That the box has rigid walls. However only the eigenstates of the Hamiltonian have energies that take discrete values, when measured in appropriate units-the superpositions of eigenstates don't have definite values for their energy and the probability distribution is continuous, it's not discrete.

If the initial state is an eigenstate of the Hamiltonian, any state that’s the solution of the Schrödinger equation will have the same energy: (to simplify notation I use units where hbar=h/(2π)=1)

idψ/dt=Hψψ(x,t)=exp(iHt)ψ(x,0)

for a time-independent Hamiltonian, indeed, whatever the potential.

If ψ(x,0) is any eigenstate of H, of energy E, i.e.

Hψ(x,0)=Eψ(x,0)

then

Hψ(x,t)=Hexp(iHt)ψ(x,0)=exp(iHt)Hψ(x,0)=exp(iHt)Eψ(x,0)=Eexp(iHt)ψ(x,0)=Eψ(x,t)

So an initial eigenstate of the Hamiltonian remains an eigenstate of the Hamiltonian with the same energy for all time.

The property that the values of the energy can be labeled by integers (i.e. that the spectrum is discrete) is a general property of the eigenstates of any operator, in the present case it's the Hamiltonian, defined over a compact domain (in the present case a box, with rigid walls).

For the particle in a box, H = -d^2/dx^2 and the eigenfunctions must satisfy the boundary conditions ψ(0)=0 and ψ(a)=0. The solutions of the equation

(-d^2/x^2)ψ=Eψ

that satisfy the boundary conditions, are

ψ(x)=(2/a)1/2sin(nπ x/a)

with n=1,2,..

and there aren't any other.

The corresponding eigenvalues are

En=(nπ/a)2

(In units where hbar=1 and the mass of the particle, m=1/2, which is always possible by a choice of units.)

so the least possible value is for n=1.

So En/(π/a)2 = n2

(in units where hbar=1; it's an easy exercise to deduce that hbar and m can only appear as En/( hbar π/a/(2m)1/2 )2=n2 which shows what the ``useful'' unit for the energy is.)

It's a standard exercise to prove that no linear superposition of eigenstates can have average energy less than E1. However for any linear superposition of eigenstates, the energy will have a continuous distribution, which can be shown-another standard exercise-to have as its average and its typical value (the value corresponding to the maximum of the distribution) not less than E1.

Quantum mechanics appears as the statement that the wave function, corresponding to a state of the particle in a box, is a solution of the Schrödinger equation, not in the mathematical properties of this equation. That the spectrum is discrete is a mathematical property of the equation, independently of what is the meaning of the constants that appear in it.

Quantum mechanics further appears as the statement about how to compute probabilities, once the wavefunction is known.

If, instead of solving the Schrödinger equation for the wavefunction of a particle in a box, one would solve the classical wave equation for waves in a cavity, one would, also find that the normal modes would have discrete energies, while the superpositions wouldn't. The difference with quantum mechanics is that the quantum system describes a particle in equilibrium with a bath of quantum fluctuations, which is why a probabilistic description makes sense; the classical wave equation describes a wave in the absence of any bath, which is why it doesn't make sense to consider a histogram of the energy of different solutions of the classical wave equation as a probability distribution.

Dear Stam Nicolis

First, thank you for your mathematical clarification,

but my question still exists, what is the kind of this energy E that comes from the Hamiltonian?

as we know the potential energy in box is 0...

So is it the kinetic energy? but the kinetic energy isn't certain in box nevertheless the energy of the particle must be certain (equals E) in each measurement.

With best regards.

Dear Mazen Khoder,

The energy is the kinetic energy, since the Hamiltonian in the box is H = p2/(2m). The energy in general-and, in the present case, the kinetic energy-of eigenstates of the Hamiltonian is known with certainty, for all time, that's what the one line calculation shows. The energy of an eigenstate of the Hamiltonian is equal to the corresponding eigenvalue. So if the initial condition is an eigenstate it remains the same eigenstate for all time and its energy remains the same and its probabilty distribution is a delta function. If the initial condition is NOT an eigenstate of the Hamiltonian, it can be expressed as a linear combination of the eigenstates. In that case the energy of the particle has a probability distribution, which can be computed from the initial condition and from the Schrödinger equation it's then possible to compute the probability distribution of the energy for all time.

Dear Stam Nicolis

Thank you for your reply, but why can't say also that kinetic energy = P^2/2m,

and as we know the momentum P isn't certain in box for the ground state for example?

With best regards.

I did say it: The kinetic energy is P2/(2m) and the ground state is a state of definite kinetic energy, (hbar π/a)2/(2m). It isn't a state with definite momentum, however, since the ground state wavefunction isn't an eigenfunction of the momentum operator-and this is, indeed, consistent with the fact that the walls of the box break translation invariance in space; translation invariance in time is, however, preserved.

The tricky point here is the presence of boundaries.

Dear Stam Nicolis

Thank you for your reply, yes I understand that the ground state wavefunction isn't an eigenfunction of the momentum operator but it is an eigenfunction for the kinetic energy operator (for the particle in box), but this is the strange thing that our definition of operators leads us to it! I mean the momentum is variable but the kinetic energy is fixed!

In my opinion, that means that this kinetic energy is not for the current momentum!

With best regards.

Of course it is-the confusion arises by assuming that eigenstates of kinetic energy must be eigenstates of momentum. They need not be: Momentum generates translations in space-which aren’t symmetries, in the present case, since the walls of the box are fixed-while energy (in the present case just the kinetic energy) generates translations in time, which are symmetries of the theory, since the Hamiltonian is bounded from below and is independent of time.

Said differently: The eigenfunctions f(x), of eigenvalue λ, of P=-id/dx (in units hbar=1, m=1/2), are defined by

Pf(x)=λf(x) f(x) = Aexp(iλx),

However the boundary conditions f(0)=0 (or f(a)=0) imply that A=0, i.e. that f(x)=0, which means that the operator P doesn't have any eigenfunctions under these conditions; which implies that P and P2 can’t have common eigenfunctions, with these boundary conditions.

(The constant function, f(x)=A≠0, does satisfy Pf=0.f, P2f=0.f, but the boundary conditions imply that A=0, which disqualifies it from being an eigenfunction of either at all (that's why the lowest value for the energy is E1 > 0; n=0, would lead to E0=0, but the corresponding ``eigenfunction'' vanishes identically, so n=0 is excluded.)

The only boundary conditions that can lead to common eigenfunctions of P and P2 are periodic boundary conditions, f(0)=f(a).