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Interesting equivalent to Schrödinger equation

  1. Apr 18, 2010 #1
    While messing around with the Schrödinger equation on paper, I found an interesting, elegant way of expressing it. Let [tex]P[/tex] be the probability density [tex]|\Psi |^2[/tex], and let [tex]\vec Q[/tex] be a real-valued vector field. [tex]\vec F[/tex] is a vector field describing the forces acting on the system when in a given configuration. Then,

    [tex]\frac{\partial P}{\partial t}=-\nabla \cdot \vec Q[/tex]
    [tex]\frac{\partial \vec Q}{\partial t}=\frac{P\vec F}m-\frac{\vec Q\nabla \cdot \vec Q + \frac{\hbar^2}{4m^2}[\nabla,P]\Delta P}P[/tex]

    Sorry if this looks obvious, but I haven't seen this mentioned in any book. Hopefully, my calculations are correct. It's evident that [tex]\vec Q[/tex] is a velocity density. The first equation just says that the probability density decreases as the wave function expands about a point. The term [tex]-\frac{\vec Q}P \nabla \cdot \vec Q[/tex] represents the flow of velocity density in the direction of the velocity itself. However, I'm unsure of how to physically interpret the last term, which is a strange looking one. Does it simply mean that the probability density tends to accelerate away from concentrations of probability density?
  2. jcsd
  3. Apr 18, 2010 #2
  4. Apr 18, 2010 #3
    Well, any interpretation that could be made of the original Schrödinger equation could also be made of these, I suppose, since they're directly derived from it. There is no notion of an "actual" configuration inherent in these equations. However, my physics is a bit rusty so I don't remember if the version of the Schrödinger equation that I've been working with works equally for any number of particles. If it doesn't, then this is for one isolated particle only.

    If someone is interested in checking the correctness of my rearranging, here is how I arrived at this. Define the variables [tex]P[/tex] and [tex]\phi[/tex] so that [tex]\Psi =\sqrt P e^{i\phi}[/tex]. Then we get:
    [tex]d\Psi =\left(\frac{dP}{2\sqrt P}+id\phi\sqrt P\right)e^{i\phi}[/tex]
    [tex]d^2\Psi =\left(\frac{d^2 P}{2\sqrt P}-\frac{\left(dP\right)^2}4 P^{-\frac 3 2}-\left(d\phi\right)^2\sqrt P +\frac{idPd\phi}{\sqrt P}+id^2\phi\sqrt P\right)e^{i\phi}[/tex]
    [tex]dP=2\sqrt P\left(d\Psi e^{-i\phi}\right)_{Re}[/tex]
    [tex]d\phi =\frac{\left(d\Psi e^{-i\phi}\right)_{Im}}{\sqrt P}[/tex]
    Dividing the Schrödinger equation by [tex]i\hbar[/tex], we get:
    [tex]\frac{\partial \Psi}{dt}=\frac{i\hbar}{2m}\Delta\Psi -\frac{iV\Psi}\hbar[/tex]
    [tex]\frac{\partial P}{dt}=-\frac{\hbar}m\left(\nabla P\nabla\phi -P\Delta\phi\right)=-\frac{\hbar}m\nabla\cdot\left(P\nabla \phi}\right)[/tex]
    [tex]\frac{\partial \phi}{dt}=\frac\hbar{2m}\left(\frac{\Delta P}{2P}-\frac{\left(\nabla P\right)^2}{4P^2}-\left(\nabla\phi\right)^2\right)-\frac V\hbar[/tex]
    Now let [tex]\vec Q=\frac\hbar mP\nabla\phi[/tex]. Then we end up with the equations above.

    (In every place I have used the symbols [tex]\nabla[/tex] and [tex]\Delta[/tex] above, I only mean the derivatives in the spatial directions.)

    Okay, so after some searching it turns out that [tex]\vec Q[/tex] is apparently called the probability current or probability flux and is usually denoted by the symbol [tex]\vec j[/tex], although I couldn't find it explicitly written out in terms of [tex]P[/tex] anywhere.
    Last edited: Apr 18, 2010
  5. Apr 18, 2010 #4
    Actually, by looking at this equation, you can see that it is of the same form as any continuity equation. Therefore, the [tex]\vec Q[/tex] is equal to the probability current density. It has the physical meaning that [tex]\vec Q \cdot \hat{\mathbf{n}}[/tex] gives the probability of a particle crossing a unit surface perpendicular to the unit vector [tex]\hat{\mathbf{n}}[/tex] in unit time. It is more customary to denote [tex]\vec Q \equiv \vec J[/tex].

    By doing the standard transformations to the Schroedinger equation for a single particle in an external potential [tex]V(\vec r)[/tex], one can show that:

    \vec J = \frac{\hbar}{2 m \iota} \left( \Psi^{\ast} \nabla \Psi - \nabla \Psi^{\ast} \Psi\right)
  6. Apr 18, 2010 #5


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    Your derivation is a little odd-looking, but I am pretty sure that your Q is simply the probability current (check the derivation in the wikipedia page if you like).

    The first equation in your original post is the continuity equation, which I believe was originally derived by Madelung in the 1920's. The second equation is odd-looking, because your derivation is non-standard, but if you haven't made any errors, then it should be the deBroglie-Bohm (or dBB) equation describing the velocity of a quantum particle. The second term that you were puzzling over is called the "quantum force", and is what causes the "randomness" in the trajectories of quantum particles. I hadn't really thought about it in the terms you used (I am still kinda new to dBB), but I guess the quantum force could accurately be described as opposing the local buildup of probability density.
  7. Apr 18, 2010 #6
    If we have a fluid with a flow velocity field [tex]\vec{u}(\vec{r}, t)[/tex] and there is some scalar charactersitic (mass, charge, probability, etc) described by a density distribution [tex]\rho(\vec{r}, t)[/tex], then, the current density for the very same property is given by:

    \vec{J} = \rho \vec{u}

    Using the formulas for probability density ([tex]\rho = \Psi^{\ast} \Psi[/tex] and probability current density (see the quoted formula), we can define a flow velocity:

    \vec{u} = \frac{\vec{J}}{\rho} = \frac{\hbar}{2 m \iota} \left( \frac{\nabla \Psi}{\Psi} - \frac{\nabla \Psi^{\ast}}{\Psi^{\ast}}\right) = \frac{\hbar}{m} \Im{\frac{\nabla \Psi}{\Psi}}
    Last edited: Apr 18, 2010
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