Deriving the free particle wave function e^(2πi(kx−nt))

[Shan K]

I was studying a book on QM and found out that the wave function for a free particle of completely undetermined position traveling in positive x direction is given by

e^(2(pi)i(kx - nt)) where n is the frequency

i have been trying a lot to derive it but till now i can't . Can anyone help me

 

[coki2000]

It comes from Schrödinger Equation as you know.

When write time-independent Schrödinger Eq, we get.

$$\frac{d^2\psi }{dx^2}=-k^2\psi$$ where $k = \frac{\sqrt{2mE}}{\hbar}$

From second order differential equations, it turns out to be:

$$\psi(x)=Ae^{ikx} + Be^{-ikx}$$ It is just like the free particle in an infinite square box. But in that case, we had boundary conditions. This time, we have none so time-independent solution is this.

Since we know that time dependence term is $e^{-\frac{iEt}{\hbar}}$ and $E=\hbar \omega$

Our general time-dependent solution becomes

$\Psi (x,t)=Ae^{i(kx - wt)} + Be^{-i(kx + wt)}$

I hope this helps you.

 

[Shan K]

It comes from Schrödinger Equation as you know.

When write time-independent Schrödinger Eq, we get.

$$\frac{d^2\psi }{dx^2}=-k^2\psi$$ where $k = \frac{\sqrt{2mE}}{\hbar}$

From second order differential equations, it turns out to be:

$$\psi(x)=Ae^{ikx} + Be^{-ikx}$$ It is just like the free particle in an infinite square box. But in that case, we had boundary conditions. This time, we have none so time-independent solution is this.

Since we know that time dependence term is $e^{-\frac{iEt}{\hbar}}$ and $E=\hbar \omega$

Our general time-dependent solution becomes

$\Psi (x,t)=Ae^{i(kx - wt)} + Be^{-i(kx + wt)}$

I hope this helps you.

thanks coki . But in the book they are saying it comes for maxwell's electromagnetic theory .