# Conservation of momentum in infinite square well

This is inspired by Griffiths QM section 2.2, on the infinite square well, which is about how far I've gotten (so, sorry if this is addressed later in the book).

For any given starting wavefunction, you can express it as a sum over the solutions of the time-independent Schrödinger equation. The coefficients in the sum are constant in time, therefore the expectation value of energy is constant in time. The chapter says "this is a manifestation of conservation of energy in quantum mechanics." Ok.

Now in problem 2.5, we have a wavefunction that is an even mixture of the first two stationary states. Part D has us compute the expectation value of momentum, and the solution is sinusoidal in time. So momentum is not conserved.

How can momentum be not conserved? Why is energy conserved but not momentum?

• A classical analogy would be something like a puck bouncing back and forth on an air hockey table. However, in the square well, we do have states with nonzero energy that conserve momentum (stationary states), which we don't have in the classical case. So I'm not sure how far we can take the analogy. Commented Jan 31, 2016 at 8:09

This subtlety is related to the fact that the momentum operator $$\hat{P}$$ (unlike the Hamiltonian $$\hat{H}=\frac{\hat{P}^2}{2m}$$) has no eigenfunctions compatible with the Dirichlet boundary conditions, and $$\hat{P}$$ is not a self-adjoint operator. This is essentially Example 4 in F. Gieres, Mathematical surprises and Dirac’s formalism in quantum mechanics, arXiv:quant-ph/9907069, see p. 6, 39, 44-45. In particular, we cannot choose a common set of eigenfunctions for $$\hat{P}$$ and $$\hat{H}$$. See also e.g. this related Phys.SE post and links therein.
Noether's theorem says that if the Lagrangian of your system is invariant under translations then momentum is conserved. Since the potential $$U(x)$$ is not translation invariant the Lagrangian isn't either. So we shouldn't expect momentum to be conserved. For a classical particle moving in a potential $$U(x)$$ this is true as well. The particle continuously loses and gains momentum as it moves up and down in the potential.