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Mathematical signature of electronmagnetic field and electron's deBroglie functions

  1. Jul 10, 2007 #1
    I asked a similar question ihttps://www.physicsforums.com/showthread.php?t=171030", but got the suggestion to come here:

    The electromagnetic field, as solution of the Maxwell equations, has the signature [itex]\def\R{\mathbb{R}}f:\R^4\to \R^3\times \R^3[/itex], indicating that each point in 4D spacetime gets assigned a pair of vectors for the electric and the magnetic component respectively?

    Now I wonder what the mathematical signature of the deBroglie function for an electron is? I would expect, that it also starts like [itex]\def\R{\mathbb{R}}f:\R^4\to...[/itex], but I would like to know where it is mapping to, and what the physical units of the target values are.

    If someone furthermore can provide the function or a typical, general representative, this would be great.

    Important: I am not asking for Schrödinger wave functions.

    Last edited by a moderator: Apr 22, 2017
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  3. Jul 10, 2007 #2

    Hans de Vries

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    In the most complete theory: Dirac Theory, the electron has 4 complex
    spinor components: [itex]\def\R{\mathbb{R}}f:\R^4\to \mathbb{C}^4[/itex]

    Which depend on the direction of the spin of the electron and the
    momentum/ angular momentum, for instance in the hydrogen solutions
    of Dirac's equation. Generally these values are normalized so that
    integrating over all space of [itex]{\bar \psi} \psi[/itex] gives the rest mass energy of the
    electron. The spinless electron has only one complex value.

    There's something very tricky about what a "point" is, which often
    leads to misunderstandings. Now although you may say that there are
    six independent variables (E and B) per "point" It's better to say that
    the EM (potential) field has 4 independent variables per point: [itex]V, A_x, A_y, A_z[/itex]:
    The potential + magnetic vector potential.

    There are 16 independent derivatives in time and space of these values,
    because derivatives compare independent adjacent points which have
    independent values for [itex]V, A_x, A_y, A_z[/itex]

    These 16 independent derivatives are reduced to the 6 values of the EM field:

    E_x= -\frac{\partial A_x}{\partial t}-\frac{\partial V}{\partial x} \qquad
    E_y= -\frac{\partial A_y}{\partial t}-\frac{\partial V}{\partial y} \qquad
    E_z= -\frac{\partial A_z}{\partial t}-\frac{\partial V}{\partial z}[/tex]

    B_x= \frac{\partial A_y}{\partial z}-\frac{\partial A_z}{\partial y}\qquad
    B_y= \frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z}\qquad
    B_z= \frac{\partial A_x}{\partial y}-\frac{\partial A_y}{\partial x}[/tex]

    Four of the derivatives don't occur at all in the definition of E and B while
    the remaining 12 are used in pairs. So, we are left with only 6 values.

    Now, E and B are uniquely defined from V and A but we can not uniquely
    reconstruct V and A from E and B. In order to do so we would need all 16
    independent derivatives to recover V and A by integration.

    So, although 6 values per point seems more than 4. It's actually less
    because the "size of the point" is not the same in both cases.

    Regards, Hans.
    Last edited: Jul 10, 2007
  4. Jul 10, 2007 #3
    Thanks Hans for the clearcut answer. May I ask for a few clarifications?

    What would be the SI units of the four complex (or 8 real) values?

    Do you mean that integration of each is normalized and individually gives the rest mass? :confused: Rather not, I would guess. How are the four [itex]{\bar \psi_i} \psi_i[/itex] combined? I am assuming here that the rest mass is just [itex]\in\mathbb{R}[/itex] and is not, say, a four-vector.

    Ah, not really surprising: since E and B are always orthogonal, there is hardly room for 6 degrees of freedom. But I wouldn't have known it is these four.

  5. Jul 10, 2007 #4

    Hans de Vries

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    Dirac's [tex]{\bar \psi}\psi[/tex] is very similar to Schroedinger's [tex] \psi^*\psi[/tex], now the latter is
    normalized to give a dimensionless 1 once integrated over space. The
    normalization of [tex]{\bar \psi}\psi[/tex] varies, but most commonly is done in order to
    give m (the restmass) when integrated over space.

    Actually, since c is generally set to 1 in QFT we need to look a bit closer
    to get the SI units. m is actually mc so the normalization gives SI units:

    [tex]\sqrt{\mbox{Joule meter/second}}\ \ \ [/tex]

    (Well, It's just a normalization)

    A different normalization would be to make it the charge e. Important is that
    both m and e are Lorentz invariant scalars. They are the same in any reference
    frame. (But maybe I'm going a bit to deep here)

    It's indeed a scalar value calculated from the four complex spinor components:

    [tex]{\bar \psi}\psi\ =\ \psi_1^*\psi_1 + \psi_2^*\psi_2 - \psi_3^*\psi_3 - \psi_4^*\psi_4 [/tex]

    The latter two are subtracted because they correspond to negative energy,
    positive charge components. These components are zero for an electron
    at rest, but can not be neglected at very high speeds. There are several
    reasons why these components make sense, for instance:

    The electron has an inherent magnetic dipole moment which partly transforms
    into an effective electric dipole moment when seen from other (high speed)
    reference frames. This effective electric dipole moment is proportional to the
    positively charged components [tex]\psi_3, \psi_4[/tex]

    E and B are always orthogonal in the case of radiation from a single (point)
    source. In general it doesn't need to be true and E and B can have any
    value independently of each other. So there are six degrees of freedom.
    If you read my response more carefully then you'll see what I did mean
    to say.

    Regards, Hans
    Last edited: Jul 10, 2007
  6. Jul 11, 2007 #5
    Uuum, I am lost a bit on notation here: In Priestley's 'Complex Analysis' [itex]{\bar \psi}[/itex] denotes the complex conjugate and in Treiman's 'The Odd Quantum' [itex] \psi^*[/itex] is also the complex conjugate. So I hope this is still the same here and with 'very similar' you are rather referring to the normalization than to some different meaning of bar and star.

    and in an earlier post you say

    This clearly counts to four and you say that the 6 values [itex]E_{xyz}, B_{xyz}[/itex] are uniquely determined from the four. If the 6 are uniquely determined, I don't see how there can be more degrees of freedom.

    I don't want to split hairs here. I am just confused.

    Last edited: Jul 11, 2007
  7. Jul 11, 2007 #6
    This is not splitting hairs, and your confusion is well-justified. In spite of using the same symbol [itex] \psi [/itex] in quantum mechanics and in QFT, we are talking about two completely different things. Regrettably, this reuse of the Greek alphabet creates a lot of confusion.

    In quantum mechanics we have wave function [itex]\psi(\mathbf{x}, t) [/itex], which assigns a complex number to each space point [tex] \mathbf{x}[/itex] (an eigenvalue of the position operator) at each time [itex] t [/itex]. Notation [itex]\psi^*(\mathbf{x}, t) [/itex] defines a complex-conjugated function. The fundamental law of quantum mechanics says that the product [itex]\psi^*(\mathbf{x}, t)\psi(\mathbf{x}, t) [/itex] defines the probability density for finding particle in point [tex] \mathbf{x}[/itex] at time [itex] t [/itex]. Then, by definition of probability, the integral over entire space at any time instant [itex] t [/itex] is exactly 1

    [tex] \int d^3x \psi^*(\mathbf{x}, t)\psi(\mathbf{x}, t) = 1 [/tex]

    In quantum field theory we have Dirac's quantum field [itex]\psi(\mathbf{x}, t) [/itex] which has absolutely nothing to do with probabilities. This functions assigns an operator (actually, a group of four operators, which act in the Fock space) to each point [itex] (\mathbf{x}, t) [/itex] in the Minkowski spacetime. It is debatable that Minkowski spacetime of QFT has any relationship to measured particle positions. In particular, I don't think it is possible to interpret argument [itex] \mathbf{x} [/itex] as eigenvalue of a position operator. Notation [itex]\psi^{\dag}(\mathbf{x}, t) [/itex] defines a Hermitian-conjugated field operators, and notation [itex]\overline{\psi}(\mathbf{x}, t) = \psi^{\dag}(\mathbf{x}, t) \gamma_0[/itex] defines adjoint operators, where [itex]\gamma_0[/itex] is a Dirac [itex] 4 \times 4 [/itex] matrix. One can form products like [itex]\overline{\psi}(\mathbf{x}, t) \psi(\mathbf{x}, t) [/itex], but they have nothing to do with probability densities. If you integrate these products on [itex] \mathbf{x} [/itex], do not expect to get 1.

    The true reason of having quantum fields, their adjoints, etc. in QFT is not related to the probabilistic interpretation. Fields are needed, because they are used in construction of interaction Hamiltonians (also operators in the Fock space). Then things like [itex]\overline{\psi}(\mathbf{x}, t) \psi(\mathbf{x}, t) [/itex] are often met when we are calculating S-matrix elements from these Hamiltonians. This is why such products and their properties are important in QFT.

    Note that ordinary wave functions (probability density amplitudes, whose absolute squares are normalizable to 1) can be also defined in QFT. So, there is no fundamental difference between QM and QFT.
  8. Jul 11, 2007 #7

    Hans de Vries

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    Dirac's [itex]\psi[/itex] is a four component value [tex](\psi_1,\psi_2,\psi_3,\psi_4)[/itex] and the meaning of
    [itex]{\bar \psi}[/itex] is [tex](\psi_1^*,\psi_2^*,-\psi_3^*,-\psi_4^*)[/itex] where * denotes complex conjugate.

    You'll only study this after you've studied Schroedingers equation (which
    has only one component) and which I presumed you know about. Only
    a small part of the people who learn Schroedinger's equation will get
    to actually study Dirac's relativistic Quantum Mechanical equation.

    You start with 4 values per point: [itex]V, A_x, A_y, A_z[/itex]

    A point is infinitesimal small. You need ONE + FOUR adjacent points to
    determine the SIXTEEN independent derivatives which you need to
    calculate the SIX components of the E and B fields.

    If one naively says that the EM field has SIX independent values per "point"
    then you must realize that you have actually used a small environment of
    multiple points instead of only a single point.

    Therefor, the SIX values of E and B represent a LOWER "information density"
    as the FOUR values of V and A. If you know A and V, then you can calculate
    E and B, This is not true the other way around.

    Regards, Hans
  9. Jul 11, 2007 #8
    It is true that many QFT textbooks, e.g., Bjorken & Drell, consider four-component Dirac's [itex]\psi[/itex] as a wave function and Dirac's equation as a relativistic analog of the Schroedinger equation. They even solve the Dirac's equation for the hydrogen atom. Later they admit that there are lot of problems with this idea: no clear probabilistic interpretation, negative energies, Klein's paradox, zitterbewegung, etc. and suggest to use Dirac's equation for quantum fields only.

    I think you can avoid a lot of frustration if you accept from the beginning that KG and Dirac's equations are *not* analogs of the Schroedinger equation and that quantum fields (that satisfy covariant field equations) have nothing to do with wave functions of particles.
  10. Jul 12, 2007 #9

    Hans de Vries

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    I would say that Dirac's equation provides a very significant improvement
    over Schroedinger's equation in terms of the atomic spectra. Gold would
    have a silver color according to Schroedinger's equation.

    See for instance table 10.5 on page 359 in Paul Strange's "Relativistic
    Quantum Mechanics" :

    Code (Text):

    One electron energy eigenvalues of the mercury atom:

    n   l   j             NRHFS   DHFS      DHF      EXPERIMENTAL

    1   0   1/2          5535.8  6130.2    6148.5     6107.9
    2   0   1/2          932.0   1090.3    1100.5     1090.6
    2   1   1/2          896.9   1047.8    1053.7     1044.8
    2   1   3/2          896.9    903.0     910.3      903.0

    NRHFS: non relativistic Hartree-Fock with Slater exchange
    DHFS:  relativistic Hartree-Fock with Slater exchange
    DFH:   relativistic Hartree-Fock without Slater
    The normal and anomalous Zeeman effect are directly accounted for
    by the Dirac equation as well as more recently studied effects such
    as Magnetic Dichroism: The dependence of the probability of some
    electronic transitions on the state of polarization of the incident photon.

    Klein Paradox was one of the issues that let to the interpretation of
    of the wave function as a charge, current and spin density rather than
    a probability:

    [tex] {\bar \psi} \gamma_\mu \psi = j_\mu = \mbox{charge, current density} [/tex]

    In the rest frame we have for the spin density: (using the upper bi-spinor)

    [tex] {\psi^\dagger \sigma_k \psi} = \mbox{spin density} [/tex]

    Now the calculation of atomic spectra and molecular modeling show that
    we should indeed consider the charge and spin to be spread out all over
    the wave-function and that electrons actually interact in this way. Also
    with such a static state of distributed charge/current there isn't any EM
    radiation according to classical Electro magnetics.

    Regards, Hans
  11. Jul 12, 2007 #10

    I agree with that.

    This looks suspicious to me. I am not ready to abandon quantum mechanics and its probabilistic interpretation. There is a way to perform relativistic calculations of atoms without the use of Dirac's equation. It is called the "Breit Hamiltonian". You can find a description of this method in sections 83-84 of

    V. B. Berestetskii, E. M. Liv****z, and L. P. Pitaevskii, "Quantum electrodynamics"

    More detailed derivations can be found in section 9.3 of http://www.arxiv.org/physics/0504062 [Broken]

    I am not sure if this method is more or less accurate than Dirac's equation, but I like it more, because it doesn't require any twists in the standard formalism of quantum mechanics.
    Last edited by a moderator: May 3, 2017
  12. Jul 12, 2007 #11
    In fact the [itex]\psi[/itex] was not my problem. I was aware that it does not represent the Schroedinger wave function for the simple reason that I explicitly noted that I was not asking for the wave equation:smile: The \bar was my problem, because from math courses and, as mentioned, from Priestley's book, I thought it denotes the complex conjugate.

    I learned that the \bar has a quite specific meaning in QFT. And I learned that the seemingly innocent question

    What is the field function (now I avoid 'wave') for an electron in a similar way as the E and B fields represent the photon?​

    is either naive or not specific enough. I learned that [itex]V, A_x, A_y, A_z[/itex] may also be taken to represent what gives rise to a photon or more. And from your discussions, it seems that the field function for the electron is even less clear cut.

    In fact, I had the hope that if [itex]\def\R{\mathbb{R}}f:\R^4\to\R^3\times R^3[/itex] represents the (field function of a) photon and [itex]\def\C{\mathbb{C}}\def\R{\mathbb{R}}g:\R^4\to\C^4[/itex] represents the electron, then the interaction of a photon and an electron, where the electron "swallows" the photon to gain in energy should be some function [itex]T(f,g)\mapsto h[/itex], mapping electron and photon to a new photon. Of course [itex]T[/itex] would have to be defined on a suitable function space that can accomodate both of the slightly different signatures of the functions involved.

  13. Jul 12, 2007 #12

    Hans de Vries

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    The probabilistic interpretation was never abandoned with relativistic quantum
    mechanics. I would say it needs extra considerations.

    I can think of two ways to reconcile the distributed charge density picture
    with the probability picture. (I strongly prefer the second). Here we go
    speculating into the interpretations of QM...

    1) Often people presume that the electron, as a particle, jumps randomly
    through the wave function and thereby creates the illusion of a charge
    density. This has no chance to explain classically why orbits don't radiate.

    2) One can assume that what we call a single electron is in fact a
    combination of N+1 particles plus N 'anti'-particles. So there is a surplus
    of 1 particle guaranteeing unitarity. It is however undetermined which
    of the N+1 particles is THE surplus particle. This might be a different
    particle all the time.

    The fact that there is a surplus particle causes the other N pairs to
    be slightly separated, creating a distributed charge density. The chance
    to actually 'detect' the surplus particle is highest in the location where the
    charge density is highest.

    Regards, Hans.
    Last edited: Jul 12, 2007
  14. Jul 12, 2007 #13
    This looks very complicated to me. Why can't we describe a single electron by a wave function (in the position space or momentum space) whose square is the probability density. For two electrons this would be a wavefunction depending on two arguments, etc. The same stuff as in ordinary non-relativistic quantum mechanics. I haven't seen a good argument why we can't keep the same formalism for relativistic electrons. Before modifying quantum mechanics I want to see a solid proof that this is absolutely necessary.

    You are trying to save interpretation of Dirac's [itex] \psi [/itex] as some kind of wave function by changing the definition of what wave function actually is (by introducing charge and current densities which have not been present in the original definition of the wave function). For me it is difficult to swallow. I think that there is no need to insist on Dirac's [itex] \psi [/itex] as a generalization of wave function. This object becomes a "quantum field" in the full-blown QFT, and as I tried to argue elsewhere quantum fields and wavefunctions have nothing in common, except (unfortunately) similar notation.
  15. Jul 12, 2007 #14

    Hans de Vries

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    In practice, most people who use non-relativistic QM consider the wave
    function as a distributed charge denity/ spin density. That is Chemist,
    Molecular modelers, Solid State physicist. et-cetera. Simply because
    electrons interact with other electrons as if these were distributed charge/
    spin densities plus Pauli's exclusion principle.

    Almost all physicist consider the probability element to be there as well.
    It's the "Shut up and calculate approach" You adhere to what works.

    The early arguments for the distributed charge density picture came from
    relativistic QM equations which seemed to suffer from negative probabilities.

    I wouldn't want to dismiss the Klein Gordon and Dirac equations. The first
    becomes the Schroedinger equation in the non relativistic case, while the
    latter becomes the Pauli spin equation, the Schroedinger equation + spin.

    Well, I presume I'm telling you noting new here.

    Regards, Hans
  16. Jul 13, 2007 #15
    I think that Klein-Gordon and Dirac equations should be used for quantum fields only. A relativistic analog of the Schroedinger equation for the state vector [itex] | \Psi \rangle [/itex] of a free particle should look like this:

    [tex] i \hbar \frac{\partial}{\partial t} | \Psi (t)\rangle = \sqrt{\mathbf{P}^2 c^2 + m^2c^4}| \Psi (t)\rangle [/tex]

    where [itex] \mathbf{P} [/itex] is the momentum operator, [itex]m[/itex] is particle's mass, and [itex] \sqrt{\mathbf{P}^2 c^2 + m^2c^4} [/itex] is the usual expression for the free particle Hamiltonian. In the position representation, in this formula one should use wave functions [itex] \Psi (\mathbf{r}, t) [/itex] in place of [itex] | \Psi \rangle [/itex] and usual space derivatives in place of [itex] \mathbf{P} [/itex].

    This equation is first-order in t, and it allows the same probabilistic interpretation for the wave function [itex] \Psi (\mathbf{r}, t) [/itex] as in non-relativistic quantum mechanics. For example, the normalization

    [tex] \int \limits_{V} |\Psi (\mathbf{r}, t)|^2 d^3r = 1 [/tex]

    remains valid at all times t. Moreover, using formulas for boost transformations of the Newton-Wigner position operator one can show that the wave function in the moving reference frame is normalized too.
  17. Jul 13, 2007 #16

    Hans de Vries

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    This is a reasonable proposal, but the problem with this equation is that it
    is non-local. The time derivative at a position x will depend on far away
    points instead of the direct environment.

    It is the square root of the operator which causes this. The momentum
    operator is basically a diagonal matrix where each diagonal element is
    (-1, +1). That is, it takes the difference of two adjacent points in space.

    The square of the momentum operator is also a diagonal matrix. Its
    diagonal elements are (1,-2,1), the second order differences. Add to
    this the Unity matrix times the mass term to get the total diagonal
    operator from which the square root has to be taken.

    It turns out that the square root isn't a diagonal matrix anymore, but
    has non-zero elements far away from the diagonal.

    Regards, Hans
  18. Jul 13, 2007 #17
    When you take the time derivative of both sides of the equation

    i\hbar\partial_t |\Psi\rangle = E_p |\Psi\rangle

    the result is

    -\hbar^2\partial^2_t |\Psi\rangle = i\hbar E_p \partial_t |\Psi\rangle

    and when you substitute the previous equation into the right hand side you get

    -\hbar^2\partial^2_t |\Psi\rangle = E^2_p |\Psi\rangle

    and substituting the usual position representations here gives the Klein-Gordon equation. So isn't defining time evolution with this merely equivalent to defining it with the KG equation and adding a condition that only positive frequency solutions are accepted?

    To me it seems, that the Shrodinger equation doesn't produce the usual conserving probability current, because the Hamiltonian is too strange. Or have we just found the solution to the conserving probability current with Klein-Gordon equation?
  19. Jul 13, 2007 #18
    Sorry, I failed to follow your logic. Momentum and energy are commuting operators. So, they are both diagonal in the momentum representation. They are both non-diagonal in the position representation, because they do not commute with the position operator. The situation is the same as is non-relativistic quantum mechanics. I don't see any problem.
  20. Jul 13, 2007 #19
    The most fundamental equation describing the time evolution of a state vector is

    | \Psi(t) \rangle = \exp{\frac{i}{\hbar} Ht} |\Psi(0) \rangle
    [/tex] (1)

    where I denoted the Hamiltonian as [itex]H[/itex] instead of your [itex]E_p[/itex]. This equation can be rewritten as

    | \Psi(t) \rangle = (1 + \frac{i}{\hbar} Ht - \frac{1}{2 \hbar^2} H^2t^2 + \ldots) |\Psi(0) \rangle

    then the second time derivative of [itex] | \Psi(t) \rangle [/itex] at t=0 is

    \frac{\partial^2}{\partial t^2}| \Psi(t=0) \rangle = - \frac{1}{ \hbar^2} H^2 |\Psi(0) \rangle

    from which your "Klein-Gordon" equation follows

    - \hbar^2 \frac{\partial^2}{\partial t^2}| \Psi(t=0) \rangle = H^2 |\Psi(0) \rangle

    Similar formulas can be written for the 3rd and higher time derivatives. But I am not sure what additional insight you can get from that. If you are interested in finding the time evolution, then use eq. (1).

    What do you mean by saying "the Shrodinger equation doesn't produce the usual conserving probability current"? The integral of the probability density (or the norm of the state vector [itex] \langle \Psi | \Psi \rangle [/itex]) remains equal to 1 at all times and in all reference frames, because the time evolution (1) is unitary as long as the Hamiltonian [itex] H [/itex] is Hermitian.
  21. Jul 13, 2007 #20

    Hans de Vries

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    [tex] i \hbar \frac{\partial}{\partial t} | \Psi (t)\rangle = \sqrt{\mathbf{P}^2 c^2 + m^2c^4}| \Psi (t)\rangle [/tex]

    This equation is discussed in the books and literature and is generally
    dismissed because of it's non-local solutions. See for instance:

    Bjorken and Drell, "Relativistic Quantum Mechanics chapter 1.2"
    Paul Strange, "Relativistic Quantum Mechanics chapter 3.1"

    If you work it out you get an infinite series in higher spatial derivatives:

    [tex] \sqrt{1 + P^2} \ =\ 1 + \frac{1}{2}P^2 - \frac{1}{8}P^4 + \frac{1}{16}P^6 - \frac{5}{128}P^8 + \frac{7}{256}P^{10} - \frac{21}{1024}P^{12} + \frac{33}{2048}P^{14} - \frac{429}{16384}P^{16} + ....[/tex]

    Regards, Hans
    Last edited: Jul 13, 2007
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