}y .$ Notice that the first term vanishes as $(H-z) \psi_- = 0$. Doing the same computation for the region $\{x > y\}$ and combining the results we see that $\iint G(x,y;z) \xi(y) (H-z) \eta(x) ~\mathrm{d}y ~\mathrm{d}x = \int \xi(y) \eta(y) ~\mathrm{d}y$ and so $(H-z) \int G(x,y;z) \xi(y) ~\mathrm{d}y = \xi(x)$ in the sense of distributions.

Therefore, when $W_H \neq 0$, we can define $(H-z)^{-1} \xi(x) = \int G(x,y;z) \xi(y) ~\mathrm{d}y$ for every $\xi$ that is bounded and continuous.

Now what happens when $z$ approaches a root of $W_H$? When this happens, necessarily the coefficient $c_{--}$ in $\eqref{decomplefteigen}$ of the discussion above tends to zero, and as a result both $a(y)$ and $b(y)$ will blow-up. And so we see that we get corresponding poles of $(H-z)^{-1} \xi$ (for fixed $x$, as a function of $z$).

Resonance at zero

Of particular interest to us is what happens as $z \to 0$. Recall that the coefficients $c_{-\pm}$ in $\eqref{decomplefteigen}$ are dependent on $z$. One easily calculates that $W_H(z) = 2 i \sqrt{z} c_{--}(z)$ and is the inverse of the coefficient of $b(y)$ in $\eqref{expby}$.

Now, as $z \to 0$, we have that $\psi_+(x;z)$ converges uniformly on compact sets to the solution of $H\psi = 0$ that is identically 1 for all sufficiently large positive $x$, and similarly $\psi_-(x;z)$ converges uniformly on compact sets to the solution of $H\psi =0$ that is identically 1 for all sufficiently large negative $x$. When the limit $W_H(0) = 0$, this signifies that $\psi_\pm$ are linearly dependent, and that $\psi_-(x;z)$ approaches a constant value for all sufficiently large positive $x$ too; this value cannot be zero since the 0 solution is unique. Hence we conclude that if $W_H(z) \to 0$, and hence $b(y) \to \infty$, then $a(y)$ must also blow-up at the same rate.

So if we define $G(x,y;z)$ by the expression $\eqref{defgreen}$, we see that the behavior of the Green function at zero is determined by how $W_H(z)$ approaches $0$. The possibilities are:

1. $W_H(0) \neq 0$. In this case, both $a(y)$ and $b(y)$ have regular limits as $z\to 0$. For all sufficiently large positive $y$, $b(y)$ is a constant and $a(y) = c_a - y$ for some constant $c_a$. Similarly for $y$ large and negative. For fixed $y$, the function $G(x,y;0)$ is bounded in $x$. For fixed $x$, the function $G(x,y;0)$ is bounded in $y$.
2. $W_H(z) \to 0$ at the rate $\sqrt{z}$, which implies that $c_{--}$ approaches a regular value. In this case we say that the system has a resonance at zero.
3. $W_H(z) \to 0$ at rate $z$ or faster (here we use that $c_{--}$ is a meromorphic function of $\sqrt{z}$ by construction, and so in this case $c_{--}$ approaches $0$). In this case we say that the system has a pole at zero.

It is interesting to note that the free evolution as treated in the previous installments actually has $c_{-+} = 0$ and $c_{--} = 1$, and thus we have that $W_H(z) = 2 i \sqrt{z}$ and exhibits a resonance at zero. This resonance at zero is what is responsible for the $O(t^{-1/2})$ generic rate of decay of solutions.

Returning to the inverse Laplace transform, we see that if we plug in $z = ip$ as above, this gives $$\phi(t,x) = \frac{1}{2\pi i} e^{\gamma t} \int_{-\infty}^\infty e^{-iw t} (H - (w+i\gamma))^{-1} \phi_0(x) ~\mathrm{d}w.$$ (Here we wrote $z = w + i\gamma$.) When $V$ is real valued, the resolvent set must contain both the upper and lower half (open) planes. (Suppose not, then $(H - z) \psi = 0$ has a non-trivial exponentially decaying solution. Multiply by $\bar{\psi}$ and integrate by parts, we have that $\int \bar{\psi} H \psi = \int |\psi'|^2 + V |\psi|^2$ is real, by $\int z |\psi|^2$ has a non-vanishing imaginary part, so the sum of the two cannot equal 0.) This implies that we can compute $\phi(t,x)$ by taking the limit as $\gamma \to 0$.

Assuming now that there are no eigenvalues on the negative real line, then by contour integration we can rewrite the integral as $\phi(t,x) = \lim_{\gamma \to 0} \frac{1}{2\pi i} e^{\gamma t} \int_{0}^\infty e^{-iw t} (H - (w+i\gamma))^{-1} \phi_0(x) ~\mathrm{d}w - \frac{1}{2\pi i} e^{-\gamma t} \int_{0}^\infty e^{-iw t} (H - (w-i\gamma))^{-1} \phi_0(x) ~\mathrm{d}w.$ Using finally that we expect the resolvent $(H-z)$ to be meromorphic in $\sqrt{z}$ (by our construction!), we can formally suppress that limit in $\gamma$ and write $\phi(t,x) = \frac{1}{\pi i} \int_{-\infty}^\infty e^{-i \alpha^2 t} (H-\alpha^2)^{-1} \phi_0(x) \alpha ~\mathrm{d}\alpha$ where we understand that $(H-\alpha^2)$ to mean $\lim_{\gamma \to 0} (H - \alpha^2 - i \mathrm{sgn}(\alpha) \gamma)$.

This now is an oscillatory integral, and the leading contribution comes from the value of $\alpha (H - \alpha^2)^{-1} \phi_0(x)$ as $\alpha \to 0$. In the case where there is a resonance at zero, $(H - \alpha^2)^{-1}$ behaves like $(z)^{-1/2} = \alpha^{-1}$. So that $\alpha (H - \alpha^2)^{-1}$ is regular. And this gives a $t^{-1/2}$ decay from stationary phase arguments. In the case where $W_H(0) \neq 0$, we have that $(H-\alpha^2)^{-1}$ behaves boundedly provided $\phi_0$ is integrable. And in this case stationary phase actually gives a improvement to $t^{-3/2}$ decay.

Explicit example

Let's implement the above heuristic discussion by doing an explicit computation. For this example we will let $V(x)$ be the characteristic function of the interval $[-1,1]$. We can then solve very explicitly for $\psi_{\pm}(x)$. By symmetry, $\psi_-(x) = \psi_+(-x)$.

We require $\psi_-(x;z)$ to be $\exp(-i \sqrt{z} x)$ for $x < -1$. In the interval $[-1,1]$, the solution has to be a linear combination of $\exp(\pm i \sqrt{z-1} x)$: note that if $z\in \mathbb{C}\setminus \mathbb{R}_+$ then so is $z - 1$. Continuity at $x = -1$ requires then the coefficients satisfy $e^{i \sqrt{z}} = c_+(z) e^{- i \sqrt{z-1}} + c_- e^{i \sqrt{z-1}}$ and $- \sqrt{z} e^{i \sqrt{z}} = \sqrt{z-1} c_+(z) e^{- i \sqrt{z-1}} - \sqrt{z-1} c_-(z) e^{i \sqrt{z-1}}.$ This can be solved by \begin{align} c_+ &= \frac12 \left( 1 - \frac{\sqrt{z}}{\sqrt{z-1}} \right) e^{i \sqrt{z}} e^{i \sqrt{z-1}} \newline c_- &= \frac12 \left( 1 + \frac{\sqrt{z}}{\sqrt{z-1}} \right) e^{i \sqrt{z}} e^{-i \sqrt{z-1}} \end{align} and evaluating the Wronskian at the origin $x = 0$ we find $$W_H(z) = 2i \sqrt{z-1} ( c_-^2 - c_+^2)$$ and hence $$\lim_{z\to 0} W_H(z) = \sinh(-2) \neq 0.$$ In particular, we see that for this system (like most perturbations of the free Schrödinger equation with a bounded, compactly supported potential) there is no resonance at zero.

(One can also check that there are no poles in $\mathbb{C}\setminus \mathbb{R}_+$ for this $W_H$.)

Asymptotics

Now, let us apply this heuristic and study the asymptotics of the initial value problem where the initial data which we denote by $\Xi_k(x)$ is vanishing on $(-\infty,1]$ and equals $e^{ikx}$ on $(1,\infty)$. To study the asymptotics, we again use the Laplace transform method. We can define $\hat{\phi}(p,x)$ to be the unique bounded solution to $(H - ip) \psi = -i \Xi_k$. As in the previous installment, we will take the branch cut for $\sqrt{p}$ to be the negative real axis. (Note, this is different from what we used in the previous discussion, but is more convenient for the Laplace transform argument, as this moves the branch cut away from the poles.)

We will also take $\omega = (1 - i) / \sqrt{2}$ so that $\omega^2 = -i$.

Solving for $\hat{\phi}(p,x)$

For $p$ with positive real parts, the solution can be given by the ansatz $$\hat{\phi}(p,x) = \begin{cases} \lambda \exp( \omega \sqrt{p} (x+1)) & x < -1 \newline \lambda [ c_+ \exp( \omega \sqrt{p+i} x) + c_- \exp(-\omega \sqrt{p+i} x) ] & -1 \leq x \leq 1 \newline \eta \exp( -\omega \sqrt{p} (x-1)) + \frac{-i}{k^2 - ip} e^{ikx} & x > 1 \end{cases}$$ Continuity at $x = -1$ means \begin{align} c_+ = \frac12 \left( 1+ \frac{\sqrt{p}}{\sqrt{p+i}} \right) e^{\omega \sqrt{p+i}} \newline c_- = \frac12 \left( 1- \frac{\sqrt{p}}{\sqrt{p+i}} \right) e^{-\omega \sqrt{p+i}} \end{align}

Next, denote by $$\Omega = \omega \left[ (\sqrt{p} + \sqrt{p+i}) c_+ e^{\omega \sqrt{p+i}} + ( \sqrt{p} - \sqrt{p+i}) c_- e^{-\omega \sqrt{p+i}} \right].$$ This $\Omega$ is essentially the Wronskian. We claim

Lemma
$\Omega$ has no roots with $\Re p \geq 0$.
Proof:

Simplifying the formula for $\Omega$ we see that $\Omega = \frac{\omega}{2 \sqrt{p+i}} \left[ (\sqrt{p} + \sqrt{p+i})^2 \exp( 2\omega \sqrt{p+i}) - (\sqrt{p} - \sqrt{p+i})^2 \exp(-2\omega \sqrt{p+i}) \right] .$ Near $p = -i$, denote by $\alpha = \sqrt{p+i}$, we have $\Omega = \frac{\omega}{2 \alpha} \left[ ( \sqrt{\alpha^2 - i} + \alpha)^2 \exp (2 \omega \alpha) - (\sqrt{\alpha^2 - i} - \alpha)^2 \exp(-2\omega \alpha) \right].$ Taylor expanding we see that the limit as $\alpha \to 0$ of $\Omega$ is finite (it takes value $-2 \sqrt{2} i$).

Away from $p = -i$, a root can only happen if $(\sqrt{p} + \sqrt{p+i})^2 \exp( 2\omega \sqrt{p+i}) = (\sqrt{p} - \sqrt{p+i})^2 \exp(-2\omega \sqrt{p+i}).$ Noting that $(\sqrt{p} + \sqrt{p+i})(\sqrt{p} - \sqrt{p+i}) = -i$ this requires $(\sqrt{p} + \sqrt{p+i})^4 \exp(4 \omega \sqrt{p+i}) = -1.$ Now, when $\Re p \geq 0$, we have that $|\sqrt{p} + \sqrt{p+i}|^2 \geq |p| + |p+i|$ from the law of cosine, and our choice of branch cut. By the triangle inequality, we have that $|p| + |p+i| \geq |p - (p+i)| = 1$ with equality only when $p, p+i, 0$ are collinear (in fact we need $p$ to be on the negative imaginary axis and $p+i$ on the positive imaginary axis for this to hold).

Furthermore, when $\Re p \geq 0$ we also have $\Re (\omega \sqrt{p+i}) \geq 0$, with equality only if $p+i$ is on the negative imaginary axis. Together this means that when $\Re p \geq 0$, both $|\sqrt{p} + \sqrt{p+i}|^2 \geq 1$ and $|\exp(4\omega \sqrt{p+i})| \geq 1$ must hold, with at least one of the two inequalities being strict. Therefore $\Omega$ has no root.

Remark
In fact, following the same method of proof, we see that the same argument also shows that $\Omega$ has no root for $p$ in the second quadrant. We do not rule out, however, the possibility of roots in the third quadrant (outside the branch cut $p = -i$). Roots of $\Omega$, being essentially roots of the Wronskian, correspond to poles of $p \mapsto \hat{\phi}(p,x)$. To properly account for the asymptotics it would be necessary to account for all these poles, and make sure their total number is either finite or that they contribute summably to the contour integration. In the discussion below we omit this analysis; taking on faith that, since by Lemma 1 the Wronskian doesn't contribute additional poles to the right half plane, any remaining poles will give exponentially decaying contributions to the evolution, and can therefore be neglected.
Remark
It may also be worthwhile to observe that near $p = 0$, $\Omega$ approaches $\sinh(2)$.

Solving the compatibility condition (continuity of both $\hat{\phi}(p,x)$ and its first derivative) at $x = 1$ we find \begin{align} \lambda &= \Omega^{-1} \frac{k - i \omega \sqrt{p}}{k^2 - ip} e^{ik} \newline \eta &= \Omega^{-1} \frac{e^{ik}}{k^2 - ip} \underbrace{\left[ ( k + i \omega\sqrt{p+i} ) c_+ e^{\omega \sqrt{p+i}} + (k - i \omega \sqrt{p+i}) c_- e^{-\omega \sqrt{p+i}} \right]}_{\Upsilon} \end{align}

Observation
The quantity $\Upsilon$ is related to the quantity $\Omega$: one sees that when $k = i \omega \sqrt{p}$ the two are equal up to multiplication by $i$. The condition that $k = i\omega \sqrt{p} \iff \sqrt{p} = \omega k$ can only be satisfied, given our branch cut choice and that $k\in \mathbb{R}$, when $k > 0$. In this case $p = -i k^2$. So we can summarize by saying that when $k \geq 0$, $\Upsilon(-ik^2) = i \Omega(-ik^2)$.

Putting everything together we have the following expressions for $\hat{\phi}(p,x)$ $\tag{x < -1} \hat{\phi}(p,x) = \frac{e^{ik}}{\Omega} \frac{k-i \omega \sqrt{p}}{k^2 - ip} e^{\omega\sqrt{p}(x+1)},$ \begin{multline} \tag{$x\in[-1,1]$} \hat{\phi}(p,x) = \frac{e^{ik}}{\Omega} \frac{k - i\omega \sqrt{p}}{k^2 - ip} \Big[ \cosh(\omega \sqrt{p+i}(x+1))\newline + \frac{\sqrt{p}}{\sqrt{p+i}} \sinh(\omega\sqrt{p+i}(x+1)) \Big],\qquad\qquad\qquad\quad \end{multline} $\tag{x > -1} \hat{\phi}(p,x) = \frac{i e^{ik}}{k^2 - ip} \left[ \frac{\Upsilon}{i\Omega} e^{-\omega \sqrt{p}(x-1)} - e^{ik(x-1)} \right].$

Particles moving to the right: $k > 0$

Intuitively we expect in this case, seeing the particles all started out to the right of the barrier, the solution will converge toward vacuum. Let's see if this intuition is verified.

To the left of the barrier $x \lt -1$

Here the formula for $\hat{\phi}(p,x)$ is very simple. The only obstructions to the contour integration are the branch cuts at $\mathbb{R}_-$ and $\mathbb{R}_- - i$, and the potential pole at $k^2 = ip$. (As discussed in Remark 2, here and below we will ignore any additional roots of $\Omega$ in the third quadrant and just pretend they don't exist.)

We first look at the pole: for $k^2 = ip$ we need $p = -ik^2$ which gives $i\omega\sqrt{p} = k$ (using that $k > 0$). Therefore the zero in the denominator $k^2 - ip$ is exactly cancelled by the zero in the numerator $k - i\omega \sqrt{p}$. Hence we see that the solution to the left of the barrier must decay to zero.

For the branch cuts, we notice that $\Omega^{-1}$ is a bounded continuous function in neighborhoods of $p = 0$ and $p = -i$ (with the two branch cuts removed). Around the two branch cuts we proceed as in the previous installment. Near $\mathbb{R}_-$ we reparametrize by $\sigma = -i \sqrt{p}$. The contour integral can be rewritten as $$\label{basiccontour} \int_{\mathbb{R}} e^{-\sigma^2 t} \sigma \cdot \text{ (smooth function of \sigma) }~\mathrm{d}\sigma$$ and we see that by method of stationary phase the leading order term to the decay is given by $t^{-3/2} \hat{\phi}(0,x) \approx k^{-1} t^{-3/2}$ (independent of $x$). Near $\mathbb{R}_- - i$ we parametrize by $\sigma = -i \sqrt{p+i}$, and again the contour integral can be written in the above form and we see that the leading order term to the decay is given by $t^{-3/2} \hat{\phi}(-i,x)$. Expanding this term we see $\hat{\phi}(-i, x) \approx \frac{1}{k+1} e^{-i (x+1)}$ so this term also gives decay that is independent of $x$.

On the barrier $x \in [-1,1]$

Again we see that there can be no poles: the apparent pole at $k^2 = ip$ being cancelled by the zero in the numerator. So we are down to the integration around the branch cuts.

Using the same change of variables as the previous step, we see that near $\mathbb{R}_-$ the contour integral takes the form $\eqref{basiccontour}$ and hence the leading order contribution is $O(k^{-1}t^{-3/2})$. Near $\mathbb{R}_- - i$ we need to use additionally the fact that $\frac{1}{\sigma} \sinh(\omega \sigma(x+1))$ is a smooth (analytic) function in $\sigma$, and we again conclude that the leading order contribution is also $O((k+1)^{-1}t^{-3/2})$. (The rate is automatically uniform in $x$ since we are working on a bounded region.)

To the right of the barrier $x \gt 1$

When $x > 1$, again we need to worry about a potential pole at $p = -ik^2$. Here, however, we will use Observation 4 above, which shows that in a neighborhood of $p = -ik^2$ the function $\frac{\Upsilon}{i\Omega} e^{-\omega \sqrt{p}(x-1)} - e^{ik(x-1)}$ is holomorphic and vanishes at $p = -ik^2$. Hence there is no pole to worry about.

(The definition of $\Upsilon$ depends on $c_\pm$ and has $\sqrt{p+i}$ in the denominator; one however sees that the same argument as the regularity of $\Omega$ shows that $\Upsilon(0) = k\cosh(2) - i \sinh(2)$ and $\Upsilon(-i) = k(1-2i)$ and are regular on the domain of definition.)

A similarly argument with the branch cut shows that pointwise the solution decays like $O(k^{-1} t^{-3/2})$.

Particles with zero momentum $k = 0$

Next we treat the case when the particle has zero momentum. Here our intuition is less clear: in the case without barriers, the particles tend to spreadout. What happens now? Will the particles leak through the barrier and form a constant distribution to the left, just as it did in the free case?

Left of the barrier

Here $\hat{\phi}(p,x)$ simplifies to $\frac{1}{\Omega} \frac{\omega}{\sqrt{p}} e^{\omega \sqrt{p}(x+1)} .$ We see that to the left of the barrier, instead of having a pole at the origin, we now have a zero resonance. This means that while the solution still decays to 0 on the left of the barrier, it will only do so at the rate $t^{-1/2}$ (as oppose to $t^{-3/2}$ before), coming from the contour integral around $\mathbb{R}_-$. The integral around the other branch cut decays faster.

On the barrier

Similar to the situation to the left of the barrier, we have a zero resonance on the branch cut along $\mathbb{R}_-$, and the decay rate we see here is again $t^{-1/2}$.

To the right of the barrier

We can simplify $\hat{\phi}(p,x) = - \frac{1}{p} \left[ \frac{\Upsilon}{i\Omega} e^{-\omega \sqrt{p}(x+1)} - 1 \right].$ and of particular interest is what happens as $p \to 0$. Noting that as $p \to 0$ (since $k =0$) we have that $\frac{\Upsilon}{i\Omega} \to 1$, we see that the terms in the brackets approach $0$. So here we do not have a pole anymore. But do we still have a resonance? Setting $\beta = \sqrt{p}$ and Taylor expanding near zero, we have that $\Upsilon = i \sinh(2) + i \omega \beta \cosh(2) + \ldots$ while $i\Omega = i \sinh(2) + 2 i\omega \beta \cosh(2) + \ldots$ and so $\frac{\Upsilon}{i\Omega} = 1 - \coth(2) \omega \beta + \ldots$ so this means that near $p = 0$ we have the expansion $\hat{\phi}(\beta^2,x) = \frac{\omega(x-1 + \coth(2))}{\beta} + \ldots$ Since in our domain $x-1 + \coth(2) > 0$, this means that to the right of the barrier there is a resonance at zero, and that the leading order decay rate is given by $t^{-1/2} (x-1 +\coth(2)) ,$ seeing as the lack of resonance at $p = -i$ gives only $O(t^{-3/2})$ terms.

Particles moving to the left $k \lt 0$, except $k = -1$

The case $k = -1$ is isolated, because any putative poles from the formulas for $\hat{\phi}(p,x)$ will sit at $p = -i$, and we need to be careful checking whether this gives a pole or a resonance.

For general $-1 \neq k < 0$, there will be a bona fide pole at $p = -i$. Let's examine the properties in detail.

Left of the barrier

As $k - i \omega \sqrt{p} \neq 0$ now, $\hat{\phi}(p,x)$ has a pole at $p = -ik^2$. This being on the imaginary axis, contributes a non-decaying "stationary" state from the residue theorem. Evaluating the contour integral we need to know the value of $\Omega(-ik^2)$, which we can explicitly compute. The result depends on whether $|k|$ is bigger or less than 1. \begin{align} k & < -1 & \Omega(-i k^2) &= -\frac{2k^2 - 1}{\sqrt{k^2 - 1}} \sin(2 \sqrt{k^2 - 1}) + 2ki \cos(2 \sqrt{k^2 - 1}) \newline k & \in (-1,0) & \Omega(-ik^2) &= - \frac{2k^2 - 1}{\sqrt{1-k^2}} \sinh(2 \sqrt{1-k^2}) + 2ik \cosh(2\sqrt{1-k^2}) \newline k & = -1 & \Omega(-i) &= -2 - 2i \end{align} Then the solution $\phi(t,x)$ should approach the plane-wave $$\frac{2k}{i \Omega(-ik^2)} e^{-i k^2 t} e^{ik(x+2)}.$$ The amplitude of this wave is $| 2k / \Omega(-ik^2)|$. We see that $$|\Omega(-ik^2)|^2 = \begin{cases} 4k^2 + \frac{1}{k^2 - 1} \sin^2(2\sqrt{k^2 - 1}) & |k| > 1 \newline 4k^2 + \frac{1}{1-k^2} \sinh^2(2 \sqrt{1-k^2}) & |k| < 1 \newline 8 & |k| = 1 \end{cases}$$ In particular, as $|k|$ increases from $0$ to $1$, the transmitted amplitude $| 2k / \Omega(-ik^2)|$ is strictly increasing, starting from $0$ when $k = 0$ and rising to $1 / \sqrt{2}$ when $k = -1$. Furthermore, when $k \searrow -\infty$ we see that the transmitted amplitude approach 1.

Such behavior is physically expected: the effect of a potential barrier of finite height and finite extent is larger on low momentum particles (which are less likely to tunnel through the barrier) than on high momenta ones. (Thinking of the classical picture: there is a strict cut-off value of momentum below which a classical particle cannot make it over a given hill, while above which all particles make it over.)

The rate at which the solution converge toward this steady state is governed by the integration around the branch cuts. When $k \neq 0,-1$, we see the same behavior as the $k > 0$ picture will hold, with rate of convergence given by $t^{-3/2}$.

On the barrier

The analysis on the barrier is largely the same as to the left, except we get a slightly different profile: When $|k| < 1$ the limiting profile is $$\frac{2k e^{ik}}{i \Omega(-ik^2)} e^{-ik^2 t} \left[ \cosh(\sqrt{1-k^2}(x+1)) - \frac{k}{i \sqrt{1-k^2}} \sinh(\sqrt{1-k^2} (x+1)) \right]$$ while for $|k| > 1$ the limiting profile is $$\frac{ 2k e^{ik}}{i \Omega(-ik^2)} e^{-ik^2 t} \left[ \cos( \sqrt{k^2 - 1} (x + 1)) + \frac{ik}{\sqrt{k^2 - 1}} \sin(\sqrt{k^2 - 1}(x+1)) \right].$$

The only feature of note is that when $|k| < 1$ the profile decays exponentially across the barrier, exactly as expected from standard examinations of the quantum tunneling effect.

Right of the barrier

To the right of the barrier, we see something new. The pole at $p = -ik^2$ contributes a limiting profile $$- e^{-ik^2 t} e^{ik} \left[ \frac{\Upsilon(-ik^2)}{i\Omega(-ik^2)} e^{-ik(x-1)} - e^{ik(x-1)} \right].$$ Here, as $k < 0$, the ratio $\Upsilon / i\Omega$ does not equal $1$. We can evaluate explicitly: when $k < 0$ $\Upsilon(-ik^2) = \begin{cases} - \frac{\omega}{\sqrt{1-k^2}} \sinh(2 \sqrt{1-k^2}) & |k| < 1 \newline - \frac{\omega}{\sqrt{k^2 - 1}} \sin(2 \sqrt{k^2 - 1}) & |k| > 1 \newline 2 \omega & |k| = 1 \end{cases}.$ Clearly then $\left| \frac{\Upsilon(-ik^2)}{\Omega(-ik^2)} \right| < 1$ when $k \in (-\infty,0)$. Furthermore, as $k \to 0$ the ratio approaches 1, while as $k \to -\infty$ the ratio approaches 0.

Returning to the profile, we see that we have again an easy, physical interpretation. The portion given by $e^{ik x-ik^2 t}$ is the incoming plane wave moving in from the right. The portion with the $\Upsilon/\Omega$ coefficient is now the portion that gets reflected by the barrier. As the momentum goes down to zero, the reflected proportion approaches unity; while as the momentum goes to infinity, the reflected proportion approaches zero.

The branch cuts again contribute only a $t^{-3/2}$ decay rate, there being no resonances at the origin or at $p = -i$.

When $k = -1$

When $k = -1$ the would-be pole coincides with $p = -i$, so we have to examine whether additional resonances occur. Using that $\sqrt{p}$ is holomorphic in a neighborhood of $p = -i$, we can see that in fact the singularities at $p = -i$ are poles and not resonances. And hence the solution decays at rate $t^{-3/2}$ toward the steady state.

Furthermore the analysis of the previous section largely applies to yield the steady state solution. The only additional item to observe is that on the barrier, when $p = -i$ and $k = -1$, the limiting solution reduce to the linear function $$\frac{ e^{-i}}{i -1} e^{-it} \left( 1 - i -ix \right).$$

1. This question as stated does not make sense; to ask whether an operator is invertible one has to specify what one expects as the domain and range of the operator. We leave this deliberately vague for now. ^
2. This is a special, easy case of what is called Abel's identity. ^
Willie WY Wong
Assistant Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.