# Classical analog of the statement "$E$ must exceed the minimum value of $V(x)$

Overall question:

Griffiths problem 2.2 states that $$E$$ must exceed the minimum value of $$V(x)$$ for every normalizable solution to the time-independent Schrodinger equation. Then, it asks for a proof and what the classical analog of this statement is. I understand the proof for this, but I am very confused about the classical analog.

Approach:

I was initially thinking that this statement means that the energy must be greater than the potential energy at this $$x$$ value. I assumed $$E$$ was the total energy, so this would mean that there must be kinetic energy always. However, I remain confused about many things:

Specific Questions:

1. What is kinetic energy in quantum mechanics?

2. Is it proper to say that $$E$$ here is total energy?

3. Is this the proper classical analog? I don't need an answer, but a hint would be nice. An answer is okay as well but I'm not sure it's allowed.

4. Why can we use the concepts potential and potential energy interchangeably? Here, we are saying that potential has the same units as energy, which means it's potential energy I gathered?

Update: I have also thought about the classical analog being underdamped motion, but I am still curious about the above questions and interpretation.

1. Typically, the kinetic energy operator is formally defined to be $$\hat T = \frac{1}{2m} \hat P^2$$.
2. Yes
3. If $$E\not\geq V_{min}$$, then $$E at every point in space. What would this tell you about the kinetic energy?
4. $$\hat V$$ is the potential energy operator. Some sources may drop the word "energy" from this description but, to be blunt, if you're studying quantum mechanics you should be able to comfortably interpret this rather mild abuse of terminology (which, I would note, you have). You are free to avoid this abuse and refer to the "finite potential energy well" and the "delta function potential energy" and the "potential energy step" and the "Coulomb potential energy" (so on and so forth) for your entire physics career, but I suspect this will quickly grow tiresome and you will quickly join the club.

(1)In quantum mechanics, there is no essentially no such thing as Kinetic energy of a particle, rather it is the expectation value of the $$\hat{T}$$ operator, given as: $$\langle\hat{T}\rangle=\left\langle\frac{\hat{p}^2}{2m}\right\rangle$$
(2)E represents, the total energy or the eigenvalue obtained when the wave function $$\psi_n$$ is operated on the total Hamiltonian of the system. $$\hat{H}\psi_n=E_n\psi_n$$
(3)This can be thought classically as, $$T+V=E$$ $$\frac{m}{2}v^2=E-V=\alpha<0$$ which implies, $$v^2=\frac{2\alpha}{m}<0$$ which is impossible.