Time Dependence of Ring States

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Section 22.8 Time Dependence of Ring States

We know, from the theory of Fourier series, that we can write any initial probability distribution, which is necessarily periodic, as a sum of the energy eigenstates:

\begin{equation} \Phi(\phi) = \sum_{m=-\infty}^\infty c_m \Phi_m(\phi) = \sum_{m=-\infty}^\infty c_m \left( \frac{1}{\sqrt{2\pi r_0}} e^{im\phi} \right)\tag{22.8.1} \end{equation}

where, for the probability distribution to be normalized, we must have:

\begin{equation} \sum_{m=-\infty}^\infty |c_m|^2 = 1\tag{22.8.2} \end{equation}

To find the time evolution of the eigenstates \(\Phi_m(\phi)\text{,}\) we must solve the \(t\) equation (22.4.3). Since, for each \(\Phi_m\text{,}\) we have now found the value of the constant \(E=E_m\text{,}\) given by (22.5.5), we can solve (22.4.3) trivially:

\begin{equation} T(t) = e^{-\frac{\hbar}{i}E_m t}\tag{22.8.3} \end{equation}

A deep theorem in the theory of partial differential equations states that if you have found an expansion of the initial probability density in terms of the eigenstates of the Hamiltonian, then the time evolution of that probability density is simply obtained by multiplying each eigenstate individually by the appropriate time evolution:

\begin{equation} \Phi(\phi,t) = \sum_{m=-\infty}^\infty c_m \Phi_m(\phi) e^{-\frac{\hbar}{i}E_m t}\tag{22.8.4} \end{equation}

BE CAREFUL! There are an infinite number of different values for the energy, depending on the eigenstate of the Hamiltonian. It is incorrect to multiply the initial state (22.8.1) by a single over-all exponential time factor. Each term in the series gets its own time evolution.