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

[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 .