Recent content by feynwomann
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Solve Finite Potential Well: Schrödinger Eqn. & k=qtan(q*a)
Oh it totally makes sense now. I forgot that my teacher said there was a mistake in the exercise. It's supposed to be B = 0. I didn't think of dividing the equations (I had actually done the first step, but didn't see how I could take it further). Anyway, thanks for the help!- feynwomann
- Post #3
- Forum: Advanced Physics Homework Help
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Solve Finite Potential Well: Schrödinger Eqn. & k=qtan(q*a)
' I've got these solutions to the Schrödinger equation (##-\frac{\hbar} {2m} \frac {d^2} {dx^2} \psi(x) + V(x)*\psi(x)=E*\psi(x)##): x < -a: ##\psi(x)=C_1*e^(k*x)## -a < x < a: ##\psi(x)=A*cos(q*x)+B*sin(q*x)## x > a: ##\psi(x)=C_2*e^(-k*x)## ##q^2=\frac {2m(E+V_0)} {\hbar^2}## and ##k^2=\frac...- feynwomann
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- Finite Potential Potential well
- Replies: 2
- Forum: Advanced Physics Homework Help