Question on discrete commutation relation in QFT

user1139
Messages
71
Reaction score
8
Homework Statement
The statement is in title
Relevant Equations
The equations are given below
Given the commutation relation

$$\left[\phi\left(t,\vec{x}\right),\pi\left(t,\vec{x}'\right)\right]=i\delta^{n-1}\left(\vec{x}-\vec{x}'\right)$$

and define the Fourier transform as

$$\tilde{\phi}(t,\vec{k})=\frac{1}{\sqrt{L^{n-1}}}\int_{\,0}^{\,L}\phi(t,\vec{x})e^{-i\vec{k}\cdot\vec{x}}\,\mathrm{d}^{n-1}\vec{x}$$

$$\tilde{\pi}(t,\vec{k})=\frac{1}{\sqrt{L^{n-1}}}\int_{\,0}^{\,L}\pi(t,\vec{x})e^{-i\vec{k}\cdot\vec{x}}\,\mathrm{d}^{n-1}\vec{x}$$

Is it then correct to say the following?

$$\left[\tilde{\phi}(\vec{k}),\tilde{\pi}(\vec{k}')\right]=i\delta_{\vec{k},-\vec{k}'}=i\delta_{-\vec{k},\vec{k}'}$$

i.e. can I use both Kronecker deltas interchangeably?
 
Physics news on Phys.org
Have you tried to derive the commutator? You also should destinguish the operators ##\phi(t,\vec{x})## from ##\tilde{\phi}(\vec{k})##, because otherwise it leads to confusion. If so, where is your problem. Of course ##\delta_{\vec{k},-\vec{k}'}=\delta_{-\vec{k},\vec{k}'}##.
 
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