# Energy of a Free Particle starting in an Infinite Square Well

A particle is in the ground state of an infinite square well with walls in the range x=[0,a]. At time t=0, the walls are removed suddenly and the particle becomes free. What is the energy of the free particle?

What I know: $$\begin{split} V(x) &= 0, \; \; \; 0\le x \le L \\ &= \infty, \; \; \;otherwise \end{split}$$ $$\begin{split} \psi(x,0)&=\sqrt{\frac{2}{a}} sin(\frac{\pi x}{a}) \\ E_1 &= \frac{\pi^2 \hbar^2}{2ma^2} \end{split}$$

I've found the wave function in momentum space $\phi(k)$ by taking the Fourier transform of the initial wavefunction. $$\begin{split} \phi(k) &= \frac{1}{2\pi\hbar} \int_{0}^{a} dx e^{ikx} \psi(x,0) \\ &=\frac{1}{a\pi\hbar} \frac{-\pi L (1+e^{-ikL)}}{k^2L^2 - \pi^2} \end{split}$$

I know $<E>=\frac{<p^2>}{2m}$, so I need to find $$<p^2>= \int_{0}^{a} k^2 \mid \phi(k) \mid^2 dk$$ Unfortunately, when I evaluate this integral, it diverges. Is there another way can I find energy of a free particle that yields an appropriate answer?

Note: I've also tried to evaluate the momentum space Schroedinger equation $i\hbar \frac{\partial \phi(p)}{\partial t} = H \phi(p)$. However, $\phi(p)$ is not time dependent from my evaluation, so the answer it yields is $0$.

• The potential is not yet introduced in your calculations. The energy should be the same as the regular infinite square well. Also your final answer is not time dependent because all your calculations are done explicitly at $t=0$. You can reintroduce time dependence with $\psi(x,t)=\int\phi(x,0)exp(i(kx-E(k)t/\hbar))dk$. Mar 13, 2018 at 0:31
• If I want to find the energy at some later time t, I should take the equation you suggested to reintroduce time dependence and then solve the Schroedinger equation? Mar 13, 2018 at 0:48
• Edit: Isn't energy conserved in the type of system where a wall is suddenly removed? So wouldn't that mean the free particle would still have the same amount of energy as it did before the wall was removed? Mar 13, 2018 at 1:07

You have to solve the time dependent Schrödinger equation on the unlimited x-axis with the initial condition $$\psi(x, t=0)=\psi(x,0)=\sqrt{\frac{2}{a}} sin(\frac{\pi x}{a}) \tag 1$$ for 0≤x≤a and $$\psi(x,0)=0 \tag 1$$ for x≤0 and x≥a (in the ground state with L=a). For the initial value problem, you first consider the time-dependent Schrödinger equation for t≥0 which has only a Hamiltonian with the kinetic energy term because the potential energy is zero: $$i\hbar \frac {\partial \psi(x,t)}{\partial t}= -\frac{\hbar^2}{2m}\frac{\partial^2 \psi(x,t)}{\partial x^2} \tag2$$ You assume a Fourier representation of the x-dependence of the solution $$\psi (x,t)=\frac {1}{\sqrt{2 \pi}} \int_{-\infty}^{+\infty} \hat\psi(k,t)\exp (ikx)dk \tag 3$$ giving you the linear first order ordinary differential equation in time $$i\hbar \frac {\partial \hat \psi(k,t)}{\partial t}=\frac{\hbar^2k^2}{2m}\hat \psi(k,t) \tag4$$ which is the Schrödinger equation in momentum representation and has the solution $$\hat \psi(k,t)=\hat \psi(k,t=0)\exp(-i\frac{\hbar k^2}{2m}t) \tag 5$$ where $$\hat\psi(k,t=0)=\frac {1}{\sqrt{2 \pi}}\int_0^a \sqrt{\frac{2}{a}} sin(\frac{\pi x}{a})\exp(-ikx)dx=\frac {1}{\sqrt{a\pi}}\frac {\frac{\pi}{a}(\exp(-ika)+1)}{(\frac{\pi}{a})^2-k^2} \tag 6$$ is the Fourier transform of the initial condition given by equations (1).
As mentioned before, equation (5) is the time dependent solution of the Schrödinger equation in momentum representation (k-space). The expectation value for the energy is given by the scalar product $$<E>=\int_{-\infty}^{+\infty}\hat\psi^*(k,t)\frac{\hbar^2k^2}{2m}\hat\psi(k,t)dk=\int_{-\infty}^{+\infty}\hat\psi^*(k,0)\frac{\hbar^2k^2}{2m}\hat\psi(k,0)dk \tag7$$ It can be seen that there is no time dependence of the expectation value of the energy which stays the same as at $t=0$. An evaluation of the integral (7) is not necessary. Thus the expectation value of the energy stays constant at the expectation value at time $t=0$ which should be the same as the lowest energy eigenvalue of the Schrödinger equation for the infinite potential square well given in the question: $$E_1=\frac{\pi^2 \hbar^2}{2ma^2} \tag 8$$