Derivation of current density in quantum mechanics

[Lojzek]

I read some derivations of current density from the quantum equations of motion (like Scrödinger's and Klein-Gordon's). They derive an equation with the same form as continuity equation:

div(A)+dB/dt=0

Then they conclude that A=current density and B=density.

However there are non-zero vector fields that have zero divergence, so we could add any of them to A and the continuity equation would still be true. How can we know that A is the true expression for current density?

 

[genneth]

You don't really. In the case of Schroedinger, you can treat it as a classical complex field and apply Lagrangian methods. The field, being complex, has global phase symmetry, which by Noether's theorem identifies a pair of charge and currents. The charge is of the form \psi^* \psi which we postulated is the probability density over position, and so we identify the current identified as the probability current.

 

[lbrits]

I read some derivations of current density from the quantum equations of motion (like Scrödinger's and Klein-Gordon's). They derive an equation with the same form as continuity equation:

div(A)+dB/dt=0

Then they conclude that A=current density and B=density.

However there are non-zero vector fields that have zero divergence, so we could add any of them to A and the continuity equation would still be true. How can we know that A is the true expression for current density?

The Lagrangian treatment is done here: http://www.physics.thetangentbundle.net/wiki/Quantum_mechanics/Lagrangian_formulation

To answer your question, one normally derives the continuity equation from saying \mathbf{J} is a current density and \rho is a density, and then from using flux arguments. So when you derive an equation like \boldsymbol\nabla\cdot\mathbf{J} + \tfrac{\partial\rho}{\partial t} = 0, you simply recognize it as the continuity equation and make the identification.

Note, the continuity equation isn't a constitutive equation. For example, in the heat equation, the continuity equation must be supplemented by \mathbf{J} \propto \boldsymbol\nabla T. In the case of the wavefunction, your constitutive equation is the thing you derive from the S.E.:

\mathbf{J} = \frac{\hbar}{2mi}\left(\Psi^* \vec \nabla \Psi - \Psi \vec \nabla \Psi^*\right) = \frac\hbar m \mbox{Im}(\Psi^*\vec\nabla\Psi)=\mbox{Re}(\Psi^* \frac{\hbar}{im} \vec\nabla \Psi)

 

[TriKri]

We have

\mathbf{\hat{p}}=m\mathbf{\hat{v}} \Leftrightarrow \mathbf{\hat{v}}=\frac{\mathbf{\hat{p}}}{m} = \frac{-i\hbar\mathbf{\nabla}}{m}=\frac{\hbar\mathbf{\nabla}}{im}

and if we assume that the wave function is normalized, we will get the expectation value of the particle's velocity by

\left<\mathbf{\hat{v}}\right> = \left<\Psi\left|\mathbf{\hat{v}}\right|\Psi\right> = \left<\Psi\left|\frac{\hbar}{im}\mathbf{\nabla}\right|\Psi\right> = \int\limits_{}^{}\Psi^*\frac{\hbar}{im}\mathbf{\nabla}\Psi\, dV.

where dV is a small volume element. If we define

\mathbf{J_\mathbb{C}} = \Psi^*\frac{\hbar}{im}\mathbf{\nabla}\Psi

so that

\mathbf{J} = \mbox{Re}(\mathbf{J_\mathbb{C}})

we get the simple relation

\left<\mathbf{\hat{v}}\right> = \int\limits_{}^{}\mathbf{J_\mathbb{C}}\, dV.

I think this is a fundamental property that you want to have.

Let's continue with looking at how the expectation value of the position changes with time. It is well known that the time derivative of the expectation value of a time-independent operator \hat{A} easily can be obtained using the Hamiltonian, namely

\frac{d}{dt}\left<\hat{A}\right> = (i\hbar)^{-1}\left<\left[\hat{A},\,\hat{H}\right]\right>

so in our case we have

\frac{d}{dt}\left<\hat{x}\right> = (i\hbar)^{-1}\left<\left[\hat{x},\,\hat{H}\right]\right> = (i\hbar)^{-1}\left<\left[\hat{x},\,\frac{\mathbf{\hat{p}}^2}{2m}+V(\mathbf{r},\,t)\right]\right>

Since \hat{x} commutes with V(\mathbf{r},t) and also with \hat{p}_y and \hat{p}_z, we have

\frac{d}{dt}\left<\hat{x}\right> = \frac{1}{2mi\hbar}\left<\left[\hat{x},\,\hat{p}_x^2}\right]\right> = \frac{1}{2mi\hbar}\left<\left[\hat{x},\,\hat{p}_x}\right]\hat{p}_x + \hat{p}_x\left[\hat{x},\,\hat{p}_x}\right]\right>

We know that \left[\hat{x},\hat{p}_x\right]=i\hbar, so we get

\frac{d}{dt}\left<\hat{x}\right> = \frac{1}{2mi\hbar}\left<i\hbar\hat{p}_x + \hat{p}_x i\hbar\right> = \frac{\left<\hat{p}_x\right>}{m}

Since we will get the same result with the expectation values of y and z, we obtain

\frac{d}{dt}\left<\mathbf{\hat{r}}\right> = \frac{\left<\mathbf{\hat{p}}\right>}{m} = \left<\mathbf{\hat{v}}\right>

Using the result from above, we get

\frac{d}{dt}\left<\mathbf{\hat{r}}\right> = \int\limits_{}^{}\mathbf{J_\mathbb{C}}\, dV.

Or, if we would so like

\frac{d}{dt}\left<\mathbf{\hat{r}}\right> = \left<\mathbf{\hat{v}}\right> = \int\limits_{}^{}\mathbf{J_\mathbb{C}}\, dV.

 

[andremelzi]

\mathbf{J} = \frac{\hbar}{2mi}\left(\Psi^* \vec \nabla \Psi - \Psi \vec \nabla \Psi^*\right) = \frac\hbar m \mbox{Im}(\Psi^*\vec\nabla\Psi)=\mbox{Re}(\Psi^* \frac{\hbar}{im} \vec\nabla \Psi)

How can I show this?

 

[tom.stoer]

See post #3; just use the Lagrangian specified in http://www.physics.thetangentbundle.net/wiki/Quantum_mechanics/Lagrangian_formulation

From this you can derive
a) the S.E. by applying the Euler-Lagrange equations and
b) probability and current density by applying Noether's theorem for the global U(1) symmetry of the wave function

 

[andremelzi]

My doubt is more basic. How do I show the equalities?

 

[preposterous]

please correct latex equation

 

[tstin]

My doubt is more basic. How do I show the equalities?

If z is a complex number, then z=z_{1}+iz_{2} and its complex conjugate is z^*=z_{1}-iz_{2} Then it is easy to prove z+z^*=z_{1}+iz_{2}+z_{1}-iz_{2}=2z_{1} and from this is is obvious that z_{1}=\mbox{Re}\left(z\right)=\frac{z+z^*}{2}

Similarly, z_{2}=\mbox{Im}\left(z\right)=\frac{z-z^*}{2}

Now consider a complex-valued function \psi and some operator whose result is also complex valued: \hat{O}\psi. Now let our complex number z be z=\psi^*\left(\hat{O}\psi\right). Then from the above identity, \frac{1}{2}\left[\psi^*\left(\hat{O}\psi\right)+\psi\left(\hat{O} \psi \right)^*\right]=\mbox{Re}\left[\psi^*\left(\hat{O}\psi\right)\right]

Now we substitute the general operator for the quantum mechanical momentum operator, \hat{O}=\frac{\hbar}{i}\nabla

Then we find \hat{O}\psi=\frac{\hbar}{i}\nabla\psi and \left(\hat{O}\psi\right)^*=-\frac{\hbar}{i}\nabla\psi^*

Then it follows that \mbox{Re}\left[\psi^*\left(\hat{O}\psi\right)\right]=\mbox{Re}\left[\psi^*\frac{\hbar}{i}\nabla\psi\right]=\frac{1}{2}\left[\psi^*\frac{\hbar}{i}\nabla\psi-\psi\frac{\hbar}{i}\nabla\psi^*\right]=\frac{\hbar}{2i}\left[\psi^*\nabla\psi-\psi\nabla\psi^*\right]

Now if z=\left(\psi^*\nabla\psi\right)

then z^*=\psi\nabla\psi^*

So we get the second identity (which has a typo in the previous post): \frac{\hbar}{im}\mbox{Im}\left[\psi^*\nabla\psi\right]=\frac{\hbar}{2mi}\left[\psi^*\nabla\psi-\psi\nabla\psi^*\right]

Put it all together and you get the equality
\frac{\hbar}{2mi}\left[\psi^*\nabla\psi-\psi\nabla\psi^*\right]=\frac{\hbar}{im}\mbox{Im}\left[\psi^*\nabla\psi\right]=\mbox{Re}\left[\psi^*\frac{\hbar}{im}\nabla\psi\right]