Probability current density in E.M. field

[aa]

Derive the probability current density for a particle
in an electromagnetic field.

(I previously posted this on StackExchange. Please pardon,
but I have been spending a lot of time on this and if anyone
knows exactly what the subtle trick involved is, I
would really appreciate it.)

http://physics.stackexchange.com/qu...lity-current-density-factors-of-2-discrepancy

Timaeus at StackExchange said that I need to be less
cavalier with moving operators around (since everything is
technically an operator and operators do not commute in
general). Yet after trying to be more rigorous
about preserving order, it still doesn't work for me.

We begin:

$$\dfrac{\partial \rho}{\partial t}<br /> =<br /> \dfrac{\partial}{\partial t} (\Psi^* \Psi)<br /> =<br /> \dfrac{\partial \Psi^*}{\partial t} \Psi<br /> +<br /> \Psi^* \dfrac{\partial \Psi}{\partial t}$$

H is, if we substitute in $-i\hbar \nabla$ for $\vec{p}$:

$$H = \frac{1}{2m}(\vec{p} - \frac{q}{c} \mathbf{A}) \cdot<br /> (\vec{p} - \frac{q}{c} \mathbf{A}) + q\phi \\<br /> = \frac{1}{2m}(-i\hbar \nabla - \frac{q}{c} \mathbf{A}) \cdot<br /> (-i\hbar \nabla - \frac{q}{c} \mathbf{A}) + q\phi \\<br /> = \frac{1}{2m}(i\hbar \nabla + \frac{q}{c} \mathbf{A}) \cdot<br /> (i\hbar \nabla + \frac{q}{c} \mathbf{A}) + q\phi$$

Schrödinger equation and its complex conjugate:

$$\dfrac {\partial \Psi}{\partial t} = \dfrac{H\Psi}{i\hbar}$$

$$\dfrac {\partial \Psi^*}{\partial t} = \dfrac{(H \Psi)^*}{-i\hbar}$$

Substitute in:

$$\dfrac {\partial \rho}{\partial t}<br /> =<br /> \dfrac{-1}{i\hbar} [(H\Psi)^* \Psi - \Psi^* (H\Psi)]$$

Substitute in H:

$$\dfrac {\partial \rho}{\partial t}<br /> = \dfrac{-1}{i\hbar}<br /> \{ ( [\frac{1}{2m}(+i\hbar \nabla + \frac{q}{c} \mathbf{A})<br /> \cdot (+i\hbar \nabla + \frac{q}{c} \mathbf{A}) + q\phi]\Psi )^* \Psi \\<br /> - \Psi^*( [\frac{1}{2m}(+i\hbar \nabla + \frac{q}{c} \mathbf{A})<br /> \cdot (+i\hbar \nabla + \frac{q}{c} \mathbf{A}) + q\phi] \Psi)<br /> \}$$

Apply complex conjugate:

$$\dfrac {\partial \rho}{\partial t}<br /> =\dfrac{-1}{i\hbar}<br /> \{ ( [\frac{1}{2m}(-i\hbar \nabla + \frac{q}{c} \mathbf{A})<br /> \cdot (-i\hbar \nabla + \frac{q}{c} \mathbf{A}) + q\phi]\Psi^* ) \Psi \\<br /> - \Psi^*( [\frac{1}{2m}(+i\hbar \nabla + \frac{q}{c} \mathbf{A})<br /> \cdot (+i\hbar \nabla + \frac{q}{c} \mathbf{A}) + q\phi] \Psi)<br /> \}$$

FOIL:

$$\dfrac {\partial \rho}{\partial t}=\dfrac{-1}{i\hbar}<br /> \{ ( [\frac{1}{2m}(i\hbar i\hbar \nabla^2 + (-i\hbar) \nabla \cdot (\frac{q}{c} \mathbf{A})<br /> + (\frac{q}{c} \mathbf{A}) \cdot (-i\hbar) \nabla + \frac{q^2}{c^2} \mathbf{A}^2)<br /> + q\phi]\Psi^* ) \Psi \\<br /> - \Psi^*( [\frac{1}{2m}(i\hbar i\hbar \nabla^2 + i\hbar \nabla \cdot (\frac{q}{c} \mathbf{A})<br /> + (\frac{q}{c} \mathbf{A}) \cdot (i\hbar \nabla) + \frac{q^2}{c^2} \mathbf{A}^2)<br /> + q\phi] \Psi)<br /> \}$$

Multiply everything out:

$$\dfrac {\partial \rho}{\partial t}<br /> =\frac{-i\hbar}{2m}(\nabla^2 \Psi^*) \Psi<br /> + \frac{1}{2m} (\nabla \cdot \frac{q}{c} \mathbf{A}) \Psi^* \Psi<br /> + \frac{1}{2m} (\frac{q}{c} \mathbf{A}) \cdot (\nabla \Psi^*) \Psi<br /> + \frac{-1}{i\hbar} \frac{1}{2m} \frac{q^2}{c^2} \mathbf{A}^2 \Psi^* \Psi \\<br /> + \frac{-1}{i\hbar} \frac{1}{2m} q \phi \Psi^* \Psi \\<br /> + (\Psi^*) \frac{1}{2m} (i\hbar)(\nabla^2 \Psi)<br /> + (\Psi^*) \frac{1}{2m} \nabla \cdot (\frac{q}{c} \mathbf{A}) \Psi<br /> + (\Psi^*) \frac{1}{2m} (\frac{q}{c} \mathbf{A}) \cdot (\nabla \Psi)<br /> + \frac{1}{i\hbar} (\Psi^*) \frac{1}{2m} \frac{q^2}{c^2} \mathbf{A}^2 \Psi \\<br /> + \frac{1}{i\hbar} \frac{1}{2m}(\Psi^*)q\phi \Psi$$

The terms containing $\phi$ and $\frac{q^2}{c^2} \mathbf{A}^2$
cancel and there's a fact that $\Psi \nabla^2 \Psi^* - \Psi^* \nabla^2 \Psi<br /> = \nabla \cdot(\Psi \nabla \Psi^* - \Psi^* \nabla \Psi)$, so

$$\dfrac {\partial \rho}{\partial t}<br /> = \frac{-i\hbar}{2m} \nabla \cdot (\Psi \nabla \Psi^* - \Psi^* \nabla \Psi) \\<br /> + \frac{1}{2m} (\nabla \cdot \frac{q}{c} \mathbf{A}) \Psi^* \Psi<br /> + \frac{1}{2m} (\frac{q}{c} \mathbf{A}) \cdot (\nabla \Psi^*) \Psi<br /> + (\Psi^*) \frac{1}{2m} \nabla \cdot (\frac{q}{c} \mathbf{A}) \Psi<br /> + (\Psi^*) \frac{1}{2m} (\frac{q}{c} \mathbf{A}) \cdot (\nabla \Psi)$$

Note that of the 5 terms, the 2nd and 4th are the same, so

(1) $$\dfrac {\partial \rho}{\partial t}<br /> = \frac{-i\hbar}{2m} \nabla \cdot (\Psi \nabla \Psi^* - \Psi^* \nabla \Psi) \\<br /> + \dfrac{q}{mc} (\nabla \cdot \mathbf{A}) \Psi^* \Psi<br /> + \dfrac{q}{2mc} \mathbf{A} \cdot (\Psi \nabla \Psi^*)<br /> + \dfrac{q}{2mc} \mathbf{A} \cdot (\Psi^* \nabla \Psi)$$

https://en.wikipedia.org/wiki/Probability_current tells us
that the final result should be $\dfrac{\partial \rho}{\partial t}<br /> = - \nabla \cdot \mathbf{j}$ and that

$$\mathbf{j} = \dfrac{1}{2m} [(\Psi^* \mathbf{\hat{p}} \Psi<br /> - \Psi \mathbf{\hat{p}} \Psi^* ) - 2 \frac{q}{c} \mathbf{A} |\Psi|^2]$$

or using $\mathbf{\hat{p}} = -i\hbar \nabla$,

$$\mathbf{j} = \dfrac{1}{2m} [(\Psi^* (-i\hbar \nabla) \Psi<br /> - \Psi (-i\hbar \nabla) \Psi^* ) - 2 \frac{q}{c} \mathbf{A} |\Psi|^2]$$

$$\mathbf{j} = \dfrac{-i\hbar}{2m} (\Psi^* \nabla \Psi<br /> - \Psi \nabla \Psi^* ) - \dfrac{1}{2m} 2 \frac{q}{c} \mathbf{A} |\Psi|^2$$

$$\mathbf{j} = \dfrac{-i\hbar}{2m} (\Psi^* \nabla \Psi<br /> - \Psi \nabla \Psi^* ) - \dfrac{q}{mc} \mathbf{A} |\Psi|^2$$

Applying $\dfrac{\partial \rho}{\partial t} = - \nabla \cdot \mathbf{j}$,

$$\dfrac{\partial \rho}{\partial t}<br /> = - \nabla \cdot [\dfrac{-i\hbar}{2m} (\Psi^* \nabla \Psi<br /> - \Psi \nabla \Psi^* ) - \dfrac{q}{mc} \mathbf{A} |\Psi|^2]$$

$$\dfrac{\partial \rho}{\partial t}<br /> = \dfrac{i\hbar}{2m} \nabla \cdot (\Psi^* \nabla \Psi<br /> - \Psi \nabla \Psi^* ) + \dfrac{q}{mc} \nabla \cdot<br /> (\mathbf{A} |\Psi|^2)$$

Apply an identity about $\nabla$ operating on a scalar times a vector
at https://en.wikipedia.org/wiki/Vector_calculus_identities
$\nabla \cdot (\phi \mathbf{B}) =<br /> \mathbf{B} \cdot \nabla \phi +<br /> \phi (\nabla \cdot \mathbf{B})$ (I changed the letters),

$$\dfrac{\partial \rho}{\partial t}<br /> = \dfrac{i\hbar}{2m} \nabla \cdot (\Psi^* \nabla \Psi<br /> - \Psi \nabla \Psi^* )<br /> + \dfrac{q}{mc} (\mathbf{A} \cdot \nabla (\Psi^* \Psi)<br /> + (\Psi^* \Psi) (\nabla \cdot \mathbf{A}))$$

Product rule:

$$\dfrac{\partial \rho}{\partial t}<br /> = \dfrac{i\hbar}{2m} \nabla \cdot (\Psi^* \nabla \Psi<br /> - \Psi \nabla \Psi^* )<br /> + \dfrac{q}{mc} (\mathbf{A} \cdot (\Psi^* \nabla \Psi)<br /> + \mathbf{A} \cdot (\Psi \nabla \Psi^*)<br /> + (\Psi^* \Psi) (\nabla \cdot \mathbf{A}))$$

$$\dfrac{\partial \rho}{\partial t}<br /> = \dfrac{i\hbar}{2m} \nabla \cdot (\Psi^* \nabla \Psi<br /> - \Psi \nabla \Psi^* )<br /> + \dfrac{q}{mc} \mathbf{A} \cdot (\Psi^* \nabla \Psi)<br /> + \dfrac{q}{mc} \mathbf{A} \cdot (\Psi \nabla \Psi^*)<br /> + \dfrac{q}{mc} (\Psi^* \Psi) (\nabla \cdot \mathbf{A})$$

Rearrange some terms so that we can compare with equation (1):

$$\dfrac{\partial \rho}{\partial t}<br /> = \dfrac{i\hbar}{2m} \nabla \cdot (\Psi^* \nabla \Psi<br /> - \Psi \nabla \Psi^* )<br /> + \dfrac{q}{mc} (\Psi^* \Psi) (\nabla \cdot \mathbf{A})<br /> + \dfrac{q}{mc} \mathbf{A} \cdot (\Psi \nabla \Psi^*)<br /> + \dfrac{q}{mc} \mathbf{A} \cdot (\Psi^* \nabla \Psi)$$

Now we're very close to equation (1),

(1) $$\dfrac {\partial \rho}{\partial t}<br /> = \frac{-i\hbar}{2m} \nabla \cdot (\Psi \nabla \Psi^* - \Psi^* \nabla \Psi) \\<br /> + \dfrac{q}{mc} (\nabla \cdot \mathbf{A}) \Psi^* \Psi<br /> + \dfrac{q}{2mc} \mathbf{A} \cdot (\Psi \nabla \Psi^*)<br /> + \dfrac{q}{2mc} \mathbf{A} \cdot (\Psi^* \nabla \Psi)$$

but are off by some factors of 2.
Do you see the error in my steps?

 

[TSny]

In your "FOIL" step (see below), think about whether or not the Del operator circled in red should operate on $\Psi^{*}$ as well as $\mathbf{A}$.

 

[aa]

It works! "Incorporating the $\Psi$ and $\Psi^*$ first", before doing
the $\nabla$, creates a "6th term", that fills in the missing piece(s).
THANK YOU TSny!