small insight into TISE

18 Nov 2025

When solving the TISE, you seem to keep ending up with the second deriv wrt x of φ equals some constant times V(x) - E. Solving this ODE gives e to the power of the square root of whatever is there. Two insights from this as I'm studying to help see the pattern of what the time-independent wave function for that energy will look like.

First: wherever V(x) - E is positive (the potential is larger than the wave energy), you get the e to the power of the square root of a positive quantity, which is a real number. This is why you get exponential decay in these regions which can lead to tunneling to the other side of a potential barrier. And elsewhere, you get the square root of a negative quantity, which is imaginary; this is why you get sinusoidal solutions.

Second: for sinusoidal solutions, the difference between V(x) and E controls the wavelength. Larger V(x) leads to longer wavelength. Super fun. Awesome. Amazing. Cool. Okay.

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