Continuity Equation in an Electromagnetic Field

[Rubiss]

Homework Statement

Derive the continuity equation for a charged particle in an electromagnetic field

Homework Equations

The time-dependent Schrödinger equation and its complex conjugate are

$$i\hbar\frac{\partial \psi}{\partial t}=\frac{1}{2m}(-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})^{2}\psi+e\phi\psi$$

$$i\hbar\frac{\partial \psi^{*}}{\partial t}=\frac{1}{2m}(+i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})^{2}\psi^{*}+e\phi\psi^{*}$$

The Attempt at a Solution

I proceed in much the same way I would when deriving the continuity equation without a magnetic field. I multiply the top equation by psi-star, the bottom by psi and subtract the bottom equation from the top equation to obtain

$$\frac{\partial \rho}{\partial t} = \frac{-\hbar}{2mi}(\psi^{*}\vec{\nabla}^{2}\psi - \psi \vec{\nabla}^{2} \psi^{*})+\frac{e}{2mc}(2|\psi|^{2}\vec{\nabla} \cdot \vec{A}+\psi^{*}\vec{A} \cdot \vec{\nabla}\psi + \psi \vec{A} \cdot \vec{\nabla}\psi^{*})$$

Now I pull a divergence out of the first quantity in the parentheses on the right, and that becomes the the probability current when there is no magnetic field. Then I use the fact that the divergence of A is zero. This leaves me with

$$\frac{\partial \rho}{\partial t} = -\vec{\nabla} \cdot \vec{j} + \frac{e}{2mc} (\psi^{*}\vec{A} \cdot \vec{\nabla}\psi + \psi \vec{A} \cdot \vec{\nabla}\psi^{*})$$

Now I pull the A out of parentheses:

$$\frac{\partial \rho}{\partial t} = -\vec{\nabla} \cdot \vec{j} + \frac{e}{2mc} \vec{A} \cdot (\psi^{*}\vec{\nabla}\psi + \psi\vec{\nabla}\psi^{*})$$

This becomes

$$\frac{\partial \rho}{\partial t} = -\vec{\nabla} \cdot \vec{j} + \frac{e}{2mc} \vec{A} \cdot (\vec{\nabla}|\psi|^{2})$$

and I can pull the gradient out because del dot A is zero:

$$\frac{\partial \rho}{\partial t} = -\vec{\nabla} \cdot \vec{j} + \frac{e}{2mc} \vec{\nabla} \cdot (\vec{A}|\psi|^{2})$$

Now pull the divergence out of both terms:

$$\frac{\partial \rho}{\partial t} = -\vec{\nabla} \cdot (\vec{j} + \frac{e}{2mc}\vec{A}|\psi|^{2})$$

Now I am very close to the correct answer (I know because the result is on the page "probability current" on Wikipedia). My only problem is that there should NOT be a 2 in the denominator. I have spent a long time trying to find out why this 2 is there. Any help would be appreciated.

 

[TSny]

The time-dependent Schrödinger equation and its complex conjugate are

$$i\hbar\frac{\partial \psi}{\partial t}=\frac{1}{2m}(-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})^{2}\psi+e\phi\psi$$

$$i\hbar\frac{\partial \psi^{*}}{\partial t}=\frac{1}{2m}(+i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})^{2}\psi^{*}+e\phi\psi^{*}$$

I multiply the top equation by psi-star, the bottom by psi and subtract the bottom equation from the top equation to obtain

$$\frac{\partial \rho}{\partial t} = \frac{-\hbar}{2mi}(\psi^{*}\vec{\nabla}^{2}\psi - \psi \vec{\nabla}^{2} \psi^{*})+\frac{e}{2mc}(2|\psi|^{2}\vec{\nabla} \cdot \vec{A}+\psi^{*}\vec{A} \cdot \vec{\nabla}\psi + \psi \vec{A} \cdot \vec{\nabla}\psi^{*})$$

See if you can show that you are missing a couple of factors of 2 above. I think it should be

$$\frac{\partial \rho}{\partial t} = \frac{-\hbar}{2mi}(\psi^{*}\vec{\nabla}^{2}\psi - \psi \vec{\nabla}^{2} \psi^{*})+\frac{e}{2mc}(2|\psi|^{2}\vec{\nabla} \cdot \vec{A}+2\psi^{*}\vec{A} \cdot \vec{\nabla}\psi + 2\psi \vec{A} \cdot \vec{\nabla}\psi^{*})$$

 

[Rubiss]

See if you can show that you are missing a couple of factors of 2 above. I think it should be

$$\frac{\partial \rho}{\partial t} = \frac{-\hbar}{2mi}(\psi^{*}\vec{\nabla}^{2}\psi - \psi \vec{\nabla}^{2} \psi^{*})+\frac{e}{2mc}(2|\psi|^{2}\vec{\nabla} \cdot \vec{A}+2\psi^{*}\vec{A} \cdot \vec{\nabla}\psi + 2\psi \vec{A} \cdot \vec{\nabla}\psi^{*})$$

I think you're right, but I have rewritten my steps many times, and do not get that factor of 2 to show up. Very frustrating.

 

[TSny]

I think you're right, but I have rewritten my steps many times, and do not get that factor of 2 to show up. Very frustrating.

For

$$i\hbar\frac{\partial \psi}{\partial t}=\frac{1}{2m}(-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})^{2}\psi+e\phi\psi$$

try writing it as

$$i\hbar\frac{\partial \psi}{\partial t}=\frac{1}{2m}(-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})\cdot (-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})\psi+e\phi\psi$$

First write out $(-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})\psi$ and then operate on the result with the other $(-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})$. If you encounter $\vec{\nabla}\cdot(\vec{A}\psi)$ then remember that $\vec{\nabla}$ operates on both $\vec{A}$ and $\psi$ according to the product rule.

 

[Rubiss]

For

$$i\hbar\frac{\partial \psi}{\partial t}=\frac{1}{2m}(-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})^{2}\psi+e\phi\psi$$

try writing it as

$$i\hbar\frac{\partial \psi}{\partial t}=\frac{1}{2m}(-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})\cdot (-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})\psi+e\phi\psi$$

First write out $(-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})\psi$ and then operate on the result with the other $(-i\hbar \vec{\nabla} - \frac{e}{c} \vec{A})$. If you encounter $\vec{\nabla}\cdot(\vec{A}\psi)$ then remember that $\vec{\nabla}$ operates on both $\vec{A}$ and $\psi$ according to the product rule.

Ah, yes! That is the crucial part I was missing - del acts on both psi and A.

Thanks so much!