danny.

Aug 17, 2018

A Free Particle

A solution to the Schrodinger equation for a Free Particle.

The time independent Schrodinger equation has different solutions depending on the potential that the particle feels. As we find the first solution it would make sense if it was the simplest possible potential, or, V=V0V=V_0. When we put this potential into the Schrodinger equation it becomes

Eψ=−ℏ22m∇2ψ+V0ψE\psi=\frac{-\hbar^2}{2m}\nabla^2\psi + V_0\psi

In one dimension

(V0−E)ψ=ℏ22m∂2ψ∂x2(V_0-E)\psi=\frac{\hbar^2}{2m}\frac{\partial^2\psi}{\partial x^2}

This form of second order differential equation has a general solution

ψ=Aeαx(2.2.1)\psi = Ae^{\alpha x}\tag{2.2.1}

This is proven over in the mathematics section. To find the coefficients AA and α\alpha that the Schrodinger equation produces we sub this general solution into it. By doing so we have

(V0−E)(Aeαx)=ℏ22m∂2∂x2(Aeαx)(V_0-E)\left(Ae^{\alpha x}\right)=\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\left(Ae^{\alpha x}\right)

Taking the second derivative

(V0−E)(Aeαx)=ℏ22m(Aα2eαx)(V_0-E)\left(Ae^{\alpha x}\right)=\frac{\hbar^2}{2m}\left(A\alpha^2e^{\alpha x}\right)

From this we can see that AA can have any value and by dividing through by ψ\psi

(V0−E)=ℏ2α22m(2.2.2)(V_0-E)=\frac{\hbar^2\alpha^2}{2m}\tag{2.2.2}

Solving for α\alpha we have if (V0−E)<0(V_0-E) < 0

α=±i2m(V0−E)ℏ=±ik(2.2.3)\alpha =\pm i \frac{\sqrt{2m(V_0-E)}}{\hbar}=\pm ik \tag{2.2.3}

And for (V0−E)>0(V_0-E) > 0

α=±2m(V0−E)ℏ=±k(2.2.4)\alpha =\pm \frac{\sqrt{2m(V_0-E)}}{\hbar} =\pm k \tag{2.2.4}

Knowing this means that we can now write out the general solution for a free particle's wavefunction

For (V0−E)<0(V_0-E) < 0

ψ=Ae−ikx+Beikx(2.2.5) \psi=Ae^{-ikx}+Be^{ikx}\tag{2.2.5}

And for (V0−E)>0(V_0-E) > 0

ψ=Ae−kx+Bekx(2.2.6) \psi=Ae^{-kx}+Be^{kx}\tag{2.2.6}

For the case of the free particle we can rule out the (2.2.6)(2.2.6) as its derivative is discontinuous and does not meet the requirements of a wavefunction. However, for the (2.2.5)(2.2.5) case this corresponds to a plane wave with frequency kk and is a valid wavefunction. Knowing that kk is the spatial frequency we can sub back into (2.2.2)(2.2.2) and find the expression for the energy with α=ik\alpha=ik

E−V0=ℏ2k22m(2.2.7)E-V_0=\frac{\hbar^2k^2}{2m}\tag{2.2.7}

To finish, we can combine our wavefunction with (2.1.3)(2.1.3) to arrive at

ψ=(Ae−ikx+Beikx)exp⁡(−iℏk22mt)(2.2.8)\psi=\left(Ae^{-ikx}+Be^{ikx}\right)\exp{\left(\frac{-i\hbar k^2}{2m}t\right)} \tag{2.2.8}

[![]([data:figure221])](https://www.desmos.com/calculator/eee88rqsyc)

~

Danny wray 2025