Noether Current Derivation Issue: Why Don't the Last Two Terms Vanish?

[CAF123]

Sounds ok to me, except that I wouldn't call the 2nd piece "internal". A better name might be "intrinsic", since it's relevant to the intrinsic spin indices. Then the term "internal piece" could be used later for gauge tuplet indices.

The reason that I was unclear why $\phi'(x) - \phi(x)$ should be equal to the full generator is because the coordinates in $\phi'(x) - \phi(x)$ are not shifted at all, so it appears there is no orbital piece having any effect.

Thanks for the suggestion about Ballentine - would you recommend his book for introducing the path integral formulism/ Feynman path integrals and time dependent perturbation theory. I also have Sakurai and Griffiths.

I have a few more questions with regard to material from earlier on in the book by FMS, if that is ok.
- P.40 eqn (2.133). I was trying to understand the factor of 1/2 appearing on the LHS of that equation. Intuitively, I think the 1/2 is to compensate for exactly half of the entries in the $\omega_{\rho \nu}$ matrix being not independent, so the number of independent generators is also halved. However, I was looking to try to get the half via a more mathematical analysis: Given $$i \omega_{\rho \nu}L^{\rho \nu}\Phi = \omega_{\rho \nu} \left(\frac{\delta x^{\mu}}{\delta \omega_{\rho \nu}} \frac{\partial \Phi}{\partial x^{\mu}} - \frac{\delta F}{\delta \omega_{\rho \nu}}\right) \Rightarrow -i \omega_{\nu \rho}L^{\rho \nu}\Phi = -\omega_{\nu \rho} \left(-\frac{\delta x^{\mu}}{\delta \omega_{\nu \rho}} \frac{\partial \Phi}{\partial x^{\mu}} + \frac{\delta F}{\delta \omega_{\nu \rho}}\right) $$ and then relabel gives $$-i \omega_{\rho \nu}L^{\nu \rho}\Phi = \omega_{\rho \nu} \left(\frac{\delta x^{\mu}}{\delta \omega_{\rho \nu}} \frac{\partial \Phi}{\partial x^{\mu}} - \frac{\delta F}{\delta \omega_{\rho \nu}}\right)$$ then I tried to add this to the first equation, but it didn't give (2.133). Do you have any ideas?

-In another P.SE thread here,http://physics.stackexchange.com/questions/119381/spin-matrix-for-various-spacetime-fields I obtain the generator of rotations of the SO(2) rotation group for an infinitesimal rotation of 2D vectors, collectively comprising a vector field. I now tried to relate this to the spin-1/2 electron system, but it appears vectors representing states for that system transform under the Pauli matrices instead. Is there an underlying reason for this? I also noticed that $$\text{Id} + \omega \begin{pmatrix} 0&1\\-1&0 \end{pmatrix} = \text{Id} + i \omega \begin{pmatrix} 0&-i \\i&0 \end{pmatrix},$$ so I seemed to have made contact with one of the Pauli matrices. What is so special about this particular Pauli matrix showing up here?

Thank you, strangerep.

 

[strangerep]

Thanks for the suggestion about Ballentine - would you recommend his book for introducing the path integral formulism/ Feynman path integrals and time dependent perturbation theory.

Ballentine is a modern development of QM, not QFT. The reason I keep pushing Ballentine in your direction is that I sense your proficiency in ordinary QM needs improvement.

For path-integral stuff,... well,... Greiner & Reinhardt develop QFT by both the canonical method, and by path integrals.

I have a few more questions with regard to material from earlier on in the book by FMS, if that is ok. - P.40 eqn (2.133). I was trying to understand the factor of 1/2 appearing on the LHS of that equation. [...]Do you have any ideas?

Not really. It's a double-counting thing. There is some freedom in how one defines generators and parameters (up to a scale factor), so one chooses the factor to make subsequent calculations more convenient.

[...] so I seemed to have made contact with one of the Pauli matrices. What is so special about this particular Pauli matrix showing up here?

Any 2x2 matrix can be expressed as a linear combination of the Pauli matrices and the unit matrix, so that much is unremarkable. To get more insight, you could try working out the effect of the other Pauli matrices on an arbitrary 2D vector. What transformation of the 2D vectors do they generate? IOW, what matrices do you get when you exponentiate $a \sigma_x$ and $b \sigma_z$ (where $a,b$ are real parameters) ?

And what is the most general linear transformation of a 2D vector space?

 

[CAF123]

The reason that I was unclear why $\phi'(x) - \phi(x)$ should be equal to the full generator is because the coordinates in $\phi'(x) - \phi(x)$ are not shifted at all, so it appears there is no orbital piece having any effect.

Did you have any comments with regard to what I wrote above? There is a sketch on P.67 Ballentine that I thought may help, but I couldn't get much from it in terms of answering my question.

Ballentine is a modern development of QM, not QFT. The reason I keep pushing Ballentine in your direction is that I sense your proficiency in ordinary QM needs improvement.

You are right, I have only done a single Griffiths level QM course. More comes after the summer break, before the following year where I then do the QFT courses. I should also mention that the reason I am inclined to pursue FMS (although I accepted Greiner's derivation for the Noether derivation) is because this book was the book the professor I am working with assigned to aid me in the project I am doing. Despite not having done any QFT, the professor is keen for me to do a little bit in correlation functions and ward identities. I believe these have connections to the quantum field theory counterparts to the conserved currents arising from the classical field theory analysis. Would you be able to tell me if the coverage of this in FMS looks reliable? (pp.42-45, 104-109.) Thanks very much. After the project, I will study the more general treatment from other texts, e.g some of which you talked about.

Any 2x2 matrix can be expressed as a linear combination of the Pauli matrices and the unit matrix, so that much is unremarkable.

Is it correct to say that the spin-1/2 electron spin states transform under the quantity $$1 + i\omega \begin{pmatrix} 0&-i\\i&0 \end{pmatrix}?$$ I just wondered because, from what I have read, the spin1/2 electron system transforms under the fundamental rep of SU(2) (the rescaled Pauli matrices). I suppose those states can be mapped to vectors in a 2D Euclidean plane (since j=1/2 => 2 values for m) in which case they would transform. Or if I understand Ballentine, P.172 eqn (7.50) correctly, to make contact with what I wrote, $a \cdot \sigma = i \omega \sigma_y$, with vanishing coefficients for the other Pauli matrices in the linear combination. However, this would involve making one of the coefficients complex = iω.

 

[strangerep]

Did you have any comments with regard to what I wrote above?

Well,... (deep breath...), this involves some rather advanced concepts in field theory, but I'll try to give a sketch...

Have you ever heard the saying that "elementary particle types can be classified according to the unitary irreducible representations of the Poincare group" ?

A related, but perhaps easier, concept is that the values of total spin, and spin projection, are determined by finding the unitary irreducible representations of the rotation group. The latter is exactly what Ballentine performs in section 7.1. I get the feeling many people kinda gloss over that section, eager to move onward, but it contains incredibly important foundational material, that should be studied -- and then meditated upon.

In the case of the full Lorentz group (and hence the Poincare group_, one finds that there are no finite-dimensional unitary irreducible representations, but only infinite-dimensional reps. Hence they are necessarily field representations.

Now comes the big insight: the Lagrangians and the fields they're built from work in concert to yield a representation of the Poincare algebra(!). This is a deep and crucial insight, essentially responsible for "why field theory is the way it is" -- to quote Weinberg. It means that we can find certain expressions built from the fields which satisfy the Poincare commutation relations. In the classical case, this is implemented via Poisson brackets and functional derivatives. You can study this further in Greiner, section 2.5. The quantities corresponding to each continuous symmetry of the Lagrangian also generate that symmetry transformation -- in the sense of Poisson brackets.

This principle -- of building a field representation of the Poincare group -- then guides the choice of possible interactions between the free fields. The fields in the interacting theory must still give a representation of the Poincare group, though it is a different representation from that given by the free fields. Weinberg shows how this usefully restricts the possible choice of interaction terms in the Lagrangian.

Ballentine does a similar thing for the non-relativistic Galilean case in his section 3.4, case (iii). It's only the relatively easy case of a spinless particle interacting with an external field, but the guiding principle is that he's still trying to ensure that the net result gives a representation of the Galilean algebra. This criterion severely restricts the possible forms of the interaction, but it turns out that this covers a vast number of cases.

Anyway,... getting back to the classical field case... the orbital part of the generator is still in there, though slightly disguised. See, e.g., Greiner's eq(2.70). (BTW, did you ever look at the MTW reference I mentioned earlier? It's relevant here.)

All this stuff is essentially why I suggested you study that whole chapter of Greiner carefully, right to the end (rather than just stopping at Noether's thm). A physicist needs a deep understanding of the field representations of symmetry groups.

You are right, I have only done a single Griffiths level QM course. More comes after the summer break, before the following year where I then do the QFT courses. I should also mention that the reason I am inclined to pursue FMS (although I accepted Greiner's derivation for the Noether derivation) is because this book was the book the professor I am working with assigned to aid me in the project I am doing. Despite not having done any QFT, the professor is keen for me to do a little bit in correlation functions and ward identities. I believe these have connections to the quantum field theory counterparts to the conserved currents arising from the classical field theory analysis.

That's quite an advanced topic. Have you studied Green's functions yet? (They're related to the simplest 2-point correlation functions.) I'm not sure about the wisdom of trying to study these before a basic course in QFT, but heh, maybe I'm wrong.

Would you be able to tell me if the coverage of this in FMS looks reliable? (pp.42-45, 104-109.)

I don't think I can give you reliable advice about that, since I don't know what was in your professor's mind. (Did he give you a written statement of the project, or just some vague waffle?)

Tbh, I think it's all a bit advanced for where you are right now, and you're kinda being thrown in the deep end. But you might be able to get a more intuitive understanding of correlations functions (and path integrals) from Zee's QFT book.

Is it correct to say that the spin-1/2 electron spin states transform under the quantity $$1 + i\omega \begin{pmatrix} 0&-i\\i&0 \end{pmatrix}?$$ I just wondered because, from what I have read, the spin1/2 electron system transforms under the fundamental rep of SU(2) (the rescaled Pauli matrices).

The usual rotation group is represented in the case of spin-1/2 particles as $SU(2,C)$, i.e., 2x2 complex unitary matrices.

I suppose those states can be mapped to vectors in a 2D Euclidean plane (since j=1/2 => 2 values for m) in which case they would transform.

Except that 2-complex-dimensional, not 2-real-dimensional.

Or if I understand Ballentine, P.172 eqn (7.50) correctly, to make contact with what I wrote, $a \cdot \sigma = i \omega \sigma_y$, with vanishing coefficients for the other Pauli matrices in the linear combination. However, this would involve making one of the coefficients complex = iω.

Can you work out this problem: what are the (matrix) generators and Lie algebras for the groups $SU(2,C)$, and $SL(2,R)$ ?

 

[CAF123]

The usual rotation group is represented in the case of spin-1/2 particles as $SU(2,C)$, i.e., 2x2 complex unitary matrices.

Except that 2-complex-dimensional, not 2-real-dimensional.

Is there a reason why the state vectors in two dimensional space do not transform under the group of 2D real matrices SL(2,R)? (..or is that what you are getting me to see below?)

Can you work out this problem: what are the (matrix) generators and Lie algebras for the groups $SU(2,C)$, and $SL(2,R)$ ?

SU(2) is locally isomorphic to SO(3) which means it shares the same Lie algebra as SO(3), satisfying commutation relations $[T_a, T_b] = i\epsilon_{abc}T_c$. In two dimensions, suitable representations of the generators are $T_a = 1/2 \sigma_a$ where $\sigma_a$ are the Pauli matrices.

For $SO(2,R)$, the generator would be the 2x2 rotation matrix. For SL(2,R), from this document, it appears the Lie algebra is the same up to a sign in the last commutation relation.http://infohost.nmt.edu/~iavramid/notes/sl2c.pdf

 

[strangerep]

Is there a reason why the state vectors in two dimensional space do not transform under the group of 2D real matrices SL(2,R)? (..or is that what you are getting me to see below?)

Partly, yes.

SU(2) is locally isomorphic to SO(3) which means it shares the same Lie algebra as SO(3), satisfying commutation relations $[T_a, T_b] = i\epsilon_{abc}T_c$. In two dimensions, suitable representations of the generators are $T_a = 1/2 \sigma_a$ where $\sigma_a$ are the Pauli matrices.

For $SO(2,R)$, the generator would be the 2x2 rotation matrix. For SL(2,R), from this document, it appears the Lie algebra is the same up to a sign in the last commutation relation.http://infohost.nmt.edu/~iavramid/notes/sl2c.pdf

(I presume you meant "...same as for su(2,C) up to a sign...".)

OK, so when you asked earlier about a Pauli matrix that turned up in something you were doing, herein lies the reason: when you're working in 2D, the Pauli matrices are always floating around somewhere. Depending on whether you multiply (some of) them by $i$, you get different algebras. So you need to be clear up front about which group is applicable to the scenario you're considering.

The important insight is that the state vectors are only of secondary importance. What matters most is the dynamical group one is trying to represent as Hilbert space operators. E.g., many physical scenarios involve the rotation group, and some involve $SL(2,R)$, not to mention various other stuff.

So... first one must determine the dynamical group applicable to a physical scenario, then find all the unitary irreducible representations thereof (along the lines of what Ballentine does in sect 7.1 for the rotation group). The structure of the group's spectrum (Casimir values, and other eigenvalues) determines the dimension and structure of the Hilbert space(s) suitable for modelling that scenario.

 

[CAF123]

Thanks strangerep, please tell me if this is about right:

So we know the spin states of the spin-1/2 electron system transform under representations of SU(2) because those (matrix) reps yield the correct observables (I.e eigenvalues) when they act on spin states belong to the spin-1/2 electron system. The eigenvalues of the Pauli matrices are $\pm 1$ and so to obtain the correct values of spin measured along an arbritary axis, ($\pm \hbar/2$ - the $m$ quantum number) we necessarily multiply the Pauli matrices by this factor. That is why the Pauli matrices are used in this case.

An arbritary ket $|\alpha\rangle$ transforms under a finite rotation like $|\alpha\rangle \rightarrow |\alpha ' \rangle = D(\hat n, \phi) |\alpha\rangle = \exp (i\phi \hat n \cdot J/\hbar)$. For states in the spin 1/2 electron system, (that is states of a spin 1/2 system that are linear combinations of some basis vectors, e.g could choose the eigenvectors of one of the Pauli matrices.) $J \rightarrow S \rightarrow \hbar/2 \sigma_i$ and all states (or spinors) transform under $\exp(i\phi \hat n \cdot S/\hbar) = \exp(i \phi \hat n \cdot \sigma/2)$

What is the reason for the spin states of this system transforming under a generator comprising three matrices?, i.e why is $\mathbf S = \frac{\hbar}{2}\sigma = \frac{\hbar}{2}(\sigma_1, \sigma_2, \sigma_3)$? What is special about three?

Is there any reason why, when I computed the spin matrix for a 2 dimensional vector field, I obtained the generator of SO(2) and not the Pauli matrices? It looks like the result I got: $$\text{Id} + i \omega \begin{pmatrix}0&-i\\i&0 \end{pmatrix}$$ is the infinitesimal version of $\exp(i\omega \hat n \cdot \sigma)$ with $$\hat n \cdot \sigma = \sigma_2 = \begin{pmatrix} 0&-i \\ i&0 \end{pmatrix}.$$ which seems to mean $\hat n = (0,1,0)$.

 

[strangerep]

Sorry, I've learned that it's an inefficient use of my time to try and deconstruct posts like that.

Study Ballentine ch7 carefully, at least up to and including section 7.6.

Then decide whether any followup questions remain.

 

[CAF123]

Do you mean to say that what I wrote is incorrect?
A representation of the spin operators S_x, S_y, S_z can be found by applying the (Casimir of SU(2)) operator S² and S_z onto the basis states $| 1/2\rangle$ and $|-1/2 \rangle$. and using the fact that $S_z| \pm 1/2\rangle = \pm \hbar/2 | \pm 1/2 \rangle$ and similar result for $S^2$. The results obtained match the Pauli matrices. I think that should be more accurate than what I wrote previously.

However, in a two dimensional Hilbert space (i.e one in which the spin 1/2 electron states live), why do we mention the z component of spin?

From reading the relevant chapters in Ballentine, I am not seeing an answer to this question:

An arbritary ket $|\alpha\rangle$ transforms under a finite rotation like $|\alpha\rangle \rightarrow |\alpha ' \rangle = D(\hat n, \phi) |\alpha\rangle = \exp (i\phi \hat n \cdot J/\hbar)$. For states in the spin 1/2 electron system, (that is states of a spin 1/2 system that are linear combinations of some basis vectors, e.g could choose the eigenvectors of one of the Pauli matrices.) $J \rightarrow S \rightarrow \hbar/2 \sigma_i$ and all states (or spinors) transform under $\exp(i\phi \hat n \cdot S/\hbar) = \exp(i \phi \hat n \cdot \sigma/2)$

...

Is there any reason why, when I computed the spin matrix for a 2 dimensional vector field, I obtained the generator of SO(2) and not the Pauli matrices? It looks like the result I got: $$\text{Id} + i \omega \begin{pmatrix}0&-i\\i&0 \end{pmatrix}$$ is the infinitesimal version of $\exp(i\omega \hat n \cdot \sigma)$ with $$\hat n \cdot \sigma = \sigma_2 = \begin{pmatrix} 0&-i \\ i&0 \end{pmatrix}.$$ which seems to mean $\hat n = (0,1,0)$.

Sorry for prolonging this example, but I wanted to try to relate the spin matrix I found earlier on (by trying out the generic spin matrix for a vector field obtained by the transformation theory of fields) and apply it to a physical example. The first one that popped into my head was the spin 1/2 electron system and I now want to see if the connection above is a sensible one.
Thanks.

 

[strangerep]

Do you mean to say that what I wrote is incorrect?

I meant that it would be too much work (for me) to disentangle what was correct from what was subtly incorrect or backwards.

(Recall the oath sworn in a court: one swears to tell the truth, the whole truth, and nothing but the truth. The point is that as soon as one injects something not quite right, the whole is no longer the truth, even though an ignorant person or liar might insist that it's still "mostly true".)

A representation of the spin operators S_x, S_y, S_z can be found by applying the (Casimir of SU(2)) operator S² and S_z onto the basis states $| 1/2\rangle$ and $|-1/2 \rangle$. and using the fact that $S_z| \pm 1/2\rangle = \pm \hbar/2 | \pm 1/2 \rangle$ and similar result for $S^2$. The results obtained match the Pauli matrices. I think that should be more accurate than what I wrote previously.

Unfortunately, it's kinda backwards. See below.

However, in a two dimensional Hilbert space (i.e one in which the spin 1/2 electron states live), why do we mention the z component of spin?

Where did the 2D Hilbert space come from and how? (If you had studied Ballentine, instead of just skim-reading it, you'd be able to answer that better than you have so far.)

From reading the relevant chapters in Ballentine, I am not seeing an answer to this question:

That's because you're skim-reading rather than studying carefully. By "study" I mean: put your current question aside temporarily, and study those chapters in and of themselves -- without reference to your other stuff. Then come back and try to relate.

Sorry for prolonging this example, but I wanted to try to relate the spin matrix I found earlier on (by trying out the generic spin matrix for a vector field obtained by the transformation theory of fields) and apply it to a physical example. The first one that popped into my head was the spin 1/2 electron system and I now want to see if the connection above is a sensible one.

You need to be clear whether you're considering a classical case, or a quantum case. There is no such thing as quantized spin-1/2 in the classical case.

Maybe you need to do a reset. I.e., compose a new question in a new thread?

 

[CAF123]

Thanks strangerep, I will come back to Ballentine later. But for the time being, the matrix that I derived acts on space-time and therefore not on the space where the spin states live (that being the 2D Hilbert space (to answer your question above, it is 2D because a spin 1/2 system has two independent directions or orbital directions, which give rise to the term spin up/ spin down - I am not sure if that is the answer you were looking for)) so there really is no connection.

I will maybe make a new thread later, but I have a final question in relation to this thread title topic if that is okay. I understand that to every continuous (global) symmetry of the action, meaning it remains invariant under the transformation of the fields, we may associate a conserved current to it. My question is: what constitutes a symmetry transformation?

Given $S = \int_{\mathcal D}\,\text{d}^d x \mathcal L(\phi, \partial \phi)$ we have that $$S' = \int_{\mathcal D}\,\text{d}^d x \left|\frac{\partial x'}{\partial x}\right| \mathcal L(F(\phi(x)), \partial_{\mu}' F(\phi(x))),$$ so this would suggest to me that provided,
a) the Jacobian factor $|\partial x'/\partial x|$ is unity,
b)$F(\phi(x)) = \phi(x)$,
c)$\partial_{\mu}' F(\phi(x)) = \partial_{\mu}\phi$

then the transformation is deemed to be a symmetry. Would that be right?

I suppose those conditions a)- c) would always make $S'=S$ and therefore always invoke a symmetry, however those conditions are not wholly self contained (In other words, a symmetry transformation need not imply a)-c)). Other symmetries could arise depending on the form and structure of the lagrangian. To illustrate what I mean, under the transformation $\phi'(x) = e^{i\theta}\phi(x)$ we have an invariant action for a lagrangian given here, bottom of first page, http://www.itp.phys.ethz.ch/research/qftstrings/archive/12HSQFT1/Chapter04.pdf . So even though the field transformed non-trivially, the action may still be invariant because the lagrangian was overall unaffected.

Does this all seem accurate?

 

[strangerep]

[...] I am not sure if that is the answer you were looking for [...]

It's not.

what constitutes a symmetry transformation?

That's answered in the 3rd paragraph of Beisert notes you linked to.

Given $S = \int_{\mathcal D}\,\text{d}^d x \mathcal L(\phi, \partial \phi)$ we have that $$S' = \int_{\mathcal D}\,\text{d}^d x \left|\frac{\partial x'}{\partial x}\right| \mathcal L(F(\phi(x)), \partial_{\mu}' F(\phi(x))),$$ so this would suggest to me that provided,
a) the Jacobian factor $|\partial x'/\partial x|$ is unity,
b)$F(\phi(x)) = \phi(x)$,
c)$\partial_{\mu}' F(\phi(x)) = \partial_{\mu}\phi$

then the transformation is deemed to be a symmetry.

Yes, but it's a rather trivial symmetry.

[...] under the transformation $\phi'(x) = e^{i\theta}\phi(x)$ we have an invariant action for a lagrangian given here, bottom of first page, http://www.itp.phys.ethz.ch/research/qftstrings/archive/12HSQFT1/Chapter04.pdf . So even though the field transformed non-trivially, the action may still be invariant because the lagrangian was overall unaffected.

Yes.

More generally, we might have a symmetry in which the Lagrangian is changed, by the change is compensated by the Jacobian term, hence leaving the overall action invariant. Or the extra bit might boil down to being a total derivative, which doesn't affect the equations of motion.

There's actually a hierarchy of symmetries (though the terminology for each type seems nonuniform among different authors). If you really want to know more about this, see my posts in this thread and this one .