Ballentine on construction of the (Galilean) symmetry generators

In summary, the author is saying that the unitary operator ##U(v) = exp(iv·G)## describes the instantaneous ##(t = 0)## effect of a transformation to a frame of reference moving at the velocity ##v## with respect to the original reference frame. Its effects on the velocity and position operators are: ##UVU^{−1} =v−vI, UQU^{−1}=Q−vtI##. Find an operator ##G_t## such that the unitary operator ##U(v, t) = exp(iv·G_t)##
  • #1
EE18
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In Problem 3.7, Ballentine says:
The unitary operator ##U(v) = exp(iv·G)## describes the instantaneous ##(t = 0)## effect of a transformation to a frame of reference moving at the velocity ##v## with respect to the original reference frame. Its effects on the velocity and position operators are:
##UVU^{−1} =v−vI, UQU^{−1}=Q##. Find an operator ##G_t## such that the unitary operator ##U(v, t) = exp(iv·G_t)## will yield the full Galilei transformation: ##UVU^{−1} =V−vI, UQU^{−1} =Q−vtI##. Verify that ##G_t## satisfies the same commutation relation with P, J, and H as does G.

This seems to imply that the development in the text itself (Chap 3.4) was fine. But why should this be? Surely we've just shown that G in general actually takes a different form and G=MQ is wrong?

Edit: I am having some trouble using MathJax -- isn't ## the delimiter here?
 
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  • #2
EE18 said:
In Problem 3.7, Ballentine says:
The unitary operator ##U(v) = exp(iv·G)## describes the instantaneous ##(t = 0)## effect of a transformation to a frame of reference moving at the velocity ##v## with respect to the original reference frame. Its effects on the velocity and position operators are:
##UVU^{−1} =v−vI,##
Note the typo: it should be ##UVU^{−1} = V−vI##. (Disappointingly, this typo is still present in the 2nd edition.)

EE18 said:
##UQU^{−1}=Q##. Find an operator ##G_t## such that the unitary operator ##U(v, t) = exp(iv·G_t)## will yield the full Galilei transformation: ##UVU^{−1} =V−vI, UQU^{−1} =Q−vtI##. Verify that ##G_t## satisfies the same commutation relation with P, J, and H as does G.

This seems to imply that the development in the text itself (Chap 3.4) was fine. [...]
One should place Problem 3.7 in the context where it was first mentioned -- see the text near the bottom of p79 and top of p80. It's only later that he moves on to different forms of the symmetry operators corresponding to more specific cases.

EE18 said:
Edit: I am having some trouble using MathJax -- isn't ## the delimiter here?
Yes. Sometimes you need to reload the page to make MathJax to do its thing.
 
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  • #3
I think the logic is the other way, i.e., in the approach to construct non-relativistic QM from the symmetries of the Newtonian spacetime model, you have the generators ##\hat{H}##, ##\hat{\vec{P}}##, ##\hat{\vec{J}}##, and ##\hat{\vec{K}}## for the Galilei algebra. As it turns out this algebra has a non-trivial central charge, ##\hat{M}##, which is crucial to get a physically useful ray representation of the Galilei group. Physically, ##\hat{M}## represents the total mass of the system.

After eliminating all "trivial central charges" the relevant part of the "quantum Galilei algebra" (also known as "Bargmann algebra") reads
$$[\hat{H},\hat{P}_j]=0, \quad [\hat{H},\hat{K}_j]=-\mathrm{i} \hat{P}_j, \quad [\hat{K}_j,\hat{P}_k]=\mathrm{i} \hat{M} \delta_{kl} \hat{1}, \quad [\hat{K}_j,\hat{K}_k]=0, \quad [\hat{P}_j,\hat{P}_k]=0.$$
From Noether's theorem you get that ##\hat{P}_k## and ##\hat{H}## are all not explicitly time-dependent, but that
$$\mathring{\hat{K}}_j=0=\frac{1}{\mathrm{i}} [\hat{K}_j,\hat{H}]+\partial_t \hat{K}_j = \hat{P}_j+\partial_t\hat{K}_j.$$
From this you get
$$\hat{K}_j=\hat{M} \hat{X}_j-\hat{P}_j t,$$
where ##\hat{X}_j## are explicitly time-independent operators, which necessarily fulfill the commutation relations for the usual position operators with all the generators of the Galilei group, which shows that there are always position operators (for the "center of mass position" of the considered quantum system).
 
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  • #4
strangerep said:
Note the typo: it should be ##UVU^{−1} = V−vI##. (Disappointingly, this typo is still present in the 2nd edition.)One should place Problem 3.7 in the context where it was first mentioned -- see the text near the bottom of p79 and top of p80. It's only later that he moves on to different forms of the symmetry operators corresponding to more specific cases.Yes. Sometimes you need to reload the page to make MathJax to do its thing.
Thank you for your response. I had two quick followup questions if that's OK.
1) So you are saying that the ##G_t## in Problem 3.7 is the most generally correct one. So that I understand correctly -- the ##G## which is constructed using the Schur's Lemma method on 80-81 is NOT correct in general because it used that ##G## commutes with ##Q## which is not in general true for ##G_t##? It's baffling to me that Ballentine would have gone ahead in the discussion with ##G## if so seeing as the result is so materially different.

2) I want to establish the RHS for the ##UVU^{−1} =V−vI## and ##UQU^{−1} =Q−vtI## expressions. For the first, do we consider the action of some ##V'## and some ##V## on ##| v \rangle## eigenstates (by analogy to the development in terms of ##| x \rangle## near (3.42))? And then we appeal to ##V## being self-adjoint so that the ##| v \rangle## are complete (in the rigged Hilbert space sense) and thus we get the operator equality ##V−vI##?
 
  • #5
vanhees71 said:
I think the logic is the other way, i.e., in the approach to construct non-relativistic QM from the symmetries of the Newtonian spacetime model, you have the generators ##\hat{H}##, ##\hat{\vec{P}}##, ##\hat{\vec{J}}##, and ##\hat{\vec{K}}## for the Galilei algebra. As it turns out this algebra has a non-trivial central charge, ##\hat{M}##, which is crucial to get a physically useful ray representation of the Galilei group. Physically, ##\hat{M}## represents the total mass of the system.

After eliminating all "trivial central charges" the relevant part of the "quantum Galilei algebra" (also known as "Bargmann algebra") reads
$$[\hat{H},\hat{P}_j]=0, \quad [\hat{H},\hat{K}_j]=-\mathrm{i} \hat{P}_j, \quad [\hat{K}_j,\hat{P}_k]=\mathrm{i} \hat{M} \delta_{kl} \hat{1}, \quad [\hat{K}_j,\hat{K}_k]=0, \quad [\hat{P}_j,\hat{P}_k]=0.$$
From Noether's theorem you get that ##\hat{P}_k## and ##\hat{H}## are all not explicitly time-dependent, but that
$$\mathring{\hat{K}}_j=0=\frac{1}{\mathrm{i}} [\hat{K}_j,\hat{H}]+\partial_t \hat{K}_j = \hat{P}_j+\partial_t\hat{K}_j.$$
From this you get
$$\hat{K}_j=\hat{M} \hat{X}_j-\hat{P}_j t,$$
where ##\hat{X}_j## are explicitly time-independent operators, which necessarily fulfill the commutation relations for the usual position operators with all the generators of the Galilei group, which shows that there are always position operators (for the "center of mass position" of the considered quantum system).
I think this answer is a little (a lot) above my head mathematically though I appreciate it's precision -- unfortunately Ballentine does not use the language of algebraic structures in this way (though I am trying to read Symmetry and Quantum Mechanics by Corry right now so hopefully this comes up). Thank you for your answer at any rate!
 
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  • #6
EE18 said:
1) So you are saying that the ##G_t## in Problem 3.7 is the most generally correct one. So that I understand correctly -- the ##G## which is constructed using the Schur's Lemma method on 80-81 is NOT correct in general because it used that ##G## commutes with ##Q## which is not in general true for ##G_t##? It's baffling to me that Ballentine would have gone ahead in the discussion with ##G## if so seeing as the result is so materially different.
The important thing here is to distinguish (1) abstract representation of the Galilei group (i.e., representing the Galilei operators as unitary operators on some as-yet-unspecified Hilbert space), and (2) a concrete Hilbert space that has an extra notion of "position" operators ##Q_\alpha##. That's what Ballentine does in eq(3.36). The rest of section 3.4 is just an exercise in requiring that the abstract Galilei operators continue to make physical sense on this concrete Hilbert space, in the way that they work with the new operator ##Q_\alpha##.

There's a fudge a bit later in eq(3.37) where he introduces a velocity operator via a time derivative of ##\langle Q \rangle## though he didn't introduce a "time" operator. (This is because a "time" operator in QM is problematic and the theory must be constructed with time remaining just a parameter, but not an operator, -- unlike spatial position.)

So it's not one version of ##G## being "correct" or not -- it's about what happens when you take an abstract representation of the Galilei operators and plonk them onto the concrete Hilbert space built from position kets ##|x_\alpha\rangle##, and then analyze the enlarged set of commutation relations that now include ##Q_\alpha## and adjust things so that the whole concrete representation make physically intuitive sense.

Btw, this distinction between abstract and concrete representations (Hilbert spaces) arises again later, in ch7, where one deals with quantum angular momentum. If you haven't read ch7 yet, keep this distinction in mind when you get there. :oldsmile:

EE18 said:
2) I want to establish the RHS for the ##UVU^{−1} =V−vI## and ##UQU^{−1} =Q−vtI## expressions. For the first, do we consider the action of some ##V'## and some ##V## on ##| v \rangle## eigenstates (by analogy to the development in terms of ##| x \rangle## near (3.42))? And then we appeal to ##V## being self-adjoint so that the ##| v \rangle## are complete (in the rigged Hilbert space sense) and thus we get the operator equality ##V−vI##?
You seem to be overthinking this.
Hint: do problem 3.3 (Baker-Campbell-Hausdorff identity) if you haven't yet, and then try to apply the result here.
 
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  • #7
strangerep said:
The important thing here is to distinguish (1) abstract representation of the Galilei group (i.e., representing the Galilei operators as unitary operators on some as-yet-unspecified Hilbert space), and (2) a concrete Hilbert space that has an extra notion of "position" operators ##Q_\alpha##. That's what Ballentine does in eq(3.36). The rest of section 3.4 is just an exercise in requiring that the abstract Galilei operators continue to make physical sense on this concrete Hilbert space, in the way that they work with the new operator ##Q_\alpha##.

There's a fudge a bit later in eq(3.37) where he introduces a velocity operator via a time derivative of ##\langle Q \rangle## though he didn't introduce a "time" operator. (This is because a "time" operator in QM is problematic and the theory must be constructed with time remaining just a parameter, but not an operator, -- unlike spatial position.)

So it's not one version of ##G## being "correct" or not -- it's about what happens when you take an abstract representation of the Galilei operators and plonk them onto the concrete Hilbert space built from position kets ##|x_\alpha\rangle##, and then analyze the enlarged set of commutation relations that now include ##Q_\alpha## and adjust things so that the whole concrete representation make physically intuitive sense.

Btw, this distinction between abstract and concrete representations (Hilbert spaces) arises again later, in ch7, where one deals with quantum angular momentum. If you haven't read ch7 yet, keep this distinction in mind when you get there. :oldsmile:You seem to be overthinking this.
Hint: do problem 3.3 (Baker-Campbell-Hausdorff identity) if you haven't yet, and then try to apply the result here.
With respect to the first line of questioning -- you're saying (I think) that Ballentine's concrete representation of ##G## given in the body of the text is the representation valid at ##t=0## only, no?

With respect to the second line of questioning, I'm not sure I agree that BCH is at play. For example, just below (3.43) it's noted that we equate (3.43) and (3.41) to first order (with the latter involving a tacit appeal to BCH) in order to develop the commutator (3.44). I would have expected that the equality (3.49) is arrived at by analogy to (3.41) for the LHS and (3.43) for the RHS. But then I don't see us develop a commutator from it analogous to (3.44)?

Lastly, would it be OK to PM you some questions about Ballentine? I see from the blurb at the bottom of your posts that you are a fan and, as I'm self studying it, I unfortunately have no one to ask questions to about it. No worries at all if not!
 
  • #8
EE18 said:
Lastly, would it be OK to PM you some questions about Ballentine? I see from the blurb at the bottom of your posts that you are a fan and, as I'm self studying it, I unfortunately have no one to ask questions to about it.
It's better to ask such questions in this (public) forum, or the "Advanced Physics" homework forum. That way, any answers given may also benefit others, and if I'm not available there's various other people here could certainly answer your questions.

(I'll reply to your other questions later when I have more time.)
 
  • #9
EE18 said:
With respect to the first line of questioning -- you're saying (I think) that Ballentine's concrete representation of ##G## given in the body of the text is the representation valid at ##t=0## only, no?
Ballentine is saying there's no loss of generality in using that ##G## at ##t=0##, since ##t## is just a parameter. We just have to use the Hamiltonian ##H## to perform time translations consistently on all the operators.
 
  • #10
EE18 said:
With respect to the second line of questioning, I'm not sure I agree that BCH is at play. For example, just below (3.43) it's noted that we equate (3.43) and (3.41) to first order (with the latter involving a tacit appeal to BCH) in order to develop the commutator (3.44). I would have expected that the equality (3.49) is arrived at by analogy to (3.41) for the LHS and (3.43) for the RHS. But then I don't see us develop a commutator from it analogous to (3.44)?
Just try working on the LHS of (3.49), i.e.,$$ e^{iv\cdot G} V e^{-iv\cdot G}$$ using (3.39), i.e.,$$V ~=~ i[H,Q] ~,$$and some other commutation relations previously derived. If you still don't get it, you'll have to show your attempt in detail here (using Latex), else I can't see what you might be missing or doing wrong.
 
  • #11
EE18 said:
I think this answer is a little (a lot) above my head mathematically though I appreciate it's precision -- unfortunately Ballentine does not use the language of algebraic structures in this way (though I am trying to read Symmetry and Quantum Mechanics by Corry right now so hopefully this comes up). Thank you for your answer at any rate!
But isn't that more or less what's done in Ballentine? I've to check my copy when I'm back home.
 
  • #12
vanhees71 said:
But isn't that more or less what's done in Ballentine?
Ballentine doesn't use Noether's theorem in that section. He appeals to special cases to get relationships between ##V,P,M## involving external fields, if present.
 
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  • #13
strangerep said:
Just try working on the LHS of (3.49), i.e.,$$ e^{iv\cdot G} V e^{-iv\cdot G}$$ using (3.39), i.e.,$$V ~=~ i[H,Q] ~,$$and some other commutation relations previously derived. If you still don't get it, you'll have to show your attempt in detail here (using Latex), else I can't see what you might be missing or doing wrong.
Here I use BCH on the LHS of (3.49) for a particular component ##V_\alpha##:

$$e^{i\textbf{v}\cdot \textbf{G}} V_\alpha e^{-i\textbf{v}\cdot \textbf{G}} = V_\alpha + [\textbf{v}\cdot \textbf{G},V_\alpha] = V_\alpha + [v_j G_j,i[H,Q_\alpha]]$$
where I've assumed that higher order terms truncate and where I'm not sure i understand how to go further than this. But to be honest I really do not think that Ballentine uses BCH to arrive at (3.49) seeing as he has not even asked us to prove it yet (he does so at the end of chapter problems) and seeing as it has the same structure as (3.41) and (3.43), the equality between which follows from an entirely different argument.

Maybe you can set me straight here?
 
  • #14
EE18 said:
$$e^{i\textbf{v}\cdot \textbf{G}} V_\alpha e^{-i\textbf{v}\cdot \textbf{G}} = V_\alpha + [\textbf{v}\cdot \textbf{G},V_\alpha] = V_\alpha + [v_j G_j,i[H,Q_\alpha]]$$
Use the Jacobi identity on the last commutator, then use other previously derived commutators to simplify further.

EE18 said:
where I've assumed that higher order terms truncate
You don't have to assume this. Just work on the lowest order commutator. If you can show that it is indeed a multiple of the identity then the higher order commutators necessarily vanish.

EE18 said:
But to be honest I really do not think that Ballentine uses BCH to arrive at (3.49) seeing as he has not even asked us to prove it yet (he does so at the end of chapter problems) and seeing as it has the same structure as (3.41) and (3.43), the equality between which follows from an entirely different argument.
But at that point one does not yet have eq(3.44), i.e., $$[Q_\alpha, P_\beta] = i \delta_{\alpha\beta} I ~.$$Now, however, you'll be able to use that commutator, (and at least one other already-derived commutator).
 
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  • #15
strangerep said:
Use the Jacobi identity on the last commutator, then use other previously derived commutators to simplify further.
Per your advice (from private messages) to me, I've begun a reread of Ballentine and am back here. I'm at the point where I need to use the Jacobi identity but I end up obtaining ##e^{i\textbf{v}\cdot \textbf{G}} V_\alpha e^{-i\textbf{v}\cdot \textbf{G}} = V_\alpha + [\textbf{v}\cdot \textbf{G},V_\alpha] = V_\alpha + [v_j G_j,i[H,Q_\alpha]] \implies v_\alpha I = iv_j([G_j,Q_\alpha],H] - \delta_{\alpha,j}I)## but it's not clear how to go further from here. It seems to me like to obtain (3.50) Ballentine uses perhaps a completely different argument that has nothing to do with (3.49). Perhaps Ballentine is saying that the action at ##t=0## on the position basis is of ##e^{i\textbf{v}\cdot \textbf{G}}|\textbf{x}\rangle = \textbf{x}|\textbf{x}\rangle## so that ##Q = Q'## under this transformation and thus we get that the two commute, i.e. 3.50?

Edit: I think I am correct and that there is no connection between 3.49 and 3.50 since 3.49 is used later to prove 3.58.
 
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  • #16
EE18 said:
It seems to me like to obtain (3.50) Ballentine uses perhaps a completely different argument that has nothing to do with (3.49).
It comes from the definition of the Galilei boost acting on the spacetime coordinates. See the table at the bottom of p69. That boost is $$x_\alpha ~\to~ x'_\alpha = x_\alpha + v_\alpha t ~,$$so if we restrict attention to ##t=0##, spatial position is unchanged by the boost.

Another way to see it is to construct the spacetime representation of the boost generator. The general formula for this is $$G_\beta ~=~ \left. \frac{\partial x'_\alpha}{\partial v_\beta} \right|_{v=0} \;
\frac{\partial}{\partial x_\alpha} ~.$$What does this evaluate to? What can we infer from this? And can we use this result to guess a likely solution to Problem (3.7)?
 
  • #17
strangerep said:
It comes from the definition of the Galilei boost acting on the spacetime coordinates. See the table at the bottom of p69. That boost is $$x_\alpha ~\to~ x'_\alpha = x_\alpha + v_\alpha t ~,$$so if we restrict attention to ##t=0##, spatial position is unchanged by the boost.

Another way to see it is to construct the spacetime representation of the boost generator. The general formula for this is $$G_\beta ~=~ \left. \frac{\partial x'_\alpha}{\partial v_\beta} \right|_{v=0} \;
\frac{\partial}{\partial x_\alpha} ~.$$What does this evaluate to? What can we infer from this? And can we use this result to guess a likely solution to Problem (3.7)?
Gotcha, agreed with this approach. Thank you! If you get the chance, I just asked another question re: Ballentine and the construction in Chap 3.4. Seriously, thank you as always for your very considered and helpful suggestions/discussion around this awesome book (which sometimes needs some details filled in, at least for me)!
 
  • #18
EE18 said:
If you get the chance, I just asked another question re: Ballentine and the construction in Chap 3.4.
Huh? Where? I'm starting to lose track of your questions. :oldfrown:
 

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