Q.Given in Fig. 5.11 are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle must have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.

1. Since $V(x)=0$ for $x<a$ and $V(x)=V_0>E$ for $x\ge a$, the kinetic energy $K=E-V(x)$ would be negative for $x\ge a$ — impossible. So the particle is confined to $x<a$ and can never be found for $x\ge a$. The lowest possible total energy for which the particle can exist at all is $E_{\min}=0$ (the lowest value $V(x)$ ever takes). This is the classic single-step potential: like a ball rolling on a flat frictionless track that meets a step/ledge higher than its kinetic energy can climb, so it is turned back at $x=a$ and can never cross over.
2. The marked energy $E$ lies below $V_0$, which is itself the lowest step of the rising staircase — so $V(x)\ge V_0>E$ everywhere the staircase is defined, and $K=E-V(x)<0$ there. The particle cannot be found anywhere along the staircase at this energy. Since even the first, smallest step already exceeds $E$, the particle's minimum energy to exist there at all is $E_{\min}=V_0$. Physically this is like a ball sitting at the base of a long flight of ever-higher stairs — it doesn't even have enough energy to climb onto the very first step, let alone the higher ones beyond.
3. Here $V(x)=V_0>E$ for $x<a$ and $x>b$ (the two outer walls), but $V(x)=-V_1$ (below zero) in between — so the particle is confined entirely to the well, $a<x<b$, and cannot be found outside it. Its minimum possible total energy is the floor of the well, $E_{\min}=-V_1$. This is the standard rectangular potential well — the same idealised model used for a particle (e.g. a nucleon) trapped inside a finite, attractive potential region with walls too high for it to climb over classically, such as inside a nucleus or a simple finite square well.
4. The barriers on either side rise to a peak $V_0>E$ over $a/2<|x|<b/2$, so the particle cannot be found in those two barrier regions, $-b/2<x<-a/2$ and $a/2<x<b/2$; outside the barriers ($|x|>b/2$) and inside the central well ($|x|<a/2$) it is allowed, since $V(x)\le E$ there. The lowest point of the whole curve is the central well's floor, so $E_{\min}=-V_1$. This double-hump shape is a simple model for a particle trapped in a central attractive well flanked by two barriers it cannot classically cross — e.g. a particle bound between two humps, as in a simplified model of a symmetric double-well/double-barrier system.
   ✓Final answer

   1. forbidden for $x\ge a$, $E_{\min}=0$ — like a ball unable to climb a single step higher than its energy.
   2. forbidden for all $x$ at the marked energy, $E_{\min}=V_0$ — like a ball at the base of a staircase too tall to climb even the first step.
   3. allowed only in $a<x<b$, $E_{\min}=-V_1$ — the classic rectangular potential well (e.g. a particle trapped inside a finite well).
   4. forbidden in $-b/2<x<-a/2$ and $a/2<x<b/2$, $E_{\min}=-V_1$ — a particle trapped in a central well flanked by two barriers it cannot cross.

Concept

Total mechanical energy is constant, $E=K+V(x)$, so $K=E-V(x)$. Kinetic energy can never be negative, therefore the particle is allowed only where $V(x)\le E$ and forbidden wherever $V(x)>E$. The least total energy a particle can have equals the minimum value of $V(x)$ (there $K=0$).

(a) Single upward step

For $x<a$, $V=0<E$, so the region is allowed. For $x\ge a$, $V=V_0>E$, giving $K<0$, so the particle cannot be found for $x\ge a$. The lowest potential is $0$, so the minimum total energy is $0$. Physical picture: a particle meeting a potential step, e.g. an electron approaching a metal boundary.

(b) Rising staircase

The marked energy lies below the whole curve ($E<V_0\le V(x)$ for all $x$), so $K<0$ everywhere and the particle cannot be found anywhere with this energy. To be found even in its lowest broad region it needs $E\ge V_0$, so the minimum total energy is $V_0$. Physical picture: a charge driven through a succession of rising potential steps.

(c) Rectangular well

$V=V_0>E$ for $x<a$ and for $x>b$, so those regions are forbidden; $V=-V_1<E$ for $a<x<b$, which is allowed. The particle is confined to $a<x<b$, and the minimum total energy is $-V_1$ (the floor of the well). Physical picture: a particle trapped in a box / finite square well, like a molecule bouncing between two rigid walls.

(d) Twin barriers with a central well

$V=-V_1<E$ in the central well $|x|<a/2$ (allowed) and $V=0<E$ for $|x|>b/2$ (allowed), but each hump rises to $V_0>E$. The particle is forbidden where the humps rise above $E$, i.e. in $-\tfrac{b}{2}<x<-\tfrac{a}{2}$ and $\tfrac{a}{2}<x<\tfrac{b}{2}$; a particle sitting in the central well is classically trapped. The minimum total energy is $-V_1$. Physical picture: a particle bound in a well guarded by potential barriers, e.g. an $\alpha$-particle held inside a nucleus.

Concept: Classically Forbidden Regions — $K = E - V(x) \ge 0$

Step 1: State the governing rule.

A particle can exist only where $V(x) \le E$ (so $K \ge 0$); it is forbidden wherever $V(x) > E$. Its minimum possible total energy equals the lowest value $V(x)$ reaches.

Step 2: Apply this to each potential shape.

1. Single step: allowed for $x<a$ ($V=0<E$), forbidden for $x\ge a$ ($V=V_0>E$); $E_{\min}=0$.
2. Rising staircase: $E$ lies below every part of the curve, so forbidden everywhere; $E_{\min}=V_0$ (the lowest step).
3. Rectangular well: forbidden where $V=V_0>E$ (outside $a<x<b$), allowed inside the well where $V=-V_1<E$; $E_{\min}=-V_1$.
4. Twin barriers with central well: allowed in the central well and far outside, forbidden on the barrier humps where $V$ rises above $E$; $E_{\min}=-V_1$. Step 3: Read off the minimum energy in each case as the lowest point of $V(x)$ reachable by the particle. Final Answer: (a) forbidden $x\ge a$, $E_{\min}=0$; (b) forbidden everywhere, $E_{\min}=V_0$; (c) confined to $a<x<b$, $E_{\min}=-V_1$; (d) forbidden on the barriers, $E_{\min}=-V_1$