4
$\begingroup$

The Schrodinger equation is usually written in complex form $$ i \hbar \frac{\partial\psi}{\partial t} = \left( -\frac{\hbar^2}{2m} \nabla^2 + V \right) \psi $$ with complex solution $\psi$. It's well-known that this can equivalently be written as two coupled real-valued equations by taking the polar decomposition of the solution $\psi = R \exp (i S/\hbar)$, inserting it into the Schrodinger equation above, and separating the real and imaginary parts of the equation. Defining $\rho = R^2$ and $v = (\nabla S)/m$, this results in a continuity equation $$ \partial\rho/\partial t + \nabla \cdot (\rho v) $$ and something that looks like the Hamilton-Jacobi equation $$ -\frac{\partial S}{\partial t} = \frac{\| \nabla S \|^2}{2m} + V + Q $$ except for the presence of the so-called quantum potential $$ Q = -\frac{\hbar^2}{2m} \frac{\nabla^2 R}{R} .$$ This formulation was studied by Bohm and others.

I'm interested in whether there's an analogous approach that can be taken for the Pauli equation $$ \left[ \frac{1}{2m} \left( (\hat{\mathbf{p}} - q \mathbf{A})^2 - q\hbar \mathbf{\sigma} \cdot \mathbf{B} \right) + q\phi\right] |\psi \rangle = i \hbar \frac{\partial}{\partial t} | \psi \rangle $$ where $\mathbf{\sigma}$ is the usual vector of Pauli matrices, $\mathbf{A}$, $\mathbf{B}$, $\phi$, and $q$ are the vector potential, the magnetic field, the scalar potential, and the electric charge, and the momentum operator $\hat{\mathbf{p}}$ has its usual form. Because the solutions are two-component entities and the Pauli matrices themselves introduce complex numbers, it's not obvious to me what would be the analogous approach to taking the polar decomposition of the solution as was done for the Schrodinger equation.

Is there are standard approach to this that yields something analogous to the quantum potential in the Schrodinger equation case?

$\endgroup$
3
  • $\begingroup$ If there is no magnetic field (time-reversal symmetry) you can show that all the eigenfunctions can be chosen to be real. That's not possible in the presence of magnetic fields. $\endgroup$
    – Mauricio
    Commented Nov 20 at 22:14
  • $\begingroup$ Linked for spherical coordinates, no reality concerns. $\endgroup$ Commented Nov 20 at 23:15
  • $\begingroup$ Because the Pauli equation is 2-complex-component, there would be 4 real-valued parameters, and it would be a lot more unwieldy to work with. $\endgroup$ Commented Nov 21 at 4:44

1 Answer 1

6
$\begingroup$

Is there are standard approach to this that yields something analogous to the quantum potential in the Schrodinger equation case?

You should check the book The Undivided Universe, by D. Bohm and B.Hiley, in chapter 10 they develop what you are asking for. Basically they call it Bohm Schiller Tiomno model (BST). The original reference for the model is this article.


We need to consider the bidimensional spinor

$$ \psi= \begin{pmatrix} \psi_1\\ \psi_2 \end{pmatrix} $$

The probability density and velocity are defined as

$$ P = |\psi_1|^2 + |\psi_2|^2 $$

and

$$ \vec v = { |\psi_1|^2\nabla S_1/m + |\psi_2|^2 \nabla S_2/m \over |\psi_1|^2 + |\psi_2|^2 } $$

for which the continuity equation follows

$$ {\partial P \over \partial t} + \nabla \cdot (P \vec v) = 0 $$

Now, at this point we should consider $\psi_j = R_je^{iS_j}$. There will be four real quantities $R_1,R_2,S_1,S_2$. Assuming that the particle is a spinning object with some orientation given by three Euler angles $(\theta,\phi,\chi)$, we write the spinor as

$$ \psi = \rho^{1/2} \begin{pmatrix} \cos(\theta/2)\exp[i(\chi+\phi)/2]\\ i\sin(\theta/2)\exp[i(\chi-\phi)/2]\\ \end{pmatrix} $$

By choosing $(\theta,\phi)$ to represent the orientation of the angular momentum vector and $\chi$ as the angle of rotation. For spin one-half particle, we get

$$ \vec S = \frac 12 \frac{\psi^\dagger \vec \sigma \psi}{\rho} $$

and the angle of rotation is determined by $S_1$ and $S_2$

$$ \chi = \frac{S_1+S_2}{2}. $$

At this moment, following the book I mentioned in the beggining, they say that it is possible to obtain a generalization of the Hamilton-Jacobi for $S = \chi$. They first write the velocity

$$ \vec v = {1\over 2m}\left (\nabla S +\cos \theta \nabla \phi \right) - e\vec A $$

and then they show that it is possible to get the following equation (I didn't tried to)

$$ {\partial S\over \partial t} -\cos \theta \vec v \cdot \nabla \phi + \frac{1}{2m} \left (\nabla S +\cos \theta \nabla \phi - e\vec A \right )^2 - \frac{1}{2m}{\nabla^2 \rho^{1/2} \over \rho^{1/2}} + V + \frac{1}{2m} \left [ (\nabla \theta)^2 + \sin^2 \theta (\nabla \phi)^2 \right ] + \mu \vec S \cdot \vec B = 0. $$


I personally think that the BST model is an interesing exercise of non-relativistic QM, but the interpretation of QM they propose in the book is not considered mainstream science. I'm just using this as a reference to answer the question.

In the BST paper they explore in more details what conditions should be matched in a classical theory to achieve the same behavior of the described system. They say:

We conclude then that in the Pauli theory, the angular momentum , S, is always pointing along the principal axis (1) of the body. Such a special orientation of the angular momentum cannot be maintained for the most general kind of torque that may act on the body. In the next section and in a subsequent paper, we shall see, however, that the Pauli equation implies special kind of quantum-mechanical torque that permits this condition to be maintained as a consistent subsidiary condition.

So, analogously with the "quantum potential" that naturally appears when we parametrize the Schrodinger equation with real parameters, we have a quantum torque. To explore it in details, they develop a model describing a charged classical fluid and compare it with the Pauli equation. In the papers they don't appeal to any specific interpretation of Quantum theory, but appeal to Madelung fluid model.

$\endgroup$
1
  • $\begingroup$ Thanks. I will take me a while to get a copy of the book, so I appreciate you outlining the main steps. $\endgroup$
    – Brick
    Commented Nov 21 at 14:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.