# Wavelength uncertainity for particle in a box

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1. Oct 11, 2014

### Maharshi Roy

For a particle in a box, we have been explained that the wave function is like a standing wave. Then we wrote:-
λ = 2L/n
where λ is the wavelength of the nth energy state. But a moving particle is considered as a wave packet and so does not have an unique wavelength. Then, how are we determining energy for nth state using this unique λ. There must be an uncertainity in energy too.

2. Oct 11, 2014

### Staff: Mentor

More precisely, the possible wave functions are superpositions of these standing waves.

There is. That's why the uncertainty principle is so often written as the product of the uncertainty of two observables; the smaller the uncertainty in one the greater the uncertainty in the other, but usually neither is zero.

3. Oct 12, 2014

### vanhees71

If your particle is prepared in an eigenstate of the Hamiltonian, there is no uncertainty in energy. You'll measure with 100% probability the corresponding eigenvalue of the Hamiltonian. Note that for a particle confined to a box, the energy eigenstates are true (square integrable) eigenstates. However, you should note that there is no momentum operator, and the position-momentum uncertainty relation sometimes qualitatively quoted in connection with the particle in a rigid box is questionable.

4. Oct 12, 2014

### Staff: Mentor

The standing-wave solutions (energy eigenstates) of the particle in a box are not "moving particles" in the quantum-mechanical sense. Their probability distributions do not change with time. They are "stationary states."

If you construct a superposition of some number of energy eigenstates with different values of n, it will not be a stationary state. The probability distribution will "slosh" back and forth between the walls of the box. For such a state, the energy, momentum and wavelength are all uncertain.