# How to Derive the Time Evolution Equation for Quantum Phase?

In quantum mechanics, the wavefunction $$\psi(x,t)$$ outputs a complex number that describes the probability amplitude of finding a particle in a particular place and time. The complex number can be written in the form $$Ae^{i\theta}$$ where $$\theta$$ is the "quantum phase".

Of course the time evolution of a quantum particle is given by the time-dependent Schrodinger Equation: $$i\hbar \dfrac{\partial \psi}{\partial t} = -\dfrac{\hbar^{2}}{2m}\nabla ^{2} \psi +U\psi$$

What I wanted to do is derive the time evolution equation of the quantum phase $$\theta(x,t)$$ assuming a constant $$A$$. The first step in my derivation was I made a substitution $$\psi(x,t)=Ae^{i\theta(x,t)}$$: $$i\hbar \dfrac{\partial (Ae^{i\theta(x,t)})}{\partial t} = -\dfrac{\hbar^{2}}{2m}\nabla ^{2} (Ae^{i\theta(x,t)}) + U(Ae^{i\theta(x,t)})$$ Then assuming constant $$A$$ and performing the derivatives: $$i\hbar (ie^{i\theta(x,t)})\dfrac{\partial \theta(x,t)}{\partial t} = -\dfrac{\hbar^{2}}{2m}(-e^{i\theta(x,t)})(\nabla\theta(x,t)^2-i\nabla ^{2} \theta(x,t)) + U(e^{i\theta(x,t)})$$ I then divided out the $$e^{i\theta(x,t)}$$ terms and it leaves us with: $$-\hbar \dfrac{\partial \theta(x,t)}{\partial t} = \dfrac{\hbar^{2}}{2m}(\nabla\theta(x,t)^2-i\nabla ^{2} \theta(x,t)) + U$$ But the problem is that $$\theta(x,t)$$ is supposed to be real-valued, so why is there still a term with a complex coefficient in the equation? That is, the term $$-i\nabla ^{2} \theta(x,t)$$? I must have made a mistake somewhere, but I checked and double checked all the calculus steps I did (IE, chain rule, product rule, etc.) and I can't seem to find it. Any help is greatly appreciated!

You have two equations: the real part is zero and the imaginay part is zero. You would have had more information of you allowed $$A$$ to vary, then you would be able to interpret what you are doing in physical terms. For the complete algebra see the wiki article on the Madelung equations.

What I wanted to do is derive the time evolution equation of the quantum phase $$\theta(x,t)$$ assuming a constant $$A$$. The first step in my derivation was I made a substitution $$\psi(x,t)=Ae^{i\theta(x,t)}$$:

Schrödinger's equation is actually two coupled differential equations for two real functions. You cannot reduce these to one differential equation for one real function. Assuming this would be possible leads to contradictions, as you have noticed. (So essentially you did a proof by contradiction.)

However, what you can do, is to make the substitution $$\psi(x,t)=A(x,t)e^{i\theta(x,t)}$$ Actually, it turns out to be a little bit easier to make a slightly different substitution: $$\psi(x,t)=\sqrt{\rho(x,t)}e^{iS(x,t)/\hbar}$$

$$\frac{\partial\rho}{\partial t} +\frac{1}{m}\nabla (\rho\nabla S) =0 \tag{1}$$ $$\frac{\partial S}{\partial t} +\frac{1}{2m}(\nabla S)^2 +U-\frac{\hbar^2}{2m}\frac{\Delta\sqrt{\rho}}{\sqrt{\rho}} =0 \tag{2}$$ You see that these are coupled differential equations for $$\rho(x,t)$$ and $$S(x,t)$$.
But in the limit $$\hbar\to 0$$ (the classical limit of quantum mechanics) equation (2) approximately becomes an uncoupled differential equation for $$S(x,t)$$ alone. $$\frac{\partial S}{\partial t} +\frac{1}{2m}(\nabla S)^2 +U \approx 0 \tag{3}$$ which is the Hamilton-Jacobi equation known from classical mechanics.
• Great answer, but I don't think there is a contradiction in the OP. It simply implies that $\nabla^2\theta = 0$, which is true (from the continuity equation) when $A$ is constant, which is what was assumed. Commented Jul 25 at 12:42