# Infinite square well and Heisenberg picture

The infinite square well is often a mainstay of introductory quantum physics courses. Its boundary conditions at the well-walls are easily solved to the find the Hamiltonian's eigenfunctions in the position basis.

The course then might move on to the simple harmonic oscillator, and around this time the Heisenberg picture may be introduced. It's then demonstrated that expectation values in the Heisenberg and Schrodinger pictures are equivalent. As an example of this, a quantum physics student might be asked to calculate the expectation of the position $$Q$$ in a superposition of the ground and first excited state, say $$\frac{1}{\sqrt{2}} (|0\rangle + |1\rangle)$$. The student would find that the expectation is sinusoidal in time in both the Schrodinger and Heisenberg pictures. In the Schrodinger picture, the sinusoidal time dependence falls out of the time dependence of the eigenstates of the Hamiltonian. In the Heisenberg picture, it falls out of the time dependence of the $$Q$$ and $$P$$ operators. In particular, the Heisenberg operators follow the classical equations of motion (easily seen by solving the Heisenberg time evolution equation): \begin{align} Q(t) =& Q(0) \cos(\omega t) + \frac{P(0)}{m \omega} \sin(\omega t) \\ P(t) =& P(0) \cos(\omega t) - Q(0)m \omega \sin(\omega t) \, . \end{align}

The infinite square well also demonstrates a sinusoidal position dependence when looking at a superposition of the ground and first excited state. This is easily seen in the Schrodinger case, where the sinusoidal position dependence falls out once again from the phase time dependence of the Hamiltonian's eigenstates. Even if the time dependence of Q is very complicated in the Heisenberg picture, I would like to see the sinusoidal dependence of the position expectation of a superposition of two eigenstates to see that it can be done in this picture.

This is where I become confused. For one thing, it is tricky for me to handle the abrupt walls of the potential. One might anticipate that like the simple harmonic oscillator, the infinite square well Q and P operators should follow the analogous time dependence of the classical equations of motion. However, the classical equation of motion of Q is a triangle wave, not anything sinusoidal, and so it's difficult for me to see how one could get a sinusoidal dependence for the simple case of the expectation of position for a superposition of two eignenstates. Following the comments, I expect perhaps the resolution is that Q can expanded in a Fourier-esque sum with operator-valued coefficients. I'm tempted to believe we can find such a sum starting in the Schrodinger picture by calculating the matrix elements of Q in the energy eigenbasis and then affixing the necessary phase factors in time. It would be preferable to have a method that only uses the Heisenberg picture, however.

My question is, how do we find the time dependence of Q and P in the Heisenberg picture in the infinite square well? Can we find their expectation values in the state $$\frac{1}{\sqrt{2}} (|1\rangle + |2\rangle)$$ and show they match with those found in the Schrodinger picture?

Edited to make it clear that we should not expect a simple sinusoidal time dependence of Q on time and to add the idea of a series of operator-valued coefficients.

• I do not know if it helps with your question, but I once made this animation: commons.wikimedia.org/wiki/… – Pieter Oct 9 '18 at 23:35
• The problem is the same as for the classical solution of a particle in an infinite square well: the equations of motion are singular at the boundary. You can only solve this by either considering the well as a limit of increasingly steep smooth potentials; or by imposing a reflecting boundary condition on $p$ at the edges. Either method still works in quantum mechanics, but implementing either one is extremely onerous, compared to the almost triviality of the Schrödinger-picture solution. – Buzz Oct 10 '18 at 0:39
• @Pieter That's a nice animation. I'm not sure it helps in this case, but it's a good visualization of the oscillations. – user196574 Oct 10 '18 at 0:57
• @Buzz Would you consider writing that comment up as an answer? It might be onerous, but could you show an example with increasingly steep smooth potentials or with the boundary condition? – user196574 Oct 10 '18 at 0:58
• $P:= -i\hbar \frac{d}{dx}$ is not a self-adjoint operator for the infinite square well, so "momentum" is not a well-defined observable here. – J. Murray Oct 10 '18 at 3:54