Ballentine-sections 3.2 and 3.3

[WannabeNewton]

Hi! In section 3.2 of Ballentine, the author considers the most general effect of a Galilean transformation on space-time coordinates $\{\mathbf{x}, t \}$ (3.5) and denotes such a transformation by $\tau = \tau (R,\mathbf{a},\mathbf{v}, s)$ where $R$ is a rotation matrix, $\mathbf{a}$ is a space displacement, $\mathbf{v}$ is a 3-velocity, and $s$ is a time displacement. Based on his later notation directly below, $\tau \{\mathbf{x}, t \} = \{\mathbf{x'}, t' \}$ it seems like $\tau$ is a map $\tau : \mathbb{R}^{3}\times \mathbb{R} \rightarrow \mathbb{R}^{3}\times \mathbb{R}$.

He then says that $\tau$ must have corresponding to it a unitary transformation $U(\tau)$ with $|\psi \rangle \rightarrow U|\psi \rangle$ and $A\rightarrow UAU^{-1}$ as usual (top of p.67). Furthermore, the Galilei group has ten parameters, $s_{\mu}$ (the 3 rotation angles, 3 velocity components, 3 spatial translations, and 1 temporal translation), and associated with each $s_{\mu}$ is a Hermitian generator $K_{\mu}$ such that $U(s_{\mu}) = e^{i s_{\mu} K_{\mu}}$. At the inception of section 3.3 Ballentine notes that a unitary transformation $U(\tau)$ corresponding to a space-time transformation $\tau$ will have the most general form $U(\tau) = \prod_{\mu=1}^{10}U(s_{\mu}) = \prod_{\mu=1}^{10}e^{is_{\mu}K_{\mu}}$ (3.9).

What's confusing me is the functional form of $U(\tau)$ when compared to the functional forms of the $U(s_{\mu})$. Each $U(s_{\mu})$ corresponds to a one-parameter family of unitary operators but $\tau$ seems to be an endomorphism of Galilean space-time so what does $U(\tau)$ correspond to? At first I thought it corresponded to a ten-parameter family of unitary operators $U(s_1,...,s_{10})$ since $\tau = \tau (R,\mathbf{a},\mathbf{v}, s) = \tau(s_1,...,s_{10})$ but the fact that he subsequently writes $\tau \{\mathbf{x}, t \} = \{\mathbf{x'}, t' \}$ has me confused as to what $\tau$ is as a mathematical object, what explicit functional dependence it has on the $s_{\mu}$, and if I can actually call $U(\tau)$ a ten-parameter family of unitary operators.

Thanks in advance for clearing this up!

 

[Jilang]

I took the curly brackets to mean doing the transform on the something with the coordinates of {x,t} gives you something with the coordinates of {x',t'}. But perhaps I am not reading enough into it.

 

[George Jones]

I don't have my Ballentine home with me, but it seems like

Based on his later notation directly below, $\tau \{\mathbf{x}, t \} = \{\mathbf{x'}, t' \}$ it seems like $\tau$ is a map $\tau : \mathbb{R}^{3}\times \mathbb{R} \rightarrow \mathbb{R}^{3}\times \mathbb{R}$.

Yes.

Think of each real number $s_\mu$ as corresponding to a specific $\tau$. For example: $s_7$ is the specific $\tau$ that translates by $s_7$ along $x$, while leaving all other coordinates untouched; $s_{10}$ is the specific $\tau$ that translates by $s_{10}$ temporally, while leaving all other coordinates untouched; etc.

Also, there is no sum on $s_\mu K_\mu$.

 

[bhobba]

Yes.

Its been a while since I went through that chapter but as far as I recall that's it.

The thing I recall is it took me a little while to fully understand - its a bit dense.

The other thing I recall is there is a bit of thought required about exactly why a free particle forms an irreducible set.

The other thing is the exact key assumption being made in quantitization - it took me a while to nut that one out.

Thanks
Bill

 

[Jilang]

I don't have my Ballentine home with me, but it seems like



Yes.

Think of each real number $s_\mu$ as corresponding to a specific $\tau$. For example: $s_7$ is the specific $\tau$ that translates by $s_7$ along $x$, while leaving all other coordinates untouched; $s_{10}$ is the specific $\tau$ that translates by $s_{10}$ temporally, while leaving all other coordinates untouched; etc.

Also, there is no sum on $s_\mu K_\mu$.

There is in 3.10
U = I + iƩsK summing over μ from 1 to 10

 

[George Jones]

Expanding 3.9 gives

$$\begin{align}
U &=e^{is_{1}K_{1}}e^{is_{2}K_{2}}\cdots e^{is_{10}K_{10}}\\
&=\left( 1+is_{1}K_{1}+\ldots \right) \left( 1+is_{2}K_{2}+\ldots \right) \cdots \left( 1+is_{10}K_{10}+\ldots \right) \\
&=\left(1+is_{1}K_{1}+is_{2}K_{2}+is_{3}K_{3}+is_{4}K_{4}+is_{5}K_{5}+is_{6}K_{6}+is_{7}K_{7}+is_{8}K_{8}+is_{9}K_{9}+is_{10}K_{10}+\ldots \right),
\end{align}$$

where $\ldots$ in the last line refers to terms of second order and higher.

 

[George Jones]

Oops, I originally left out a bunch of factors of $i$ in my previous post, which I have now edited and corrected. I inadvertently used standard math notation instead of standard physics notation. I will atone for my sins by using below the physicists' fast and loose version of functional analysis.

I now am in my office, and I have Ballentine at hand. There is a bit of a clash of notation between the material near equations 3.2 and 3.3, and the material near equations 3.9 and 3.10.

Think of each real number $s_\mu$ as corresponding to a specific $\tau$. For example: $s_7$ is the specific $\tau$ that translates by $s_7$ along $x$, while leaving all other coordinates untouched; $s_{10}$ is the specific $\tau$ that translates by $s_{10}$ temporally, while leaving all other coordinates untouched; etc.

This is what the chart after equation 3.13 indicates.

Let's do the $s_7$ example. Take $s_7$ to be a real number and $K_7 = -id/dx$ (suppressing all other coodinates). Then

$$\begin{align}
U\left(s_7 \right) \psi \left(x\right) &= e^{is_7 K_7} \psi \left(x\right)\\
&= \left( 1 + s_7 \frac{d}{dx} +\frac{s_7^2}{2!} \frac{d^2}{dx^2} + \dots \right) \psi \left(x\right) \\
&= \psi \left(x\right) + s_7 \frac{d\psi}{dx}\left(x\right) +\frac{s_7^2}{2!} \frac{d^2 \psi}{dx^2}\left(x\right) + \dots \\
&= \psi \left(x+s_7 \right)\\
&= U \left( \tau \right) \psi \left(x\right)
\end{align}$$

The last equality, with $\tau$ a spatial x-translation by amount $s_7$, follows from equation 3.8. I hope I got the signs right.

 

[WannabeNewton]

Oh wow, that definitely clears it up quite elegantly. Thank you so much George!

If I may ask another question: starting from (3.12) on p.69 and assuming $\omega \sim s_{\mu} \sim \epsilon$, we have $e^{i\omega}U = (1 + i\omega + O(\epsilon^2))(1 + i\sum_{\mu=1}^{10}s_{\mu}K_{\mu} + O(\epsilon^2)) = (1 + i\omega)I + i\sum_{\mu=1}^{10}s_{\mu}K_{\mu} + O(\epsilon^2)$ and separately $\epsilon^{i\epsilon K_{\mu}}\epsilon^{i\epsilon K_{\nu}}\epsilon^{-i\epsilon K_{\mu}}\epsilon^{-i\epsilon K_{\nu}} = I - \epsilon^{2}[K_{\mu},K_{\nu}] + O(\epsilon^3)$.

Now I understand the general argument for why the left hand sides of these two expressions are supposed to be equal (as elucidated by Ballentine in the paragraph directly under (3.11)) but I don't get how the right hand sides equate since one is written to 2nd order in $\epsilon$ whereas the other is written to 1st order in $\epsilon$. Furthermore, even if we ignore the error terms and just equate the two (as Ballentine says to do directly under (3.12)) we seem to get $- \epsilon^{2}[K_{\mu},K_{\nu}] = i\omega I + i\sum_{\mu=1}^{10}s_{\mu}K_{\mu}$ and I don't immediately see how this leads to $[K_{\mu},K_{\nu}] = i\sum_{\lambda}c^{\lambda}_{\mu\nu}K_{\lambda} + ib_{\mu\nu}I$ (3.13); the indices don't even match up.

Thanks again for the help.

 

[Jilang]

Oops, I originally left out a bunch of factors of $i$ in my previous post, which I have now edited and corrected. I inadvertently used standard math notation instead of standard physics notation. I will atone for my sins by using below the physicists' fast and loose version of functional analysis.

I now am in my office, and I have Ballentine at hand. There is a bit of a clash of notation between the material near equations 3.2 and 3.3, and the material near equations 3.9 and 3.10.



This is what the chart after equation 3.13 indicates.

Let's do the $s_7$ example. Take $s_7$ to be a real number and $K_7 = -id/dx$ (suppressing all other coodinates). Then

$$\begin{align}
U\left(s_7 \right) \psi \left(x\right) &= e^{is_7 K_7} \psi \left(x\right)\\
&= \left( 1 + s_7 \frac{d}{dx} +\frac{s_7^2}{2!} \frac{d^2}{dx^2} + \dots \right) \psi \left(x\right) \\
&= \psi \left(x\right) + s_7 \frac{d\psi}{dx}\left(x\right) +\frac{s_7^2}{2!} \frac{d^2 \psi}{dx^2}\left(x\right) + \dots \\
&= \psi \left(x+s_7 \right)\\
&= U \left( \tau \right) \psi \left(x\right)
\end{align}$$

The last equality, with $\tau$ a spatial x-translation by amount $s_7$, follows from equation 3.8. I hope I got the signs right.

Thanks, that really cool to see it that way round. I am less clear though why the unitary condition requires dU/ds + dU/ds dagger needs to be zero? I'm not getting the line above equation 3.3

 

[WannabeNewton]

Thanks, that really cool to see it that way round. I am less clear though why the unitary condition requires dU/ds + dU/ds dagger needs to be zero? I'm not getting the line above equation 3.3

$I = UU^{\dagger}\\ = (I + s\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} + O(s^2))(I + s\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0} + O(s^2)) \\= I + s(\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} +\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0} ) + O(s^2)$

so $\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} +\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0} = 0$ (ignoring 2nd order and higher terms in $s$).

 

[Jilang]

$I = UU^{\dagger}\\ = (I + s\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} + O(s^2))(I + s\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0} + O(s^2)) \\= I + s(\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} +\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0} ) + O(s^2)$

so $\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} +\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0} = 0$ (ignoring 2nd order and higher terms in $s$).

That's brilliant. Many thanks. I now understand mathematically why unitary transformations look like they do. That's some accomplishment on both our parts especially on a Friday!

 

[Jilang]

I just had a horrible thought though... What happens to equation 3.3 if K does not equal K dagger? What requires it to be Hermitian?

 

[George Jones]

Now I understand the general argument for why the left hand sides of these two expressions are supposed to be equal (as elucidated by Ballentine in the paragraph directly under (3.11)) but I don't get how the right hand sides equate since one is written to 2nd order in $\epsilon$ whereas the other is written to 1st order in $\epsilon$.

Do you buy Ballentine's statement "It is clear that all 11 parameters $\left\{ \omega, s_\lambda \right\}$ must be infinitesimal and of order $\epsilon^2$"? If you don't, pretend that you do for the rest of this post, and we can talk about it later.

the indices don't even match up.

As relativists everywhere know, Ballentine's (3.12) is an example really horrible notation, since $\mu$ is used there as a dummy summation index on the right. and a free index on the left. Better to use $\lambda$ as the summation index on the right in (3.12). If the left sides of (3.11) and (3.12) are to be equated, we must have (with slight abuse of notation) $s_\lambda = s_\lambda \left(\mu,\nu\right)$ and $\omega = \omega\left(\mu,\nu\right)$, because (3.11) depends on $\mu$ and $\nu$.

Now I am going to wave my hands so furiously that folks everywhere should be able to feel the breeze.

Putting everything together, write $s_\lambda \left(\mu,\nu\right) = \epsilon^2 c^\lambda_{\mu\nu}$, where $\epsilon$ is "small"', bu the $c^\lambda_{\mu\nu}$ aren't, and put this in the sum (now over $\lambda$) on the right of 3.12.

 

[George Jones]

I just had a horrible thought though... What happens to equation 3.3 if K does not equal K dagger? What requires it to be Hermitian?

Take $\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} = iK$, and, initially, don't assume anything about $K$ What do you then get for

$$\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} +\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0}?$$

 

[Jilang]

Oh wow, that definitely clears it up quite elegantly. Thank you so much George!

If I may ask another question: starting from (3.12) on p.69 and assuming $\omega \sim s_{\mu} \sim \epsilon$, we have $e^{i\omega}U = (1 + i\omega + O(\epsilon^2))(1 + i\sum_{\mu=1}^{10}s_{\mu}K_{\mu} + O(\epsilon^2)) = (1 + i\omega)I + i\sum_{\mu=1}^{10}s_{\mu}K_{\mu} + O(\epsilon^2)$ and separately $\epsilon^{i\epsilon K_{\mu}}\epsilon^{i\epsilon K_{\nu}}\epsilon^{-i\epsilon K_{\mu}}\epsilon^{-i\epsilon K_{\nu}} = I - \epsilon^{2}[K_{\mu},K_{\nu}] + O(\epsilon^3)$.

Now I understand the general argument for why the left hand sides of these two expressions are supposed to be equal (as elucidated by Ballentine in the paragraph directly under (3.11)) but I don't get how the right hand sides equate since one is written to 2nd order in $\epsilon$ whereas the other is written to 1st order in $\epsilon$. Furthermore, even if we ignore the error terms and just equate the two (as Ballentine says to do directly under (3.12)) we seem to get $- \epsilon^{2}[K_{\mu},K_{\nu}] = i\omega I + i\sum_{\mu=1}^{10}s_{\mu}K_{\mu}$ and I don't immediately see how this leads to $[K_{\mu},K_{\nu}] = i\sum_{\lambda}c^{\lambda}_{\mu\nu}K_{\lambda} + ib_{\mu\nu}I$ (3.13); the indices don't even match up.

Thanks again for the help.

I think your multiplication may be wrong in the first equation on the RHS. The later terms are missing a factor w.

 

[Jilang]

Take $\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} = iK$, and, initially, don't assume anything about $K$ What do you then get for

$$\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} +\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0}?$$

Thanks Mr Jones, I can see that it works if K is Hermitian, I am just wondering why it has to be. It seems like the introduction of imaginary numbers is necessary because K is Hermitian, but I am unclear as to why K needs to be just Hermitan, though I understand why U is. Ballentine says the text is "that we may write". But it's not saying it has to be. Sorry if I'm being a bit dense, but it's been a long week.
I think you get iK + iK dagger

 

[WannabeNewton]

Thanks Mr Jones, I can see that it works if K is Hermitian, I am just wondering why it has to be. It seems like the introduction of imaginary numbers is necessary because K is Hermitian, but I am unclear as to why K needs to be just Hermitan, though I understand why U is. Ballentine says the text is "that we may write". But it's not saying it has to be. Sorry if I'm being a bit dense, but it's been a long week.
I think you get iK + iK dagger

If $\frac{\mathrm{d} U}{\mathrm{d} s} = -\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}$ then $\frac{\mathrm{d} U}{\mathrm{d} s}$ is "purely imaginary" in the sense that $\frac{\mathrm{d} U}{\mathrm{d} s} = iK$ for some $K = K^{\dagger}$. It's analogous to how if $a$ is a complex number and $a = -a^{*}$ then $a = ib$ for some real number $b$.

 

[George Jones]

Take $\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} = iK$, and, initially, don't assume anything about $K$ What do you then get for

$$\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} +\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0}?$$

Then,

$$\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} +\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0} = iK -iK^\dagger = i\left( K - K^\dagger \right)?$$

In order for this to be zero, we must have $K = K^\dagger$.

[edit]Scooped by WannabeNewton while typing[/edit].

 

[WannabeNewton]

Do you buy Ballentine's statement "It is clear that all 11 parameters $\left\{ \omega, s_\lambda \right\}$ must be infinitesimal and of order $\epsilon^2$"? If you don't, pretend that you do for the rest of this post, and we can talk about it later.

Thanks, as a matter of fact I'm actually unsure as to why Ballentine's statement is true.

As relativists everywhere know, Ballentine's (3.12) is an example really horrible notation, since $\mu$ is used there as a dummy summation index on the right. and a free index on the left. Better to use $\lambda$ as the summation index on the right in (3.12). If the left sides of (3.11) and (3.12) are to be equated, we must have (with slight abuse of notation) $s_\lambda = s_\lambda \left(\mu,\nu\right)$ and $\omega = \omega\left(\mu,\nu\right)$, because (3.11) depends on $\mu$ and $\nu$.

Now I am going to wave my hands so furiously that folks everywhere should be able to feel the breeze.

Putting everything together, write $s_\lambda \left(\mu,\nu\right) = \epsilon^2 c^\lambda_{\mu\nu}$, where $\epsilon$ is "small"', bu the $c^\lambda_{\mu\nu}$ aren't, and put this in the sum (now over $\lambda$) on the right of 3.12.

Thanks again George, that makes perfect sense. The abuse of indices in (3.12) confused the living hell out of me.

 

[Jilang]

Then,

$$\frac{\mathrm{d} U}{\mathrm{d} s}|_{s=0} +\frac{\mathrm{d} U^{\dagger}}{\mathrm{d} s}|_{s=0} = iK -iK^\dagger = i\left( K - K^\dagger \right)?$$

In order for this to be zero, we must have $K = K^\dagger$.

[edit]Scooped by WannabeNewton while typing[/edit].

Thanks Mr Jones, it's obvious when you do it. I'll try not to bother you for a few chapters!

 

[George Jones]

Thanks, as a matter of fact I'm actually unsure as to why Ballentine's statement is true.

Let's rearrange the presentation a bit. Using (3.7) a few time gives $e^{i\omega}U = e^{i\epsilon K_\mu} e^{i\epsilon K_\nu} e^{-i\epsilon K_\mu} e^{-i\epsilon K_\mu}$. But (3.11) shows $e^{i\omega}U = e^{i\epsilon K_\mu} e^{i\epsilon K_\nu} e^{-i\epsilon K_\mu} e^{-i\epsilon K_\mu} = I - \epsilon^2 \left[ K_\mu , K_\nu] \right] + O\left( \epsilon^3 \right)$. Hence, $e^{i\omega}U = I - \epsilon^2 \left[ K_\mu , K_\nu] \right] + O\left( \epsilon^3 \right)$. Since the right side of the last equation has no first-order $\epsilon$ terms, the left side can't have any either.

Equivalently, write $s_\lambda \left(\mu,\nu\right) =\epsilon a^\lambda_{\mu\nu} + \epsilon^2 c^\lambda_{\mu\nu}$. Since the right side has epsilon times zero all the $a^\lambda_{\mu\nu}$ must be zero.

 

[WannabeNewton]

Hence, $e^{i\omega}U = I - \epsilon^2 \left[ K_\mu , K_\nu] \right] + O\left( \epsilon^3 \right)$. Since the right side of the last equation has no first-order $\epsilon$ terms, the left side can't have any either.

Equivalently, write $s_\lambda \left(\mu,\nu\right) =\epsilon a^\lambda_{\mu\nu} + \epsilon^2 c^\lambda_{\mu\nu}$. Since the right side has epsilon times zero all the $a^\lambda_{\mu\nu}$ must be zero.

Gotcha, thanks! Just to make sure I'm on the same page overall, consider the sequence of space-time transformations given by a boost $-\epsilon$ along the $x$-axis, a time translation by $-\epsilon$, and the respective inverses.

We have on the one hand $e^{i\epsilon H}e^{i\epsilon G_{x}}e^{-i\epsilon H}e^{-i\epsilon G_{x}} = I + \epsilon^{2}[G_{x},H]+O(\epsilon^{3})$ where $H$ is the generator of time translations and $G_{x}$ the generator of boosts in the $x$ direction. On the other hand the sequence of space-time transformations give us $(t,x,y,z)\rightarrow (t,x - \epsilon^2,y,z)$ which is a spatial translation along $x$ by $-\epsilon^2$ so the associated unitary transformation is $U = I + i \epsilon^2 P_{x} + O(\epsilon^3)$ (using Ballentine's sign convention for the generator of spatial translations $P_{\alpha}$) where $s_{x}(t,x) = \epsilon^2$ and $s_{\mu}(t,x) = 0$ for all $\mu \neq x$.

The two different expressions we have must be equivalent up to a phase factor $\omega(t,x) = \epsilon^2 b_{tx}$ so $[G_x,H]=i P_{x} +i b_{tx}I$. To relate back to (3.13) would it then be fair to say that $\sum c^{\lambda}_{tx}K_{\lambda} = P_{x}\Rightarrow c^{x}_{tx} = 1,c^{\mu}_{tx} = 0(\forall \mu\neq x)$? Also just to be careful, is Ballentine's "(?)" in front of the identity in (3.21) simply the factor $ib_{tx}$?

Thanks!

 

[Jilang]

Gotcha, thanks! Just to make sure I'm on the same page overall, consider the sequence of space-time transformations given by a boost $-\epsilon$ along the $x$-axis, a time translation by $-\epsilon$, and the respective inverses.

We have on the one hand $e^{i\epsilon H}e^{i\epsilon G_{x}}e^{-i\epsilon H}e^{-i\epsilon G_{x}} = I + \epsilon^{2}[G_{x},H]+O(\epsilon^{3})$ where $H$ is the generator of time translations and $G_{x}$ the generator of boosts in the $x$ direction. On the other hand the sequence of space-time transformations give us $(t,x,y,z)\rightarrow (t,x - \epsilon^2,y,z)$ which is a spatial translation along $x$ by $-\epsilon^2$ so the associated unitary transformation is $U = I + i \epsilon^2 P_{x} + O(\epsilon^3)$ (using Ballentine's sign convention for the generator of spatial translations $P_{\alpha}$) where $s_{x}(t,x) = \epsilon^2$ and $s_{\mu}(t,x) = 0$ for all $\mu \neq x$.

The two different expressions we have must be equivalent up to a phase factor $\omega(t,x) = \epsilon^2 b_{tx}$ so $[G_x,H]=i P_{x} +i b_{tx}I$. To relate back to (3.13) would it then be fair to say that $\sum c^{\lambda}_{tx}K_{\lambda} = P_{x}\Rightarrow c^{x}_{tx} = 1,c^{\mu}_{tx} = 0(\forall \mu\neq x)$? Also just to be careful, is Ballentine's "(?)" in front of the identity in (3.21) simply the factor $ib_{tx}$?

Thanks!

I'm agreeing with everything you just said. Just hit another query at 3.22 where ε seems to mean something else entirely before it switches back to being infinitesimal in the next equation. Am I being picky or just missing something obvious?

 

[George Jones]

I didn't bring Ballentine home with me, as my knapsack only has a finite volume. I might stop by and pick it up, but I'm not sure.

 

[Jilang]

I didn't bring Ballentine home with me, as my knapsack only has a finite volume. I might stop by and pick it up, but I'm sure.

Please don't go to any trouble as I am not unduly troubled by it. It just seems strange that when there are all those Greek letters to choose from you use the same one to mean different things on the same page. There might be a good reason for it though that is alluding me!

 

[WannabeNewton]

Am I being picky or just missing something obvious?

$\epsilon_{\alpha_1...\alpha_n}$ is the standard notation for the n-dimensional Levi-Civita symbol. $\epsilon$ by itself is also standard notation for expansion parameters when working with perturbed quantities; it isn't really a problem since the former has indices to indicate that it's the Levi-Civita symbol.

I didn't bring Ballentine home with me, as my knapsack only has a finite volume. I might stop by and pick it up, but I'm sure.

Thanks!

 

[Jilang]

$\epsilon_{\alpha_1...\alpha_n}$ is the standard notation for the n-dimensional Levi-Civita symbol. $\epsilon$ by itself is also standard notation for expansion parameters when working with perturbed quantities; it isn't really a problem since the former has indices to indicate that it's the Levi-Civita symbol.

Thanks! I wondered if there might be a good reason. Just 2 pages of maths left to go now before the Physics kicks in!

 

[strangerep]

Since George hasn't answered yet...

[...] To relate back to (3.13) would it then be fair to say that $\sum c^{\lambda}_{tx}K_{\lambda} = P_{x}\Rightarrow c^{x}_{tx} = 1,c^{\mu}_{tx} = 0(\forall \mu\neq x)$?

That looks right to me.

BTW, the generators of a Lie algebra are also basis elements in an abstract vector space. So in cases like this you could think of the 2 sides as vectors, and make some rapid deductions by relying on linear independence of the basis elements. (I hope that makes sense.)

Also just to be careful, is Ballentine's "(?)" in front of the identity in (3.21) simply the factor $ib_{tx}$?

More likely it's $ib_{t\alpha}$.

But yes, it denotes the coefficient of the central element $I$ in the Lie algebra.

 

[WannabeNewton]

BTW, the generators of a Lie algebra are also basis elements in an abstract vector space. So in cases like this you could think of the 2 sides as vectors, and make some rapid deductions by relying on linear independence of the basis elements. (I hope that makes sense.)

That makes perfect sense, thank you!

More likely it's $ib_{t\alpha}$.

But yes, it denotes the coefficient of the central element $I$ in the Lie algebra.

Cool. Sorry for the onslaught of questions but it's almost winter break and I figured I should finally bring myself around to working through the book in a more serious manner

 

[strangerep]

Sorry for the onslaught of questions

No need to be sorry.

but it's almost winter break and I figured I should finally bring myself around to working through the book in a more serious manner

Great idea. Maybe that will eventually force me to figure out whether certain little things later in the book are typos or just my misunderstandings. (I couldn't find an errata list back when I was going through Ballentine closely.)

Seems like Jilang is also working through it?