Poincare Algebra from Poisson Bracket with KG Action

jfy4
Messages
645
Reaction score
3

Homework Statement


Consider the Klein-Gordan action. Show that the Noether charges of the Poincare Group generate the Poincare Algebra in the Poisson brackets. There will be 10 generators.

Homework Equations


<br /> \{ A,B \}=\frac{\delta A}{\delta \phi}\frac{\delta B}{\delta \pi}-\frac{\delta A}{\delta \pi}\frac{\delta B}{\delta \phi}<br />
<br /> j_{a}^{\mu}=\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)}A_{a}^{\nu}\partial_\nu \phi - A_{a}^{\mu}\mathcal{L}<br />
<br /> \pi = \frac{\partial \mathcal{L}}{\partial (\partial_0 \phi)}<br />
<br /> \Box \phi - m^2 \phi=0<br />
<br /> Q_a = \int d^3 x j^{0}_{a}<br />

The Attempt at a Solution


Starting with the action
<br /> \mathcal{L}=\frac{1}{2}(\partial_\mu \phi \partial^\mu \phi -m^2 \phi^2)<br />
for the conserved currents for both translations and lorentz transformations I obtain:
<br /> p^\mu = \int d^3 x (\partial^0 \phi \partial^\mu \phi - g^{0\mu}\mathcal{L})<br />
<br /> m^{\alpha\beta}=\int d^3 x (\theta^{0\alpha}x^\beta - \theta^{0\beta}x^\alpha)<br />
with
<br /> \theta^{\alpha\beta}=\partial^\alpha \phi \partial^\beta \phi - g^{\alpha\beta}\mathcal{L}<br />
The for \{ p^\mu ,p^\nu \} I get
<br /> \frac{\delta p^\mu}{\delta \phi}=g^{0\mu}m^2 \phi \quad\text{and}\quad \frac{\delta p^\mu}{\delta \pi} = \partial^\mu \phi - g^{0\mu}\partial^0 \phi<br />
which gives
<br /> \begin{align}<br /> \{ p^\mu ,p^\nu \} &amp;= \int d^4 x (g^{0\mu}m^2 \phi)(\partial^\nu \phi - g^{0\nu}\partial^0 \phi)-(\partial^\mu \phi - g^{0\mu}\partial^0 \phi)(g^{0\nu}m^2 \phi) \\<br /> &amp;= \int d^4 x m^2(g^{0\mu} \phi \partial^\nu \phi - \partial^\mu \phi g^{0\nu}\phi) \\<br /> &amp;= \int d^4 x m^2 g^{0\mu}\frac{1}{2} \partial^\nu (\phi^2) - \int d^4 x m^2 g^{0\nu}\frac{1}{2} \partial^\mu (\phi^2) =0<br /> \end{align}<br />
since the field vanishes on the boundary. Now if I have done this right... to compute \{ p^\mu , m^{\alpha\beta} \} I can borrow the momentum stuff, and for the Ms I get
<br /> \frac{\delta m^{\alpha\beta}}{\delta \pi}=(\partial^\alpha \phi - g^{0\alpha}\partial^0 \phi)x^\beta - (\partial^\beta \phi - g^{0\beta}\partial^0 \phi)x^\alpha<br />
and
<br /> \frac{\delta m^{\alpha}}{\delta \phi}=g^{0\alpha}m^2 \phi x^\beta - g^{0\beta}m^2 \phi x^\alpha<br />
Now at this point I have tried to compute the poisson bracket of those above guys, however I cannot seem to retrieve the appropriate form... I know I need to get a superpostion of momenta but I can't seem to get the momentum to come back out. The things I have tried range from integration by parts of various terms, to substituting in the the EOM for the m^2 term and trying to integrate by parts from that. But no luck as of yet. Could someone give me a solid push in how to manipulate these guys into the correct end form, thanks.
 
Last edited:
Physics news on Phys.org
just looking at the term \{ p^0 , m^{0i} \} I get the following
<br /> \begin{align}<br /> &amp;=\int d^3 x m^2 \phi (-\partial^i \phi x^0) -0 \\<br /> &amp;= -\int d^3 x (m^2 \phi)\partial^i \phi x^0 = -\int d^3 x (\partial_\lambda \partial^\lambda \phi ) \partial^i \phi x^0 \\<br /> &amp;= \int d^3 x \partial^\lambda \phi \partial_\lambda (\partial^i \phi x^0) \\<br /> &amp;= \int d^3 x \partial^\lambda \phi (\partial_\lambda \partial^i \phi)x^0 + \int d^3 x \partial^0 \phi \partial^i \phi \\<br /> &amp;= \text{something}+p^i <br /> \end{align}<br />
which is the desired result except for the "something". Can someone point out why that first term is zero, which it should be. Thanks.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top