Undergrad The thermal interpretation of quantum physics

[A. Neumaier]

@A. Neumaier How does the reformulation of quantum mechanics and electroweak theory in terms of the Clifford algebra $C(\mathbb{R}^{3,1})$ (cf. https://www.researchgate.net/publication/305483161_Geometric_Algebra_as_a_Unifying_Language_for_Physics_and_Engineering_and_Its_Use_in_the_Study_of_Gravity, https://www.researchgate.net/publication/252922221_Mysteries_and_Insights_of_Dirac_Theory, https://www.researchgate.net/publication/1739641_Gauge_Gravity_and_Electroweak_Theory) mathematically fit into your coherent foundations for quantum physics?

Here is a link to a presentation by Lasenby that needs no permissions to read.

His work is unrelated to interpretation questions, so I don't understand why you ask the question. Clearly, any mathematical tool consistent with standard quantum mechanics is also compatible with my coherent foundations. This includes geometric algebra.

 

[member 673059]

Here is a link to a presentation by Lasenby that needs no permissions to read.

His work is unrelated to interpretation questions, so I don't understand why you ask the question. Clearly, any mathematical tool consistent with standard quantum mechanics is also compatible with my coherent foundations. This includes geometric algebra.

Perhaps this wasn't the right thread to put my question, but my question wasn't an interpretation question, but rather a mathematical question of how to derive the mathematics of the Clifford algebra formalism from coherent spaces and quantum spaces in section 5.2 and 5.3. I think now that they would instead probably introduce the Clifford algebra of an Euclidean or Hilbert space somewhere in chapter 4 on Euclidean spaces, and define linear operators and outermorphisms and their adjoint, definiteness, hermiticity, and positivity using Clifford algebra instead of matrices, as well as a scalar product of a Clifford algebra that corresponds to the inner product, from which the definitions of a coherent space and product and quantum space in Proposition 5.3.1 follow, but where the quantum space is usually a subspace of the Clifford algebra $C(\mathbb{H})$ instead of the Euclidean/Hilbert space $\mathbb{H}$. From then, it would probably be simply a question of changing the notation from bra-ket notation to Clifford algebra notation.

 

[A. Neumaier]

Perhaps this wasn't the right thread to put my question, but my question wasn't an interpretation question, but rather a mathematical question of how to derive the mathematics of the Clifford algebra formalism from coherent spaces and quantum spaces in section 5.2 and 5.3.

Since this is a technical question related to my book, and the answer is nontrivial, it is better to ask this at PhysicsOverflow, in the thread linked to as a comment, or as a separate question there.

 

[ftr]

Rather, the next level of naturally visualizable detail would be to think of the particle's world tube as being actually a moving cloud of mass (or energy, or charge, etc.), with mass density $\rho(z)=\langle m\delta(x_k-z)\rangle$ for a distinguishable particle $k$ inside a multiparticle system. For a single particle state, this would be $\rho(z)=m|\psi(z)|^2$, recovering Schrödinger's intuition about the meaning of the wave function. Of course, this is already the field picture, just in first quantized form.

So could you say that $\rho(z)=m|\psi(z)|^2$ ( from post #273) means that the wavefunction(squared) represents the energy contained that makes up the mass according to E=mc^2, nothing else.

 

[A. Neumaier]

So could you say that $\rho(z)=m|\psi(z)|^2$ ( from post #273) means that the wavefunction(squared) represents the energy contained that makes up the mass according to E=mc^2, nothing else.

No. The Schrödinger equation is for non-relativistic QM, while Einstein's relation refers to the relativistic case only.

Moreover, the wave function also encodes phase information, and in the multiparticle case much more.

 

[ftr]

No. The Schrödinger equation is for non-relativistic QM, while Einstein's relation refers to the relativistic case only.

But if you take the integral of both sides then the integral of rho comes out to be m, what does that mean?
Of course I know the non relativistic vs relativistic(Schrodinger vs Dirac).

 

[A. Neumaier]

But if you take the integral of both sides then the integral of rho comes out to be m, what does that mean?.

It means that rho is the mass density, nothing more.

 

[ftr]

It means that rho is the mass density, nothing more.

Mass density equals energy density, correct?

 

[A. Neumaier]

Mass density equals energy density, correct?

No, these are very different concepts, otherwise there wouldn't be two different names for them. For example, an electromagnetic field has zero mass density but usually a positive energy density.

 

[ftr]

Of course we are talking about massive particles. It just seems more intuitive to talk about energy density than mass density for a single particle. I hope some other people give us their two cents.

 

[PeterDonis]

It just seems more intuitive to talk about energy density than mass density for a single particle.

As @A. Neumaier has pointed out, if we are talking about the Schrodinger equation, we are talking about non-relativistic QM, where mass-energy equivalence does not apply, so "mass density" and "energy density" are simply different things.

If you want to talk about the relativistic case, you'd be talking about quantum field theory, and the discussion would belong in a separate thread, not this one.

 

[A. Neumaier]

It just seems more intuitive to talk about energy density than mass density for a single particle.

It is also more intuitive to talk about temperature than about energy, but this does not make these concepts equivalent. There are good reasons why physicists use different names for different concepts.

A glass of water has a different mechanical energy density depending on whether it is still or stirred, but its mass density at its bottom does not change. There is no meaningful way to equate the two concepts.

 

[ftr]

It means that rho is the mass density, nothing more.

So what is this rho mass density, is it mathematical only or is it something real scattered across the space that makes up the electron/particle ,for example , does it represent the size of the electron(assuming some cut off in the function), or what?. I don't mind a long answer, I know you are very good at one liners answers.

There is no meaningful way to equate the two concepts.

I know you don't mean that E=mc^2 is wrong. I also know that they are used as different concepts but I don't understand your "no meaningful way".

 

[A. Neumaier]

So what is this rho mass density, is it mathematical only or is it something real scattered across the space that makes up the electron/particle ,for example , does it represent the size of the electron(assuming some cut off in the function), or what?. I don't mind a long answer, I know you are very good at one liners answers.
I know you don't mean that E=mc^2 is wrong. I also know that they are used as different concepts but I don't understand your "no meaningful way".

Mass density tells you how mass is distributed in space, while energy density tells you how energy is distributed in space.

E=mc^2 gives only one of many contributions to the relativistic energy density. The latter also contains kinetic energy, potential energy, thermal energy, electromagnetic energy, etc..

 

[ftr]

Mass density tells you how mass is distributed in space, while energy density tells you how energy is distributed in space.

E=mc^2 gives only one of many contributions to the relativistic energy density. The latter also contains kinetic energy, potential energy, thermal energy, electromagnetic energy, etc..

All that is well understood.
I was trying to understand something from your "interpretation"(which I personally think it is a new theory and a good one). Ok, I will be more blunt. Do you think Schrodinger's equation is correct? because I see that you seem to hint that the equation should be changed so that Psi itself carries the units of energy and so no more "interpretation" is needed. How close is my assessment?

 

[A. Neumaier]

Do you think Schrodinger's equation is correct? because I see that you seem to hint that the equation should be changed so that Psi itself carries the units of energy and so no more "interpretation" is needed. How close is my assessment?

Totally off the mark.

Schrödinger's equation $i\hbar \dot\psi = H\psi$ is fundamental and cannot be changed since it just says that the Hamiltonian is the generator of time translations.

For an n-particle system in the position representation, psi must have units length to the power -3n/2 as its squared modulus is a probability density.

 

[PeterDonis]

This seems like a good time to close the thread.