Exploring the deBroglie Function: Understanding Its Mathematical Signature

[birulami]

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.

Thanks,
Harald.

 

[Hans de Vries]

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.

Thanks, Harald.

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:

[tex]
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]

[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.

 

[birulami]

Thanks Hans for the clearcut answer. May I ask for a few clarifications?

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]

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

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.

Do you mean that integration of each is normalized and individually gives the rest mass? 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.

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.

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.

Harald.

 

[Hans de Vries]

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?

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)

Do you mean that integration of each is normalized and individually gives the rest mass? 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.

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]

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.

Harald.

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

 

[birulami]

Dirac's [tex]{\bar \psi}\psi[/tex] is very similar to Schroedinger's [tex] \psi^*\psi[/tex]

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.

So there are six degrees of freedom.

and in an earlier post you say

It's better to say that the EM (potential) field has 4 independent variables per point: The potential + magnetic vector potential [itex]V, A_x, A_y, A_z[/itex]

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.

Cheers,
Harald.

 

[meopemuk]

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.

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

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.

 

[Hans de Vries]

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.

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.

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.

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

 

[meopemuk]

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.

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.

 

[Hans de Vries]

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.

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" :

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

 

[meopemuk]

I would say that Dirac's equation provides a very significant improvement
over Schroedinger's equation in terms of the atomic spectra.

I agree with that.

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:

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

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.

 

[birulami]

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 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 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 accommodate both of the slightly different signatures of the functions involved.

Harald.

 

[Hans de Vries]

This looks suspicious to me. I am not ready to abandon quantum mechanics and its probabilistic interpretation.

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.

 

[meopemuk]

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.

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.

 

[Hans de Vries]

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.

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

 

[meopemuk]

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.

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.

 

[Hans de Vries]

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 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

 

[jostpuur]

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.

When you take the time derivative of both sides of the equation

[tex]
i\hbar\partial_t |\Psi\rangle = E_p |\Psi\rangle
[/tex]

the result is

[tex]
-\hbar^2\partial^2_t |\Psi\rangle = i\hbar E_p \partial_t |\Psi\rangle
[/tex]

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

[tex]
-\hbar^2\partial^2_t |\Psi\rangle = E^2_p |\Psi\rangle
[/tex]

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?

 

[meopemuk]

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

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.

 

[meopemuk]

When you take the time derivative of both sides of the equation

[tex]
i\hbar\partial_t |\Psi\rangle = E_p |\Psi\rangle
[/tex]

the result is

[tex]
-\hbar^2\partial^2_t |\Psi\rangle = i\hbar E_p \partial_t |\Psi\rangle
[/tex]

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

[tex]
-\hbar^2\partial^2_t |\Psi\rangle = E^2_p |\Psi\rangle
[/tex]

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?

The most fundamental equation describing the time evolution of a state vector is

[tex]
| \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

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

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

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

from which your "Klein-Gordon" equation follows

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

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).

To me it seems, that the Shrodinger equation doesn't produce the usual conserving probability current, because the Hamiltonian is too strange.

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.

 

[Hans de Vries]

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.

[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

 

[meopemuk]

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"

Yes, this equation is often mentioned and dismissed without explanation that would satisfy me. I read on page 5 of Bjorken and Drell:

If we expand it, we obtain an equation containing all powers of the derivative operator and thereby a nonlocal theory. Such theories are very difficult to handle and present an unattractive version of the Schroedinger equation in which the space and time coordinates appear in unsymmetrical form.

I skip their "difficult to handle" and "unattractive" comments, as they are not objective. They are right that space and time coordinates appear in this equation in unsymmetrical form. Is it good or bad?

We learned in kindergarten that special relativity demands equivalence of space and time coordinates. However, it is also true that in quantum mechanics space and time are not equivalent. There is an operator of position, and there is no time operator. Who is right, quantum mechanics or special relativity? In my opinion, this is the central question of modern theoretical physics. Without answering it, there is no hope to build a quantum theory of gravity.

In http://www.arxiv.org/physics/0504062 I am trying to answer this question. Briefly, my conclusion is that the space-time equivalence is not an immediate consequence of the relativity principle. It is an additional unproved assumption. Fully relativistic (quantum) theories can be constructed which do not respect this equivalence. In order to satisfy the principle of relativity it is sufficient to make sure that the theory realizes an (unitary) representation of the Poincare group. So, in my opinion the unsymmetrical appearance of the relativistic Schroedinger equation is not a bad thing. A lot can be said about these issues. Perhaps, we will need a separate thread to cover them.

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]

Thanks for this amazing formula. I have seen only first three terms before.

Eugene.

 

[Hans de Vries]

[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}{32768}P^{16} + ...[/tex]Thanks for this amazing formula. I have seen only first three terms before.

Eugene.

staring a bit at the numbers give me this:

[tex] i \hbar \frac{\partial}{\partial t} | \Psi (t)\rangle \ =\ \sqrt{\mathbf{P}^2 c^2 + m^2c^4}| \Psi (t)\rangle\ =\ \ mc^2\left\{ 1 + \sum_{n=1}^\infty \frac{(2n-2)!}{n!(n-1)!\ 2^{2n-1}} \left( \frac{\mathbf{P}^2}{m^2c^2}\right)^{n} \right\} |\Psi (t)\rangle [/tex]

Regards, Hans

 

[masudr]

staring a bit at the numbers give me this:

I need to get back in practice.

 

[Anonym]

staring a bit at the numbers give me this

It so great pleasure to read your posts!

Regards, Dany.

 

[lightarrow]

staring a bit at the numbers give me this:

[tex] i \hbar \frac{\partial}{\partial t} | \Psi (t)\rangle \ =\ \sqrt{\mathbf{P}^2 c^2 + m^2c^4}| \Psi (t)\rangle\ =\ \ mc^2\left\{ 1 + \sum_{n=1}^\infty \frac{(2n-2)!}{n!(n-1)!\ 2^{2n-1}} \left( \frac{\mathbf{P}^2}{m^2c^2}\right)^{n} \right\} |\Psi (t)\rangle [/tex]

Regards, Hans

Of course there is a (-1)^(n+1) which is missing inside the sum.

 

[jostpuur]

[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

This series does not converge for [tex]|P|>1[/tex], and I would be suprised if it converged when an operator is subtituted in place of P. Doesn't this mean, that the series can be used to obtain high accuracy correction terms in non-relativistic limit, but it cannot be used to define a relativistic operator?

 

[jostpuur]

The most fundamental equation describing the time evolution of a state vector is

[tex]
| \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

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

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

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

from which your "Klein-Gordon" equation follows

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

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).

The first order differential equation cannot be presented in position representation. So if I want position representation, then I'm forced to use Klein-Gordon equation.

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.

It could be I made a too quick conclusion. I was looking this thing from a position representation point of view.

Can you give this operator any other definition in position representation than this:

[tex]
\psi(x) \mapsto \int\frac{d^3p}{(2\pi\hbar)^3} \psi(p)\sqrt{p^2 + m^2} e^{ip\cdot x/\hbar}
[/tex]

Where [tex]\psi(p)[/tex] is the Fourier transformation of the original [tex]\psi(x)[/tex]. This seems to be a well defined operator, altough not as nice those that can be written in terms of derivative operators.

 

[Hans de Vries]

Of course there is a (-1)^(n+1) which is missing inside the sum.

And we should also include a +/- to allow for anti-particles as well :

[tex] i \hbar \frac{\partial}{\partial t} | \Psi (t)\rangle \ =\ \sqrt{\mathbf{P}^2 c^2 + m^2c^4}| \Psi (t)\rangle\ =\ \pm \ mc^2\left\{ 1 + \sum_{n=1}^\infty \frac{(-1)^{n+1}\ (2n-2)!}{n!(n-1)!\ 2^{2n-1}} \left( \frac{\mathbf{P}^2}{m^2c^2}\right)^{n} \right\} |\Psi (t)\rangle [/tex]

Or written out as a differential operator:

[tex] i \hbar \frac{\partial}{\partial t} | \Psi (t)\rangle \ =\ \pm \ mc^2\left\{ 1 - \sum_{n=1}^\infty \frac{ (2n-2)!}{n!(n-1)!\ 2^{2n-1}} \left( \frac{\hbar^2}{m^2 c^2}\ \right)^{n} \left( \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} +\frac{\partial^2}{\partial z^2} \right)^{n}\right\} |\Psi (t)\rangle [/tex]

Including the EM interactions:

[tex] i \hbar \frac{\partial}{\partial t} | \Psi (t)\rangle \ =\ \pm \ mc^2 \left\{ 1 - \sum_{n=1}^\infty \frac{ (2n-2)!}{n!(n-1)!\ 2^{2n-1}} \left( \frac{1}{m^2 c^4}\ \sum_{i=1}^3\left( c \hbar \frac{\partial}{\partial x_i}-ieA_{x_i} \right)^2 \ \ \right)^{n}\right\} |\Psi (t)\rangle \ +\ eV\ |\Psi (t)\rangle [/tex]

This series does not converge for [tex]|P|>1[/tex], and I would be suprised if it converged when an operator is subtituted in place of P. Doesn't this mean, that the series can be used to obtain high accuracy correction terms in non-relativistic limit, but it cannot be used to define a relativistic operator?

So, it seems it would works as long as cp doesn't exceed the rest mass
energy. That would be up to [itex]v=\sqrt{1/2}\ c [/itex]. Now, if you take the squares
then all higher order terms cancel and you recover the Klein Gordon equation.
I suspect that it should give the same results as the Klein Gordon equation
if you do a Lattice simulation.

The difference is that, if you do a simulation with the Klein Gordon equation,
then the first order derivative in time is an arbitrary initial condition, Such
an arbitrary initial condition may not be the result from an actual ongoing
physical process.

It looks like that, in this case, One might actually recover the first order
derivatives, belonging to an actual physical process, by sampling in a
somewhat non-local area around each point, and this would then be the
correct interpretation of the non-locality. I don't know.

I would think that Nature itself would work in the simplest way. That is, it
uses only local points to determine what happens next. Like in the quadratic
form of the Klein Gordon equation.


Regards, Hans

 

[Haelfix]

Neat little intro exercise for a grad student. Stick this on a lattice and compute the contributions from the surrounding points. I'll have to tag this for future reference.

 

[meopemuk]

The first order differential equation cannot be presented in position representation. So if I want position representation, then I'm forced to use Klein-Gordon equation.

Why do you say this? You can just substitute the momentum operator under the square root with the operator of gradient. This doesn't look very familiar, but, really, nothing special. Functions (e.g., square roots) of operators are well-studied things.

It could be I made a too quick conclusion. I was looking this thing from a position representation point of view.

Can you give this operator any other definition in position representation than this:

[tex]
\psi(x) \mapsto \int\frac{d^3p}{(2\pi\hbar)^3} \psi(p)\sqrt{p^2 + m^2} e^{ip\cdot x/\hbar}
[/tex]

Where [tex]\psi(p)[/tex] is the Fourier transformation of the original [tex]\psi(x)[/tex]. This seems to be a well defined operator, altough not as nice those that can be written in terms of derivative operators.

Yes, this is a fine way of writing the free particle Hamiltonan in the position representation.

Eugene.

 

[meopemuk]

Originally Posted by jostpuur:

This series does not converge for [itex]| P| > 1 [/itex], and I would be suprised if it converged when an operator is subtituted in place of P.

So, it seems it would works as long as cp doesn't exceed the rest mass
energy. That would be up to [itex]v=\sqrt{1/2}\ c [/itex].

The series could be non-convergent, but the square root itself is a well-defined function for all real P or v. I think that your condition [itex]v=\sqrt{1/2}\ c [/itex] should not be interpreted as any physical limit.

Eugene.

 

[jostpuur]

Unfortunately I don't know much of square roots of operators. If I was now given a task of defining the operator [itex]\sqrt{-\nabla^2 + m^2}[/itex], I would define it like this

[tex]
\sqrt{-\nabla^2+m^2}\psi(x) := \int\frac{d^3x'\; d^3p}{(2\pi)^3} \psi(x') \sqrt{|p|^2 + m^2} e^{ip\cdot(x-x')}
[/tex]

Would other definitions agree with this one?

But this is not fully making sense. Isn't it well established fact, that Klein-Gordon equation does not conserve the probability [itex]\int d^3x\;|\psi|^2[/itex]? Or could it be, that it actually conserves it, but it is just not manifestly apparent? Now when you define the time evolution with the Shrodinger's equation (with this Hamiltonian with square root), the time evolution also satisfies the Klein-Gordon equation. So if Shrodinger's equation conserves the probability, and Klein-Gordon equation does not, this is quite paradoxical.

I would sure like some well done mathematical analysis conserning this operator. Merely noting that the Taylor series don't work for it doesn't seem very satisfactory.

 

[meopemuk]

Unfortunately I don't know much of square roots of operators. If I was now given a task of defining the operator [itex]\sqrt{-\nabla^2 + m^2}[/itex], I would define it like this

[tex]
\sqrt{-\nabla^2+m^2}\psi(x) := \int\frac{d^3x'\; d^3p}{(2\pi)^3} \psi(x') \sqrt{|p|^2 + m^2} e^{ip\cdot(x-x')}
[/tex]

Would other definitions agree with this one?

In my opinion, this definition is perfectly OK.

But this is not fully making sense. Isn't it well established fact, that Klein-Gordon equation does not conserve the probability [itex]\int d^3x\;|\psi|^2[/itex]? Or could it be, that it actually conserves it, but it is just not manifestly apparent? Now when you define the time evolution with the Shrodinger's equation (with this Hamiltonian with square root), the time evolution also satisfies the Klein-Gordon equation. So if Shrodinger's equation conserves the probability, and Klein-Gordon equation does not, this is quite paradoxical.

I think that in order to answer this it is useful to take a bit more abstract and general approach. In quantum mechanics, probabilities are preserved if the state vector gets transformed by a unitary operator [itex] U [/itex] (Wigner theorem)

[tex] |\Psi \rangle \to U |\Psi \rangle [/tex]

So, the operator of time evolution [itex] U(t) [/itex] must be unitary

[tex] |\Psi (t)\rangle = U(t) |\Psi (0)\rangle [/tex]

Time evolution is a one-parametric subgroup of transformations from the Poincare group. Then by Stone theorem, representatives of time shifts in the Hilbert space must have the form

[tex] U(t) = \exp(\frac{i}{\hbar} Ht)[/tex]

where [itex] H [/itex] is a Hermitian operator (called Hamiltonian). So, we obtain

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

which can be differentiated by t to obtain

[tex] -i \hbar \frac{\partial}{\partial t}|\Psi (t)\rangle = H|\Psi (t)\rangle [/tex]

So, it follows from very general principles (conservation of probability, group structure of inertial transformations, the principle of relativity) that the time dependence of state vectors (and/or wave functions) should be described by the above equation, where H is an Hermitian operator.

The Klein-Gordon equation does not have this form. This is the reason why KG does not conserve probability and why it is wrong to consider KG as a relativistic generalization of the Schroedinger equation.

 

[Hans de Vries]

Unfortunately I don't know much of square roots of operators. If I was now given a task of defining the operator [itex]\sqrt{-\nabla^2 + m^2}[/itex], I would define it like this

[tex]
\sqrt{-\nabla^2+m^2}\psi(x) := \int\frac{d^3x'\; d^3p}{(2\pi)^3} \psi(x') \sqrt{|p|^2 + m^2} e^{ip\cdot(x-x')}
[/tex]

Would other definitions agree with this one?

That's OK. The operator working on the wave function in configuration space
is indeed a convolution with the Fourier transform of the propagator in
momentum space.

Regards, Hans.

 

[Hans de Vries]

But this is not fully making sense. Isn't it well established fact, that Klein-Gordon equation does not conserve the probability [itex]\int d^3x\;|\psi|^2[/itex]? Or could it be, that it actually conserves it, but it is just not manifestly apparent? Now when you define the time evolution with the Shrodinger's equation (with this Hamiltonian with square root), the time evolution also satisfies the Klein-Gordon equation. So if Shrodinger's equation conserves the probability, and Klein-Gordon equation does not, this is quite paradoxical.

The whole point is, I think, that the Schrödinger wave function is defined as:

[tex]\psi \ =\ \Psi\ e^{-imc^2t/\hbar}[/tex]

Where only [itex]\Psi[/itex] can vary and [itex]\exp(-imc^2t/\hbar)[/itex] is "hardcoded" in the solution.
The same is true in QFT working with plane waves in momentum space.

Now while the Schrödinger and QFT solutions are also the solutions of
the Klein Gordon equation, it does not generate them in a lattice
simulation if not given the proper initial conditions.

[tex]\partial^2_t \psi\ =\ \nabla\psi - m^2\psi[/tex]

is a real equation without imaginary parts. If you give it a real wave function
as input, together with 0 or real initial values for [itex]\partial_t \psi[/itex] then it will stay
forever real in a lattice simulation, and the integral over space can grow
indefinitely in some cases.

Given the proper initialization for [itex]\partial_t \psi[/itex], which is entirely imaginary if the
wave function is real, then it should continue to simulate exactly as with
Schrödinger's equation. (ignoring the small terms which are thrown away if
one goes from Klein Gordon to Schrödinger)Regards, Hans