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?

 

[WannabeNewton]

This isn't a question about chapter 3 of Ballentine itself but is related to the notation of the chapter. In chapter 2 of Srednicki's QFT text, Srednicki goes about deriving the commutator relations for the generators of the proper orthochronous Lorentz group and consequently the structure coefficients of the Lorentz algebra similarly to Ballentine's derivation of these quantities for the Galilei group. However the notations differ slightly and I'm having trouble parsing Srednicki's notation having become comfortable with Ballentine's notation.

He starts with an infinitesimal Lorentz transformation $\Lambda^{\mu}{}{}_{\nu} = \delta^{\mu}{}{}_{\nu} + \delta \omega^{\mu}{}{}_{\nu}$ where $\delta \omega_{\mu\nu} = -\delta \omega_{\nu\mu}$. In expression 2.12 he writes, for a unitary transformation $U(\Lambda)$ of an infinitesimal Lorentz transformation, $U(1 + \delta \omega) = I + \frac{i}{2\hbar}\delta \omega_{\mu\nu}M^{\mu\nu}$.

I assume by $1 + \delta \omega$ he means the identity operator $1$ and the matrix $\delta \omega^{\mu}{}{}_{\nu}$ now construed as an operator $\delta \omega$, with both operators acting on Minkowski space-time. Now he's using abstract indices to represent the 2-form $\delta \omega_{\mu\nu}$ and then contracting it with the indices on $M^{\mu\nu}$ and based on his definition of the angular momentum operator $\mathbf{J}$ component-wise by $J_i = \frac{1}{2}\epsilon_{ijk}M^{jk}$ and the Lorentz boost operator $\mathbf{K}$ also component-wise by $K_{i} = M^{0i}$ it seems that the indices on $M^{\mu\nu}$ aren't abstract indices but rather just labels for Hermitian generators of the proper orthochronous Lorentz group along the different spatial axes in which case I can neither make sense of how he gets $U(1 + \delta \omega) = I + \frac{i}{2\hbar}\delta \omega_{\mu\nu}M^{\mu\nu}$ nor make sense of how to even parse it.

Thanks.

 

[George Jones]

Each each $\Lambda$ is Lorentz transformation matrix; consequently, each $\Lambda^{\mu}{}{}_{\nu}$ is a component of a matrix, i.e., a number; so, each $\delta \omega_{\mu \nu}$ is a number (infintesimal?); aach $M^{\mu \nu}$ is an operator (or matrix, if the space is finite-dimensional and a basis has been chosen); $\delta \omega_{\mu \nu} M^{\nu \mu}$ is a linear combination of operators.

 

[strangerep]

Adding a little to what George said...

I assume by $1 + \delta \omega$ he means the identity operator $1$ and the matrix $\delta \omega^{\mu}{}{}_{\nu}$ now construed as an operator $\delta \omega$, with both operators acting on Minkowski space-time.

An element of the abstract Lorentz group can be represented as 4x4 matrices acting as transformations on Minkowski spacetime, or it can be represented as a unitary operator on Hilbert space. Srednicki is obviously mapping between these 2 representations. In more general situations, however, you may need to think of $U$ as a mapping from abstract group elements to their concrete representations as unitary operators. But of course, when reading Srednicki's textbook, you got to keep these two distinctions straight, since he's doing a rep-to-rep mapping.

Hence: when he writes $U(1+\delta\omega)$, the argument is obviously a 4x4 matrix, denoting a transformation applicable on Minkowski spacetime. But the rhs of (2.12) is an operator on Hilbert space, so he uses $I$ instead of $1$.

(And now I wonder whether I've only confused the issue.)

 

[WannabeNewton]

Each each $\Lambda$ is Lorentz transformation matrix; consequently, each $\Lambda^{\mu}{}{}_{\nu}$ is a component of a matrix, i.e., a number; so, each $\delta \omega_{\mu \nu}$ is a number (infintesimal?); aach $M^{\mu \nu}$ is an operator (or matrix, if the space is finite-dimensional and a basis has been chosen); $\delta \omega_{\mu \nu} M^{\nu \mu}$ is a linear combination of operators.

Ok good that's what I figured the notation was representing. So just to tie it back to Ballentine's notation, if we take the definitions $J_{i} = \epsilon_{ijk}M^{jk}$ and $K_{i} = M^{i0}$ and note that $M^{ij} = \epsilon^{ijk}J_{k}$ as a result of the first definition, we have $\frac{1}{2}\delta \omega_{\mu\nu}M^{\mu\nu} = \delta \omega_{i0}M^{i0}+\frac{1}{2}\epsilon^{ijk}\omega_{ij}J_{k} = \delta \omega_{0x}K_{x}+\delta \omega_{0y}K_{y}+\delta \omega_{0z}K_{z} +\omega_{xy}J_{z} + \omega_{yz}J_{x} + \omega_{zx}J_{y}$
so if I wrote $U(1 + \delta \omega) = I + i\sum_{\mu = 1}^{6}s_{\mu}K_{\mu}$ as in Ballentine (units of $\hbar = 1$) and made the identifications $s_1 = \delta \omega_{0x}$, $s_{4} = \delta \omega_{xy}$ etc. and $K_{1} = K_{x}$, $K_{4} = J_{z}$ etc. would it be fair to say it's equivalent to what Srednicki writes?

Also, consider a killing field $\xi^{\mu}$ in Minkowski space-time and a constant 4-velocity field $u^{\mu}$. Letting $x^{\mu}$ be the position vector field relative to $u^{\mu}$, we can decompose $\xi^{\mu}$ as $\xi^{\mu} = 2E^{[\nu}u^{\mu]}x_{\nu} + \epsilon^{\mu\nu\alpha\beta}x_{\nu}u_{\alpha}B_{\beta}$ where $E^{\mu},B^{\mu}$ are space-like vectors. Pure Lorentz boosts correspond to the "electric" part $2E^{[\nu}u^{\mu]}x_{\nu}$ and pure rotations correspond to the "magnetic part" $\epsilon^{\mu\nu\alpha\beta}x_{\nu}u_{\alpha}B_{\beta}$ where the rotation axis is given by $B^{\mu}$.

If we go to the rest frame of $u^{\mu}$ and consider an elemental pure rotation given by $B^{\mu} = (\partial_z)^{\mu}$ for example then $\xi = x\partial_{y}-y\partial_{x}$ which is in obvious 1-1 correspondence with the generator of rotations about the $z$ axis $J_z = X P_{y} -Y P_{x}$. For an elemental pure Lorentz boost in the $x$ direction, for which $E^{\mu} = (\partial_{x})^{\mu}$, we get $\xi = x\partial_t+t\partial_x$. Can an analogous 1-1 correspondence between this and $K_z$, the generator of Lorentz boosts in the $x$ direction, be made in the same obvious manner? This is just so I can connect the Hermitian generators of the proper orthochronous Lorentz group back to the familiar notion of Minkowski space-time killing fields as the generators of isometric flows in Minkowski space-time.

Adding a little to what George said...
In more general situations, however, you may need to think of $U$ as a mapping from abstract group elements to their concrete representations as unitary operators.

Could you elucidate this point? I understand that when we write $U(1 + \delta \omega)$, we basically just saying that to each representation of the infinitesimal Lorentz transformation as an operator $1 + \delta \omega; \delta^{\mu}{}{}_{\nu} + \delta \omega^{\mu}{}{}_{\nu}$ on Minkowski space-time there is associated a representation of the same infinitesimal Lorentz transformation as a unitary operator $I + \frac{i}{2\hbar}\delta \omega_{\mu\nu} M^{\mu\nu}$ on Hilbert space, a rep to rep mapping as you put it, but I'm afraid I don't quite understand the quoted statement above.

Thanks guys!

 

[rubi]

Could you elucidate this point? I understand that when we write $U(1 + \delta \omega)$, we basically just saying that to each representation of the infinitesimal Lorentz transformation as an operator $1 + \delta \omega; \delta^{\mu}{}{}_{\nu} + \delta \omega^{\mu}{}{}_{\nu}$ on Minkowski space-time there is associated a representation of the same infinitesimal Lorentz transformation as a unitary operator $I + \frac{i}{2\hbar}\delta \omega_{\mu\nu} M^{\mu\nu}$ on Hilbert space, a rep to rep mapping as you put it, but I'm afraid I don't quite understand the quoted statement above.

The correct mathematical setting to talk about this stuff is the theory of unitary representations of Lie groups. In general, you are given a Lie group $G$, that is a group with an additional manifold structure, such that the group multiplication and the inversion are smooth maps. In your case, this group is the Poincare group (or rather it's connected component of the identity). A unitary representation of a group is a pair $(\mathcal H, \pi)$ with $\mathcal H$ being a Hilbert space and $\pi:G\rightarrow U(\mathcal H)$ is a (strongly continuous) map from the group into the unitary operators on $\mathcal H$ such that $\pi(e) = \mathrm{id}_\mathcal{H}$ and $\pi(g)\pi(h)=\pi(gh)$. The goal of representation theory is to classify all these representations (up to equivalence).

For the Poincare group, it turns out that you can classify all the (irreducible, roughly speaking "minimal") (projective) representations by two numbers, which we call mass and spin. This is due to Wigner ("On unitary representations of the inhomogeneous Lorentz group").

There is much more to be said about this. For example the role of the infinitesimal transformations is played by Lie algebras and their representations and there is a way to relate them to the representations of the Lie group (that's why we actually do it) and one has to deal with some complications. But I'm not sure how much you want to learn about it, so I stop here for now.