The role of wave function in QED

[Quantum River]

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.

I have tried to read the original paper of Tomonaga in 1946 Progress of Theoretical Physics, but just not could get the original paper. I have no access to Progress of Theoretical Physics. And this is the only paper that concerns the foundations of QED in my knowledge. Dyson's review The Radiation of Tomonaga, Schwinger, and Feynman gives a rather short description of Outline of the Theoretical Foundations(Dyson's words). Could anyone recommend me some papers in this direction.

What I am confused about is the role of wave function in QED. In QM, the wave function means the distribution probability in the space (Born)? Then what concept or quantity in QFT means such things (distribution probability in the space) or corresponds to the wave function ? Is there some correspondence principle between QM and QED? Could the basic QED equation retreat to QM Schrodinger equation or Dirac equation under some conditions?

For example, in the calculation of Lamb thift, what is the use of wave function (1s state of hydrogen)? It seems the Lamb shift has some connections with the wave function value at the r=0 point. Look at Baranger, Bethe, and Feynman's calculation. (Phys. Rev., Vol. 92, NO. 2,482) They use the wave function phi(r=0) to calculate the Lamb shift. But there is no wave function (of course no 1s wave function) in Quantum field theory? When calculating scattering, it is easier to understand the role of QFT, but when considering the Bounded states, I just could not understand how to calculate the bounded state wave function from QFT.

Quantum River

 

[Demystifier]

To understand a difference between two different quantum theories one should first understand the difference between the corresponding classical theories. While QM is a theory of particles, QED is a theory of fields. If you understand the difference between particles and fields in classical mechanics, the difference in the quantum case is essentially the same.

However, in reality, things are not that simple. QED is not only a theory of fields, it is also a theory of particles (photons and electrons). It does not have an analog in the classical case. The correspondence between particles and fields is essentially the correspondence between wave functions and field operators. Wave functions are certain matrix elements of the field operators. For textbooks where this is explained to some extent see the textbooks of Ryder or Schweber. You may also see
http://xxx.lanl.gov/abs/quant-ph/0609163

 

[Quantum River]

To understand a difference between two different quantum theories one should first understand the difference between the corresponding classical theories. While QM is a theory of particles, QED is a theory of fields. If you understand the difference between particles and fields in classical mechanics, the difference in the quantum case is essentially the same.

However, in reality, things are not that simple. QED is not only a theory of fields, it is also a theory of particles (photons and electrons). It does not have an analog in the classical case. The correspondence between particles and fields is essentially the correspondence between wave functions and field operators. Wave functions are certain matrix elements of the field operators. For textbooks where this is explained to some extent see the textbooks of Ryder or Schweber. You may also see
http://xxx.lanl.gov/abs/quant-ph/0609163

But in Quantum field theory, particle and field are one thing. Electron is a particle in classical physics and Quantum mechanics (Schrodinger Equation), but it is a field in Quantum field theory. On the first thought, QM corresponds to classical physics (such as Newtonian physics), while Quantum field theory corresponds to classical field (such as Maxwellian electromagnetic field). But in Quantum field theory we not only quantize electromagnetic field, but also electron field (Does it exist?). I think the first thought is somewhat naive.

Wave functions are certain matrix elements of the field operators.
Could you explain the sentence above more concretely?

 

[marlon]

If you understand the difference between particles and fields in classical mechanics, the difference in the quantum case is essentially the same.

That's hardly a decent answer to the OP's question. There is far more going on and you are missing out on the most essential part : how particles are connected to quantum fields in QFT.

TO THE OP :

In QFT (eg QED), particles arise due to vibrations of the quantum fields.

Think of a mattress (ie the quantum field) on which "YOU" jump in one place. The surface of the mattress vibrates and due to this vibration, there is energy coming free from the mattress (ie the vibrational energy).
If you take into account that energy is the same as mass (E=mc^2) you can see how the vibration of a quantumfield can mimic a certain particle with mass m and momentum p.

All particles (photons, electrons, etc) are born this way. For example, suppose we have two electrons (which are vibrations of a quantumfield themselves) "sitting" on the mattress. The two electrons cause the mattress to vibrate and the resulting vibrations give us the energy for a particle (a photon for example), as explained before. This is how two electrons interact with each other via a photon. Keep in mind that i am giving a simplified picture here with some loss of accuracy (like the necessary conditions for the quantum fields and conservation laws).

But essentially, in QFT, we have vibrating quantum fields that give us the particles we need. Such quantum fields are written in terms of creation and annihilation operators that acto onto a quantum state.


Finally, let's compare to QM. The main difference between QFT and QM are the basic ingredients :

1) In QM the basic ingredient are wavefunctions.
2) In QFT the basic ingredient are quantum fields.

In QFT, the "old" QM wavefunction is now the quantum field written in terms of creation and annihilation operators. In other words, the QM wavefunction has become a field operator in QFT.


regards
marlon

 

[hellfire]

You can indeed define a "wavefunction" in QFT proceeding in a similar way than for (non relativistic) QM. In QM the variables that define the configuration space are the positions. The wavefunction is a function of the positions in configuration space. In QFT the variables that define the configuration space are the fields. You can define a functional of the fields in configuration space. This "wavefunctional" would obey a Schrödinger-type equation and would be analogous to the wavefunction in QM. Take a look to section 5.2 of http://www.theorie.physik.uni-muenchen.de/~serge/T6/SS03-T6-LN5.pdf .

What you cannot do in QFT, as far as I know, is to define a wavefunction for a particle as in QM. The expression of a localized single particle \phi(x) \vert 0 \rangle "looks like" the expansion in terms of momenta of the eigenstate of postion, however, the product of two of them \langle 0 \vert \phi(x) \phi(y) \vert 0 \rangle is not equal to zero (it is equal to the propagator), which would be required in order to define a position basis and a wavefuction as such.

 

[vanesch]

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.

This is entirely correct. Only, the Schroedinger equation in QFT is quite a complicated beast, as it is an equation for a functional (and not a partial differential equation of a wavefunction over a finite-dimensional configuration space, as in QM). So nobody actually knows how to deal with it directly. Hence, the Schroedinger equation doesn't get much attention in QFT, as we don't know how to use it. In fact, the only thing we can do with it, is to write down the Schroedinger equation in integrated form, and even in an approximated way, and then even only for the case of t= - infinity to t = + infinity. The result of this is the S-matrix, which is nothing else but the solution to the Schroedinger equation for initial condition t = - infinity taken at the point t = + infinity: its elements are the individual integrals of the Schroedinger equation for specific initial conditions ("incoming particles").

What I am confused about is the role of wave function in QED. In QM, the wave function means the distribution probability in the space (Born)? Then what concept or quantity in QFT means such things (distribution probability in the space) or corresponds to the wave function ? Is there some correspondence principle between QM and QED? Could the basic QED equation retreat to QM Schrodinger equation or Dirac equation under some conditions?

The only relationship we have is that of the Born rule for the final solution of the Schroedinger equation (t = + infinity) when we have initial conditions at t = - infinity. It are hence the squares of the elements of the S-matrix, and when properly put into context, it gives you the cross sections for certain reactions.

For example, in the calculation of Lamb thift, what is the use of wave function (1s state of hydrogen)? It seems the Lamb shift has some connections with the wave function value at the r=0 point. Look at Baranger, Bethe, and Feynman's calculation. (Phys. Rev., Vol. 92, NO. 2,482) They use the wave function phi(r=0) to calculate the Lamb shift. But there is no wave function (of course no 1s wave function) in Quantum field theory?

You can use the superposition principle as well in QFT as in QM. If you have the solutions (the matrix elements of the S-matrix) for plane waves, you know that the solution to an initial state which is a superposition of plane waves will be the superposition of the corresponding final states.

When calculating scattering, it is easier to understand the role of QFT, but when considering the Bounded states, I just could not understand how to calculate the bounded state wave function from QFT.

What is usually done (as far as I know, I'm no expert) is to start with a specific bound state (usually obtained in NR QM), then use the superposition principle to "transfer" this to a superposition of final states, and eventually "project" this on other bound states (using the Born rule again, in a basis of bound states). This gives you then the transition probability from the initial bound state to the final bound state.

 

[Demystifier]

But in Quantum field theory, particle and field are one thing. Electron is a particle in classical physics and Quantum mechanics (Schrodinger Equation), but it is a field in Quantum field theory. On the first thought, QM corresponds to classical physics (such as Newtonian physics), while Quantum field theory corresponds to classical field (such as Maxwellian electromagnetic field). But in Quantum field theory we not only quantize electromagnetic field, but also electron field (Does it exist?). I think the first thought is somewhat naive.

Wave functions are certain matrix elements of the field operators.
Could you explain the sentence above more concretely?

Please, see Eq. (74) in
http://xxx.lanl.gov/abs/quant-ph/0609163
Also, read Secs. VIII and IX in it, as well as the post of vanesch above.

 

[reilly]

I'ts important to recognize that any quantum field theory in use is equivalent to a standard quantum theory -- that is, anything that is characterized in terms of particle creation and destruction operators, can be reformulated in the ordinary QM use of Fock space. This is very apparent in non-relativistic many-body theory, and is discussed in any number of publications. Note also that particle creation and destruction can occur, via transitions between subspaces with different numbers of particles.

With the imposition of relativity, life gets more difficult. While we regularly use a momentum operator in QFT, we do not use a position operator. Rather, we use the space-time parametrization (x,y,z,t) as if the spatial coordinates are equivalent to position eigenstates. But we do use wave functions in momentum space, and consider their Fourier transforms as spatial wavefunctions. Yes, we use wavefunctions and state vectors in relativistic QFT -- that's how we compute scattering crossections for example.

Regards
Reilly Atkinson

 

[CarlB]

I'ts important to recognize that any quantum field theory in use is equivalent to a standard quantum theory -- that is, anything that is characterized in terms of particle creation and destruction operators, can be reformulated in the ordinary QM use of Fock space.

I agree completely and think that this is very important. I have a couple QFT books that hint at this, but I have no textbooks or other references that spell it out. Do you know of any other sources?

 

[Truth Finder]

Hi,
The following may work;

http://scienceworld.wolfram.com/physics/QuantumElectrodynamics.html.

It is simple but detailed and summed up.
Doesn't it include the Dirac's Equation?!



------------------------------------------------
Either to work correctly as required, or to leave it.

 

[vanesch]

Hi,
The following may work;

http://scienceworld.wolfram.com/physics/QuantumElectrodynamics.html.

It is simple but detailed and summed up.
Doesn't it include the Dirac's Equation?!

I should look at it in a more detailled way, but this looks like a very garbled way of writing down QED, if not totally wrong...

 

[Truth Finder]

Wrong!
These equations have been absolutely believed of since feynmann... Thanks, to revise your knowledge about QFTs.

Thanks ... and Brgrds.
Nuclear Scientist!

 

[dextercioby]

The eqns need to be checked, as they are written in noncovariant language. But the assertion "Quantum electrodynamics is a generalization of quantum mechanics to include special relativity" is definitely wrong.

Daniel.

 

[Truth Finder]

Dexter,
You are right. Dirac included SR in QM. The other 2 equations included Field Quantization (May be:{DUE TO QUNIZATION OF SOURCE CHARGES}).
You can check them as much you can. But a question arises: What is the original? Is this form or the other or another one "if there is no equivalence"?
By the way, I was just asking about the meaning of its implication for Dirac's Equation for a nonconservative field. But I am sure that they are right and already believed of. I already studied this long before and discussed with many professors. And I think that it is time now for me to understand its meaning and how QED founders made it like this.

Schwartz Vandslire
------------------------------------------------
Either to work correctly as required, or to leave it.

 

[Truth Finder]

Of Course Dexter, Lagrangian Form and Hamiltonian Form are different faces to the same Theory; QED.

Schwartz Vandslire
------------------------------------------------
Either to work correctly as required, or to leave it.

 

[Amr Morsi]

T. Finder,

That's right, these equations are absolutely right.
But the problem is that many some scietists now are talking about the origin of field quantization to be due to the probability of photon itself. What do you think?

No doubt, the wave function in QFTs are that of the charges (and particles in general). But, isn't strange to have Dirac's Equation put explicitly like this (in its general forum)? Got me!

I think, as you said below, this is a very good starting point to reach the whole picture, sir.

But, isn't it some what strange that QM and QFTs in general don't speak about individual events, but about a number of same-system events?
Got you! Yaaiih!


Engineer\ Amr Morsi.

 

[Truth Finder]

OOPS!

The issue of quantum field theories is a puzzle for me till now. But, it is quite apparent that there is some differences in scientists' complete consideration for it. However many of them is talking about the theory as a total, but still away from the details. Why not and it was not till Born who explained the right meaning of the wave function... I think you pointed to that in one of your previous answers.

The last fact is absolutely right. It cannot describe only one trial. But, this doesn't have to do with the uncertainty principle? Does it! Scientists are investigating this problem in the mean time. But, we cannot deny that Quantum Mechanics still get more exact solutions? But still can't be applied to one experiment!


Schwartz Vandslire

------------------------------------------------
Either to work correctly as required, or to leave it.

 

[Anonym]

CarlB:” I agree completely and think that this is very important. I have a couple QFT books that hint at this, but I have no textbooks or other references that spell it out. Do you know of any other sources?”

I think that one may be helpful too:
Herman Hesse's book "The Glass Bead Game".

 

[reilly]

Once again, I appeal to history and note that this issue, that QED or QFT can be written in terms of ordinary (relativistic) QM was pretty much recognized some 80 years ago. First, recall that Heisenberg's matrix solution to the harmonic oscillator problem, used so-called step operators, now generally called creation and destruction operators. The rest is history.

I presume that Dirac can be cited as a reasonable authority in this matter. His groundbreaking paper on QED (1927) basically uses wavefunctions. Many papers on QED and QFT proior to WWII used wavefunction, albeit in creative ways. See, for example, Wiesskopf's paper on the electron's self energy of 1939. (Many of the important papers in QED, from inception to renormalization, can be found in Schwinger's Dover Books collection, Quantum Electrodynamics. (I'm no expert, but I suspect wave functions may be used in Stat Mech problems in which the chemical potntial is required.)

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.

Regards.
Reilly Atkinson

 

[Haelfix]

There is far more to field theory than wavefunctions, so I am of the point of view that the Dirac equation and say the relativistic generalization of the Schrodinger equation are *not* correct. For instance they miss radiative corrections like the lamb shift, that only field theory will see.

Of course they are retrieved in suitable limits or regimes where such things are not important, and for instance in looking at leading order terms in the dynamics.

Indeed there are many examples where we have absolutely no idea what the wavefunction even *is*, yet we can still calculate most quantities of interest. Take QCD as an example.

 

[Anonym]

wave functions may be used in Stat Mech problems in which the chemical potntial is required

Indeed, for example J.S. Langer, Phys.Rev., 167,183 (1968).

Present formulation of relativistic QM do not have any predictive power. The adequate QFT should contain naturally electroweak and QCD (at least).

Once again, I appeal to history and note that this issue, that QED or QFT can be written in terms of ordinary (relativistic) QM was pretty much recognized some 80 years ago. First, recall that Heisenberg's matrix solution to the harmonic oscillator problem, used so-called step operators, now generally called creation and destruction operators. The rest is history.

You are wrong. The rest is future.

 

[Amr Morsi]

Buddies,

This small file may help more about the concept of QFTs. It also inlcudes the link to the source.

Haelfix, lamb shift is not a correction in QFT or QM, it is only a consequence. No doubt between these days, or before that DIRAC'S IS THE RELATIVISTIC SHRODINGER EQUATION. Forgive me Haelfix, I am not able to discuss the facts of scientific fundamental. At least, we should have a background to stand on.

Anonym, I agree with you that QM don't account for emitting radiation,but account for interaction with radiation, through the non-conservative Schrodinger Equation, by A and phi, thanks to refer to Basics of Quantum Mechanics, for the author "White".
But, with respect to electroweak and strong fields,these of course must be get into account to get AN INFINITLY EXACT SOLUTION. But, to a very great extent, we are not in need for strong field (QCD) in chemical properties. Weak effects of course are much more present here than strong-field ones.

Reilly, I agree that True Theories (or interpretations), about the same issue, must coincide with each others,if each is 100% exact ... but, how to get the the correct idea into mind ... through which way!; "Some ways are better than others" Said Reilly ...in ...the ...other... Thread..

I totally agree with Schwartz

Amr Morsi.

 

[Haelfix]

"DIRAC'S IS THE RELATIVISTIC SHRODINGER EQUATION."

Thats not really morally true either. The Dirac equation treats spin 1/2 particles, but for instance is completely unable to treat spin 0 (Klein-Gordon).
The KG equation is really the relativistic version of the Schrodinger equation.

However, there is far more to QED than those equations, even though they were the historical starting point. There are no creation or annihilation operators present in those naive treatments (something relativity demands) and they cannot account for the plurality of experiments, ergo they are incorrect in the same way Newtons law is incorrect outside its realm of applicability.

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

 

[Anonym]

Amr Morsi:” This small file may help more about the concept of QFTs”

Thank you. It help me a lot, especially with “AN INFINITLY EXACT SOLUTION” and Haelfix notion of “that field theory is morally equivalent to some sort of relativistic Schrodinger equation”

Guys, may be better that you will spend your time studying physics?

 

[Amr Morsi]

Thanks Anonym for this advise. You are right... One has to revise many many things.

You are absolutely right: "The adequate QFT should contain naturally electroweak and QCD (at least)". Thanks for the alert. I may continue in your way.

See you.
Amr Morsi.

---------------------------------------------------------------
Know a thing about everything and know everything about a thing.

 

[Truth Finder]

Morsi,

Have you heared about the sea of particles and antiparticles?!
I adopt your way of thinking: Whatever was the sayings of the correct equations, TRUE LOGIC IS THE TRUE LOGIC.

It is totally wrong to say that QED is built upon particles' creation or annihilation operators. And, I challenge any QM / QFT Professor that may say so. If it is only naming then it can be passed, but creation or annihilation have never been the building block of QFTs, they are only included in it. And, as per Amr Morsi; they can be included in these 2 forms of equations by step-functions (or similar) in time, in the wave function. Got you, Morsi! Yaih!


Schwarz VANDSDLIRE.

-------------------------------------------------------------
Either to begin it as correctly as required, or pass it, as required.

 

[reilly]

Morsi,

Have you heared about the sea of particles and antiparticles?!
I adopt your way of thinking: Whatever was the sayings of the correct equations, TRUE LOGIC IS THE TRUE LOGIC.

It is totally wrong to say that QED is built upon particles' creation or annihilation operators. And, I challenge any QM / QFT Professor that may say so. If it is only naming then it can be passed, but creation or annihilation have never been the building block of QFTs, they are only included in it. And, as per Amr Morsi; they can be included in these 2 forms of equations by step-functions (or similar) in time, in the wave function. Got you, Morsi! Yaih!Schwarz VANDSDLIRE.

-------------------------------------------------------------
Either to begin it as correctly as required, or pass it, as required.

I doubt very seriously that any professional working with QED or QFT or the Standard Theory, or many-body physics will claim that their basic theory is built on creation and destruction operators. The mini paper cited in #22 gives a standard, vanilla formulation of QED, one that can be found in virtually an infinite number of books, and does not involve creation and destruction operators.

As Dirac showed many years ago, these creation and destruction operators proved to be very convenient in working with QED. As I pointed out above, one can choose not to use these operators, but at the cost of making many arguments and computations horrendously complex.

Do recall that Dirac generalized the notion of wave function in his transformation theory. which in many ways is a generalization of generalized coordinates in classical mechanics. It's all in the representation you choose -- recall that the use of the interaction representation was key for making practical computations in QED in the late 1940s and early 1950s.

QM, QT, QED, and QFT are, fundamentally, the same, whether you choose to work with wave functions, or creation/ destruction operators or not.
Regards,
Reilly Atkinson

 

[vanesch]

I doubt very seriously that any professional working with QED or QFT or the Standard Theory, or many-body physics will claim that their basic theory is built on creation and destruction operators.

I can be wrong on this, but I thought that in a QFT, the annihilation and creation operators span, as an algebra, the entire set of operators (maybe with some qualifier...). My reasoning goes as follows: given that in Fock space, we can reach any basis state from any other by a finite application of creation and destruction operators, the image under an operator of a basis vector can be constructed using a linear combination of creation and destruction operators, and as such, any linear operation action which maps any basis vector onto a linear combination of basis vectors can thus be constructed.
I know this is a bit a naive reasoning, and some mathematical difficulties might arise, but this shows nevertheless, that the algebra of creation and destruction operators is quite rich.

 

[Kevin_spencer2]

But, the problem is let be a Lagrangian of a QFT theory:

\int d^{4}x \mathcal L [\phi]

then you could calculate the momenta of the field

\frac{\partial L}{\partial \dot \phi}=\pi

With the Legendre trasform you get the Hamiltonian and then the Schröedigneur equation would be:

\mathcal H |\Psi (\phi) >=i\hbar \frac{\partial |\Psi>}{\partial t}

from this equation above you could get the wave function of the system and all info you need, it's simpler than working with perturbation theory.

 

[dextercioby]

Nope, generally the Schroedinger's eqn for the state vector cannot be solved in a quantum theory of interacting fields.

 

[Quantum River]

When we use the Heisenberg representation in QFT, there is no need to discuss wave function anymore. In Heisenberg's matrix mechanics of QM in the early 1920s, Heisenberg didn't need the wave function to calculate the atomic spectrum. And now we do not need to know the wave function in QFT to calculate a QFT process. We just use the Heisenberg representation in QFT.
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.

 

[Kevin_spencer2]

Nope, generally the Schroedinger's eqn for the state vector cannot be solved in a quantum theory of interacting fields.

Not even when you consider perturbative approach?? so \mathcal L =\mathcal L _{0} +\lambda V(\phi)

 

[dextercioby]

Pertubative means "approximate". You "solve" an equation by finding approximate solutions.

 

[o1911]

i am a freshmen in QM...,and my words might be wrong.

my opinon
In QM, the wave function is just one of the represence of the state Bra(ket),
-- the represence of the R operator,and all the Bras form a Bra Space -- an abstract Hilbert Space.
and in QED ,a field is discribed as a Operator defineded in the Space (but i know nothing else , i know almost nothing about QED :-( )

 

[reilly]

More Dirac and Fock -General Covariance

I can be wrong on this, but I thought that in a QFT, the annihilation and creation operators span, as an algebra, the entire set of operators (maybe with some qualifier...). My reasoning goes as follows: given that in Fock space, we can reach any basis state from any other by a finite application of creation and destruction operators, the image under an operator of a basis vector can be constructed using a linear combination of creation and destruction operators, and as such, any linear operation action which maps any basis vector onto a linear combination of basis vectors can thus be constructed.
I know this is a bit a naive reasoning, and some mathematical difficulties might arise, but this shows nevertheless, that the algebra of creation and destruction operators is quite rich.

vanesch -- You have it right. However, there are other approaches,
Schwinger's for example which do not use creation and destruction operators, but fields instead. It is strictly a matter of taste, and, sometimes, expediency.

Dirac's transformation theory is, for all practical purposes, a formulation of general covariance in QM, based on unitary transformations. So, wave functions -- yes or no -- are a matter of taste. However, when it comes to compute transition amplitudes and bound state properties, we almost always use wave functions -- as in spinors where appropriate, plane wave representations, Ylms for angular momentum and the like. No mystery, no problem; most texts will have more than enough, directly or indirectly -- which might be a referene to Dirac -- to dispell any concerns a student might have. Not to worry.

That being said, approaches to the relativistic bound state or many body problem, with multi-time wave functions are difficult.

Regards,
Reilly Atkinson

 

[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:

\phi^* \phi \ \ \ \mbox{and} \ \ \ \ i\left(\phi^*\frac{\partial \phi}{\partial t} - \phi\frac{\partial \phi^*}{\partial t} \right) [/itex] <br /> <br /> Note that these are the SAME (!) for stationary solutions except for a <br /> normalization factor of 2E/\hbar and as long a E is real. The KG probability<br /> 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.

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.

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