In the free particle case, the general solution to the time-dependent Schrödinger equation is:

$$\Psi (x, t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \phi(k) e^{i(kx-\frac{\hbar k^2t}{2m})} dk$$

I do not understand why $\phi(k)$ does not depend on time. I have the idea that $\phi(k)$ is related to the 'envelope' of the wave packet, which travels at the group velocity. So the first logic idea that came to my mind was that the envelope should change as time runs, having $\phi(k)$ a time dependence. Why am I wrong?

I am aware of the fact that $\phi(k)$ can be computed using Fourier analysis and that if $\phi(k)$ was time-dependent it would not be possible to compute it using this method. But I am interested in the physical explanation of it.

  • $\begingroup$ $\phi (k)$ is the amplitude of a pure ideal sinusoid, how could that change? $\endgroup$
    – hyportnex
    Oct 8, 2018 at 18:37
  • $\begingroup$ But the envelope does change in function of time right? $\endgroup$
    – JD_PM
    Oct 8, 2018 at 18:46
  • 1
    $\begingroup$ Have you studied the Gaussian wavepacket? $\endgroup$ Oct 8, 2018 at 18:52

3 Answers 3


I do not understand why ϕ(k) does not depend on time.

The short answer is:

$$\Phi(k,t) = \phi(k)\,e^{-\frac{i\hbar k^2}{2m}t}$$

The long(er) answer is that, for a free particle, momentum eigenstates are also stationary states. That is, a state vector $|\psi_k(t)\rangle$ with definite momentum $p = \hbar k$ has 'trivial' time evolution

$$|\psi_k(t)\rangle = |k\rangle\, e^{-\frac{i\hbar k^2}{2m}t}$$

where $|k\rangle$ is a momentum eigenket with eigenvalue $k$. Assume a general free particle state vector $|\Psi(t)\rangle$ and express it in terms of the momentum eigenstate basis:

$$|\Psi(t)\rangle = \int\mathrm{d}k\,|k\rangle\,\phi(k)\, e^{-\frac{i\hbar k^2}{2m}t}$$


$$\phi(k) \equiv \langle k |\Psi(0)\rangle$$

So, to be clear, $\phi(k)$ is the amplitude for finding the particle in the momentum eigenket $|k\rangle$ at time $t = 0$. Clearly, $\phi(k)$ is not a function of time since we evaluating the amplitude at a particular time.

Now, the final step is project all this onto the position basis:

$$\Psi(x,t) = \langle x|\Psi(t)\rangle = \int\mathrm{d}k\,\langle x|k\rangle\,\phi(k)\, e^{-\frac{i\hbar k^2}{2m}t} = \frac{1}{2\pi}\int\mathrm{d}k\,\phi(k)\,e^{ikx}\, e^{-\frac{i\hbar k^2}{2m}t}$$

where we have used

$$\langle x | k \rangle = \frac{e^{ikx}}{2\pi}$$

and thus arrive at your equation.


The solution to the time-independent Schrodinger equation for the free particle is $e^{ikx}$ and the full time-dependent form is the (unormalized) function $$ \Psi_k(x,t)=e^{ikx}e^{-iE_k t/\hbar} $$ where $E_k=\hbar^2 k^2/2m$.

If the possible energies $E_k$ were discrete, the general solution would be a sum over these energies the coefficient $\phi(k)$ giving the amplitude of each $\Psi_k(x,t)$ in the solution. Thus, in the case of a harmonic oscillator, $$ \Phi(x,t)=\sum_{k=0}^\infty c_k \Psi_k(x,t) \tag{1} $$ where $\Psi_k(x,t)$ has energy $(k+\frac{1}{2})\hbar\omega$.

In the case of the free particle, there is no restriction of the possible values of $E$, and the possible values are continuous: the sum over discrete values of energy (as examplified in (1)) must be replaced by a continuous sum, i.e an integral $$ \Phi(x,t)=\int dk \phi(k)\Psi_k(x,t) $$ with coefficients $\phi(k)$ chosen so that the function $\Phi(x,t)$ is normalizable and indeed properly normalized


The time dependence is already encapsulated in $e^{-i\epsilon_k t /\hbar}$.

The $\phi(k)$ are given by the overlap of your state $\psi$ at time $t=0$ with the momentum eigenstates $|k\rangle$, and by this definition they don't depend on time.


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