okay so i need some help interpreting some of the results,(adsbygoogle = window.adsbygoogle || []).push({});

so (-ħ^{2}/2m)Ψ''=E-V_{0}Ψ;

So i set k^{2}= 2m*(E-V_{0})/ħ^{2}

and so : Ψ''=-k^{2}Ψ

so if V_{0}=0 or is smaller than E, k^{2}is positive;

*need for help starts here*

Ψ=Ae^{ikx}+Be^{-ikx};

another result for this would also be only e^{ikx}so is the second term only there to represent the possibility of the wave to travel in both directions?

now for the trignometric result is the same, are both the sines and cosines there to represent both directions?

ok so when V_{0}is bigger than E i can just set k^{2}=-2m*(E-V_{0})/ħ^{2}and so

Ψ=Ae^{kx}+Be^{-kx}

there is no trig result correct?

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# I Time independent Schrodinger equation results (1D)

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