The role of wave function in QED

[samalkhaiat]

I am just learning QED and could not understand the role of wave function. Is the basic equation in QED the Schrodinger Equation? Is the difference between Quantum mechanics and QED just they have different Hamiltonians.


Quantum River

Relativistic & non-relativistic quantum description of any physical object (particle/field) can be achieved by at least 2 equivalent methods:
1)Schrodinger: the time-evolution of the system is given by the state vector. operators are time-independent.

So, in the coordinate representation of:

i) point particle;

$$\hat{X}\rightarrow x$$
$$\hat{P}\rightarrow -i \frac{\partial}{\partial x}$$

we have (depending on the Hamiltonian) the usual Schrodinger/Dirac equation:

$$i\frac{\partial}{\partial t}\Psi (x,t) = H(x, -i\partial_{x}) \Psi (x,t)$$

Here, the wavefunction (probability amplitude) is the component of the state $$|\Psi (t)\rangle$$ in the direction |x> :

$$|\Psi (t)\rangle = \int dx \Psi (x,t) |x\rangle$$

ii) continuous object (field);

$$\hat{\varphi}(\vec{x}) \rightarrow \phi (\vec{x})$$
$$\hat{\pi}(\vec{x})\rightarrow -i \frac{\delta}{\delta \phi (\vec{x})}$$

we have the Schrodinger representation of (relativistic/nonrelativistic) field theory:

$$i\frac{\partial}{\partial t} \Psi [\phi ,t] = H(\phi (\vec{x}), -i\frac{\delta}{\delta \phi (\vec{x})})\Psi [\phi ,t]$$

Notice that this equation is a functional differential equation for the wave functional $$\Psi [\phi ,t]$$ of the quantum field. As in the case of point particle, this number (probability amplitude) reoresents the component of the state $$|\Psi (t)\rangle$$ in the direction $$|\phi \rangle$$ :

$$|\Psi (t)\rangle = \int \cal{D}\phi \Psi [\phi,t] |\phi \rangle$$

and the square of the modulus of the amplitude is proportional to the probability that the field has the configuration $$\phi (\vec{x})$$ at t (i.e., the field is at the point $$\phi$$ in the function space).

Functional differential can be thought of as an infinite set of coupled partial differential equations. So you can imagine the "fun" when trying to solve them. Indeed, very few methods are known to solve these equations. This is why textbooks avoid the Schrodinger representation in field theory. Instead, they work with the following (equivalent) operator representation:

2) Heisenberg: the dynamics is controlled by operators. the state vectors are time-independent.

For QM of particle, you have

$$\frac{d}{dt}\hat{X}(t) = [iH , \hat{X}(t)]$$

and similar equation for $$\hat{P}(t)$$ .
In the theory of fields, the (operator) Heisenberg equations are:

$$\partial_{t} \hat{\varphi}(\vec{x},t) = [iH , \hat{\varphi}(\vec{x},t)]$$

$$\partial_{t} \hat{\pi}(\vec{x},t) = [iH , \hat{\pi}(\vec{x},t)]$$


Now, consider (for example) a massless scalar field theory;

$$H = 1/2 \int d^{3}x (\hat{\pi}^{2} + |\nabla \hat{\varphi}|^{2})$$

In the operator representation, Heisenberg's equations give the operator equation of motion;

$$\partial_{\mu}\partial^{\mu} \hat{\varphi}(\vec{x},t) = 0$$

As you might know, this equation is (very) easy to solve (naturally) in terms of crreation & annihilation operators. (I assume, you know what to do next with the solution).

In the Schrodinger representation, we can work with a coordinate base $$|\phi \rangle$$ for Fock space where the (time-independent) operator $$\varphi$$ is diagonal;

$$\hat{\varphi}(\vec{x}, 0) |\phi \rangle = \phi (\vec{x}) |\phi \rangle$$
(the eigenvalue $$\phi$$ is an ordinary function of $$\vec{x}$$ )

This equation together with

$$\hat{\pi}(\vec{x}, 0) = -i \frac{\delta}{\delta \phi (\vec{x})}$$

turn the Schrodinger equation

$$i\partial_{t} |\Psi \rangle = H |\Psi \rangle$$

into a functional differential equation;

$$i\partial_{t} \Psi [\phi ,t] =1/2 \int d^{3}x (-\frac{\delta^{2}}{\delta \phi^{2}(\vec{x})} + |\nabla \phi |^{2}) \Psi [\phi ,t]$$

where
$$\Psi [\phi ,t] \equiv \langle \phi | \Psi (t) \rangle$$

If you have the "required skills", then you can solve this equation and reproduce all the results of the operator representation.In the Schrodinger representation we do not need to speak about (eventhough we can) creation and annihilation operators.

To summarize:
In a given classical system [such system is any 3-word selection from the pairs; (particle,field),(free,interactive),(relativistic,non-relativistic)]:

1)the operator representation is possible if, and only if, the Schrodinger representation is possible.

2) the corresponding quantum theory (QM or QFT) can be solved in any operator representation if, and only if, it can be solved by an appropriate Schrodinger equation.

So, It is not true to say that the functional Schrodinger equation does not solve interacting field theories. Indeed, the Rayleigh-Schrodinger perturbation theory (which is familliar from QM) can (always) be used to calculate the S-matrix. However, it is a lot easier to determined the S-matrix in terms of the Green's functions of the operator representation.
Remember that (apart from overly simplified mathematical models) interacting field theories do not have exact solutions. This fact (naturally) does not depend on the representation(see 2 above). How the S-matrix is calculated does vary from representation to representation.

regards

sam

 

[CarlB]

Remember that (apart from overly simplified mathematical models) interacting field theories do not have exact solutions.

Actually, this sentence is a statement of your understanding of interacting field theories, at least until you've gone to the trouble of examining all possible interacting field theories and their solutions.

 

[Demystifier]

Samalkhaiat, very good post.

If someone is intersted in methods for finding solutions of the functional Schrodinger equation, see:
http://arxiv.org/abs/hep-th/9605004

 

[samalkhaiat]

Actually, this sentence is a statement of your understanding of interacting field theories,

My understanding is that of all mathematicians and physicists.

at least until you've gone to the trouble of examining all possible interacting field theories and their solutions

Sorry to disappoint you, I have done that.

 

[samalkhaiat]

]"DIRAC'S IS THE RELATIVISTIC SHRODINGER EQUATION."

Thats not really morally true either.

1) Dirac and Schrodinger equations both have the same general form;

$$i\partial_{t}\Psi = H\Psi$$

i.e., they say the same thing about the time evolution of the state.

2) The derivation of dirac's equation is based on the assumption that the general form of Schridinger equation is valid in the relativistic domain.
Indeed, this is how Dirac arrived at his equation.
So, in this sense, it is perfectly ok to regard Dirac's equation as the relativistic version of Schridinger's.
The fact that Dirac's describes a spin 1/2 particle is consistent with the known fact (from the representation theory of Poincare group) that intrinsic spin is a relativistic degree of freedom (spin is a relativistic concept).

The Dirac equation treats spin 1/2 particles, but for instance is completely unable to treat spin 0 (Klein-Gordon).

This is not true because each component of Dirac spinor does satisfy the K-G equation.
Please note that I am talking about Dirac's equation not theory! Dirac theory (= Dirac's equation + the sea postulate) is field theory not QM of single particle.

The KG equation is really the relativistic version of the Schrodinger equation.

I have good reasons to say it is not:

1) In QM, it is postulated that the time evolution of the system is given by a first order (in time) differential equation. Schrodinger and Dirac equations are linear in the time derivative, while K-G's is not.

2) as you said Dirac (and Schrodinger) equations describe (the atomic) electrons very well. Klien-Gordon equation does not provid such description.

3) and most importantly, in contrast to Schrodinger and Dirac equations, the probability density derived from the K-G equation is not necessarily positive definite.

So I object to stating that field theory is morally equivalent to some sort of relativistic Schrodinger equation, its not.

I do not know the "moral" of the subject. However I know that QFT can be formulated in terms of an appropriate Schrodinger equation.


regargs

sam

 

[samalkhaiat]

But the equations in QFT of Heisenberg representation do have the analogous forms as the Dirac equaiton or Schrodinger equation in QM of Schrodinger representation. This is interesting but I can't find a concise explanation why at the very minute now.

The relavent symmetry group determines the form of the Lagrangian which in turn determines the form of the (operator) equation of motion. However, in the full QFT (as in quantum gauge theories) it is not guaranteed that the field operator $$\hat{A}(x)$$ always satisfies the same differential equations as does the "classical" field function $$A(x)$$ .

regards

sam

 

[Quantum River]

Gross Nobel lecture on QFT and QCD.
http://nobelprize.org/nobel_prizes/physics/laureates/2004/gross-lecture.pdf

 

[Hans de Vries]

Haelfix: "The KG equation is really the relativistic version of the Schrodinger equation."

samalkhaiat: I have good reasons to say it is not:

1) In QM, it is postulated that the time evolution of the system is given by a first order (in time) differential equation. Schrodinger and Dirac equations are linear in the time derivative, while K-G's is not.

To go from Dirac to Schroedinger you apply it twice to:

-1- Get the Laplacian (2nd order) which leads to the atomic orbits.
-2- Square the covariant derivative in t which gives 3 terms containing V.
-3- Throw away the term containing $V^2$ because it's very small.
-4- Throw away the term containing $\partial V /\partial t$ because V doesn't change in time
-5- Maintain the term which has V in 1st order and a factor 2 (it's a cross term)
-6- Presume there is no magnetic vector potential A.

Further throw away:

-6- Pauli's spin term which defines Dirac's inherent magnetic dipole moment.
-7- Throw away the effective electric dipole moment term.

(The latter term is a consequence of the inherent magnetic dipole moment:
An electron moving in a pure E field has an extra energy term because the
E field transforms partly into a B field in the electron's frame. This translates
to an effective electric dipole moment in the rest frame which has an E field
only.)

And, then, nastiest of all:

-8- Replace

$$\ \ \ \frac{\partial^2 \psi}{\partial t^2}\ + \ \frac{m^2c^4}{\hbar^2}\psi\ \ \ \ \mbox{with}\ \ \ \ -2\frac{imc^2}{\hbar} \ \frac{\partial \Psi}{\partial t}$$

(see next post for details)

The latter is correct in stationary cases which describe, for instance,
the atomic orbits, where E(nergy) is constant and the same everywhere.
Some care has to be taken with the interpretation of the time-dependent
Schroedinger equation. The approximation is actually OK unless the
shape of the wave function changes very fast.

So, the first order derivative in t of both the Dirac and Schroedinger
equations have a very different origin and a very different meaning.

2) as you said Dirac (and Schrodinger) equations describe (the atomic) electrons very well. Klien-Gordon equation does not provid such description.

Klein Gorden leads just as well to Schroedingers equation if you make sure
that you apply the covariant derivative squared and again:

-8- Replace

$$\ \ \ \frac{\partial^2 \psi}{\partial t^2}\ + \ \frac{m^2c^4}{\hbar^2}\psi\ \ \ \ \mbox{with}\ \ \ \ -2\frac{imc^2}{\hbar} \ \frac{\partial \Psi}{\partial t}$$

Which in case of stationary solutions (where E is constant and the same
everywhere) and a close enough approximation in all but wave functions
with a very fast changing shape.

3) and most importantly, in contrast to Schrodinger and Dirac equations, the probability density derived from the K-G equation is not necessarily positive definite.
sam

We have for the probability density for Schroedinger's and Klein Gordon's
equation:

[tex]\phi^* \phi \ \ \ \mbox{and} \ \ \ \ i\left(\phi^*\frac{\partial \phi}{\partial t} - \phi\frac{\partial \phi^*}{\partial t} \right) [/itex]

Note that these are the SAME (!) for stationary solutions except for a
normalization factor of $2E/\hbar$ and as long a E is real. The KG probability
density is positive for electrons and negative for positrons.Regards, Hans

 

[Hans de Vries]

-8- Replace

$$\ \ \ \frac{\partial^2 \psi}{\partial t^2}\ + \ \frac{m^2c^4}{\hbar^2}\psi\ \ \ \ \mbox{with}\ \ \ \ -i\frac{2mc^2}{\hbar} \ \frac{\partial \Psi}{\partial t}$$

I should elaborate a bit on this, let:

$$\psi \ =\ \Psi\ e^{-iE_o/\hbar\ t}$$

Where $\Psi$ is the Schroedinger wavefunction and $\psi$ is the relativistic one, with $E_o$
representing the rest-mass energy, then the second order derivative in time becomes:$$\frac{\partial^2 \psi}{\partial t^2}\ \ = \ \ \left(\frac{\partial^2 \Psi}{\partial t^2}\ -\ i\frac{2mc^2}{\hbar}\ \frac{\partial \Psi}{\partial t}\ -\ \frac{m^2c^4}{\hbar^2}\ \Psi \right)\ e^{-imc^2/\hbar\ t}$$

- The first term (2nd order derivative) is thrown away. Presumed very small.
- The middle term becomes the 1st order derivative in time in Schroedingers equation.
- The third term cancels with the $m^2$ term from the Klein Gordon and Dirac equations.

The high frequency part $e^{-imc^2/\hbar\ t}$ is removed throughout the equation by replacing $\psi$ by $\Psi$

Interesting is that the factor 2m in the 1st order derivative term leads to the classical
non-relativistic equation, (as it should be):

$$E\ =\ \frac{p^2}{2m}$$ The time dependent Schroedinger equation is written as:

$$i\hbar\ \frac{\partial \Psi}{\partial t}\ \ =\ \ -\frac{\hbar^2}{2m}\ \ \frac{\partial^2 \Psi}{\partial x^2}\ +\ V\Psi$$

Notice that there is also a factor 2 in the potential energy term $V\Psi$. This because
it is a cross product as well, coming from the square of the covariant derivative.



Regards, Hans

 

[Haelfix]

I was going to write a long reply but Hans beat me to it.

"The relavent symmetry group determines the form of the Lagrangian which in turn determines the form of the (operator) equation of motion. However, in the full QFT (as in quantum gauge theories) it is not guaranteed that the field operator always satisfies the same differential equations as does the "classical" field function "

Bingo. It is this fact, along with mathematical subtelties (I really don't want to leave physicist lvl of rigor here) that separates conventional relativistic qms from field theory. They really, truly are cousins and not clones.

People have tried to make 'generalized' finite degrees of freedom Shroedinger equations for years now, that approximates the field theoretic solutions in aomw limit, usually by introducing nonlinearities and so forth. That program hasn't led to anything interesting.

Keep in mind, there is already an ambiguity in the interacting case simply between different 'pictures'. It is absolutely NOT guarenteed that the Heisenberg picture equals the Schroedinger picture, though its often presented that way in intro textbooks. In general, you require the axioms of the Stone-Von Neumann theorem to hold for this to be the case. Further, the unitary transformation that links the two has to be defined within your space.
Many papers have shown explicit examples where this fails to be the case. One can usually massage it to something close, but be sure its ambiguos at a mathematicians lvl of rigor (the Haag-Kastler axioms tend to exclude Schrodinger mechanics for noncompact manifolds for instance, but not the Heisenberg picture).

But back to physics lvl of rigor, truth be told, I tend to agree that the KG equation is *already* quite distinct from the Schrodinger equation. Its even worse for the Dirac equation, where you are trying to generalize the Schrodinger eqn + Pauli spinors to describe fermions. What you end up with are Dirac spinors. Clearly, apples and oranges, already at the algebraic level.

Field theory further departs from the dirac equation in rather deep senses as well, as you now have degrees of freedom that are no longer described by the latter.

 

[Anonym]

In QM, it is postulated that the time evolution of the system is given by a first order (in time) differential equation. Schrodinger and Dirac equations are linear in the time derivative.

The latter is correct in stationary cases which describe, for instance,the atomic orbits, where E(nergy) is constant and the same everywhere.Some care has to be taken with the interpretation of the time-dependent Schroedinger equation. The approximation is actually OK unless the shape of the wave function changes very fast.

So, the first order derivative in t of both the Dirac and Schroedinger equations have a very different origin and a very different meaning.

May you explain what do you mean from group theoretical point of view?

 

[samalkhaiat]

To go from Dirac to Schroedinger you apply it twice to:

-1- Get the Laplacian (2nd order) which leads to the atomic orbits.
-2- Square the covariant derivative in t which gives 3 terms containing V.
-3- Throw away the term containing $V^2$ because it's very small.
-4- Throw away the term containing $\partial V /\partial t$ because V doesn't change in time
-5- Maintain the term which has V in 1st order and a factor 2 (it's a cross term)
-6- Presume there is no magnetic vector potential A.

Further throw away:

-6- Pauli's spin term which defines Dirac's inherent magnetic dipole moment.
-7- Throw away the effective electric dipole moment term.

(The latter term is a consequence of the inherent magnetic dipole moment:
An electron moving in a pure E field has an extra energy term because the
E field transforms partly into a B field in the electron's frame. This translates
to an effective electric dipole moment in the rest frame which has an E field
only.)

And, then, nastiest of all:

-8- Replace

$$\ \ \ \frac{\partial^2 \psi}{\partial t^2}\ + \ \frac{m^2c^4}{\hbar^2}\psi\ \ \ \ \mbox{with}\ \ \ \ -2\frac{imc^2}{\hbar} \ \frac{\partial \Psi}{\partial t}$$

(see next post for details)

The latter is correct in stationary cases which describe, for instance,
the atomic orbits, where E(nergy) is constant and the same everywhere.
Some care has to be taken with the interpretation of the time-dependent
Schroedinger equation. The approximation is actually OK unless the
shape of the wave function changes very fast.

-9- throw away the above irrelevant garbage.:grumpy:

So, the first order derivative in t of both the Dirac and Schroedinger
equations have a very different origin and a very different meaning.

Time derivative has different meaning in different equations? WOW, Is this a new discovery?

The KG probability
density is positive for electrons and negative for positrons.

Negative probability? This must be another new discovery. So, howmany time do I need to throw a coin to get (-1) probability?

Sir, K-G is a quation for BOSONS not FERMIONS (electron or positron). the energy eigenvalues of the K-G operator are
$$E=\pm (p^{2} + m^{2})^{1/2}$$

i.e., in addition to the E>0 solutions, we have negative energy solutions. These E<0 solutions leads to negative probability. Such thing as Negative Probability is not acceptable.


regards
sam

 

[samalkhaiat]

May you explain what do you mean from group theoretical point of view?

$$\frac{\partial \Psi}{\partial t} = \lim_{t \rightarrow 0} \frac{U(t)-1}{t} \Psi$$

where

$$U(t) = \exp(-it\hat{H})$$

is the 1-parameter group of time translation which is generated by the Hamiltonian (the flow of H).

regards

sam

 

[Anonym]

Thank you, samalkhaiat. But I ask Hans de Vries and only refer to you for comparison.

 

[Hans de Vries]

-9- throw away the above irrelevant garbage.:grumpy:

If you want to compare the Schroedinger, Dirac and Klein Gordon equations
then it helps very much if you know how they relate precisely mathematically.
If you call that irrelevant than it shows you don't care. If you call that garbage
than it would show that you don't know.

Negative probability? This must be another new discovery. So, howmany time do I need to throw a coin to get (-1) probability?

Word play.

Charge density becomes probability density after normalization and
"negative probability" becomes ordinary positive probability by changing
the sign depending on the charge being positive or negative:

$$\mbox{Charge density:} \qquad J^t\ =\ i(\phi^* \partial^t \phi\ -\phi \partial^t \phi^*)$$

Sir, K-G is a quation for BOSONS not FERMIONS (electron or positron).

Spin-0 bosons, yes of course. Because it misses the two terms handling
the inherent spin 1/2 magnetic moment under relativistic transformation.
See for example: Weinberg (1.1.26)

In this regard one should call the Schroedinger equation an equation for
Spin-0 bosons as well because it misses the same two terms. The first
of these two terms was artificially added by Pauli to the Schroedinger
equation to obtain a non relativistic Spin 1/2 equation. Both terms are
necessary for a relativistic equation.


Regards, Hans

 

[Quantum River]

I am not quite familiar with strong interaction physics. So my judgment may be wrong. Besides his contribution to asymptotic freedom, Gross first derived the heterotic string theory. So Gross may be quite open-minded. Gross’s Nobel lecture is quite surprising about the QFT was nearly discarded by physicists. To a student who has no real impression of the particle physics in 1960s and 1970s, Gross’s words are quite startling. It makes me feel there may be not too many natural principles that are unshakable.

1. Gell-Mann and Weinberg’s attitude to quarks as fictitious devices may be right. If a quark exists, why can’t we separate it? Color confinement may not be the end of the story. I do not quite understand the meaning of “absence of the physical intuition”. What is physical intuition? Is Einstein’s “God does not play dice” physical intuition? Is Newton’s absolute time physical intuition? Is many physicists’ faith the four forces must be unified physical intuition? Gross said “many non-asymptotically-free theories, such as QED, are inconsistent, at very high energies. In the case of QED this is only an academic problem, since the trouble shows up at enormously high energy.”

2. Yang-Mills field theory was not very promising to lots of physicists in a period, so Gross just followed the general attitude of physicists. It may have no relation with Gross’s understanding of differential geometry.

3. Gross first proved “no renormalizable field theory that consists of theories with arbitrary Yukawa, scalar or Abelian gauge interactions could be asymptotically free.” He second showed non-Abelian gauge theories asymptotically free. So Gross is quite right. Could functional analysis provide these answers?

So Gross gave us a quite good teaching about the history of physics. I can’t see Gross’s “problems” with mathematics. I don’t believe QCD is the end story of the strong interaction. But Gross did do a good work on strong interaction and deserved a Nobel.

Physics and mathematics have different judgment principles. Two rivers will not always run parallel with each other. Two trees on the different ground will not grow into one. If Mathematics is part of the absolute truth, why can it be discovered by purely thinking of human beings? If mathematics is purely artifact of mind, why could it be used in physics? Anthropic principle is just part of the Very problem?
As for education, it is designed to stop people thinking about the hard and important problems.

 

[Anonym]

QR, our managers consider that issue not suitable here in QP of PF and I agree with them. They perhaps will improve software to remove particular posts of “criminals”.

You missed my point. I used D.J.Gross lecture as example to discuss with you interconnections between physics, mathematics and education. I never mentioned his name and personally I think that the lecture does not represent his point of view, it is kind of MWI (I do mentioned name of F. Wilczek, I am sorry, but we are humans and I have open account with Chief Editor of AOP).

To avoid any misinterpretation, I do not know “the absolute truth” and I consider myself the last pretending to have that knowledge.

The lecture require from you to be “quite open-minded”. “The early 1960’s were a period of expirimental supremacy and theoretical impotence.” … “There was hardly any theory to speak of.” In my view this is how things looks for phenomenologist that have no idea what is the theoretical physics. Mostly, from the point of small and relatively insignificant corner of theor physics called the methods of statistical physics in QFT.

Let start from WWII. We start with most dark age in physics. The discovery of 3 neutrons emitted in nuclear division must be hermetically hidden from the outsiders. Instead, the best minds enthusiastically raise the engineer projects which have nothing to do neither with the experimental physics nor with the science. After WWII slowly start Renaissance.

“Yang-Mills theory, which had appeared in the mid 1950’s was not taken seriously”.

I consider the paper by C.N. Yang and R.L. Mills, Phys. Rev. 96, 191(1954) the most important achievement in physics of 20 century. It concluded the development started from Galua, Abel, Lie, Lorentz, Poincare, Schrödinger, Weyl and Wigner. The paper present the formulation of the Principal Physical Postulate: Principle of Local Gauge Invariance. First time after Galileo the new symmetries of Nature were introduced into the Game theoretically.

“Symmetries were all the rage”. The progress and triumph was tremendous. AB paper in 1959, the connection between the internal and the external symmetries, M. Gell-Mann and Y. Neeman SU(3) in strong interactions, currents and charges, etc. And, finally, the theoretical prediction of new fundamental fermions (M.Gell-Mann, G.Zwieg and Y.Neeman).

Quantum River:“Gell-Mann and Weinberg’s attitude to quarks as fictitious devices may be right.”

Don’t believe to the liar. The atmosphere M.Gell-Mann worked in it is described by G.Zwieg (“Origins of the Quark Model”, CALT-68-805, DOE RESEARCH AND DEVELOPMENT REPORT): “When the physics department of a leading University was considering an appointment for me, their senior theorist, one of the most respected spokesmen for all of theoretical physics, blocked the appointment at a faculty meeting by passionately arguing that the ace model was the work of a “charlatan” “.

I do not know who was that “senior theorist”, but I have no doubt that he was charlatan.

In addition, at the same time enormous progress in electrodynamics was made by R.J. Glauber et al. And this is only part of the story of 1950’s and 1960’s.

Quantum River:“What is physical intuition?”

I already gave you my definition: just every day sense of reality. So, what is every day sense of reality? Here you are:

A. Einstein:”the remembering of what we are really observe and what we do not, has probably some heuristic value. However, from the principal point of view, the attempt to formulate the theory based only on observable quantities is completely nonsense. Because in the reality everything that happens are just an opposite. Only the theory itself can decide what is and is not observable. You see, the observation, generally speaking, is very complicated notion…”

This also provide explanation why quarks are real objects. If you feel that you do not have “physical intuition” yet, take blindly all A. Einstein statements. I have no example when he was wrong.

Quantum River:“Is Einstein’s “God does not play dice” physical intuition?”

No. A.Einstein refer to the foundations of the Classical Physics. Any new postulate introduced must be in compliance with all what was done during last 350 years.That is a reason why the Principle of Local Gauge Invariance provide the ultimate truth. That is a reason why it use the incredible beauty of differential geometry.

Quantum River:“Gross first proved “no renormalizable field theory that consists of theories with arbitrary Yukawa, scalar or Abelian gauge interactions could be asymptotically free.” He second showed non-Abelian gauge theories asymptotically free. So Gross is quite right. Could functional analysis provide these answers?”

I know only one notion of the physical theory: the adequate description of the natural phenomena. All rest are either preliminary models or irrelevant garbage.

The functional analysis is the mathematical framework suitable for the physical applications iff there exist the asymptotically free region either at long or at short distances. The functional analysis is the adequate mathematical framework used in non-relativistic limit of QT.

It is written “I would show that there existed no asymptotically free field theory… I knew that if the fields themselves had canonical dimensions, then for many theories this implied that the theory was trivial, i.e. free…The second part of the argument was to show that there were no asymptotically free theories at all.”

I have no idea what the writer is talking about.

Quantum River:“Two rivers will not always run parallel with each other.”

Quantum River:“But we may find a more fundamental duality which should encompass the particle-wave one as a special case. In mathematics, there is Langlands program. Fermat's last theorem is just a very small part of the program. I will not say more.”

The Fermat's last theorem is to my taste very strange story. Fermat was one of the greatest physicists of all times. He first presented the formulation of the Principle of Least Action. But he was also the outstanding mathematician and the judge. The court is the best place where you may study a human nature. My guess that the Fermat's last theorem is a joke. It starts with n=2 solution. And it is “no go” theorem. Fermat should know that to prove “no go” theorem is virtually impossible. My guess that Fermat predicted that the attempts will be made and this will take hundreds of years. Now, suppose one find the solution after years of hard work. Bravo! So what?

Be careful, you may choose a wrong river.

Quantum River:”As for education, it is designed to stop people thinking about the hard and important problems.”

QR, you have no idea what you are talking about:

Reilly:” When I taught this stuff, I used a class to discuss this very issue, just to show that there are many ways to frame a problem, but that some are more equal than others.”

Hans de Vries:” If you want to compare the Schroedinger, Dirac and Klein Gordon equations then it helps very much if you know how they relate precisely mathematically. If you call that irrelevant than it shows you don't care. If you call that garbage than it would show that you don't know.”

Reilly and Hans de Vries are obviously the teachers. As extreme example, you was not lucky to attend course of lectures given by Y. Aharonov on paradoxes in QM.

 

[CarlB]

As extreme example, you was not lucky to attend course of lectures given by Y. Aharonov on paradoxes in QM.

I would love to hear more about this. The reason I started messing around with physics, after so many years outside, was to write a book on the paradoxes of QM.

 

[Quantum River]

Mathematics is mainly from mind, while physics is mainly from experiments. There is a vast gap here. In the real life, these philosophical things are actually of least importance for lots of people including me. I may use them as a guide of treasure finding, but will not take any of these things too serious. Reading other people's work as a laborer is a main thing for me before some real construction.
As for the education, I have habituated myself to think alone, read alone and calculate alone. I am not lucky enough to have some famous men as teachers. But this is life and I have habituated myself to it.

 

[Quantum River]

Anonym, sorry for my rudeness.
Your posts provide many resources to think. Although I don't agree with some of your argument, I do think they pose the right and important questions.
I think physics is at the verge of another revolution of thoughts. The concepts of time, space, QM could be seriously changed in the coming revolution. Although Einstein is greatly influenced by Mach, general relativity is not Machian. General relativity may contain the same problems and paradoxes that are possessed by the Newtonian gravity. Quantum mechanics is a good theory, but it contains the bizarre problem of measurement. For a long time, general relativity could not be quanized, which could not be accidental. Are the present concepts of time, space right in QM?
Quantum River

 

[CarlB]

Quantum mechanics is a good theory,
but it contains the bizarre problem of measurement.

Yes. Without measurement, QM becomes a simple wave theory.

Are the present concepts of time, space right in QM?

I suspect that our notion of time is very simple compared to what is needed. Rather than try to understand time, I've been interested in a version of QM that doesn't have time. It's Schwinger's measurement algebra (SMA). The objects are quantum states, sort of like qubits, but more general. While one doesn't have time explicitly in the SMA, one does have order, as in the difference between initial and final states.

 

[Anonym]

Anonym, sorry for my rudeness.

?

Are the present concepts of time, space right in QM?

QR, name of your thread is “The role of wave function in QED”. You discuss relativistic QM. Perhaps, I will be able to answer your question if you will explain to me what are the present concepts of time, space in relativistic QT.

Dany.

 

[samalkhaiat]

If you want to compare the Schroedinger, Dirac and Klein Gordon equations
then it helps very much if you know how they relate precisely mathematically.
If you call that irrelevant than it shows you don't care.

The nonrelativistic limit of Dirac and/or K-G equations is an IRRELEVANT issue for this thread in general and my post in particular. This was the reason for using the word "irrelevant" to describe your post. I also used "garbage" to describe your post, because you "explained", very badly, the non-relativistic limit by some 13 confused steps!

If you call that garbage
than it would show that you don't know.

Well, I can do the job (you tried to do), and get Schrodinger equation from the K-G equation, by TWO very very easy steps. And I do that without any "throw away" business;

In the K-G equation

$$\frac{1}{c^{2}} \frac{\partial^{2}\phi}{\partial t^{2}} - \nabla^{2} \phi + (\frac{mc}{\hbar})^{2}\phi = 0$$

1) put

$$\phi = \exp\{\frac{i}{\hbar}(S - m c^{2}t)\}$$

and get

$$\frac{1}{2mc^{2}}\{i \hbar \frac{\partial^{2}S}{\partial t^{2}} - (\frac{\partial S}{\partial t})^{2}\} e^{iS/ \hbar} + \frac{\hbar}{i} \frac{\partial}{\partial t} (e^{iS/ \hbar}) = \frac{\hbar^{2}}{2m} \nabla^{2} (e^{iS/ \hbar})$$

2) take the non-relativistic limit: $$c \rightarrow \infty$$ , put

$$\Psi = e^{iS/ \hbar}$$

and get

$$\frac{\hbar}{i} \frac{\partial}{\partial t}\Psi = \frac{\hbar^{2}}{2m} \nabla^{2} \Psi$$

What do you call this equation? Did I need two posts and some 14 "steps" to explain this rather simple matter?

Spin-0 bosons, yes of course. Because it misses the two terms handling
the inherent spin 1/2 magnetic moment under relativistic transformation.
See for example: Weinberg (1.1.26)

Sorry, I have to use the word "garbage" again. You need to read about the representation theory of Lorentz group SO(1,3). When you do that, you will see that:
1) K-G equation is the ONLY equation satisfied by the spin-0 (scalar) representation of SO(1,3).
2) Dirac's is the only suitable equation for the bispinor representation of so(1,3).
3) Maxwell's are for the massless spin-1 (vector) representation.
4) the linear Einstein equation is satisfied by the massless spin-2(tensor) representation of so(1,3).

regargs

sam

 

[Hans de Vries]

Well, I can do the job (you tried to do), and get Schrodinger equation from the K-G equation, by TWO very very easy steps. And I do that without any "throw away" business;

In the K-G equation

$$\frac{1}{c^{2}} \frac{\partial^{2}\phi}{\partial t^{2}} - \nabla^{2} \phi + (\frac{mc}{\hbar})^{2}\phi = 0$$

1) put

$$\phi = \exp\{\frac{i}{\hbar}(S - m c^{2}t)\}$$

and get

$$\frac{1}{2mc^{2}}\{i \hbar \frac{\partial^{2}S}{\partial t^{2}} - (\frac{\partial S}{\partial t})^{2}\} e^{iS/ \hbar} + \frac{\hbar}{i} \frac{\partial}{\partial t} (e^{iS/ \hbar}) = \frac{\hbar^{2}}{2m} \nabla^{2} (e^{iS/ \hbar})$$

2) take the non-relativistic limit: $$c \rightarrow \infty$$ , put

$$\Psi = e^{iS/ \hbar}$$

and get

$$\frac{\hbar}{i} \frac{\partial}{\partial t}\Psi = \frac{\hbar^{2}}{2m} \nabla^{2} \Psi$$

What do you call this equation? Did I need two posts and some 14 "steps" to explain this rather simple matter?

May I remind you that I described how to go from Dirac, including the
EM interactions, to Schroedinger. That's a whole different story as going
from the zero interaction Klein Gordon equation to Schroedinger.

If you would have actually read my post then you should have seen that
what you have done is simply a copy of my post here:
https://www.physicsforums.com/showpost.php?p=1232057&postcount=44
Only somewhat more complex.
Regards, Hans

 

[Anonym]

$$\frac{\partial \Psi}{\partial t} = \lim_{t \rightarrow 0} \frac{U(t)-1}{t} \Psi$$

where

$$U(t) = \exp(-it\hat{H})$$

is the 1-parameter group of time translation which is generated by the Hamiltonian (the flow of H).

regards

sam

Sorry, I have to use the word "garbage" again. You need to read about the representation theory of Lorentz group SO(1,3). When you do that, you will see that:
1) K-G equation is the ONLY equation satisfied by the spin-0 (scalar) representation of SO(1,3).
2) Dirac's is the only suitable equation for the bispinor representation of so(1,3).
3) Maxwell's are for the massless spin-1 (vector) representation.
4) the linear Einstein equation is satisfied by the massless spin-2(tensor) representation of so(1,3).

You need to read about the representation theory of Lorentz group SO(1,3). When you do that, you will see that time is operator (matrix). Your definition above is correct iff t is c(or r)-number. For relativistic QM you should define matrix derivative (first order indeed).

Try to be consistent. If Hans de Vries presentation is a "garbage", then it will remain garbage even if you present it “by TWO very very easy steps”. In addition, try to avoid the NY Times style, since it present you as idiot.

QR, this may be interesting for you: J. Hilgevoord, “Time in quantum mechanics: a story of confusion”.

 

[samalkhaiat]

You need to read about the representation theory of Lorentz group SO(1,3). When you do that, you will see that time is operator (matrix).

I need to be an i***t to be able to make such a super-garbage statement.

Your definition above is correct iff t is c(or r)-number.

Be useful and prove your statement
If you can't then Avoid using "iff" because it would present you as an i***t.

Try to be consistent. If Hans de Vries presentation is a "garbage", then it will remain garbage even if you present it “by TWO very very easy steps”. In addition, try to avoid the NY Times style, since it present you as idiot.

How about Try to understand what you read. I stated the reason for using the word "garbage". So go and do consistency check.

QR, this may be interesting for you: J. Hilgevoord, “Time in quantum mechanics: a story of confusion”

NOT INTERESTED. Maybe the "confusion" part is suitable for you.

regaeds

 

[Hans de Vries]

Mr, samalkhaiatFirst you say that Dirac's equation is capable of treating spin-0 particles:

Haelfix: The Dirac equation treats spin 1/2 particles, but for instance is completely unable to treat spin 0 (Klein-Gordon).

samalhiaiat: This is not true because each component of Dirac spinor does satisfy the K-G equation.

This of course is ignoring:

1) Spin Statistics.
2) The diplole moment interaction terms.

Now when I point out (2), referring to Weinberg (1.1.26)

Dirac also iterated Eq.(1.1.23) obtaining a second-order equation,
which turned out to have just the same form as the Klein-Gordon
equation (1.1.4) except for the presence on the right-hand-side
of two additional terms

$$\left[ -e\hbar c\ {\vec \sigma}\cdot{\vec B}\ -\ e\hbar c\ {\vec \alpha}\cdot{\vec E} \right]\ \psi \qquad \qquad \mbox{(1.1.26)}$$

For a slowly moving electron, the first term dominates, and represents
a magnetic moment in agreement with the value (1.1.8) found by
Goudsmit and Uhlenbeck

Then you insult me for a second time without reason by calling this "garbage".

Now for the careful reader of this thread like Anonym this kind of insulting
behavior gives a very strange impression. He told you what this sort of
behavior makes you look like by bluntly using the i-word.

From your response above I do no get the impression that you have
understood this as an appeal to change your behavior and to discuss
in a more civilized manner, unfortunately.Regards, Hans

 

[Anonym]

NOT INTERESTED. Maybe the "confusion" part is suitable for you.

regaeds

QR mean Quantum River. The suggestion adressed to him and related to his post #55 and my answer # 57.

Regaeds.

 

[Anonym]

Now when I point out (2), referring to Weinberg (1.1.26)

In my copy of S.Weinberg it is i also present in the second term which make that operator non self-adjoint. P.A.M. Dirac throw it out without explanation. It seems to me that it is discussed in J.J. Sakurai “Advanced QM” also. What is your comment on it?

 

[Hans de Vries]

In my copy of S.Weinberg it is i also present in the second term which make that operator non self-adjoint. P.A.M. Dirac throw it out without explanation. It seems to me that it is discussed in J.J. Sakurai “Advanced QM” also. What is your comment on it?

The effective Electric moment of the (moving) electron which Dirac
presented in one line with the inherent Magnetic moment in his original
paper seems almost absent in the textbooks.
(See the list I made scanning my books)

Here's Dirac's notation (which uses a comma for the dot product
and the matrix rho which turns the matrix sigma into alpha):

$$\frac{eh}{c}(\sigma,H) + i\frac{eh}{c}\rho_1(\sigma,E)$$

Dirac initially dismissed it but it goes straight into QED. It should be right.
A moving electron in an electric field E should have an extra energy term
depending on E since it sees E partly transformed to a magnetic field B
in its rest frame.

Most books don't mention it at all, and if they do, they mostly do so
without discussion. Feynman's book has it and he also gives an alternative
form: (Ninth Lecture)

$$\left((i\nabla_\mu-eA_\mu)^2\ -\ \frac{i}{2}e\gamma_\mu\gamma_\nu F_{\mu\nu}\right)\Psi\ =\ m^2 \Psi$$

Which is the twice iterated form of the Dirac equation with the dipole
moments hidden in the F term now. You need to evaluate this term
first before seeing that there is a dependency on the E field and how.

I found this version in Zee (III-6-2)

$$\left(D_\mu D^\mu - \frac{e}{2}\sigma^{\mu\nu} F_{\mu\nu} +m^2\right)\psi\ =\ 0$$

I think it is a very basic property and should be discussed in every text
book, and, I think its behavior under rotation and Lorentz transformation
should be discussed. (Which I was planning to study)

Regards, HansThe effective Electric Dipole Moment of the moving electron in the literature:

Aitchison,Hey           Gauge Theories in Particle Physics    No
Berestetskii,Lifshìts   Relativistic Quantum Theory           Yes §32
Bjorken & Drell         Relativistic Quantum Mechanics        No
Chaichian & Nelipa      Intr. to Gauge Field Theories         No
Cottingham & Greenwood  Intr. To the Standard Model           No
Dirac (paper)           The Quantum Theory of the Electron    Yes
Itzykson & Zuber        Quantum Field Theory                  Yes  2-2-3 
Feynman                 Lectures on Physics III               No
Feynman                 Quantum Electrodynamics               Yes 12th lecture 
Feynman                 The Theory of Fundamental Processes   No
Griffits                Intr. to Elementary Particles         No
Halzen & Martin         Quarks & leptons                      No
Heitler                 The Quantum Theory of radiation       No 
Maggiore                A modern introduction to QFT          No
Peskin & Schroeder      Introduction to QFT                   No
Sakurai                 Advanced Quantum Mechaninics          Yes (3.81)
Sakurai                 Modern Quantum Mechanics              No 
Schwinger               Quantum Kinematics and Dynamics       No
Schwinger (editor)      Selected papers on QED                Yes Oppenheimer
Ramond                  Field Theory: A modern Primer         No
Ryder                   Quantum Field Theory                  No
Straeter,Wightman       PCT,Spin,Statistics and all that      No
Strange                 Relativistic Quantum Mechanics        Yes 4.6
Tomonaka,               The story of Spin                     No
Veltman                 Diagrammatica                         No
Weinberg                The Quantum Theory of Fields          Yes (1.1.26)
Zee                     QFT in a nutshell                     (F) (III-6-2)

 

[Anonym]

A moving electron in an electric field E should have an extra energy term depending on E since it sees E partly transformed to a magnetic field B in its rest frame.

-7- Throw away the effective electric dipole moment term.

The latter term is a consequence of the inherent magnetic dipole moment:
An electron moving in a pure E field has an extra energy term because the
E field transforms partly into a B field in the electron's frame. This translates
to an effective electric dipole moment in the rest frame which has an E field
only.

It should be right.

I doubt. What you say, it how it should be (as in classical ED). But H-term is hermitian and E-term is antihermitian. Now here I refer only to P.A.M. Dirac “Principles of QM, 4 Edition, Ch. 11, Par. 70 “Existence of spin”. P.A.M. Dirac do not use the Kronecker (direct) products of underlined Clifford algebra. His presentation mathematically is obscure side by side with the brilliant physical consideration. In addition he is not familiar with C.N. Yang and R.L. Mills paper. If I understand correctly, he obtain Hamilton operator not hermitian in the Schrödinger picture and hermitian in the Heisenberg picture. In addition, if it is the classical ED limit, you expect the electron magnetic moment to be measurable (observable). But N.Bohr said it is not.

I consider this point the main “souvenir” of QED using Vanesch terminology.
I have unfinished business with the coherent states now. I am planning to study next this point also. Thank you very much.

Regards, Dany.

 

[Hans de Vries]

But H-term is hermitian and E-term is antihermitian. Now here I refer only to P.A.M. Dirac “Principles of QM, 4 Edition, Ch. 11, Par. 70 “Existence of spin”.

Indeed,

$$\left[-e\hbar c ({\vec \sigma} \cdot {\vec B})\ -\ ie\hbar c \rho_1 ({\vec \sigma} \cdot {\vec E}) \right] \psi\ \ = \nonumber$$

$$-e\hbar c \left( \begin{array}{cc|cc} B_z & B_x-iB_y & iE_z & iE_x+E_y \\ B_x+iB_y & -B_z & iE_x-E_y & -iE_z \\ \hline iE_z & iE_x+E_y & B_z & B_x-iB_y \\ iE_x-E_y & -iE_z & B_x+iB_y & -B_z \end{array} \right)\ \left( \begin{array}{c} \psi_1 \\ \psi_2 \\ \psi_3 \\ \psi_4 \end{array} \right)$$But in the standard gamma matrix presentation, which inverts the signs
in the lower half, the entire matrix becomes Hermitian:

$$-e\hbar c \left( \begin{array}{cc|cc} B_z & B_x-iB_y & iE_z & iE_x+E_y \\ B_x+iB_y & -B_z & iE_x-E_y & -iE_z \\ \hline -iE_z &-iE_x-E_y & -B_z & -B_x+iB_y \\ -iE_x+E_y & iE_z & -B_x-iB_y & B_z \end{array} \right)\ \left( \begin{array}{c} \psi_1 \\ \psi_2 \\ \psi_3 \\ \psi_4 \end{array} \right)$$Regards, Hans

 

[Anonym]

I just quoted P.A.M. Dirac. I lost you. I should reproduce your calculations.

Regards, Dany.