Understanding the 1D Wave Equation for Free Particles in Quantum Mechanics

Shan K
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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
 
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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.
 
coki2000 said:
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 .
 
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