Understanding Schur's Lemma in the Context of Free Particles

[WiFO215]

I'm reading Ballentine. He states (page 80) when introducing a free particle that the set of position and momentum operators together form an irreducible set and goes ahead to apply Schur's lemma. How is it that they form an irreducible set? How exactly does one deduce that?

 

[dextercioby]

This is the classical result of the Stone-von Neumann's theorem of 1930-1931. The original proof of this theorem is given by John von Neumann and can be found, for example, in Ed. Prugovecki's <Quantum Mechanics in Hilbert Space> (1970).

The theorem was later derived as a result of group theory by George Mackey and I think Mackey's work on this appears in Bratelli and Robinson's book on operator algebras.

EDIT: I just realized (afer reading post 3) that my original post actually refers to the irreducibility of the Schrödinger's representation of the commutation relations of Born and Jordan. The algebra of operators in its abstract form has no proper bilateral ideals. The only bilateral ideal is the algebra itself, which is course not proper.

 

[strangerep]

[Ballentine] states (page 80) when introducing a free particle that the set of position and momentum operators together form an irreducible set and goes ahead to apply Schur's lemma. How is it that they form an irreducible set? How exactly does one deduce that?

The basic idea of an irreducible set in this case (i.e., for the set of operators {Q,P})
is that there is no subalgebra (other than 0 and the whole algebra) which is invariant
under the action of the algebra. I.e., one cannot find a nontrivial subalgebra which maps
into itself under the action of (i.e., under commutation with) Q and P.
In this simple case, it's pretty easy to see, after one understands the definition.

OTOH, for case (ii) on p82 with internal (spin) degrees of freedom, the algebra is
larger since there's also spin operators which commute with Q and P. Thus, in that
case, this larger set of operators is not irreducible.

 

[WiFO215]

The basic idea of an irreducible set in this case (i.e., for the set of operators {Q,P})
is that there is no subalgebra (other than 0 and the whole algebra) which is invariant
under the action of the algebra. I.e., one cannot find a nontrivial subalgebra which maps
into itself under the action of (i.e., under commutation with) Q and P.
In this simple case, it's pretty easy to see, after one understands the definition.

OTOH, for case (ii) on p82 with internal (spin) degrees of freedom, the algebra is
larger since there's also spin operators which commute with Q and P. Thus, in that
case, this larger set of operators is not irreducible.

I don't know much about algebras (and subalgebras). Is there a simpler explanation or some reference material you suggest? I'd like to learn group theory now as often times, my googling finds articles which are written through and through in the language of algebra and symmetry groups. So any reference material along that line is most welcome.

 

[strangerep]

I don't know much about algebras (and subalgebras). Is there a simpler explanation or some reference material you suggest? I'd like to learn group theory now as often times, my googling finds articles which are written through and through in the language of algebra and symmetry groups. So any reference material along that line is most welcome.

Ballentine already gives a basic explanation of the Galilean group early in ch3 (which
I presume you've already studied since your question was about stuff later in that
chapter).

You could also try Greiner & Muller's "QM -- Symmetries", which introduces Lie
group theory in QM context and gives lots of explicit calculations.

Also, Sexl & Urbantke's "Relativity, Groups, Particles" which deals with relativistic
situations has received good reviews around here -- but I haven't studied it in detail
myself.

But for the case Ballentine is considering, look at the commutation relations for
the generators of the Galilean group, tabulated in his eq(3.35). Now consider just
the J's. The commutation relation between them is:

$$\left[ J_\alpha \,,\, J_\beta \right] ~=~ i \, \epsilon_{i\alpha\beta\gamma}\, J_\gamma$$

I.e., the J's by themselves constitute a subalgebra because their commutation relations
don't yield other generators besides J's. I.e., the set of J's is "closed" under commutation.

In contrast,

$$\left[ G_\alpha \,,\, H \right] ~=~ i \, P_\alpha ~,$$

which shows that the set of G's and H are not a subalgebra, since their commutator
yields something which is neither a G nor H.

Now consider the J subalgebra again. We've established that it's a subalgebra but
what can we say about irreducibility? The (commutator) action of other generators
such as P, G on J's does not yield J's. E.g.,

$$\left[ J_\alpha \,,\, P_\beta \right] ~=~ i \, \epsilon_{i\alpha\beta\gamma}\, P_\gamma$$

which shows that the J subalgebra is not invariant under the action of the whole
Galilei algebra (since the action of P on the J subalgebra takes us out of the J
subalgebra).

(BTW, by "subalgebra" in the above, I mean of course a Lie subalgebra.)


Now back to Ballentine p80. Let's consider just the set of operators {Q, P, 1} (I added
the "1" for completeness here, since [Q,P] is proportional to 1). Can you find a
subalgebra in this set which remains invariant under the commutator action
of all the other elements?

 

[WiFO215]

Ballentine already gives a basic explanation of the Galilean group early in ch3 (which
I presume you've already studied since your question was about stuff later in that
chapter).

...

Now back to Ballentine p80. Let's consider just the set of operators {Q, P, 1} (I added
the "1" for completeness here, since [Q,P] is proportional to 1). Can you find a
subalgebra in this set which remains invariant under the commutator action
of all the other elements?

Oh! Now I get it. When you choose Q and P, under the action of any other operator, you'll still land up somewhere inside the set Q,P. Thank you so much!

As for books, I'll look into those you recommended.

Thanks once again.

 

[WiFO215]

Hold on. I have another question. In the following section, he introduces the spin operator S which as independent degree of freedom such that [Q,S] = [P,S] = 0. Thus, the action of S on {Q,P} does not take it outside of itself, since it contains zero. How come it isn't irreducible now? Or is it wrong that 0 is contained within {Q,P}? If so, then I'll consider the set {Q,P,0}. Shouldn't that be irreducible?

 

[vanhees71]

Sure you can add the three spin-component operators to the position and momentum operators. This describes you a particle with spin. If you don't add spin operators, your particles have spin 0.

In general, the observable algebra of a quantum system is determined by the ray representations of its underlying symmetry group. Position, momentum, and spin come from the ray representation of the Galilei group (or from the Lorentz group for the special case of massive particles).

 

[strangerep]

Hold on. I have another question. In the following section, he introduces the spin operator S which as independent degree of freedom such that [Q,S] = [P,S] = 0. Thus, the action of S on {Q,P} does not take it outside of itself, since it contains zero. How come it isn't irreducible now? Or is it wrong that 0 is contained within {Q,P}? If so, then I'll consider the set {Q,P,0}. Shouldn't that be irreducible?

0 need not be considered here, since the commutator of zero with anything is zero again.
The general definition of irreducibility usually has some extra words saying that the
subset must be nontrivial. In the present case, Lie algebras are also vector spaces,
and the zero element is regarded as a trivial element.

Perhaps its best to try and understand the meaning of reducibility/irreducibility
in terms of "decomposing" sets, algebras, etc, into unions of simpler/smaller things...

Here's a quote from Wikipedia,
http://en.wikipedia.org/wiki/Irreducibility_(mathematics )

A topological space is irreducible if it is not the union of two proper closed subsets.

In the present case, substitute "topological space" with "Lie algebra", and interpret
"closed" in terms of the commutator.

Applying this, we see that the set {1,Q,P,S} is reducible because it can be expressed
as a union of {1,Q,P} and {S}. Both of the latter are subalgebras which are
separately closed under the commutator action, and are both irreducible because
they have no smaller closed subalgebras. The fact that {1,Q,P,S} contains such
subalgebras is why we call it "reducible" -- it can be decomposed (i.e., "reduced") to
a union of two subalgebras.

HTH.

 

[WiFO215]

.

Applying this, we see that the set {1,Q,P,S} is reducible because it can be expressed
as a union of {1,Q,P} and {S}. Both of the latter are subalgebras which are
separately closed under the commutator action, and are both irreducible because
they have no smaller closed subalgebras. The fact that {1,Q,P,S} contains such
subalgebras is why we call it "reducible" -- it can be decomposed (i.e., "reduced") to
a union of two subalgebras.

HTH.

Exactly. This is what I picked up from your previous posts, and this is why I got confused. You (and therefore I) say that {Q,P} is irreducible even with spin. I quote Ballentine on page 82, "The set {Q,P} is not irreducible in this case..". Why is he stating the opposite?

 

[strangerep]

Exactly. This is what I picked up from your previous posts, and this is why I got confused. You (and therefore I) say that {Q,P} is irreducible even with spin. I quote Ballentine on page 82, "The set {Q,P} is not irreducible in this case..". Why is he stating the opposite?

The full quote is

The set {Q,P} is not irreducible in this case because an operator that commutes with
that set may be a function of the operators of the internal degrees of freedom.

(I.e., S). Here I think he's trying to apply Schur's lemma
(see http://en.wikipedia.org/wiki/Schur's_lemma )
but I suspect he's not applying it correctly, since the lemma is of "if-then" form,
not "if-and-only-if".

For the rest of Ballentine's treatment, however, I'm pretty sure you're safe if you
just ignore that one sentence, since it's not needed in what follows, afaict.

 

[WiFO215]

The full quote is
(I.e., S). Here I think he's trying to apply Schur's lemma
(see http://en.wikipedia.org/wiki/Schur's_lemma )
but I suspect he's not applying it correctly, since the lemma is of "if-then" form,
not "if-and-only-if".

For the rest of Ballentine's treatment, however, I'm pretty sure you're safe if you
just ignore that one sentence, since it's not needed in what follows, afaict.

The lemma is stated in Appendix A and seems to be of 'iff' and not 'if-then' form.

 

[strangerep]

The lemma is stated in Appendix A and seems to be of 'iff' and not 'if-then' form.

He's proving a version of Schur's lemma in a more specialized context of Hilbert
spaces, etc. He uses a more specific meaning of "irreducibility" in terms of
subspaces of the Hilbert space, rather than the more general algebraic
meaning I was using.

I had read almost all of Ballentine, but skipped his appendices.
Clearly that was a mistake. I should have at least known they existed. :-(

 

[WiFO215]

He's proving a version of Schur's lemma in a more specialized context of Hilbert
spaces, etc. He uses a more specific meaning of "irreducibility" in terms of
subspaces of the Hilbert space, rather than the more general algebraic
meaning I was using.

I had read almost all of Ballentine, but skipped his appendices.
Clearly that was a mistake. I should have at least known they existed. :-(

Well, I guess that leaves me back at square one. He uses the Schur's lemma he's quoted backwards to show that certain operators are multiples of the identity. QUESTION: How do you know {Q,P} is irreducible in the first place? What does irreducible mean?

 

[dextercioby]

Exactly. This is what I picked up from your previous posts, and this is why I got confused. You (and therefore I) say that {Q,P} is irreducible even with spin. I quote Ballentine on page 82, "The set {Q,P} is not irreducible in this case..". Why is he stating the opposite?

I think this 'irreducibility' part can be seen as a matter of convention. So the term itself could bear some ambiguity. We should use Lie-algebraic terms like: basis, center and central extension. In this case, adding S to the original Heisenberg algebra is claimed by Ballentine to <destroy> reducibility, but this is done in a trivial manner, namely by extending actually the center of the original algebra and not the whole algebra itself.

If {S} is an element of the center of the extended algebra, the expression below forms a short exact sequence

$$\{1\} \rightarrow \{S\} \rightarrow \{S,Q,P,1\} \rightarrow \{Q,P,1\} \rightarrow \{1\}$$

and the algebra in the middle is called a central extension of the algebra to its right.

My convention is that Q,P are still irreducible, if the orginal Heisenberg algebra is only centrally extended. Ballentine claims otherwise.

 

[strangerep]

[Ballentine] is proving a version of Schur's lemma in a more specialized
context of Hilbert spaces, etc [...]

Well, I guess that leaves me back at square one.

Really? I don't see why. I thought that his Appendices A & B explained
quite clearly what he's doing, and the context he's using.

I'll answer the rest of your post in reverse order...

What does irreducible mean?

Hopefully, we've discovered in this thread that the word can have subtly different
meanings in different contexts (and Schur's lemma can be a bit different between
these different contexts. For our purposes here, we should focus on Ballentine's
meaning of the term, which also happens to be the one that's most applicable to
QM, classification of elementary particles by irreducible representations, etc, etc.

Let's look more closely at what Ballentine says in his Appendix A...

To say that a set of operators is irreducible on a vector space V means that
no subspace of V is invariant under the action of all operators in the set.

The meaning of "irreducible" that he's using pertains to the vector space
(Hilbert space) on which the operators act. This is subtly different from
my earlier meaning in terms of the commutator action of the members of
an algebra on itself.

The fact that he's using the meaning in terms of invariance of subspaces
of V is the crucial point here. This meaning is much more useful for doing
QM, (and I'm quite pleased that this thread has forced me to see the subtle
distinctions of meaning between the different contexts more clearly).

In the context of QM, Schur's Lemma is indeed "if-and-only-if". But that's
only because the invariance of subspaces is in terms of subspaces of a
Hilbert space, where self-adjointness and orthogonality have meaning.
This extra structure enables a stronger version of Schur's lemma.

He uses the Schur's lemma he's quoted backwards to show that certain
operators are multiples of the identity.

I don't think he's quoted it "backwards". If we have a theorem which
says "A if-and-only-if B", then we can also say "B if-and-only-if A".

QUESTION: How do you know {Q,P} is irreducible in the first place?

Let's look at Ballentine's Appendix B. The Hilbert space he's using is
the usual space of wave functions $$\psi(x)$$. Representing Q
and P as multiplication by x, and differentiation wrt x, respectively,
he derives the result that there cannot be a nontrivial subspace of
the space of wave functions which is invariant under the action of
both these operators. He's uses the version of Schur's lemma in his
Appendix A to achieve this.

So when he says that {Q,P} are irreducible, remember that he means
"irreducible on a vector space", i.e., the meaning he made clear in
Appendix A, of which I gave an extract above.

When now (intrinsic) spin S is introduced, one is implicitly dealing with
a larger Hilbert space than the above, and the reducibility of P,Q must
be re-examined in terms of their actions on that larger space (on which
the new operator S also acts). Unfortunately, he doesn't say this very
explicitly on p82, but only vaguely in terms of "internal degrees of
freedom". I guess he can't say much more than that at this point
in his book, because angular momentum and spin are not dealt with in all
their glory until ch 7. Also remember that his main goal in section 3.4
is to obtain the correct association between generators of the Galilean
group and the familiar dynamical variables from classical mechanics.

My convention is that Q,P are still irreducible, if the orginal Heisenberg
algebra is only centrally extended. Ballentine claims otherwise.

Like I said above, there's more than one meaning for "irreducible", and
it depends on the context in which one is working. I guess that's par for the
course where mathematics is involved. Mathematicians like to generalize
concepts, and the features of such generalizations are not always a proper
superset of the original. :-)

Ballentine is consistent -- provided one is not as lazy as I was, and reads all
of what he writes instead of an incomplete subset.

 

[WiFO215]

Really? I don't see why. I thought that his Appendices A & B explained
quite clearly what he's doing, and the context he's using.

The meaning of "irreducible" ...
This extra structure enables a stronger version of Schur's lemma.

I see.

I don't think he's quoted it "backwards". If we have a theorem which
says "A if-and-only-if B", then we can also say "B if-and-only-if A".

:) I know that. I meant to he's said A iff B, and used B iff A. I wasn't contending anything, merely pointing out.

Let's look at Ballentine's Appendix B...Galilean
group and the familiar dynamical variables from classical mechanics.

Well, I saw the appendices, (that's why I referred to them above), but was hoping there'd be a proof without having to forward refer to chapter 4 to use specific forms of P as $$\frac{\partial}{\partial x}$$. Ballentine hasn't introduced that yet and still has somehow deduced that {Q,P} is irreducible.

Isn't that possible?

@Dextercioby: I have no idea what most of the terms you used even mean.

 

[strangerep]

Well, I saw the appendices, (that's why I referred to them above), but was
hoping there'd be a proof without having to forward refer to chapter 4 to use
specific forms of P as $$\partial/\partial x$$ . Ballentine hasn't
introduced that yet and still has somehow deduced that {Q,P} is irreducible.

Hmmm, it looks like you're right about that.

Let's see if possible to deduce this from stuff he's already covered...

In eq(3.40) he's already shown that P generates translations in x space:

$$|x+a\rangle ~=~ e^{-iaP} \, |x\rangle ~~~~~~(1)$$

from which we can also deduce

$$|x\rangle ~=~ e^{-ixP} \, |0\rangle ~~~~~~(2)$$

where the 0 in the rhs denotes the position origin. Then

$$\partial_x \, |x\rangle ~=~ \partial_x \, e^{-ixP} \, |0\rangle ~=~ -iP \, e^{-ixP} \, |0\rangle ~=~ -iP \, |x\rangle ~,$$

hence

$$i \partial_x \, |x\rangle ~=~ P \, |x\rangle ~. ~~~~~~ (3)$$

Since the $$|x\rangle$$ form a basis for the state space,
an arbitrary state can be expressed as

$$|\psi\rangle ~=~ \int\!dx\, \psi(x) \, |x\rangle ~.$$

So, using (3),

$$P | \psi \rangle ~=~ \int\!dx\, \psi(x) \, i \partial_x | x \rangle ~=~ - \int\!dx\, \psi'(x) \, |x\rangle ~=~ | -i\psi'\rangle$$

where an integration by parts has been performed in the 2nd step, and
$$\psi(x)$$ is assumed to vanish at infinity. (Actually, this step needs
more care when working in a rigged Hilbert space, which is necessary
here. But I skip that detail.)

Hence we can deduce the desired result, i.e., that the operator P
can be represented as $$-i\partial_x$$ when operating on a space
of wave functions.

(I've been working with $$\hbar=1$$ in the above. Ballentine
introduces it a bit later, and explains how the dynamical quantities
and associated operators are scaled wrt $$\hbar=1$$ on p85.)

So I guess it's "possible", but maybe a bit much to expect of someone
reading the book for the first time. OTOH, he does say in the preface
it's intended as a graduate text, so I guess he's assuming the
reader has already done a elementary intro QM course.

 

[WiFO215]

So I guess it's "possible", but maybe a bit much to expect of someone
reading the book for the first time. OTOH, he does say in the preface
it's intended as a graduate text, so I guess he's assuming the
reader has already done a elementary intro QM course.

Well, I agree with your proof, but I put "possible" in quotes too because this is nearly the same thing he's done in the appendix. You've just bridged the gap of having to refer to chapter 4 between now and the appendix.