# Ballentine-sections 3.2 and 3.3

1. Dec 6, 2013

### 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!

2. Dec 6, 2013

### 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.

3. Dec 6, 2013

### George Jones

Staff Emeritus
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 $s_\mu K_\mu$.

4. Dec 6, 2013

### bhobba

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

5. Dec 6, 2013

### Jilang

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

6. Dec 6, 2013

### George Jones

Staff Emeritus
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.

Last edited: Dec 6, 2013
7. Dec 6, 2013

### George Jones

Staff Emeritus
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.

8. Dec 6, 2013

### 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.

9. Dec 6, 2013

### Jilang

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

10. Dec 6, 2013

### WannabeNewton

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

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 $s$).

11. Dec 6, 2013

### Jilang

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!

12. Dec 6, 2013

### 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?

13. Dec 6, 2013

### George Jones

Staff Emeritus
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.

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.

14. Dec 6, 2013

### George Jones

Staff Emeritus
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

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

15. Dec 6, 2013

### Jilang

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

16. Dec 6, 2013

### Jilang

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

Last edited: Dec 6, 2013
17. Dec 6, 2013

### WannabeNewton

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 $b$.

18. Dec 6, 2013

### George Jones

Staff Emeritus
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 $K = K^\dagger$.

Scooped by WannabeNewton while typing[/edit].

19. Dec 6, 2013

### WannabeNewton

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

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

20. Dec 6, 2013

### Jilang

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