2
$\begingroup$

To derive the probability current density for a particle in an electromagnetic field, we calculate $\dfrac{\partial \rho}{\partial t} = \dfrac{\partial}{\partial t} (\Psi^* \Psi) = \dfrac{\partial \Psi^*}{\partial t} \Psi + \Psi^* \dfrac{\partial \Psi}{\partial t}$

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

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

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:

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

Substitute in $H$:

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

Apply complex conjugate:

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

FOIL:

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

Multiply everything out:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Product rule:

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

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

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

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

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

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

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

$\endgroup$

closed as off-topic by Sebastian Riese, ACuriousMind, Javier, Kyle Kanos, Martin Sep 17 '15 at 11:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 2
    $\begingroup$ I'm voting to close this question as off-topic because this homework-like question is of the "check-my work" type which is explicitly off-topic. $\endgroup$ – Sebastian Riese Sep 16 '15 at 17:11
  • 1
    $\begingroup$ Ask about "multiply it out" and I'd write that first term as $\Psi \partial_t (\Psi^*)$ instead of $(\partial_t (\Psi^*))\Psi$ to avoid confusing yourself about which operators act on which. Or just not throw away a parenthesis, if it is there, you need it. And learning to check your work is essential, everyone makes mistakes sometimes, but you have to be able to go through it and check it to make it right. $\endgroup$ – Timaeus Sep 16 '15 at 18:15
  • $\begingroup$ @Timaeus: I didn't edit the original question. I just got back to looking at your latest comments today. I appreciate very much your analysis. $\endgroup$ – a00 Sep 17 '15 at 13:40
1
$\begingroup$

Do you see an error in my steps?

Yes, I see a conceptual error. So I'll talk about that.

$\dfrac{\partial \rho}{\partial t} = \dfrac{\partial}{\partial t} (\Psi^* \Psi) = \dfrac{\partial \Psi^*}{\partial t} \Psi + \Psi^* \dfrac{\partial \Psi}{\partial t}$

should be $\dfrac{\partial \rho}{\partial t} = \dfrac{\partial}{\partial t} (\Psi^* \Psi) = \Psi \dfrac{\partial }{\partial t} \Psi^* + \Psi^* \dfrac{\partial }{\partial t}\Psi$

because $\partial_t$ is an operator and so we should only have things to the right of it that it acts on. Because you might remember that now but later you will forget and you will then make a mistake. Plus there are going to just be too many parenthesis anyway.

Now look at

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

which equals $$\frac{1}{2m}\sum_j\left((i\hbar \partial_j +\frac{q}{c} A_j)(i\hbar \partial_j +\frac{q}{c} A_j)\right) + q\phi.$$

And these are operators the one on the right acts on things then the next acts on the outcome of that and so on. Think of them like matrices. Don't imagine any vectors or dot products don't start thinking something (anything) is a scalar and don't start changing the order of anything. If you couldn't justify it if it were a matrix, then don't do it. Just treat them all like matrices. Even a scalar field like $A_j$ first it has to multiply something then a derivative has to act on that product (with the product rule) otherwise you are doing it wrong.

$\endgroup$
  • $\begingroup$ I tried to be more rigorous about keeping order of multiplications as you said, but could not fix the problem. I posted the question to physicsforums. TSny points out that in going from the expression after "Apply complex conjugate:" to the expression after "FOIL:" I should be applying some of the $\nabla$ operators to both the $\mathbf{A}$ as well as the following $\Psi$ or $\Psi^*$. That essentially adds 2 terms which fill in the "missing stuff". $\endgroup$ – a00 Sep 19 '15 at 13:44
  • $\begingroup$ I wouldn't be surprised if it turns out that my finished derivation turns out to be wrong when viewed rigorously as you describe. But this (TSny's fix) is expedient for a dilettante. Thank you VERY much for kindly going through this long question and for explaining the critical correction to the way I view quantum mechanics math. $\endgroup$ – a00 Sep 19 '15 at 13:46
  • 1
    $\begingroup$ @a00 When you apply from right to left then the product rule gives you the extra terms. At SE we have policies against check my work questions so I was trying to stick to the concept that I thought was causing the problem. During foil you might still have add enough parenthesis, but multiply it out you definitely don't. $\endgroup$ – Timaeus Sep 19 '15 at 15:01
  • $\begingroup$ Oh, so it's the right to left associativity of operator math that is the underlying reason. You have added the Why to the How; thank you. I understand now that check my work posts are not what this StackExchange is intended for. $\endgroup$ – a00 Sep 21 '15 at 15:59
  • 1
    $\begingroup$ @a00 And that's why I wasn't as detailed about your error as you might have hoped, but now you know an important concept when dealing with operators and particularly with differential operators. Make sure each one first acts, the one on the right acting first then the one next to it and also acting on the outcome of the brighter ones. Thus needing to use product rules. $\endgroup$ – Timaeus Sep 21 '15 at 16:05

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