in the Schrödinger Equation of the energy of particle in potential box , why the $n$ not equal to 0
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2 Answers
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If the quantum number $n=0$, it means that you have $\psi_n (x) =0$ and $E_n=0$, i.e., no particle inside the box, which doesn't fit the conditions of "a particle in a box".
In non-relativistic Quantum Mechanics, before to study systems of identical particles and second quantization, we study systems where the number of particles is fixed. There is no description of vacuum in this formalism, and no meaning to consider null solutions for the time-independent Schrodinger equation.
The assumption of "never null wave functions" is implicit when we impose normalization of the wave functions.
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Because the lowest energy which particle can have in the potential box is, $$ E_{1}={\frac {h^{2}}{8mL^{2}}} \tag 1$$
it's called Zero-Point energy,- minimum energy which any quantum system must have. And so particle in the box can't go lower than that, since $E_0$ would be $0$, which is forbidden by QM laws.
Unless you make $m \to \infty$, but then infinite mass particle is not particle at all, i.e. unphysical situation OR you make $L \to \infty$, i.e. you put particle in an infinite box, which is also unphysical since infinite box would mean no box at all, and as such it invalidates this problem.
answered Nov 7 at 9:00
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2$\begingroup$ ...or you can do a particle in a sunken box: $V(x) \rightarrow V(x) - h^2/(8mL^2)$ and then $E_1 = 0$. But no one does that. But you could. $\endgroup$– JEBCommented Nov 7 at 12:56
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$\begingroup$ @JEB This one is the most funniest way to abuse QM 😄🤣 $\endgroup$ Commented Nov 7 at 13:58
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$\begingroup$ What are you talking about? Every constant shift in the Hamiltonian leaves the physics completely unchanged. There is no axiom in QM that says something about the ground state energy. See also this PSE post. $\endgroup$ Commented Nov 7 at 15:02
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$\begingroup$ @TobiasFünke Citation from wiki: Likewise, it can never have zero energy, meaning that the particle can never "sit still" Oh yes, there is such law, check Zero point energy, which defines expectation value of resonator energy : $$ \left\langle {\hat {H}}\right\rangle =V_{0}+{\frac {\hbar \omega }{2}} $$. So every QM resonator/oscillator must have at least ${\frac {\hbar \omega }{2}}$ of energy scaled by some natural frequency. $\endgroup$ Commented Nov 7 at 15:13