Phase of a probability

"The Dirac equation is superficially similar to the Schrödinger equation for a free particle:

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

The left side represents the square of the momentum operator divided by twice the mass, which is the nonrelativistic kinetic energy. If one wants to get a relativistic generalization of this equation, then the space and time derivatives must enter symmetrically, as they do in the relativistic Maxwell equations—the derivatives must be of the same order in space and time. In relativity, the momentum and the energy are the space and time parts of a geometrical space-time vector, the 4-momentum, and they are related by the relativistically invariant relation

$$ \frac{E^2}{c^2} - p^2 = m^2c^2 $$

which says that the length of this vector is the rest mass m. Replacing E and p by derivative operators as Schrödinger theory requires, we get a relativistic equation:

$$ \left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi = \frac{m^2c^2}{\hbar^2}\phi $$

and the wave function$$ \phi\ $$, is a relativistic scalar, it is a complex number which has the same numerical value in all frames. Because the equation is second order in the time derivative, one must specify both the initial value of$$ \partial_t \phi\ $$, and not just $$ \phi\ $$,. This is normal for classical water waves, the initial conditions are the position and velocity, but in quantum mechanics the wavefunction is supposed to be the complete description, just knowing the wavefunction should determine the future.

In the Schrödinger theory, the probability density is given by the positive definite expression

$$ \rho=\phi^*\phi\, $$

and its current by

$$ J = -\frac{i\hbar}{2m}(\phi^*\nabla\phi - \phi\nabla\phi^*) $$

and the conservation of probability density has a local form:

$$ \nabla\cdot J + \frac{\partial\rho}{\partial t} = 0 $$

In a relativistic theory, the form of the probability density and the current must form a four vector, so the form of the probability density can be found from the current just by replacing$$ \nabla $$ by$$ \,\partial_t: $$

$$ \rho = \frac{i\hbar}{2m}(\phi^*\partial_t\phi - \phi\partial_t\phi^*) $$

Everything is relativistic now, but the probability density is not positive definite, because the initial values of both$$ \phi\ $$, and$$ \,\partial_t\phi $$can be freely chosen. This expression reduces to Schrödinger's density and current for superpositions of positive frequency waves whose wavelength is long compared to the compton wavelength, that is, for nonrelativistic motions.
It reduces to a negative definite quantity for superpositions of negative frequency waves only. It mixes up both signs when forces which have an appreciable amplitude to produce relativistic motions are involved, at which point scattering can produce particles and antiparticles.

Although it was not a successful description of a single particle, this equation is resurrected in quantum field theory, where it is known as the Klein–Gordon equation, and describes a relativistic spin-0 complex field. The non-positive probability density and current are the charge-density and current, while the particles are described by a mode-expansion.

In order to give the Klein–Gordon equation an interpretation as an equation for the probability amplitude for a single particle to have a given position, the negative frequency solutions need to be interpreted as describing the particle travelling backwards in time, so that they propagate into the past. The equation with this interpretation does not predict the future from the present except in the nonrelativistic limit, rather it places a global constraint on the amplitudes.
This can be used to construct a perturbation expansion with particles zipping backwards and forwards in time, the Feynman diagrams, but it does not allow a straightforward wavefunction description, since each particle has its own separate proper time."
-the wikipedists
 
It wasn't direct. You mixed two questions in the one. Even if they did potentially yield the same answer.

Please, be direct.
 
Vkothii said:
"The Dirac equation is superficially similar to the Schrödinger equation for a free particle:

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

The left side represents the square of the momentum operator divided by twice the mass, which is the nonrelativistic kinetic energy. If one wants to get a relativistic generalization of this equation, then the space and time derivatives must enter symmetrically, as they do in the relativistic Maxwell equations—the derivatives must be of the same order in space and time. In relativity, the momentum and the energy are the space and time parts of a geometrical space-time vector, the 4-momentum, and they are related by the relativistically invariant relation

$$ \frac{E^2}{c^2} - p^2 = m^2c^2 $$

which says that the length of this vector is the rest mass m. Replacing E and p by derivative operators as Schrödinger theory requires, we get a relativistic equation:

$$ \left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi = \frac{m^2c^2}{\hbar^2}\phi $$

and the wave function$$ \phi\ $$, is a relativistic scalar, it is a complex number which has the same numerical value in all frames. Because the equation is second order in the time derivative, one must specify both the initial value of$$ \partial_t \phi\ $$, and not just $$ \phi\ $$,. This is normal for classical water waves, the initial conditions are the position and velocity, but in quantum mechanics the wavefunction is supposed to be the complete description, just knowing the wavefunction should determine the future.

In the Schrödinger theory, the probability density is given by the positive definite expression

$$ \rho=\phi^*\phi\, $$

and its current by

$$ J = -\frac{i\hbar}{2m}(\phi^*\nabla\phi - \phi\nabla\phi^*) $$

and the conservation of probability density has a local form:

$$ \nabla\cdot J + \frac{\partial\rho}{\partial t} = 0 $$

In a relativistic theory, the form of the probability density and the current must form a four vector, so the form of the probability density can be found from the current just by replacing$$ \nabla $$ by$$ \,\partial_t: $$

$$ \rho = \frac{i\hbar}{2m}(\phi^*\partial_t\phi - \phi\partial_t\phi^*) $$

Everything is relativistic now, but the probability density is not positive definite, because the initial values of both$$ \phi\ $$, and$$ \,\partial_t\phi $$can be freely chosen. This expression reduces to Schrödinger's density and current for superpositions of positive frequency waves whose wavelength is long compared to the compton wavelength, that is, for nonrelativistic motions.
It reduces to a negative definite quantity for superpositions of negative frequency waves only. It mixes up both signs when forces which have an appreciable amplitude to produce relativistic motions are involved, at which point scattering can produce particles and antiparticles.

Although it was not a successful description of a single particle, this equation is resurrected in quantum field theory, where it is known as the Klein–Gordon equation, and describes a relativistic spin-0 complex field. The non-positive probability density and current are the charge-density and current, while the particles are described by a mode-expansion.

In order to give the Klein–Gordon equation an interpretation as an equation for the probability amplitude for a single particle to have a given position, the negative frequency solutions need to be interpreted as describing the particle travelling backwards in time, so that they propagate into the past. The equation with this interpretation does not predict the future from the present except in the nonrelativistic limit, rather it places a global constraint on the amplitudes.
This can be used to construct a perturbation expansion with particles zipping backwards and forwards in time, the Feynman diagrams, but it does not allow a straightforward wavefunction description, since each particle has its own separate proper time."
-the wikipedists
Click to expand...


Lovely. I actually know these coordinates off-hand... what book are you working from?
 
To try and get this tread back on topic:

1.) Saxion---Regardless of your qualifications, if you don't understand the notation that Vkothii has used (which is, I might add, something that an advanced undergraduate/average grad student should be familiar with), it is unlikely that you'll be able to contribute in a meaningful way to the thread.

2.) Vkothii---You never answered my question in the third(?) post. What does "phase of a probability" mean? Probabilities are measurable quantities, so they should give real numbers---that is, there should be no phase associated with them. So what's the deal?
 
BTM said:
What does "phase of a probability" mean?
Click to expand...
What indeed?
the example said:
Let the function $$ \theta_I $$ be the probability amplitude that the particle actually takes the path $$l $$ between A and B. It is referred to as a phase factor. If the point is held fixed, then the propagator is only a function of the location , and can simply be written as the probability wave solution of the Dirac equation for a free electron.
Click to expand...
I.E. show that both the gravitational field and electromagnetic field act upon the propagator in the same manner.
Both induce a phase shift which causes the wave function to disperse in the direction which minimizes the action, thereby decreasing the relative energy of the wave. The difference is that the gravitational field contribution is the result of the mass/energy, and the electromagnetic field contribution is a result of the electromagnetic energy. Both fields are representative of the local environment of a test particle.
Click to expand...
The title is an incomplete phrase. Perhaps "phase of a probability amplitude"? (??)
And represent, that's a tricky one when it comes to information content of an electron's spin (phase).
 
Vkothii said:
"The Dirac equation is superficially similar to the Schrödinger equation for a free particle:

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

The left side represents the square of the momentum operator divided by twice the mass, which is the nonrelativistic kinetic energy. If one wants to get a relativistic generalization of this equation, then the space and time derivatives must enter symmetrically, as they do in the relativistic Maxwell equations—the derivatives must be of the same order in space and time. In relativity, the momentum and the energy are the space and time parts of a geometrical space-time vector, the 4-momentum, and they are related by the relativistically invariant relation

$$ \frac{E^2}{c^2} - p^2 = m^2c^2 $$

which says that the length of this vector is the rest mass m. Replacing E and p by derivative operators as Schrödinger theory requires, we get a relativistic equation:

$$ \left(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right)\phi = \frac{m^2c^2}{\hbar^2}\phi $$

and the wave function$$ \phi\ $$, is a relativistic scalar, it is a complex number which has the same numerical value in all frames. Because the equation is second order in the time derivative, one must specify both the initial value of$$ \partial_t \phi\ $$, and not just $$ \phi\ $$,. This is normal for classical water waves, the initial conditions are the position and velocity, but in quantum mechanics the wavefunction is supposed to be the complete description, just knowing the wavefunction should determine the future.

In the Schrödinger theory, the probability density is given by the positive definite expression

$$ \rho=\phi^*\phi\, $$

and its current by

$$ J = -\frac{i\hbar}{2m}(\phi^*\nabla\phi - \phi\nabla\phi^*) $$

and the conservation of probability density has a local form:

$$ \nabla\cdot J + \frac{\partial\rho}{\partial t} = 0 $$

In a relativistic theory, the form of the probability density and the current must form a four vector, so the form of the probability density can be found from the current just by replacing$$ \nabla $$ by$$ \,\partial_t: $$

$$ \rho = \frac{i\hbar}{2m}(\phi^*\partial_t\phi - \phi\partial_t\phi^*) $$

Everything is relativistic now, but the probability density is not positive definite, because the initial values of both$$ \phi\ $$, and$$ \,\partial_t\phi $$can be freely chosen. This expression reduces to Schrödinger's density and current for superpositions of positive frequency waves whose wavelength is long compared to the compton wavelength, that is, for nonrelativistic motions.
It reduces to a negative definite quantity for superpositions of negative frequency waves only. It mixes up both signs when forces which have an appreciable amplitude to produce relativistic motions are involved, at which point scattering can produce particles and antiparticles.

Although it was not a successful description of a single particle, this equation is resurrected in quantum field theory, where it is known as the Klein–Gordon equation, and describes a relativistic spin-0 complex field. The non-positive probability density and current are the charge-density and current, while the particles are described by a mode-expansion.

In order to give the Klein–Gordon equation an interpretation as an equation for the probability amplitude for a single particle to have a given position, the negative frequency solutions need to be interpreted as describing the particle travelling backwards in time, so that they propagate into the past. The equation with this interpretation does not predict the future from the present except in the nonrelativistic limit, rather it places a global constraint on the amplitudes.
This can be used to construct a perturbation expansion with particles zipping backwards and forwards in time, the Feynman diagrams, but it does not allow a straightforward wavefunction description, since each particle has its own separate proper time."
-the wikipedists
Click to expand...

I hope you realise that none of the equations you have copied and pasted are the Dirac equation. That is $$ \left( i \gamma^\mu \partial_\mu -m \right) \psi = 0$$. A solution to the Dirac equation is automatically a solution to the Klein Gordon equation, since the KG equation is simply an expression of the relativistic dispersion relation. However, solutions to the Klein Gordon equation are not necessarily solutions to the Dirac equation.

Also, and a sincerely hope that wolv1 is not reading this, what you state about particles going backwards in time is not a very good physical interpretation of the solutions of KG. A better thing to say is that the density that you define; $$\rho = \frac{i\hbar}{2m}(\phi^*\partial_t\phi - \phi\partial_t\phi^*) $$, is not a probability density but a charge density. You can work out that the $$\phi$$ has a charge of +1 under the conserved quantity and $$\phi^*$$ a charge of -1 (depending on how you define things) meaning that all quantum numbers are the same except those of the conserved quantity under charge conjugation, which is the definition of particles and antiparticles.
 
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prometheus said:
the Dirac equation. That is $$ \left( i \gamma^\mu \partial_\mu -m \right) \psi = 0$$. A solution to the Dirac equation is automatically a solution to the Klein Gordon equation, since the KG equation is simply an expression of the relativistic dispersion relation. However, solutions to the Klein Gordon equation are not necessarily solutions to the Dirac equation.
Click to expand...
K.
How would you rewrite that to get it into an action integral formulation - there are at least two. One mass and one charge integral.

This "exercise" I thought has an interesting angle on the AB effect; you can do a gauge transformation that shows:
"... The gauge transformation is the gradient of a scalar function, and that gradient corresponds to the gradient in the probability of an interaction.

Thus by means of a variable index of refraction resulting from the absorption and emission of photons the probability density is transformed into a function of the local density of matter and energy, and so are the coordinates! ..."

Hint: there is a bit more to do than just derive the action for the EM field, but that's a detail of this I want to work through - see if it commutes with another view of the physical rep. (of information entropy).
Thing is, information has to be 'conveyed' in a classical sense which is necessarily irreversible.

Oh yeah, the actual Dirac form:
$$ \psi_\pm(x,p)\, = \, \varsigma^{\script l}(p) \exp [{i \over \hbar} p^{\alpha} g_{\alpha\beta} {\chi^\beta}]
$$
 
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Why don't you see if you can find a hammer and a nail somewhere?

I love the way you logical types establish "facts".
You should have taken predicate calc as well, it looks like your logic buffer has overflowed.
 
There is no topic. We have one guy who insists on copying things down from wiki and MIT notes and pretending to understand them. We have another that alludes to being a government funded physicist who does no research and doesn't know about Hamiltonian mechanics.

I have no idea why these people make up the stories they do, but I imagine that pandering to their need to sprout rubbish only serves to perpetuate their habit.
 
And we have another guy, who displays a rather interesting interaction dynamic, in which their contribution to the entropy of information isn't measured in their local frame??.

Who spouts a lot of strange-looking "logic", and can't explain where it came from because it "didn't come from them". Interesting.

The topic of this thread is not: "act like a plonker", it's "derive the action for the electromagnetic Lorentz phase shift for Dirac electrons".

Also there's a vague link to: " gauge transformations".

Ed: some of you guys might not get it. There's a geometry and a topology. QM logic ties the gauge of the field together, see?
Maxwell's quaternions explain the large view - the far field effects where we apply the geometry of normal Einsteinian/Euclidian spacetime.
The quantum algebras that are accessible to Boole's network analysis formalism and on up to multi-dimensional algebras, explain the up-close view, where we can (maybe) prepare a resonance, and measure it.
But we have to do both as we do "out here", except up close it has do be done at the same time - you measure the output by "running the program".

And measurement at both scales speaks to our view of the whole show - if you consider it's a big information processor, see?
The big information-processing world, and the big curve in a field that makes particles with spin precess along a preferred axis - a geometry.
Then we get them to line up algebraically. They perform algorithmically, we drive the algorithm - what with?

With the local gradient of the field that is the equivalent of a spin-phase change. An applied field drives it.
 
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