Solving the Schrödinger Equation with Fourier Transforms. A few questions

Question

I'm reading through a method for finding solutions to Schrödinger's equation through Fourier transforms, and I'm truly lost on the proof, hence I thought of asking for some help.

The following is from Hall's Quantum Theory for Mathematicians.

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Proposition 4.2 Suppose that $\psi_0$ is a “nice” function, for example, a Schwartz function. Let $\hat{\psi}_0$ denote the Fourier transform of $\psi_0$ and define $\psi(x, t)$ by $$\psi(x, t) := \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{\psi}_0(k)e^{i(kx-\omega(k)t)},$$ where $\omega(k)$ is defined by $$\omega(k) := \frac{k^2\hbar}{2m}.$$ Then $\psi(x, t)$ solves the free Schrödinger equation with initial condition $\psi_0$.

Proof. . Since the Fourier transform of a Schwartz function is a Schwartz function, $\hat{\psi}_0(k)$ will decay faster than $1/k^4$ as $k$ tends to $±∞$.

Meanwhile, by integrating the derivative of the function eikx, we obtain the estimate

$$\left|\frac{e^{ik(x+h)}-e^{ikx}}{h}\right| \le |k|.$$

Q1: how exactly is this bound derived?

We can then apply dominated convergence, using $|k|\hat{ψ}_0(k)$ as our dominating function, to move a derivative with respect to $x$ under the integral sign in the formula for $ψ(x, t)$. This derivative pulls down a factor of $ik$ inside the integral. The decay of $\hat{ψ}_0$ allows us to repeat this argument to move a second derivative with respect inside the integral. We can also move a derivative with respect to $t$ inside the integral, by a similar argument.

Q2: what does the above paragraph mean exactly?

Since $\exp{i(kx − ω(k)t)}$ satisfies the Schrödinger equation for each fixed $k$, differentiation under the integral shows that $ψ(x, t)$ satisfies the Schrödinger equation as well. The Fourier inversion formula shows that $ψ(x, 0) = ψ_0(x)$.

Q3: how does the Fourier inversion formula show that?

Answer 1

For Question 1, note that we can write

$$\begin{align} \left|e^{ik(x+h)}-e^{ikx}\right|&=\left|\int_x^{x+h} ike^{ikt}\,dt\right|\\\\ &\le \int_x^{x+h}\left|ike^{ikt}\right|\,dt\\\\ &\le |kh| \end{align}$$

We will use this result to answer the next question.

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For Question 2, note that we have

$$\begin{align} \frac{\partial}{\partial x}\psi(x,t)&=\lim_{h\to 0}\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \hat \psi_0(k)e^{-i\omega(k)t}\frac{e^{ik(x+h)} -e^{ikx}}{h}\,dk\tag1 \end{align}$$

Now, inasmuch as $\int_{-\infty}^\infty |k\hat \psi_0(k)|\,dk<\infty$, then the Dominated Convergence Theorem guarantees that we can bring the limit inside the integral in $(1)$ to find that

$$\begin{align} \frac{\partial}{\partial x}\psi(x,t)&=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \hat \psi_0(k)e^{-i\omega(k)t}\lim_{h\to 0}\frac{e^{ik(x+h)} -e^{ikx}}{h}\,dk\\\\ &=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \hat \psi_0(k) ik e^{i(kx-\omega(k)t)}\,dk \end{align}$$

as was to be shown!

Now repeating the process, we find that

$$-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)=i\hbar \frac{\partial }{\partial t}\psi(x,t)$$

which means that $\psi(x,t)$ as given by the $\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \hat \psi_0(k) e^{i(kx-\omega(k)t)}\,dk$ satisfies the free Schrodinger equation.

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For Question 3, note that we have

$$\psi(x,0)=\frac1{\sqrt{2\pi}}\int_{-\infty}^\infty \hat \psi_0(k) e^{ikx}\,dk\tag2$$

But we know that $\hat \psi_0$ is the Fourier Transform of $\psi_0$. The expression in $(2)$ is the inverse Fourier transform of $\hat \psi_0$. Hence, by the inversion theorem we see that

$$\psi(x,0)=\psi_0(x)$$

And we are done!