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Current in Dirac equation

  1. Jan 6, 2012 #1
    Hello,
    I have a question concerning the current in the Dirac equation and its corresponding operator. One can construct a current density that is
    [itex]\textbf{j}^{i} = \psi^{\dagger}\gamma^{i}\psi[/itex]
    If I want to have the current, I will have to integrate:
    [itex]I = \oint \textbf{j} \cdot \textbf{n} \, dA [/itex]
    With Gauss' theorem this should be:
    [itex]I = \int \text{div} \textbf{j} \, d^{3}x = \int \text{div} (\psi^{\dagger}\gamma \psi) \, d^{3}x [/itex]

    Okay, but now I've found in the literature, that the current operator is [itex]\hat{\textbf{j}} = \hat{\gamma} [/itex].

    But when I calculate an average value with this operator, this is
    [itex]I = \left\langle \psi | \hat{\gamma} | \psi \right\rangle = \int \psi^{\dagger}\gamma \psi d^{3}x [/itex]
    without this divergence.

    What's right? And why is one point wrong?

    Thank you all for comments,
    Melvin
     
  2. jcsd
  3. Jan 6, 2012 #2

    Avodyne

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    What you "found in the literature" is wrong.
     
  4. Jan 6, 2012 #3
    I assume, by "current", that you mean probability current density. I think you firstly need some [itex]\gamma^0[/itex]s in there. Then, if you're planning to sandwich it between states [itex]\left\langle \psi | j | \psi \right\rangle[/itex], then [itex] \gamma^0 \hat{\gamma}[/itex] is the right operator. [itex] \psi^{\dagger}\gamma^0 \gamma^i \psi [/itex] is the corresponding operator whose vacuum expectation value is that of the gammas between two psis (because that's basically what it is).*

    As for the question of the divergence, remember what that current means: the integral of the flux of the current through a closed surface gives the rate of change of probability that a particle in the state psi is found in the volume bounded by that surface.

    Hope that helps.

    *Edit: I'm not sure exactly if the [itex] \gamma^0[/itex] belongs in the operator or in the bra [itex] \left \langle \psi \right|[/itex]- I've only ever thought about these things in the context of quantum field theory, where you invariably just think about operators acting on the vacuum- but I hope the important points are clear.
     
    Last edited: Jan 6, 2012
  5. Jan 6, 2012 #4
    @muppet: You are right with the missing [itex]\gamma^{0}[/itex]. So the probability current density is [itex]\textbf{j}^{i} = \psi^{\dagger}(\textbf{r})\gamma^{0}\gamma^{i}\psi(\textbf{r})[/itex].

    The operator for the probability current density (or one component) should be something with Delta distributions in my opinion to get rid of the integrals:
    [itex]\hat{\textbf{j}^{i}} = \hat{\gamma^{0}\gamma^{i}} \delta(\textbf{r}-\textbf{r'})\delta(\textbf{r'}-\textbf{r''})[/itex]

    With that one could write
    [itex]\textbf{j}^{i}(\textbf{r}) = \left\langle \psi(\textbf{r'}) | \hat{\gamma^{0}\gamma^{i}} \delta(\textbf{r}-\textbf{r'})\delta(\textbf{r'}-\textbf{r''}) | \psi(\textbf{r''}) \right\rangle = \int \psi^{\dagger}(\textbf{r'})\gamma^{0}\gamma^{i} \psi(\textbf{r''}) \delta(\textbf{r}-\textbf{r'})\delta(\textbf{r'}-\textbf{r''}) \, d^{3}\textbf{r'} \, d^{3}\textbf{r''} = \psi^{\dagger}(\textbf{r})\gamma^{0}\gamma^{i} \psi(\textbf{r}) [/itex]
    and find again the expression from above. Is this right? Or am I doing total crap?

    Perhaps it's a problem of interpretation, but I think of this probability current density as of an electrical current density. But when I integrate this electrical current density over a closed surface I get the total current "I" (ingoing or outgoing depending on the sign).
    The question is, if there is something like this total current in the electrical case for the Dirac equation.
    I would say yes and this total current "I" should be, analog to the EM case,
    [itex]I = \oint \textbf{j} \cdot \textbf{n} \, dA [/itex].

    The next question arising from this was, if there is an operator, sandwiched between two spinors, whose expectation value is exactly this total current "I"? I wasn't able to find one because of this appearing divergence mentioned in my original post.
     
  6. Jan 7, 2012 #5
    Firstly, you don't need to put the delta functions in by hand. (If a bit of thought doesn't make this clear, let us know if you're studying this from the perspective of relativistic quantum mechanics or field theory, and what you're reading?)

    There's nothing wrong with relating this back to other current densities, mentally swapping "charge" for "probability". (Caveat: You need to be careful what you're talking about in a relativistic theory, owing to the possibility of producing new particle-antiparticle pairs.) I don't understand what operator you're trying to find, though. What's wrong with [itex]I = \oint \textbf{j} \cdot \textbf{n} \, dA[/itex]? You'll never be able to write down such an operator without specifying the closed surface S, except in trivial cases where your probability density is time independent and everything vanishes.
     
  7. Jan 7, 2012 #6
    Thank you for your new answer.
    I'm studying this from the perspective of relativistic quantum mechanics. I'm reading a German book from Wachter called "Relativistische Quantenmechanik" and there isn't given an operator neither for current density nor for total current (that there is none for the total current is plausible from your argumentation). So I searched in the WWW and found this:
    http://physics-quest.org/Book_Chapter_Dirac_Operators.pdf
    On page 5 there is derived a total (charge) current operator. Because it's not the kind of total current I searched for, because it's still a vector. It's an integration over the complete space of every single component of the current density. So what does this mean? Or is this already field theory?

    The delta functions were my own idea to reproduce the current density by using an operator. I thought about it why it's not necessary to put in them by hand. My idea: The free solution of the Dirac equation has this plane wave factor [itex]e^{i\textbf{k}\textbf{r}}[/itex]. Multiplying the states would give [itex]e^{i\textbf{k}(\textbf{r'}-\textbf{r})}[/itex]. Integrating this would yield one delta function, which makes the second integration trivial and also gives the right resultat. Right?
     
  8. Jan 7, 2012 #7

    vanhees71

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    I'm not sure what this book is about since I don't know it, but if it's a German book with the title "Relativistische Quantenmechanik" (relativistic quantum mechanics, which means something different than "Relativistische Quantentheorie", relativistic quantum theory), I fear it's one written in this old fashioned ways as if one could understand relativistic quantum theory in a heuristic way as in the non-relativistic quantum theory as "wave mechanics", i.e. describing a single interacting particle with complex-valued "wave function". This is what in German tradition is called "Relativistische Quantenmechanik", but it's old fashioned and in a way unnecessarily complicated since the interpretation of this paradigm is not appropriate beyond situations where the non-relativistic description is still a good approximation.

    The reason for this is that in the opposite case of relativistic interacting particles (i.e., particles with energies greater than the mass of these particles) there is always some probability to create new particles (or photons, which is particularly easy since photons are massless). Thus, a one-particle description is inappropriate, and from this all the trouble with "relativistic quantum mechanics" results.

    The good thing is that this is well understood nowadays and there are excellent books on quantum field theory available. My favorite is S. Weinberg, The Quantum Theory of Fields, Cambridge University Press (3 vols.). There it is explained in detail, how one constructs relativistic many-body theory as a local quantum field theory from the underlying symmetry principles of relativistic space-time.

    It turns out that the Dirac equation is valid for a fermionic four-component field operator, describing particles and anti-particles with spin 1/2, and that the current [itex]\bar{\psi} \gamma^{\mu} \psi[/itex] is an operator, describing a charge current (e.g., the electromagnetic current of particles carrying electric charge like elektrons). The time component of this current operator is not positive definite (because of the Fermion nature of these particles) and thus cannot be given the meaning of a probability current. There's of course no problem in interpreting it as a current (e.g., the electromagnetic current of charged particles in QED). At the same time, interpreting the Dirac equation as an equation for field operators of a relativistic many-body theory, solves another problem of the naive one-particle interpretation in terms of c-number fields: It makes the Hamilton operator of the theory bounded from below, i.e., it ensures the existence of a stable ground state.

    So, I'd suggest to take another book on relativistic quantum field theory and learn the modern way right from the beginning. It's, of course, a difficult subject, but for sure less difficult than the awkward way of trying to interpret the c-number Dirac field as a relativistic one-particle wave mechanics. This awkward way has been gone by the pioneers of quantum mechanics, mostly by Dirac himself, and it lead to a very complicated form of quantum electrodynamics, the so called hole theory. All these troubles are avoided by the equivalent quantum field theory point of view.
     
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