A free particle ($m\ne 0$) is moving in one dimension, with non-relativistic velocity. At $t=0$, the particle is a quantum state described by $$ \Psi(x,t=0)=\begin{cases} 0 & |x|>|a|\\ A(|x|-a)& -a\le x \le a\end{cases}.$$

I need to find the wave function $\Phi(p_x,t)$ in momentum space.

My attempt:

$$\Phi(p_x,t)=\frac1{\sqrt{2\pi\hbar}}\int_{-a}^a\Psi(x,t)e^{\frac i{\hbar} p_x x}dx.$$

I've already found that $A= e^{i\alpha}\sqrt{\frac 3{2a^3}}$. Now, I need to find the wave function $\Psi(x,t)$ at a random time $t$. Can I just say that $\Psi(x,t)=\exp{\left(-\frac{i}{\hbar} Et\right)}\Psi(x,t=0)$, with $E$ the (kinetic) energy of the particle? Then, $$ \Phi(p_x,t)=\frac{A}{\sqrt{2\pi\hbar}}\exp{\left(-\frac{i}{\hbar} Et\right)}\int_{-a}^a(|x|-a)\exp{\left( \frac i{\hbar} p_x x\right)}dx = \dots$$ Is this a correct approach?

Suppose that we measure the energy of the particle at a random time $t$. What values can we find, and what are their probabilities?

My attempt:

This is a free particle, thus its energy would be its kinetic energy $T$. I feel like I'd need some information about $\Pi(p_x,t)=|\Phi(p_x,t)|^2$. Suppose that the approach to my first question was correct, then we have $\Pi(p_x,t)$. The possible values of the energy are the eigenvalues of the kinetic energy operator $\hat{T}=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$. Do I just solve $\hat{T}\Phi(p_x,t)=E\Phi(p_x,t)$? How can I find the corresponding probabilities?


  • $\begingroup$ Let me ask you a few things: Is this wavefunction an eigenfunction of the Hamiltonian? What is the potential here? What is the advantage of working in momentum basis if $\forall x: V(x)=0$? Is this a bound state? If not, what kind of energy values can one find by observing this system? $\endgroup$
    – acarturk
    Jan 5 '20 at 10:04
  • $\begingroup$ $H=-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}$ and thus $H\Psi(x,t=0)=0$: this wavefunction is not an eigenfunction of $H$. The potential is 0 (free particle). This implies that $H$ is only a function of $p_x$ (= advantage of working in momentum space). I would guess it is bound, since the motion is restricted to $-a\le x\le a$, but the 'if not' part makes me think that I'm wrong. $\endgroup$
    – Zachary
    Jan 5 '20 at 11:44
  • 1
    $\begingroup$ Alright here's the thing: In Schrödinger picture, the wavefunctions that evolve as $\psi(x;t)=\exp\left(-i\frac{E}{\hbar}t\right)\psi(x;t=0)$ are the eigenfunctions of the unitary time evolution operator (that happen to have the same eigenfunctions as the Hamiltonian). If your initial state is not an eigenfunction, you need to expand it in a eigenbasis of $H$ in order to get a meaningful time evolution. In this case, the eigenbasis happen to be the Fourier basis (i.e. $p_x$ basis), and that's why it's the first part of the question. $\endgroup$
    – acarturk
    Jan 5 '20 at 12:03
  • $\begingroup$ This is how I would approach the problem now: $\Phi(p_x,t=0)=\frac{1}{\sqrt{2\pi\hbar}}\int_{-a}^a \Psi(x,t=0)\exp{\left(-\frac{i}{\hbar}p_x x\right)}dx$ by definition. And then, $\Phi(p_x,t)=\exp{\left(- \frac{i}{\hbar}Et\right)}\Phi(p_x,t=0)$, since $\Phi(p_x,t)$ is expressed in an eigenbasis (= $p_x$-basis) of $H$. $\endgroup$
    – Zachary
    Jan 5 '20 at 13:02

Okay here I'll try to give an introductory summary of time evolution.

Let's assume we have an arbitrary initial state $\psi(x,0)$ and a Hamiltonian operator $\hat H$. If we know the set of eigenfunctions of $\hat H$, we can expand this arbitrary $\psi(x,0)$ as a linear combination of them. The easiest case is when we have a bound state and thus only countably many energy eigenfunctions. However, the eigendecomposition for a free particle is not hard either.

For a free particle with mass $m\ne0$, we have the time dependent Schrödinger equation: $$\hat H\psi(x,t)=i\hbar\frac{\partial}{\partial t}\psi(x,t),\ \text{ with }\ \hat H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}.$$

In quantum mechanics, we usually try to find solutions to the time independent Schrödinger equation first and then apply a unitary time evolution operator (that amounts to the same thing as solving the time evolution equation $E\psi=i\hbar\partial_t\psi$ for all different $E$ values). In a similar vein, let's first try to solve $$\hat H\psi_E(x)=E\psi_E(x),\ \text{ with }\ \hat H=-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}.$$ $$\therefore\ \left(\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+E\right)\psi_E(x)=0$$

This equation is has the basic ansatz solution $\psi_E(x)=e^{\pm i x\sqrt{2mE}/\hbar}$. If we define $p:=\sqrt{2mE}$, it becomes $\psi_E(x)=e^{\pm i px/\hbar}$, which is the expected result.

At this stage, we ask, "How does one decompose a function in a basis of functions $e^{\pm i px/\hbar}$?" The answer is Fourier transform: $$\phi(p):=\mathcal F\{\psi(x)\}=\frac1{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty\psi(x)\,e^{\frac i{\hbar}px}dx.$$

The reason we did this is not visible in forward transform. However, if we write the backward transform, $$\psi(x)=\mathcal F^{-1}\{\phi(p)\}=\frac1{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty\phi(p)\,e^{-\frac i{\hbar}px}dp.$$ we see that we've represented any arbitrary $\psi(x)$ as a linear combination of the momentum (and thus energy) eigenstates.

So just let's plug in $\psi(x)=\psi(x,0)$. $$\phi(p,0):=\mathcal F\{\psi(x,0)\}=\frac1{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty\psi(x,0)\,e^{\frac i{\hbar}px}dx.$$

The function $\psi(p,0)$ in this case stands for the weighting function for the momentum eigenfunctions at time $t=0$.

But, how does these eigenfunctions evolve with time? Answer is $$E\psi_E(t)=i\hbar\frac{\partial}{\partial t}\psi_E(t)$$ with obvious solution $$\psi_E(t)=\psi_E(0)\,e^{-i\frac{E}{\hbar}t}.$$

So, $\phi(p,t)$ is pretty straightforward (notice $E_p=p^2/2m$): $$\boxed{\phi(p,t)=\phi(p,0)\,e^{-i\frac{p^2}{2m\hbar}t}.}$$

So, we can just say that $\psi(x,t)$ is $$\begin{align} \psi(x,t)&=\mathcal F^{-1}\{\phi(p,0)\,e^{-i\frac{p^2}{2m\hbar}t}\}=\frac1{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty\phi(p,0)\,e^{-i\frac{p^2}{2m\hbar}t}\,e^{-\frac i{\hbar}px}dp\\ &=\frac1{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty\phi(p,0)\,e^{-i\frac p\hbar (pt/2m+x)}dp. \end{align}$$ which is as good as it gets without plugging in the functions.

Answer to your second question is simply $|\phi(p,t)|^2=|\phi(p,0)|^2$. You can get any energy that is included in the spectrum (with probability of its weighting function squared as usual).

  • $\begingroup$ This is an amazing explanation! Thank you so much! Basically, the first thing to do when given $\Psi(x,t=0)$ and $H$ is to check if the wave function is an eigenfunction of $H$. If not, solve the TISE to find a set of stationary (?) eigenstates. After that, we can consider the time operator to construct the time dependent wave function. $\endgroup$
    – Zachary
    Jan 5 '20 at 14:05
  • 1
    $\begingroup$ Actually no matter what you should first find the energy eigenstates of $\hat H$ since e.g. $\hat H \psi = 0$ can also be an eigenstate (with energy 0). You cannot say anything before solving TISE. In the free particle case we find plane waves. In bound cases they are called "stationary states". The common point is that they are energy eigenstates and thus also stay the same in time evolution, i.e. $\psi(x,t)=\hat U_t\psi(x,0)=\alpha_t\psi(x,0)\ \exists \alpha_t\in\Bbb C^\times$. So, yes. First decompose the initial function then apply time evolution operator. $\endgroup$
    – acarturk
    Jan 5 '20 at 14:26

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