Spin Explained for the "Wikipedia Physicist" - No QM Needed!

[peteratcam]

Just my two cents:

Spin is intrinsic angular momentum - in some regards a completely classical concept, although very hard to imagine because it is very unfamiliar in the classical context. (For the experts, nothing prevents you doing a classical field theory with dirac spinor fields - the representation structure of the Poincare symmetry is the same as ever)

Because the world is quantum mechanical, the possible values of this intrinsic angular momentum are quantised. The quantisation structure is as follows:
The total intrinsic angular momentum squared is always found to be $$s(s+1)\hbar^2$$. (hbar has units of angular momentum) where the value of s depends on which particle you are studying, but is always an integer or half integer.
For the electron, proton, neutron, s = 1/2. People will say 'the electron is spin 1/2' which means, the total intrinsic angular momentum is
$$\hbar\sqrt{\frac{1}{2}(\frac{1}{2}+1)}$$
Obviously it is easier to refer the the value 's', and just say spin 1/2 (since physicists always know how to relate the value of 's' back to the measured angular momentum). The value 's' is called the spin quantum number.

(A good quantum number is a symbol (not necessarily a number) which labels quantum mechanical stationary states according to the value of a conserved quantity).

Charge is also a quantum number, but it is complicated to explain why.

If something has a 'magnetic moment' it means it behaves like a tiny bar magnet. If you know about solenoids, hopefully you'll remember that when charge moves in a circle, it creates a magnetic field - when a charge has some angular momentum it creates a magnetic field.

It's not obvious, but this property is true of intrinsic angular momentum also. The intrinsic angular momentum of the electron creates a magnetic field - the magnetic moment of the electron.

In general, there will be some relation between the intrinsic angular momentum of some particle, and the magnetic moment it has, but this is complicated to work out. The relation is the g-factor.

 

[Char. Limit]

This also explains it to me very well. Thanks.

The property is true of intrinsic angular momentum because the particle itself has a charge, right?

 

[dextercioby]

Just my two cents:

Spin is intrinsic angular momentum - in some regards a completely classical concept, although very hard to imagine because it is very unfamiliar in the classical context. (For the experts, nothing prevents you doing a classical field theory with dirac spinor fields - the representation structure of the Poincare symmetry is the same as ever)

There's no reason to talk of spin in the context of classical physics. Of all the fields we know, only 2 (em and gravitational) have a classic-level existence (they are the only interactions with infinite range). They're interaction fields to be more precise.

If you apply the Noether theorem to the e-m Lagrangian for the restricted Lorentz invariance, you get the angular momentum tensor composed of 2 parts: the orbital one and the "intrinsic" one. The "intrinsic" can be thought of "spin" only when discussing the quantization of the e-m field. Without the quantization, one doesn't have the <particle> interpretation any classical field (as I said above, there are only 2 classical fields altogether), thus can't talk about the spin of a particle, or total spin of a system of particles.

As I said in another thread, there's no spin outside flat space relativity, there's no spin outside quantum mechanics.

 

[alxm]

The property is true of intrinsic angular momentum because the particle itself has a charge, right?

No, charge is a separate property. Neutrons have no charge and half spin. So do neutrinos. Photons have no charge and integer spin.

 

[peteratcam]

This also explains it to me very well. Thanks.

The property is true of intrinsic angular momentum because the particle itself has a charge, right?

I think this is the state of affairs, although high energy physics is not my thing, so I hope someone will correct me if they know otherwise:
If a particle has charge, and has spin, then it will have a magnetic moment.
If a particle has charge and no spin, then it will not have a magnetic moment.
If a particle has no charge, but does have spin, then it might have a magnetic moment.

 

[peteratcam]

There's no reason to talk of spin in the context of classical physics. Of all the fields we know, only 2 (em and gravitational) have a classic-level existence (they are the only interactions with infinite range). They're interaction fields to be more precise.

As I said in another thread, there's no spin outside flat space relativity, there's no spin outside quantum mechanics.

In the context of classical physics, I agree. But in the context of understanding classical field theory in Minkowski space (which in principle, a mathematician in 1905 might have done, well before the Stern-Gerlach experiment), then the spinor representations of the Lorentz symmetry are still interesting.

 

[Char. Limit]

No, charge is a separate property. Neutrons have no charge and half spin. So do neutrinos. Photons have no charge and integer spin.

What I mean is that if a charged object is spinning, it exerts a magnetic field, right? So an electron with angular momentum would have a magnetic moment, because it is charged.

Does a neutron have a magnetic moment?

 

[watcher]

I don't understand electron spin. What is it? Does spin have units? Does it do anything like electric charge or gravitation does? What does it represent? Any help for a "wikipedia physicist" (as I call myself) would really help. Just keep in mind: I haven't taken a class in QM.

according to milo wolff. in an all wave model, electron spin is as a spherical rotation...

http://www.youtube.com/watch?v=uKrM...2E390874&playnext=1&playnext_from=PL&index=42


Spherical Rotation

Rotation of the inward quantum wave at the center to become an outward wave is an absolute requirement to form a particle structure. Rotation in space has conditions. Any mechanism that rotates (to creates the quantum "spin") must not destroy the continuity of the space. The curvilinear coordinates of the space near the particle must participate in the motion of the particle. Fortunately, nature has provided a way - known as spherical rotation - a unique property of 3-D space. In mathematical terms this mechanism, according to the group theory of 3-D space, is described by stating that the allowed motions must be represented by the SU(2) group algebra which concerns simply-connected geometries.

Spherical rotation is an astonishing property of 3-D space. It permits an object structured of space to rotate about any axis without rupturing the coordinates of space. After two turns, space regains its original configuration. This property allows the electron to retain spherical symmetry while imparting a quantized "spin" along an arbitrary axis as the inward waves converge to the center, rotate with a phase shift to become the outward wave, and continually repeat the cycle.

The required phase shift is a 180o rotation that changes inward wave amplitudes to become those of the outward wave. There are only two possible directions of rotation, CW or CCW. One choice is an electron with spin of +h/4pi, and the other is the positron with spin of -h/4pi.

It is an awesome thought that if 3-D space did not have this geometric property of spherical rotation, particles and matter as we know them could not exist.

 

[ytuab]

according to milo wolff. in an all wave model, electron spin is as a spherical rotation...

http://www.youtube.com/watch?v=uKrM...2E390874&playnext=1&playnext_from=PL&index=42

Spherical Rotation

Rotation of the inward quantum wave at the center to become an outward wave is an absolute requirement to form a particle structure. Rotation in space has conditions. Any mechanism that rotates (to creates the quantum "spin") must not destroy the continuity of the space. The curvilinear coordinates of the space near the particle must participate in the motion of the particle. Fortunately, nature has provided a way - known as spherical rotation -
After two turns, space regains its original configuration. This property allows the electron to retain spherical symmetry while imparting a quantized "spin" along an arbitrary axis as the inward waves converge to the center, rotate with a phase shift to become the outward wave, and continually repeat the cycle.

The required phase shift is a 180o rotation that changes inward wave amplitudes to become those of the outward wave.
It is an awesome thought that if 3-D space did not have this geometric property of spherical rotation, particles and matter as we know them could not exist.

I think it is quite natural for us to dream of a real particle and a real spinning of the electron. (These are inseparable.)
Your statement is very interesting, but I'm afraid it is imposibble to treat the spin as a real spinning in the quantum mechanics(QM).

You said the electron spin is "a spherical rotation". But how fast is the electron rotating(spinning) ?

Foundations of Quantum Physics by Charles E.Burkhardt (in page 264)
------------------------------
They imagined that the electron is a spherical shell having total charge e uniformly smeared over its surface, reminiscent of the model used to derive the classical radius of the electron in Section 1.2.5.
This spinning sphere creates a magnetic moment identical with that of a bar magnet.
Is this model consistent with the classical radius of the electron? No-- as can be seen by
equating the angular momentum of the spinning sphere to 1/2 hbar. Solving for the speed of a point on the sphere leads to a speed that is roughly 100 times the speed of light.
--------------------------------

So If the spinning speed does not exceed the speed of light, the electron must be 100 times bigger than the classical radius size(2.8 x 10^-15 m) or an proton (10^-15 m).

But of course, by the scattering experiment or the fact of the electron capture, the electron size must be much smaller than that. (In any state and any process, the electron always has spin 1/2, doesn't it?)

And in your model, after two turn, suddenly the electron field seems to regain its original configuration by returning the fully-twisted field to the untwisted original field artificially. This does not seem to be changing continuously like $$e^{i\phi/2}$$ . Is this contradictry to the experimental results of rotating the spinning neutron in this thread ?

Do I misunderstand something?

I think if we can consider an electron as a real particle with real spinning in QM, we could have already done this a long time ago (in 1920's ~1930's). If you use some "new instruments" which could not be made at that time, this is a different matter.

 

[Fredrik]

according to milo wolff. in an all wave model, electron spin is as a spherical rotation...
...
Spherical rotation is an astonishing property of 3-D space. It permits an object structured of space to rotate about any axis without rupturing the coordinates of space. After two turns, space regains its original configuration. This property allows the electron to retain spherical symmetry while imparting a quantized "spin" along an arbitrary axis as the inward waves converge to the center, rotate with a phase shift to become the outward wave, and continually repeat the cycle.

You should be more careful about what sources you're referring to at Physics Forums. This Milo Wolff doesn't seem to be an actual physicist, and more importantly, he doesn't seem to publish in peer reviewed journals. Also, the things you said about electron spin and rotations are very misleading.

 

[Naty1]

Nobody knows EXACTLY whan quantum mechanical spin is, anymore than we know exactly what "charge" is...both do have real world macroscopic effects we can observe/measure. In general spin is a degree of freedom of a particle. The polarization of light is a physical manifestation of quantum mechanical spin.

Roger Penrose notes:

Photons are indeed particles that possesses spin, but being massless, their spin behaves in a way that is a little different from the more usual spin of a massive particle (electron or proton) as necessarily spinning about its direction of motion.

From his book, THE ROAD TO REALITY, where he also goes into a lot of math related to spin ...too much of which is over my head...

 

[watcher]

I think it is quite natural for us to dream of a real particle and a real spinning of the electron. (These are inseparable.)
Your statement is very interesting, but I'm afraid it is imposibble to treat the spin as a real spinning in the quantum mechanics(QM).

what do you mean by real? you probably meant non-classical spin.

You said the electron spin is "a spherical rotation". But how fast is the electron rotating(spinning) ?

sorry if my post mislead you. let me try again.
space is what rotates in a spherical way. spin is the result of phase shift ( from up spin to down spin, vv) due to the meeting of the so called advanced and retarded emf waves (feynman),
the electron is the particle effect at the center of these waves or the amplitude of these quantum waves.

I think if we can consider an electron as a real particle with real spinning in QM, we could have already done this a long time ago (in 1920's ~1930's). If you use some "new instruments" which could not be made at that time, this is a different matter.

yes i think it is misleading to think of electron spin as a classical spin, that is a spinning spherical particle. but i thought i mentioned that the model i am referring is an all wave model as proposed by Schrodinger. whereas the particle does not spin but space spins and in turn "create" a particle ...

During this period Schrödinger turned from mainstream quantum mechanics' definition of wave-particle duality and promoted the wave idea alone causing much controversy. - wiki


The particle can only appear as a limited region in space in which the field strength or the energy density are particularly high. (Albert Einstein, Metaphysics of Relativity, 1950)

.

 

[watcher]

You should be more careful about what sources you're referring to at Physics Forums. This Milo Wolff doesn't seem to be an actual physicist, and more importantly, he doesn't seem to publish in peer reviewed journals. Also, the things you said about electron spin and rotations are very misleading.

fredrick and folks,

i don't like to appear like i am pitching for milo but his biography is in the internet, judge for yourselves if his "not so mainstream science idea of all wave model of matter" is not worthy of physics forum.

perhaps my reply to ytuab clarify some of the confusion.

 

[Fredrik]

i don't like to appear like i am pitching for milo but his biography is in the internet, judge for yourselves if his "not so mainstream science idea of all wave model of matter" is not worthy of physics forum.

I sympathize with this to some degree, but the policy here at PF is to only discuss things that have already been judged, by well-known and respected science journals. So this is actually against the forum rules. The James Randi Educational Foundation has a good forum for those who want to discuss material that isn't allowed here.

 

[watcher]

I sympathize with this to some degree, but the policy here at PF is to only discuss things that have already been judged, by well-known and respected science journals. So this is actually against the forum rules. The James Randi Educational Foundation has a good forum for those who want to discuss material that isn't allowed here.

understood

 

[ytuab]

sorry if my post mislead you. let me try again.
space is what rotates in a spherical way. spin is the result of phase shift ( from up spin to down spin, vv) due to the meeting of the so called advanced and retarded emf waves (feynman),
the electron is the particle effect at the center of these waves or the amplitude of these quantum waves.

During this period Schrödinger turned from mainstream quantum mechanics' definition of wave-particle duality and promoted the wave idea alone causing much controversy. - wiki

The particle can only appear as a limited region in space in which the field strength or the energy density are particularly high. (Albert Einstein, Metaphysics of Relativity, 1950)

I have seen some comments in which the spin is caused by the waves around the electron.
But it is difficult, I think.

The electron also has the charge.
To create the spin magnetic moment of the electron (for example, in hydrogen) the charge of one electron of the hydrogen must be spreading over the very large space, if it actually rotates. (due to spin g-factor = 2)

Of course, the EM fields can not cause this spin magnetic moment because the EM fields has spin 1 (not 1/2).

If the charge of one electron is so spreading over, when the electron was captured by the nucleus, suddenly all this charge was taken into the nucleus?

 

[passingthru]

A study of particle physics reveals that electrons have internal structure, if I understand it correctly, and that's a big if, three quarks. I don't know anything about how the quarks are coupled inside the electron. The classical concept of angular momentum of a rotating bead is the simplest way of visualizing it, with more and more sophisticated explanations growing out of continued research and very hard work. A study of quantum mechanics in an introductory course suggests that mass particles are more like standing waves in an organ pipe, than the hard little spheres we are used to picturing in our minds. Spin that, and it might be kind of like spinning a water balloon. I don't know for sure, but I really appreciate all the thought that has gone into all these explanations. Thank you for helping.

 

[Vanadium 50]

I think you are misunderstanding something. There are no quarks inside an electron, and to the extent of our ability to measure, an electron seems to have no substructure.

 

[DrChinese]

A study of particle physics reveals that electrons have internal structure, if I understand it correctly, and that's a big if, three quarks.

As Vanadium50 says: You are referring to the proton, rather than the electron.

 

[planck42]

In Dr. Stephen Hawking's famous book A Brief History of Time, he defines spin as the reciprocal of the number of revolutions of an image necessary for that image to look identical to itself e.g. keyboards have a "spin" of 1, a clean sheet of paper has a "spin" of 2, etc. The intent of the book was merely to encourage interest in physics, and not to explain advanced concepts to significant depth, so it is possible that Dr. Hawking butchered the details intentionally.

 

[Galap]

OK, I think I might be getting a (very) basic idea now. One thing confuses me though: how can electron "spin" be synonymous with charge, if an electron has one possible value for charge, and two for "spin"?

Also, in the equation for the post above, what are those strange | and > symbols?

One good way of thinking of spin up and spin down particles is to think of them as mirror images of each other. a +1/2 spin electron will be itentical to a -1/2 spin electron, except it is a 'mirror image' of sorts.

 

[haael]

I have a question: what is the relationship between "standard" spin (say, electron has spin 1/2, photon has 1) and tensor-like approach, where electron is a spinor and photon is a vector? Why do spinors yield 1/2 and vectors give 1?

Also, how spin number translates into vector direction? Photons, as vectors, point into some direction, right? I thought that spin states -1, 0, 1 are "base vectors", but the numbers don't quite match.

 

[conway]

Also, how spin number translates into vector direction? Photons, as vectors, point into some direction, right? I thought that spin states -1, 0, 1 are "base vectors", but the numbers don't quite match.

The three spin states are all along the z axis, but you can transform the basis so you get them along the x,y, and z axes. Maybe the best way to see this is to think about the way you see p orbitals (spin 1) of hydrogen drawn in textbooks. The chemists tend to use the vector representation with three dumbbells oriented along the three axes. The physicists use the z-axes symmetry with two swirling clouds and one dumbbell.

Notice that the swirling clouds have a complex phase and maybe you can see that if you add together the physicists +1 and -1 states you get the chemists "vector state" along the x or y-axis (depending if you add or subtract).

CRANK ALERT: I am a crank.

 

[haael]

Notice that the swirling clouds have a complex phase and maybe you can see that if you add together the physicists +1 and -1 states you get the chemists "vector state" along the x or y-axis (depending if you add or subtract).

OK, thanks, but what about spin 2 tensors? There are 9 numbers that build up a tensor, but only 5 spin states.

BTW, happy Easter :).

 

[conway]

The nine tensors correspond in my picture to the nine d-orbitals: a single spin-zero (with two spherical nodes), three spin-one (with a single spherical node), and five spin-two elements. You've seen the weird pictures of the electron clouds in chemistry textbooks. The peculiar thing is that these are usually shown with an implied z-axis symmetry, including the funny vertical dumbbell with the ring around the outside. I don't know if I will succede in describing in words how this lines up to the vector picture I talked about for spin-one, but I can try.

First of all, let me ask if you will agree with my picture for the p-orbitals. Let me first ignore the "physicist's" orbitals with their swirlling complex amplitudes, and concentrate on the "chemist's" orbitals, the three dumbbells aligned along the x,y, and z axis. I am going to ask you to consider what happens if we add a small component of p orbitals in any combination to the ground state, e.g:

$$0.955{|s>} + 0.2{|p_x>} + 0.2{|p_y>} + 0.1|p_z>$$

I wonder if you will agree that the effect of this superposition in, say, the hydrogen atom, is essentially to displace the ground state a small distance in the direction (2,2,1)? So basically, to a good approximation, the whole cloud just moves a little bit in that direction?

(The cloud can do more things if you allow complex values for the coefficients, but I am just thinking about the real-valued cases for now.)

 

[SpectraCat]

The nine tensors correspond in my picture to the nine d-orbitals: a single spin-zero (with two spherical nodes), three spin-one (with a single spherical node), and five spin-two elements.

There are five d-orbitals, not 9 ... d-orbitals correspond to l=2, the degeneracy is 2l+1=5. If you consider pure Cartesian functions (xx, xy, xz, yy, yz, and zz), there you can get 6 basic functions that look like d-orbitals, but this set has a linear dependency since the combination xx + yy + zz has spherical symmetry. I have no idea what you are talking about with 9 functions here.

You've seen the weird pictures of the electron clouds in chemistry textbooks. The peculiar thing is that these are usually shown with an implied z-axis symmetry, including the funny vertical dumbbell with the ring around the outside. I don't know if I will succede in describing in words how this lines up to the vector picture I talked about for spin-one, but I can try.

The vector picture for orbital angular momentum of 2 (i.e. d-orbitals) is simple. There are 5 degenerate states (the d-orbitals), each corresponding to an angular momentum vector of length sqrt[l(l+1)]=sqrt(6). Each orbital has a different m_l quantum number describing the projection of each angular momentum vector on an arbitrarily chosen space-fixed axis. By convention the z-axis is chosen because the differential operator L_z has a very simple form .. it is just:

$$-i\hbar\frac{\partial}{\partial\phi}$$, where $$\phi$$ is the polar angle in spherical polar coordinates. It is trivial to solve for the eigenstates of this operator ... they are just $$e^{i m_l \phi}$$, where m_l can take values from -l to +l. These are the complex orbitals conway was referring to ... they cannot be easily visualized due to the complex phase. Therefore, we generally take linear combinations of them to create pure-real functions that can visualized. They are eigenfunctions of the L2[/SUB] operator (any linear combination of degenerate eigenstates is also an eigenstate), but they are not eigenstates of L_z, however they can be plotted in 3-D. We take balanced symmetric and anti-symmetric combinations of the m_l=+/-2 orbitals to make the d_xy and d_x²-y² orbitals, we construct the d_xy and d_xy by combining the m_l=+/-1 orbitals, and the m_l=0 orbital and d_z² orbital are identical.

First of all, let me ask if you will agree with my picture for the p-orbitals. Let me first ignore the "physicist's" orbitals with their swirlling complex amplitudes, and concentrate on the "chemist's" orbitals, the three dumbbells aligned along the x,y, and z axis. I am going to ask you to consider what happens if we add a small component of p orbitals in any combination to the ground state, e.g:

$$0.955{|s>} + 0.2{|p_x>} + 0.2{|p_y>} + 0.1|p_z>$$

I wonder if you will agree that the effect of this superposition in, say, the hydrogen atom, is essentially to displace the ground state a small distance in the direction (2,2,1)? So basically, to a good approximation, the whole cloud just moves a little bit in that direction.

This is most definitely not correct ... the cloud does not move in the (2,2,1) direction in space. What happens is mixing in the p-character in the way you describe would create an oblate distortion of the spherically symmetric s-cloud in the x-y plane. The linear combination:

$$0.2(p_x + p_y)$$ is just $$0.2\sqrt(2)p_1$$, which is a toroid with complex phase around the z-axis. If you added $$0.2sqrt(2)p_z$$, then you would exactly balance that toroidal contribution and recover the spherical symmetry ... the fact that only 0.1 was added means that there will be a slight bulge in the x-y plane.

 

[SpectraCat]

I have a question: what is the relationship between "standard" spin (say, electron has spin 1/2, photon has 1) and tensor-like approach, where electron is a spinor and photon is a vector? Why do spinors yield 1/2 and vectors give 1?

Also, how spin number translates into vector direction? Photons, as vectors, point into some direction, right? I thought that spin states -1, 0, 1 are "base vectors", but the numbers don't quite match.

Here it looks like you are confusing the spins themselves with the spin operator. A spinor describing a spin-1/2 particle is a 2-element vector. Photons are a little bit funny, because although they have spin-1, relativistic QM tells us that because they are mass-less particles, the m_s=0 projection cannot exist. They they also can be described by a two-element vector that is mathematically similar to a spinor, except the basis states for the vector are m_s=+/-1 instead of m_s=+/- 1/2.

The spin *operators* are matrices (c.f. the wiki page on Pauli spin matrices for more insight), which have a similar mathematical form to tensors. (I have to admit I am a bit shaky on tensor math .. it has been ages since I looked at it in any detail). The dimension of the matrix corresponding to the operator is the degeneracy of the angular momentum. That is, spin matrices for spin 1/2 have dimension of 2, for spin 1 they have a dimension of 3, and so on.

 

[conway]

First of all, let me ask if you will agree with my picture for the p-orbitals. Let me first ignore the "physicist's" orbitals with their swirlling complex amplitudes, and concentrate on the "chemist's" orbitals, the three dumbbells aligned along the x,y, and z axis. I am going to ask you to consider what happens if we add a small component of p orbitals in any combination to the ground state, e.g:

$$0.955{|s>} + 0.2{|p_x>} + 0.2{|p_y>} + 0.1|p_z>$$

I wonder if you will agree that the effect of this superposition in, say, the hydrogen atom, is essentially to displace the ground state a small distance in the direction (2,2,1)? So basically, to a good approximation, the whole cloud just moves a little bit in that direction?

This is most definitely not correct ... the cloud does not move in the (2,2,1) direction in space. What happens is mixing in the p-character in the way you describe would create an oblate distortion of the spherically symmetric s-cloud in the x-y plane. The linear combination:

$$0.2(p_x + p_y)$$ is just $$0.2\sqrt(2)p_1$$, which is a toroid with complex phase around the z-axis. If you added $$0.2sqrt(2)p_z$$, then you would exactly balance that toroidal contribution and recover the spherical symmetry ... the fact that only 0.1 was added means that there will be a slight bulge in the x-y plane.

I have to point out that my interpretation of the function is the correct one, not Spectracat's. What he is doing is basically squaring amplitudes and adding probabilities; what you are supposed to do is add amplitudes first and then square to get the probabilities.
He should know that my picture of vector displacement is correct because it's just what we've been talking about in the other thread (decoherence). When the s and p orbitals mix to form a pure state (e.g. Stark effect) then the fixed displacement persists. When the s and p orbitals represent two different energy levels, then the displacement osillates back and forth with the changing relative phase of the s and p levels. To get an oblate distortion of the s orbital, as SpectraCat suggests here, you need to mix in some d orbitals, not p orbitals.

 

[SpectraCat]

I have to point out that my interpretation of the function is the correct one, not Spectracat's. What he is doing is basically squaring amplitudes and adding probabilities; what you are supposed to do is add amplitudes first and then square to get the probabilities.
He should know that my picture of vector displacement is correct because it's just what we've been talking about in the other thread (decoherence). When the s and p orbitals mix to form a pure state (e.g. Stark effect) then the fixed displacement persists. When the s and p orbitals represent two different energy levels, then the displacement osillates back and forth with the changing relative phase of the s and p levels. To get an oblate distortion of the s orbital, as SpectraCat suggests here, you need to mix in some d orbitals, not p orbitals.

I assure you that this is not the case. I didn't bring it up because I didn't want to mix the two threads, but I was assuming you were referring to the mixing of non-degenerate s- and p-orbitals (e.g. 1s and 2p). In that case, then the oblate distortion is the correct picture for the time-averaged probability density. There will also be a time-oscillating term in the probability density, which will have a more complicated shape, but it is periodic motion with a definite phase, so it will be averaged to zero over an integer number of cycles ... and I was ignoring it in the context of your current remarks. There is definitely no "fixed displacement" as you are claiming in that case. All of the orbitals in the expansion are centered on the origin (i.e. <r>=0) so I can't understand how you would think that could be the case. Write out the expansion for yourself and you will see that there can be no such displacement of the *time-independent* (i.e. fixed), probability density.

However, if you are talking about degenerate s- and p-orbitals (i.e. 2s and 2p), then in that case you are correct, and there can be a net displacement of charge away from the origin (i.e. polarization). This is because the linear combination of those degenerate states is also an eigenstate of the zero-order Hamiltonian, and so is time invariant. However, such a state is not an eigenstate of the L_^2 operator, so the angular momentum quantum number is not a good quantum number, thus it is somewhat unclear how your analysis can be useful in the context of this thread.

EDIT: I have clarified this description to distinguish between what happens for degenerate and non-degenerate s and p orbitals.

 

[haael]

I have to point out that my interpretation of the function is the correct one, not Spectracat's

You two should fight :).

The nine tensors correspond in my picture to the nine d-orbitals: a single spin-zero (with two spherical nodes), three spin-one (with a single spherical node), and five spin-two elements.

Yeah, looking at http://en.wikipedia.org/wiki/Atomic_orbital" i see nine n=3 orbitals you are probably referring to.

However. If you count all lower-spin "orbitals" or spin states, then why again spin 1 have only 3 states? Shouldn't spin 0 be counted into spin 1 as well?

I'm starting to understand: there is as many spin N states, as independent range N polynomials (minus lower symmetric ones?). This should give 2l+1 states, that's OK.

The spin *operators* are matrices (c.f. the wiki page on Pauli spin matrices for more insight), which have a similar mathematical form to tensors. (I have to admit I am a bit shaky on tensor math .. it has been ages since I looked at it in any detail). The dimension of the matrix corresponding to the operator is the degeneracy of the angular momentum. That is, spin matrices for spin 1/2 have dimension of 2, for spin 1 they have a dimension of 3, and so on.

Wait. My understanding of tensors if quantum physics is as this: There are not-quite-matrices, namely operators, that only their commutation relations matter. There are also actual matrices (tensors), that have dimension related to our space dimension count. Each Pauli matrix is an operator, and three of them form a vector.

However, where does spin tensor approach come from? As you say, scalars spin operators have dimension of 1, so there is only one spin state. Vectors have spin operators represented by 3x3 matrices, so there are 3 vectors and 3 spin states. But what with spin 2? There are only 5 spin states, but 9 independent tensors. Then why does spin 2 particle can be described by tensor?

BTW, could you please tell me what is the right English word for vector/tensor "element" :)? The little number that gets packed up into a tuple?

 

[conway]

I assure you that this is not the case. I didn't bring it up because I didn't want to mix the two threads, but the oblate distortion is the correct picture for the time-averaged probability density. There will also be a time-oscillating term in the probability density, which will have a more complicated shape, but it is periodic motion with a definite phase, so it will be averaged to zero over an integer number of cycles ... and I was ignoring it in the context of your current remarks.

It is precisely the "context" of my current remarks which you have chosen to ignore:

"When the s and p orbitals mix to form a pure state (e.g. Stark effect) then the fixed displacement persists. When the s and p orbitals represent two different energy levels, then the displacement osillates back and forth with the changing relative phase of the s and p levels."

Where do you see me talking about a time-averaged probability density? I identify two cases: one, where the displacement is fixed in time, and the other where it osillates back and forth. Nowhere have I remotely suggested taking the time average of a periodic system.

There is definitely no "fixed displacement" as you are claiming. All of the orbitals in the expansion are centered on the origin (i.e. <r>=0) so I can't understand how you would think that could be the case. Write out the expansion for yourself and you will see that there can be no such displacement of the *time-independent* (i.e. fixed), probability density.

Once again you appear to me to be simply wrong. When you say "time-independent (i.e. fixed)" you seem to be talking about the same case as I intended: how can you not see that adding a little bit of "vector p-state" ($$p_x, p_y,$$ or $$p_z$$) has the effect of displacing the wave function from the origin? If this weren't the case, then where in the world would we get those tiny oscillating dipoles we talked about in the other thread?

The phrase "s and p orbitals mix to form a pure state" is nonsensical ... if you are mixing the orbitals, then you get a superposition, not a pure state.

I gave an example of just how they mix to form a pure state: the Stark effect. In the limit of a very weak electric field, so you can ignore all higher order terms, perturbation theory gives the ground state of the hydrogen atom (a pure state) as precisely the sum of the s orbital plus a little bit of p orbital.

 

[Fredrik]

I have a question: what is the relationship between "standard" spin (say, electron has spin 1/2, photon has 1) and tensor-like approach, where electron is a spinor and photon is a vector? Why do spinors yield 1/2 and vectors give 1?

What you call "standard spin" is the number j from when we write the eigenvalue of the operator [itex]\vec J^2=J_x^2+J_y^2+J_z^2[/tex] as j(j+1). This number is the same in all inertial frames, and that makes it appropriate to use it as one of the labels that identify a particle species. That stuff about "spinors" and "vectors" refers to how the components of the quantum field changes from one inertial frame to another.

Also, how spin number translates into vector direction? Photons, as vectors, point into some direction, right? I thought that spin states -1, 0, 1 are "base vectors", but the numbers don't quite match.

Those numbers are eigenvalues of $$J_z$$. The eigenvalues are always -j, -j+1,..., j-1, j. (Photons have j=1, electrons j=1/2). These numbers have very little to do with directions in space.

There are not-quite-matrices, namely operators, that only their commutation relations matter. There are also actual matrices (tensors), that have dimension related to our space dimension count. Each Pauli matrix is an operator, and three of them form a vector.

The Pauli matrices are the matrices of the spin operators $J_x,J_y,J_z$, in the basis $(\left|\uparrow\rangle,\left|\downarrow\rangle)$ for the (2-dimensional) Hilbert space of spin states. See this post for the relationship between linear operators (on finite-dimensional vector spaces) and matrices.

BTW, could you please tell me what is the right English word for vector/tensor "element" :)? The little number that gets packed up into a tuple?

"Component". And it's usually not a number. In other contexts it's usually a real-valued function. When we're talking about quantum fields, it's an operator-valued distribution.

 

[conway]

I gave an example of just how they mix to form a pure state: the Stark effect. In the limit of a very weak electric field, so you can ignore all higher order terms, perturbation theory gives the ground state of the hydrogen atom (a pure state) as precisely the sum of the s orbital plus a little bit of p orbital.

I should have added that the physics of the system in question make it all the more obvious that my description is the correct one. You apply an electric field to a hydrogen atom in the ground state: what is more natural than you should expect to get a small net displacement of the proton from the electron cloud? This is exactly the effect of mixing in a little bit of p orbital to the ground state.

(PS Thanks to Frederik for returning to the orignial question. My side issue has drifted pretty far off topic although I don't think I started out that way. The things I'm talking about are actually the physical pictures that help me keep straight the vector/tensor business as you deal with different spin states.)

 

[SpectraCat]

I should have added that the physics of the system in question make it all the more obvious that my description is the correct one. You apply an electric field to a hydrogen atom in the ground state: what is more natural than you should expect to get a small net displacement of the proton from the electron cloud? This is exactly the effect of mixing in a little bit of p orbital to the ground state.

(PS Thanks to Frederik for returning to the orignial question. My side issue has drifted pretty far off topic although I don't think I started out that way. The things I'm talking about are actually the physical pictures that help me keep straight the vector/tensor business as you deal with different spin states.)

Ok .. I admit that I got confused by the similarity of this case with our other argument, and assumed that you were talking about non-degenerate s- and p-states. You are correct for the case of degenerate s- and p- states in the first-order Stark effect; there will be a net polarization of the charge density in the field direction.

I will go back and amend post #64 to clarify this point.

Note: I just deleted another post which I accidentally submitted when I just meant to error-checking the TeX code before doing some more editing ... it contains errors, so please disregard it if you happened to see it before I deleted it.

 

[SpectraCat]

It is precisely the "context" of my current remarks which you have chosen to ignore:

"When the s and p orbitals mix to form a pure state (e.g. Stark effect) then the fixed displacement persists. When the s and p orbitals represent two different energy levels, then the displacement osillates back and forth with the changing relative phase of the s and p levels."

Where do you see me talking about a time-averaged probability density? I identify two cases: one, where the displacement is fixed in time, and the other where it osillates back and forth. Nowhere have I remotely suggested taking the time average of a periodic system.
Once again you appear to me to be simply wrong. When you say "time-independent (i.e. fixed)" you seem to be talking about the same case as I intended: how can you not see that adding a little bit of "vector p-state" ($$p_x, p_y,$$ or $$p_z$$) has the effect of displacing the wave function from the origin? If this weren't the case, then where in the world would we get those tiny oscillating dipoles we talked about in the other thread?

See my post above ... I did not realize that you were talking about degenerate s- and p-orbitals (you did not specify the n quantum number) when I wrote my analysis, which is correct for the non-degenerate case.

I gave an example of just how they mix to form a pure state: the Stark effect. In the limit of a very weak electric field, so you can ignore all higher order terms, perturbation theory gives the ground state of the hydrogen atom (a pure state) as precisely the sum of the s orbital plus a little bit of p orbital.

Yes, this picture is correct for the induced polarization arising from a weak electric field giving rise to the Stark effect. Sorry for the confusion.