Fourier Transform in wavefunction of Free particle

Question

I was reading Introduction to Quantum Mechanics by David Griffiths and I am at Chapter 2, page 45. He says that

The general solution to the time-dependent Schrodinger equation is still a linear combination of separable solutions (only this time it's an integral over the continuous variable $k$, instead of a sum over the discrete index).

This means that we can write free particle with general wavefunction with $\Psi(x,t)$ as Fourier transform of eigenfunction of Schrodinger equation (I think). i.e., \begin{equation} \Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{i(kx-\frac{\hbar k^2}{2m}t)}dk \end{equation} where $\phi(k)$ is eigenfunction of Schrodinger Equation for free particle. (I think)
But then he writes \begin{equation} \phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \Psi(x,0)e^{-ikx}dx \end{equation} from Plancherel's theorem. This clearly means that eigenfunction of the Schrodinger Equation will depend on the initial wave function of the particle.
Is my interpretation correct? I am not sure because this is not very intuitive to me.

Also, does this mean that Fourier Transform of any general free particle wavefunction $\Psi(x,t)$ is eigenfunction of Schrodinger equation?

Answer 1

$$\Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{i(kx-\frac{\hbar k^2}{2m}t)}dk$$ where $\phi(k)$ is eigenfunction of Schrodinger Equation for free particle.(I think)

No, you misunderstood your text book. $\phi(k)$ is a completely arbitrary function. For any function $\phi(k)$ this $\Psi(x,t)$ will be a solution of Schrödinger's equation $$i\hbar\frac{\partial}{\partial t} \Psi(x,t)= -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \Psi(x,t)$$ just because $$e^{i(kx-\frac{\hbar k^2}{2m}t)}$$ for any $k$ is a solution of Schrödinger's equation.

When you take the above solution $\Psi(x,t)$ at time $t=0$, then you have $$\Psi(x,0) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \phi(k) e^{ikx}dk$$

This means that $\phi(k)$ is the Fourier transform of $\Psi(x,0)$. You can invert this transformation and get $$\phi(k) = \frac{1}{\sqrt{2\pi}}\int_{-\infty} ^{\infty} \Psi(x,0) e^{-ikx}dx$$

Answer 2

Your problem lies in this sentence:

where $\phi (k)$ is eigenfunction of Schrodinger Equation for free particle. (I think)

Your interpretation is wrong. The eigen function is $e^{ikx}$ and $\phi (k)$ is just a number which tells you the coefficient of this eigen function in "constructing" the wave function. It might be a function of $k$ (each eigen function has a different amount in constructing the wave-function) but not a function of $x$, meaning it is not an eigen function. The way to calculate this number is by using the Fourier transform, which you wrote in your second equation (and that does depend on the initial wave function, but the eigen function themselves do not change.)