I am going through Shankar's Principles of Quantum Mechanics and am having trouble finding the free particle propagator $U(t)$ that satisfies $$\lvert\psi (t)\rangle = U(t)\lvert\psi (0)\rangle$$ due to the degeneracy of the $E$ eigenkets. Shankar says that if the Hamiltonian has degenerate eigenvalues, we change the propagator equation from $$U(t) = \sum_{E} \lvert E\rangle\langle E\rvert e^{-iEt/{\hbar}}\tag{4.3.13}$$ to $$U(t) = \sum_{E} \sum_\alpha \lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar} $$ where $\lvert E, \alpha \rangle $ are the orthonormal eigenkets for the $E$ eigenspace. This makes sense to me, because it seems like whenever we deal with degeneracy to use a projection. He also says for a Hamiltonian with no degeneracy to change the sum into an integral, so I imagine that for a Hamiltonian with a continuous, degenerate spectrum we would have this propagator: $$ U(t) = \sum_\alpha\int_\Re dE\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}$$

I tried applying this to the free particle and had some trouble. In the book, after solving for the eigenkets and eigenvalues, $\lvert E\rangle = \lvert p\rangle$ and $p = \pm\sqrt{2mE}$, he chooses the propagator as a function of the momentum eigenvalues $p$ so that $$U(t) = \int_\Re dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m).\tag{5.1.9}$$ I understand how he got to this, but I'm having some trouble because this doesn't appear to be the same expression that I have above. Splitting the integral up, we have $$U(t) = \int_{-\infty}^0dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ The left integral can changed to an integral over $-p$, and the bounds switch from $0$ to $\infty$ such that: $$U(t) = \int_{0}^\infty d(-p)\lvert -p\rangle\langle -p\rvert \exp(-ip^2t/2m) + \int_0^\infty dp\lvert p\rangle\langle p\rvert \exp(-ip^2t/2m)$$ Now we can substitute and interchange $\lvert p\rangle = \lvert E, + \rangle$, $\lvert -p\rangle = \lvert E, - \rangle$, $E = p^2/2m$, and $dp = \pm mdE/\sqrt{2mE}$ to get: $$U(t) = \sum_{\alpha = \pm} \int_0^\infty dEm/\sqrt{2mE}\lvert E, \alpha\rangle\langle E, \alpha\rvert e^{-iEt/\hbar}. $$ This is the result that we are asked to prove in Exercise 5.1.1. Why is this not the same as the first expression for the propagator? Where did the extra $ m/\sqrt{2mE} $ come from? Am I just wrong in assuming that that is the general expression for a Hamiltonian with continuous and degenerate eigenvalues?


Your first expression should have a additional factor in the integrand called a density of states (DOS) and is usually denoted $\rho(E)$, i.e., $$U(t)=\sum\int\rho(E)dE|E,\alpha\rangle\langle E,\alpha|e^{-iEt/\hbar}.$$ The density of states is the number of states in the energy range $(E,E+dE)$. Look at the parabolic dispersion section of the density of states on wiki.

  • $\begingroup$ I think your answer would be improved if you were to expand on where this density of states comes from and why it is necessary. I was confused on this subject myself recently and didn't find your answer terribly helpful. $\endgroup$ – Time4Tea Jan 27 '20 at 22:10

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