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Evolution of quantum state in time can be obtained from the time-dependent Schrodinger equation $$\hat{H} \psi(x,t) = i \frac{\partial}{\partial t} \psi(x,t).$$ For time-independent Hamiltonian, the solution of the Schrodinger equation can be written in the following form $$\psi(x,t)=\exp{(-i \hat{H}t)} \psi(x,0).$$ I am wondering if it is possible to extract the phase of the propagated wavefunction. I mean if we write the wavefunction in polar representation $\psi(x,t)=R(x,t)\exp(i S(x,t))$ whether it is possible to write some expression for propagation of the phase factor $S(x,t)$ only.

If the calculation is already performed then it is not problem to take a logarithm of the wavefunction, so $$\ln \psi(x,t) = \ln R(x,t) + i S(x,t),$$ and taking imaginary part one can obtain the phase easily. However, this algorithm completely fails numerically due to the necessity to divide imaginary part of the wavefunction on the real one when transforming from coordinate to polar representation.

Thus, my question is if it is possible to connect the phase $S(x,t)$ with, let us assume real, initial state $\psi(x,0)$ by some expression which avoids explicit conversion between coordinate and polar representation of the complex numbers.

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  • $\begingroup$ Are you unhappy with the quantum HJ equation? $\endgroup$ – Cosmas Zachos Aug 29 '18 at 15:10
  • $\begingroup$ Bohmian dynamics has its own problems so I try to avoid solving HJ equations by any means. $\endgroup$ – QuantumNik Aug 29 '18 at 15:14
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If $$ \psi(x,0)=\sum_n c_n \psi_n(x) $$ where $\psi_n(x)$ is solution to the Schrodinger equation with energy $E_n$, then $$ \psi(x,t)=\sum_n c_n e^{-i E_n t} \psi_n(x) \tag{1} $$ but that's not an overall phase. You could rewrite $$ \psi(x,t)= e^{i\varphi t} \left(\sum_n c_n e^{-i (E_n-\varphi) t} \psi_n(x)\right) $$ but that overall phase has no physical meaning and is otherwise arbitrary.

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  • $\begingroup$ Thanks for your answer. This is not the same phase which I am talking about since it is coordinate-independent in your case. In my example the phase is defined uniquely (up to 2$\pi$). Otherwise, it is clear that you can add arbitrary non-topological phase to the wavefunction which will not affect observables. $\endgroup$ – QuantumNik Aug 29 '18 at 13:28
  • $\begingroup$ @QuantumNik you should clarify your question accordingly. $\endgroup$ – ZeroTheHero Aug 29 '18 at 13:42
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This problem has a simple answer if $\hat H$ has a discrete spectrum. In such a case, one has $$ {\hat H}\phi_n=E_n\phi_n, $$ with $E_n$ the eigenvalues and $\phi_n$ the eigenfunctions. Then, the arbitrary phases can always be chosen to have the $\phi_n$ real.

Then, we use the standard expansion $$ \psi(x,t)=\sum_nc_ne^{-iE_nt}\phi_n, $$ where $c_n=(\phi_n,\psi(x,0))$, and take $$ c_n=\rho_n e^{i\chi_n}. $$ Therefore, one has $$ \psi(x,t)=\sum_n\rho_n\phi_n\cos\left(E_nt+\chi_n\right)+i\sum_n\rho_n\phi_n\sin\left(E_nt+\chi_n\right) $$ that has the standard form of a complex number $Z=X+iY$. The phase is obtained straightforwardly as $$ S(x,t)=\arctan\left(\frac{\sum_n\rho_n\phi_n\sin\left(E_nt+\chi_n\right) }{\sum_n\rho_n\phi_n\cos\left(E_nt+\chi_n\right)}\right). $$ E.g., if $\psi(x,0)$ coincides with one of the eigenvectors, the above expression turns out to give for the phase $E_nt+\chi_n$ as it should.

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  • $\begingroup$ Thanks for your answer! Indeed, the analytical solution is simple. The problem appears when one solves it numerically. In your final expression for $S(x,t)$, the spatial dependence comes from eigenstates $\phi_n(x)$. For discrete spectrum, eigenstates become small on borders of the potential well. Thus, summation in both numerator and denominator is numerically poorly defined. $\endgroup$ – QuantumNik Aug 30 '18 at 14:29
  • $\begingroup$ I suspect that one can find some satisfying algorithm for such summations. These are quite common in quantum mechanics problems and I would be somewhat surprised if somebody did not think about this. $\endgroup$ – Jon Aug 30 '18 at 14:42
  • $\begingroup$ Probably you are right. Unfortunately, I did not see it anywhere in the literature. I am wondering if there is another way do to it. I bet that the phase $S(x,t)$ in regions where $\psi(x,t)$ is close to zero should depend on the properties of the potential but not on the properties of the initial state. So, my assumption is that when $\psi(x,t) \to 0$, the phase $S(x,t)$ is a function of $V(x)$ only. Question is how to proof or deny this statement and what are the concrete expression. $\endgroup$ – QuantumNik Aug 30 '18 at 14:56
  • $\begingroup$ Did you check scattering theory and the optical theorem? $\endgroup$ – Jon Aug 30 '18 at 14:57
  • $\begingroup$ No, I didn't try it. It is hard to see how it can help. As far as I know, phases in scattering theory are numbers (phase shifts etc). In my case I am looking for coordinate-resolved property which is a function. Probably, I am missing something obvious. Please, correct me if I am wrong. $\endgroup$ – QuantumNik Aug 30 '18 at 15:09

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