Griffiths QM Problem 2.2 Solution: Proving that Energy has to be Greater than Potential

[@abdussamifarooqui5109]

Love the video sir! Please keep making these videos we really appreciate this.

[@BANGATAN_ARMY_7]

Your Video Helps Me Alot... Thank You So Much Sir... The Way You Explain...It Is So Easy To Understand...☺️

[@sairanadeem385]

Thankw sir🥰

[@RishabhKhatri-rt9pf]

Wouldn't the same thing happen if E is greater than Vmax?

[@Apalion41]

I don't see how this is a proof, considering for instance that exp(-|x|) is also always concave up and yet it is normalizable.

[@harryallen5392]

It, in fact, is not always concave up. For x0, the function mirrors itself around the y-axis as the mod throws out a positive x so -|x|=-x, tending to 0 as x tends to positive infinity and thus, we see that the function is, in fact, normalisable.

[@strat0caster124]

In fact, you can construct a phi(x) = (pi)^(-1/2)*(r^2 - x^2)^(-1/4), x∈(-r, r), which is normalizable even though phi(x) -> infinity as x approaches ±r, you can verify this by integrating phi^2 over (-r, r), replace x = -rcos(theta) and integrate over theta∈(0, pi) instead. ...Yet phi(x) and phi(x)'' have the same sign (they're always positive). So, I think this problem is fundamentally wrong, at least mathematically. You need to add more constraints to phi(x) than just normalizable (in the ∫phi(x)^2dx= 1 context)