Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.

Nicholas · 2008-04-17 22:47:22

They say the quantum property of spin is not the form of angular momentum we are familiar with in the macreo world; then what is this purely quantum phenomenon they name spin?

Mitch Raemsch

M@Man · 2008-04-22 00:19:36

For any description of quantum mechanics that precedes the relativistic theory, "spin" can be dealt with, but only if you put in in by hand. It is only within a relativistic theory that the quantity we call "spin" occurs naturally. In a nutshell:

The usual (time-dependent) Schrodinger equation

$i \hbar \frac{\partial}{\partial t} \psi = (\frac{-\hbar^2}{2m} \nabla^2 + V)\psi$

is fundamentally incompatible with relativistic theory, because relativity tells us that time and space are equivalent and can be interchanged (according to the laws of the Lorentz transformation), but the time-dependent Schrodinger equation already has space and time on an unequal footing: the time coordinate occurs as a first-derivative, but the spatial coordinates (x,y,z) all occur as second-derivatives. Therefore, in order to even attempt a theory that is consistent with both quantum mechanics and special relativity, we must find a way to "balance" the space and time coordinates in the Schrodinger equation. The Einstein relation from special relativity gives us a clue as to how to proceed:

$E^2 = (pc)^2 + (mc^2)^2$

One way to balance the Schrodinger equation for use in relativity is to effective "square it", transforming the usual Schrodinger equation

$i \hbar \frac{\partial}{\partial t} \psi = H \psi$

into

$- \hbar^2 \frac{\partial^2}{(\partial t)^2} \psi = H^2 \psi$

or

$- \hbar^2 \frac{\partial^2}{(\partial t)^2} \psi = ((pc)^2+(mc^2)^2) \psi$

This is called the Klein-Gordon equation, and is one possible generalization of the Schrodinger equation to a form that is consistent with special relativity. In particular, the Klein-Gordon equation describes particles that have no "spin" and are therefore bosons.

An alternative way to balance the Schrodinger equation is, instead of trying to make the time-part second order to catch up to the space part, trying to make the space part first-order to catch up to the time part. This formulation was first proposed by Dirac, and consists of attempting to find coefficients $\vec{\alpha}$ and $\beta$ that linearize the energy:

$E = \vec{\alpha} \cdot \vec{p}c + \beta mc^2$

If you attempt this solution, you find that the coefficients $\vec{\alpha}$ and $\beta$ cannot be ordinary numbers. In order for this method to work, they must be matrices. Therefore, if the Dirac method is going to work, the particles must have some extra "dimensions" (more precisely, some kinds of "internal degrees of freedom"). If you find the simplest matrices for $\vec{\alpha}$ and $\beta$ that will work, you can solve some simple problems in the Dirac equation (as it is now called) and compare with the results from the Schrodinger equation to see what effects these new degrees of freedom have had.

If you solve the problem of a Dirac particle in a magnetic field, you find that the new degrees of freedom behave like a magnetic moment (and in fact, you predict that the electron's g-factor is 2 - a feat that cannot be reproduced without such a treatment). Thus these new degrees of freedom behave in a manner similar to angular momentum - but they clearly are not ordinary angular momenta. If you solve the problem of the hydrogen atom, you find that these new degrees of freedom, which behave like angular momenta, "combine" with the ordinary (orbital) angular momentum of the electron to produce "spin-orbit coupling" - an effect that has long been observed, and can be treated using nonrelativistic quantum mechanics, but cannot be explained by it.

These and other properties suggest that this "internal angular momentum" behaves analogously (in some senses) to the angular momentum of the electron if it were spinning on its axis. This explanation is only an analogy, however, and it does not hold up to even classical scrutiny. Thus we must content ourselves with explaining these internal degrees of freedom as "more-or-less" spin angular momenta, and stick to the concrete descriptions of their properties provided by the Dirac equation. Since the Dirac equation has these new degrees of freedom built into it from the beginning, it can only describe fermions.

Astro · 2010-05-08 05:54:50

M@Man wrote:
For any description of quantum mechanics that precedes the relativistic theory, "spin" can be dealt with, but only if you put in in by hand. It is only within a relativistic theory that the quantity we call "spin" occurs naturally. In a nutshell:

The usual (time-dependent) Schrodinger equation

$i \hbar \frac{\partial}{\partial t} \psi = (\frac{-\hbar^2}{2m} \nabla^2 + V)\psi$

is fundamentally incompatible with relativistic theory, because relativity tells us that time and space are equivalent and can be interchanged (according to the laws of the Lorentz transformation), but the time-dependent Schrodinger equation already has space and time on an unequal footing: the time coordinate occurs as a first-derivative, but the spatial coordinates (x,y,z) all occur as second-derivatives. Therefore, in order to even attempt a theory that is consistent with both quantum mechanics and special relativity, we must find a way to "balance" the space and time coordinates in the Schrodinger equation. The Einstein relation from special relativity gives us a clue as to how to proceed:

$E^2 = (pc)^2 + (mc^2)^2$

One way to balance the Schrodinger equation for use in relativity is to effective "square it", transforming the usual Schrodinger equation

$i \hbar \frac{\partial}{\partial t} \psi = H \psi$

into

$- \hbar^2 \frac{\partial^2}{(\partial t)^2} \psi = H^2 \psi$

or

$- \hbar^2 \frac{\partial^2}{(\partial t)^2} \psi = ((pc)^2+(mc^2)^2) \psi$

This is called the Klein-Gordon equation, and is one possible generalization of the Schrodinger equation to a form that is consistent with special relativity. In particular, the Klein-Gordon equation describes particles that have no "spin" and are therefore bosons.

An alternative way to balance the Schrodinger equation is, instead of trying to make the time-part second order to catch up to the space part, trying to make the space part first-order to catch up to the time part. This formulation was first proposed by Dirac, and consists of attempting to find coefficients $\vec{\alpha}$ and $\beta$ that linearize the energy:

$E = \vec{\alpha} \cdot \vec{p}c + \beta mc^2$

If you attempt this solution, you find that the coefficients $\vec{\alpha}$ and $\beta$ cannot be ordinary numbers. In order for this method to work, they must be matrices. Therefore, if the Dirac method is going to work, the particles must have some extra "dimensions" (more precisely, some kinds of "internal degrees of freedom"). If you find the simplest matrices for $\vec{\alpha}$ and $\beta$ that will work, you can solve some simple problems in the Dirac equation (as it is now called) and compare with the results from the Schrodinger equation to see what effects these new degrees of freedom have had.

If you solve the problem of a Dirac particle in a magnetic field, you find that the new degrees of freedom behave like a magnetic moment (and in fact, you predict that the electron's g-factor is 2 - a feat that cannot be reproduced without such a treatment). Thus these new degrees of freedom behave in a manner similar to angular momentum - but they clearly are not ordinary angular momenta. If you solve the problem of the hydrogen atom, you find that these new degrees of freedom, which behave like angular momenta, "combine" with the ordinary (orbital) angular momentum of the electron to produce "spin-orbit coupling" - an effect that has long been observed, and can be treated using nonrelativistic quantum mechanics, but cannot be explained by it.

These and other properties suggest that this "internal angular momentum" behaves analogously (in some senses) to the angular momentum of the electron if it were spinning on its axis. This explanation is only an analogy, however, and it does not hold up to even classical scrutiny. Thus we must content ourselves with explaining these internal degrees of freedom as "more-or-less" spin angular momenta, and stick to the concrete descriptions of their properties provided by the Dirac equation. Since the Dirac equation has these new degrees of freedom built into it from the beginning, it can only describe fermions.

That is a major simplification of the Klein-Gorden Equation. So simple, its not even right. It's a form of a premature derivation of the Klein-Gorden equation. At best all we have from $- \hbar^2 \frac{\partial^2}{(\partial t)^2} \psi = ((pc)^2+(mc^2)^2) \psi$ is a mere energy-momentum relation.

The Klein-Gorden equation is:

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

And it describes spinless particles. Your equation has failed to rearrange terms and replace the momentum with its respective quantized operator.

Astro · 2010-05-10 07:37:51

Sorry - also - what is this supposed to mean ''particles without spin, and therefoe should be bosons?''

That's simply rubbish. Photons are bosons with spin 1. Explain how both idea's are compatible, because it sounds like you're making this stuff up.

Astro · 2010-05-10 08:05:46

According to this, http://www.springerlink.com/content/q7l6464m52354367/ states that spin zero-particles must be bosons... it's something which is new to me.

Physics education research, electronic materials, and the musings of Christopher Moore, Ph.D.

#1 2008-04-17 22:47:22

Quantum spin

#2 2008-04-22 00:19:36

Re: Quantum spin

#3 2010-05-08 05:54:50

Re: Quantum spin

M@Man wrote:

#4 2010-05-10 07:37:51

Re: Quantum spin

#5 2010-05-10 08:05:46

Re: Quantum spin

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