# Solving the Schrödinger Equation with Fourier Transforms. A few questions

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.

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?

• (It's Schrödinger, not Schröedinger. Sometimes it's written as Schroedinger, where oe is used as a substitute för ö, but with öe it just looks weird.) Commented Dec 3, 2022 at 18:24

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.

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.

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!

• Hi @sam . Pleased to see that this was useful. Please let me know if there are any questions. Commented Dec 5, 2022 at 21:10