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

[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 \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 &quot;spinors&quot; and &quot;vectors&quot; refers to how the components of the quantum field changes from one inertial frame to another.<br /> <br /> <blockquote data-attributes="" data-quote="haael" data-source="post: 2651651" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> haael said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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 &quot;base vectors&quot;, but the numbers don&#039;t quite match. </div> </div> </blockquote>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.<br /> <br /> <br /> <blockquote data-attributes="" data-quote="haael" data-source="post: 2653320" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> haael said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> 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. </div> </div> </blockquote>The Pauli matrices are the matrices <i>of</i> 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 <a href="https://www.physicsforums.com/showthread.php?p=2514428" class="link link--internal">this post</a> for the relationship between linear operators (on finite-dimensional vector spaces) and matrices.<br /> <br /> <blockquote data-attributes="" data-quote="haael" data-source="post: 2653320" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> haael said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> BTW, could you please tell me what is the right English word for vector/tensor &quot;element&quot; :)? The little number that gets packed up into a tuple? </div> </div> </blockquote>&quot;Component&quot;. And it&#039;s usually not a number. In other contexts it&#039;s usually a real-valued function. When we&#039;re talking about quantum fields, it&#039;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.

 

[haael]

What you call "standard spin" is the number j from when we write the eigenvalue of the operator J^2=J_x^2+J_y^2+J_z^2 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.

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.

Thanks, guys, that's what I wanted to know. So spin states basically carry less information than the whole spin tensor. That's why there are fewer spin states than tensor components.

 

[conway]

Thanks, guys, that's what I wanted to know. So spin states basically carry less information than the whole spin tensor. That's why there are fewer spin states than tensor components.

Actually, I think I was going to show, using the case of atomic orbitals, how to reconstruct the tensor picture from the spin states before I got sidetracked. Just as the three p states (1, 0, and -1) can be transformed to convey the vector information, the five d states (2, 1, 0, -1, and -2) can be transformed to convey what I would consider the tensor information. Just as the vector p-states, taken in superposition with the ground state, give you static displacements from the origin, the "tensor" d states allow you to make arbitrary small deformations of the s state into a general ellipsoid. That's the same thing the stress tensor does in a solid.

I can follow up on this if anyone is interested.

 

[Nick R]

Quick question - is it productive to ever think about spin as something other than a thing that describes the "intrinsic magnetic field" of a particle?

 

[conway]

Quick question - is it productive to ever think about spin as something other than a thing that describes the "intrinsic magnetic field" of a particle?

Well, I think so. The atomic orbitals have integer-valued spin numbers that don't come from the intrinsic magnetic field, they come from the motion of the electron wave around the atom. Those are the states I was proposing to analyze in my last post.

 

[Fredrik]

Quick question - is it productive to ever think about spin as something other than a thing that describes the "intrinsic magnetic field" of a particle?

Yes, especially when the particle is uncharged.

The spin operators are defined as the generators of the SU(2) subgroup of an irreducible representation of the universal covering group of the symmetry group of spacetime (which can be either the Galilei group or the Poincaré group), and unfortunately there's no easy way to really understand what that means.

A good start is to realize that the spin component operators are (by definition) the ones that appear in the first order terms when you Taylor expand a rotation operator around the identity:

U(\theta)=1-i\sum_k\theta_kJ_k+\mathcal O(\theta^2)

(The sign depends on how you define the parameters, and I don't remember if + or - is preferred). And the operator whose eigenvalue can be interpreted as one of the labels that identifies the particle species, is defined by

J^2=\sum_k J_k^2

 

[passingthru]

As others have said, spin is intrinsic angular momentum, and is assigned discreet values (unlike continuous values for orbits or rotation). The electron, eg, has two discreet (key word) values: 1/2 and -1/2 spin.

Please understand that the guy who coined the intrinsic angular momentum of a particle as "spin" is a complete imbicile. The term is misleading.

For the electron, "spin" is synonymous with "charge." You have a positive charge (same as +1/2 spin) and a negative charge (same as -1/2 spin).

That 'imbicilie' was P.A.M. Dirac. Name-calling is very stupid.

 

[Char. Limit]

That 'imbicilie' was P.A.M. Dirac. Name-calling is very stupid.

You must admit, however, that "spin" is as good a way to describe that particular property of a particle as "color" is to describe quarks.

And that's really not good.

 

[jtbell]

Quick question - is it productive to ever think about spin as something other than a thing that describes the "intrinsic magnetic field" of a particle?

It contributes to the total macroscopic angular momentum of a system. See for example the Einstein - de Haas effect.

 

[haael]

Actually, I think I was going to show, using the case of atomic orbitals, how to reconstruct the tensor picture from the spin states before I got sidetracked. (...)

I can follow up on this if anyone is interested.

I am interested.

 

[conway]

Okay. Did it make sense to you when I talked about how you can move the s cloud in any direction from the origin by adding in a small component of p_x, p_x, and p_z?

 

[SpectraCat]

Well, I think so. The atomic orbitals have integer-valued spin numbers that don't come from the intrinsic magnetic field, they come from the motion of the electron wave around the atom. Those are the states I was proposing to analyze in my last post.

Just a quibble .. it will be less confusing if you stick with the convention and refer to "orbital angular momentum" by that name, rather than by saying that atomic orbitals have "spin". In the context of atomic physics, "spin" is reserved for the intrinsic angular momentum.

 

[conway]

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.

This is not the first time you have gone back and changed your posts after seeing my response.

When I post on Physicsforums, I take the risk of exposing myself to ridicule if my ideas are stupid. It is a risk I am willing to take. Apparently you are not.

The PhysicsForums archives are a permanent record of what was said in these discussions. By going back and altering that record to your benefit, you not only make yourself look smarter than you actually are, you make me look worse because I am seen to be responding inappropriately to things you apparently never said.

I have benefited a great deal from these discussions we’ve had over the last few weeks, and I will no doubt miss having these arguments with you. But I am not willing to continue on an unlevel playing field with an opponent who does not respect or even understand the basic principles of fair play. I will therefore not be responding to your posts in the future.

 

[SpectraCat]

This is not the first time you have gone back and changed your posts after seeing my response.

When I post on Physicsforums, I take the risk of exposing myself to ridicule if my ideas are stupid. It is a risk I am willing to take. Apparently you are not.

The PhysicsForums archives are a permanent record of what was said in these discussions. By going back and altering that record to your benefit, you not only make yourself look smarter than you actually are, you make me look worse because I am seen to be responding inappropriately to things you apparently never said.

That is something I have never done ... as far as I am aware it is expressly against the rules of PF, and is a bannable offense.

I have benefited a great deal from these discussions we’ve had over the last few weeks, and I will no doubt miss having these arguments with you. But I am not willing to continue on an unlevel playing field with an opponent who does not respect or even understand the basic principles of fair play. I will therefore not be responding to your posts in the future.

Conway .. get real. I *amended* my post .. that means I added clarifications to it .. and in the case of post 64 I removed one incorrect statement that I admitted was wrong in the context of your remarks. What more could you possibly want? If you go back and look at what I wrote, you see that I noted that post had been edited, and I said what I had changed. In fact, I am pretty sure you quoted the statement in question in another post, which means it is still in the public record. Go post a big all-caps announcement that you were right and I was wrong if you like.

This is not a game, so the "level playing field" analogy is nonsense. It is also not about you and me, or any other individual .. and you take this stuff WAY too personally. As I have said about 50 times in my posts to you, what I am concerned with is getting everything correct so that future readers can follow what we have done.

I could frankly care less what you or anyone else thinks about how smart or dumb I am ... I also am not afraid to make mistakes and own up to them. I try very hard not to make incorrect statements when I post, but when I do make mistakes, I try to make sure that they do not lead to the confusion of other readers. You need to lighten up and get that Everest size chip off your shoulder.

 

[Frame Dragger]

You know, I came to this site to LEARN, from people who were in various fields of research, teaching, etc. I'm constantly amazed by the number of people who come here thinking it will be their launching point for their new theories! Not only is that counter to the purpose of this site, but it's contrary to the scientific method.

Conway, I don't know why you haven't been banned... presumably because Cat is too kind to report you, but cut the crap. You haven't benefited one iota from these "discussions", because they are always semi-related ramblings from you, which SpectraCat, CollinsMark and others challenge. You sidestep, and continue with some other aspect of your "theory", all the while admitting total ignorance of the physics into which you're treading (QED, general QFTs, etc).

You're deluded, or you have a very focused agenda, but don't play this sob-story crap about editing posts, and level playing fields. Would you like to know why you percieve the field between you and Cat to be so uneven?... HE knows what the hell he's talking about, and when he doesn't he admits it and doesn't speculate.

You're yet to even post a proof of your non-theory which everyone has been telling you is frankly silly to begin with. Obviously a lot of your self-esteem is wound into this theory of yours, or the notion of your level of comprehension. My advice is that instead of NOT responding to the one person who has unfailingly engaged you (Cat), would be that you 'go home' and do the research required to level the field.

You COULD have come here to learn, then see if your theory matches existing evidence, right? Instead, you do... this. The sad thing, is that you're acting this way on a site full of people who see right through you. I grant that must not be what you're used to, but my advice is to find your level, and try to rise from that point. If you're a fabulous genius, then you should do so rapidly, and begin to form theories based on KNOWLEDGE, not complete horsegarbage.

 

[conway]

Conway, I don't know why you haven't been banned... presumably because Cat is too kind to report you, but cut the crap. You haven't benefited one iota from these "discussions", because they are always semi-related ramblings from you...

You know, I might have let it pass or taken it up with SpectraCat in a private communication, but I looked at his alterations (sorry: amendments! Spectracat never "alters" his posts.) and I’m pretty sure he was taking advantage of them to re-argue the points I had made after his original post. I can’t exactly prove it...because he’s amnded his postings. I don’t expect you to understand how wrong this is but Spectracat ought to. And he has certainly amended posts in the past to make some particular arguments look less foolish (or at least less incorrect) than they turned out to be.

You profess wonderment that the moderators haved not yet banned me from the forum. I am also sensitive to this possibility, and that is one reason why it is important to me to have an accurate record of what was said by whom and when. In light of the type of invective that has been thrown back and forth in this discussion (in fact mostly forth and very little back), including your own most recent missive, I want the record of who said what to be clear and accurate.

Spectracat's excuse for amending his posts seems to be so that future visitors will not be confused by wrong science. In his mind, this gives him license to repeatedly tell me I don't know what I'm talking about, and then when it turns out he was wrong, he's entitled to go back and fix his mistakes. For the sake of "future readers".

It’s true that I am not popular in this discussion group and people frequently criticize me for the my tone and attitude, but after carefully reviewing my correspondence and comparing it to that of others, FrameDragger, I’m pretty sure I have nothing to apologize for.

 

[haael]

Okay. Did it make sense to you when I talked about how you can move the s cloud in any direction from the origin by adding in a small component of p_x, p_x, and p_z?

Yeah, go on.

 

[vanesch]

Could you guys please stop having sterile fights here ? That doesn't help anybody, and honestly, I don't think "future generations" will care anything about who was able to put up the best image of himself during a silly fight on an internet forum

The fight about angular wavefunctions and their appearance and their different possible superpositions and so on is actually off-topic here, because the thread is about *intrinsic* spin. Intrinsic spin hasn't much to do with angular wavefunctions except for the fact that they are related to representations of the rotation group.

So let us focus back on the original discussion about intrinsic spin and let us forget about the souvenir future generations might get of us ...

 

[Frame Dragger]

Could you guys please stop having sterile fights here ? That doesn't help anybody, and honestly, I don't think "future generations" will care anything about who was able to put up the best image of himself during a silly fight on an internet forum

Pfft, are you serious? I'm fairly sure that whoever leaves the best impression on the internet wins total wisdom and a tasteful gift-bag from Sears.

EDIT: *looks at post #89*

Okay, looking at the history of the thread, I see that vanesch is right. It was an old thread from last year that got revived last week, and haael's question about the relation between the z-axis representation of spin versus the vector/tensor representation was off topic. I'll start a new thread to continue that discussion.

Hint: The staff member wasn't offering you a choice, and s/he probably didn't need your added input to realize that things in the thread would ultimately trend their way. That's what being an admin MEANS.

Still, thanks for obfuscating Vanesch's clarity.

 

[conway]

The fight about angular wavefunctions and their appearance and their different possible superpositions and so on is actually off-topic here, because the thread is about *intrinsic* spin. Intrinsic spin hasn't much to do with angular wavefunctions except for the fact that they are related to representations of the rotation group.

So let us focus back on the original discussion about intrinsic spin and let us forget about the souvenir future generations might get of us ...

Okay, looking at the history of the thread, I see that vanesch is right. It was an old thread from last year that got revived last week, and haael's question about the relation between the z-axis representation of spin versus the vector/tensor representation was off topic. I'll start a new thread to continue that discussion.

 

[conway]

Pfft, are you serious? I'm fairly sure that whoever leaves the best impression on the internet wins total wisdom and a tasteful gift-bag from Sears.

EDIT: *looks at post #89*


Hint: The staff member wasn't offering you a choice, and s/he probably didn't need your added input to realize that things in the thread would ultimately trend their way. That's what being an admin MEANS.

Still, thanks for obfuscating Vanesch's clarity.

The moderator asked us to stop our little catfight (no pun intended, SpectraCat) and I have respected that. I'm not sure he was offering you a choice either, so maybe you should consider doing the same.