This time, we will actually calculate the electron motion using the Schrodinger equation.
Here, the wave function of the electron when there is no energy due to the potential is obtained. The time-independent Schrodinger equation obtained last time
-\frac{\hbar^2}{2m}\frac{d^2 \phi(x)}{dx^2}=E\phi(x)
Is used. Here, put the constants together
k=\sqrt{\frac{2mE}{\hbar^2}}
With this
\frac{d^2 \phi(x)}{dx^2}=-k^2\phi(x)
By solving the second-order differential equation, the wave function
\psi(x,t)=Ae^{ikx-i\omega t}+Be^{-ikx-i\omega t}
Here about ω
\omega=\frac{E}{\hbar}Than\\
\omega=\frac{\hbar k^2}{2m}
By substituting this into the wave function formula
\psi(x,t)=Ae^{ik[x-\frac{\hbar k}{2m}t]} + Be^{-ik[x+\frac{\hbar k}{2m}t]}
If x is x (t) so that the shoulder of e becomes 0 when A = 1 and B = 0
x(t)=\frac{\hbar k}{2m}t
This represents the velocity of a plane wave. The moving speed of the peak position of a plane wave is called the phase velocity.
v=\frac{\hbar k}{2m}
Also, regarding the wave function when A = 1 and B = 0, the real part and the imaginary part are
Re[\psi(x,t)]=cos(k\left[ x-\frac{\hbar k}{2m}t \right])\\
Im[\psi(x,t)]=sin(k\left[ x-\frac{\hbar k}{2m}t \right])
When the time of E = 1.00, 0.25 [eV] was shown by animation, the following waveform was obtained.
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