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Continuity Equation in an Electromagnetic Field

  1. Feb 20, 2013 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    The time-dependent Schrodinger equation and its complex conjugate are

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

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

    3. 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

    [tex] \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^{*})[/tex]

    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

    [tex] \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^{*})[/tex]

    Now I pull the A out of parentheses:

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

    This becomes

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

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

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

    Now pull the divergence out of both terms:

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

    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.
     
  2. jcsd
  3. Feb 20, 2013 #2

    TSny

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

    [tex] \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^{*})[/tex]
     
  4. Feb 21, 2013 #3
    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.
     
  5. Feb 21, 2013 #4

    TSny

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    For

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

    try writing it as

    [tex]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 [/tex]

    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.
     
  6. Feb 21, 2013 #5

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

    Thanks so much!
     
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