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Let us suppose I have a particle in a 1D infinite potential well of length $L$ which is in the ground state. The energy is given by

$$E=\frac{\hbar^2\pi^2}{2mL^2}.$$

Now I decrease the size of the well gradually to lets say $L-x$. This means that particle is still inside the 1D well, since it can't escape infinite potential, but the energy of the particle is less than the new ground state energy given by

$$E=\frac{\hbar^2\pi^2}{2m(L-x)^2}.$$

Which means the particle can't exist in the well. So how to explain this contradiction?

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  • $\begingroup$ "but the energy of the particle is less than the new ground state energy" - what do you mean by this? Since you have moved the walls, there no longer exists a state of definite energy with the particle's previous energy: "the energy of the particle" is either definite with the new value or indefinite, but either way, where is the contradiction? $\endgroup$
    – ACuriousMind
    Aug 16, 2020 at 11:28

1 Answer 1

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This is a classic example used to illustrate the adiabatic theorem. If you narrow the walls slowly enough, the particle will stay in the ground state of the box at all times. Therefore, its energy will increase slowly. This makes sense if you think about it. Moving the walls can cause the particle to gain energy. This can be true even in the classical case (a collision with a moving wall would add energy to a classical particle).

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  • $\begingroup$ What if it’s not done slowly (fast enough compared to energy gap)? Does the particle suddenly gain energy? $\endgroup$ Aug 15, 2020 at 22:01
  • $\begingroup$ The particle has to gain energy to stay above the ground state. If you narrow the walls quickly, you'll probably end up with a superposition of the ground state and some excited states, which means your energy will increase even more than it would in the slow-walls case. $\endgroup$ Aug 15, 2020 at 22:23
  • $\begingroup$ I have been thinking about it in classical sense like you said.I can see that since the size of the well is decreased, the number of collisions per unit time increases,and thus more work is done per unit time.But this just means that the particle also changes direction more often per unit time.So there should be no increase in momentum, right?....Also let's say the particle can gain energy by hitting moving potential wall, but there's chance that particle might not hit the walls when its moving right? $\endgroup$
    – Nishan
    Aug 15, 2020 at 22:45
  • $\begingroup$ The classical analogy isn't perfect. You could, technically, narrow the walls without adding energy to the particle in classical mechanics, but not in quantum mechanics. As for momentum, the expectation value of momentum in a stationary state of the well is always zero. $\endgroup$ Aug 15, 2020 at 22:55

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