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Results tagged with Search options user 84488
18 results

Applies to questions of primarily educational value - not only questions that arise from actual homework assignments, but any question where it is preferable to guide the asker to the answer rather than giving it away outright. Please READ THE GUIDANCE IN META before asking homework-like questions.

since P is a constant and can be taken outside of the integral There is no reason whatsoever why $p$ should be a constant, unless specified so; in particular, in your exercise the task is to find …
answered Jul 1 '15 by gented
Let $|n'\rangle$ be a basis of the Hilbert space, then $$\textrm{tr}\Big[|\alpha\rangle\langle\alpha|A\Big]=\sum_{n'}\langle n'|\alpha\rangle\langle\alpha|A|n'\rangle=\sum_{n'}\langle\alpha|A|n'\rang … answered Oct 1 '15 by gented In general one has$$ K_{\alpha\beta} = \frac{1}{2}K_{\alpha\beta} + \frac{1}{2}K_{\alpha\beta} + \frac{1}{2}K_{\beta\alpha} - \frac{1}{2}K_{\beta\alpha} = \frac{1}{2}K_{\left\{\alpha, \beta\right\ …
answered Aug 1 '16 by gented
Start from $d^4p = dp_{\mu}\,dp^{\mu}$ which is manifestly Lorentz invariant. In order to obtain the actual measure to integrate against you have to pair this up with the mass-shell condition for the …
answered Jun 27 '15 by gented
Start noticing that ${(\gamma^{\alpha})}^2 =1\cdot g^{\alpha \alpha}$ and that $$\textrm{tr} (\gamma^{\mu}\gamma^{\nu}\gamma^5)= \textrm{tr}\left(\frac{1}{g^{\alpha \alpha}}{(\gamma^{\alpha})}^2\gam … answered Sep 27 '15 by gented You are making confusion between operators and their representations on the position basis. Your hypotheses 1) and 2) are wrong (more ill-interpreted): as you can easily notice the left hand sides con … answered Jul 1 '15 by gented The exponential is already a number, so it has no components, by definition. The only vector quantity in the expression of \mathbf{E} is \mathbf{E}_0, therefore it is the only part that you have t … answered Jul 10 '17 by gented Diagonalising an operator means finding its eigenstates. Without loss of generality your Hamiltonian can be written as$$ H = c_1 a^{\dagger}a + c_2 a^{\dagger}a^{\dagger} + c_3 a a $$with a^{\dag … answered Dec 15 '15 by gented If the first measurement yields the value A_1 with certainty, this means the initial state has collapsed into u_1 after the first observation. In particular one has, inverting the above back:$$ | …
answered Sep 30 '15 by gented
Let $\square = \eta^{\mu\nu}\partial_{\mu}\partial_{\nu}$ and let $\Lambda$ be any operator leaving the metric invariant, i. e. $\Lambda\eta\Lambda^{-1}=\eta$. From the above it follows that the comp …
answered Dec 11 '15 by gented
The expression that you want to get automatically follows from the definition of the electromagnetic tensor as $F_{\mu\nu} = \partial_{\mu}A_{\nu}- \partial_{\nu}A_{\mu}$. In particular one has $$\pa … answered Nov 7 '16 by gented Notice that S_j = S^1_j\otimes 1_2 + 1_1\otimes S^2_j, with S_j living in \mathcal{H_1}\otimes\mathcal{H_2}. The Hamiltonian commutes with (S^2, S_z), therefore it can be diagonalised onto th … answered Jun 21 '17 by gented Using distributions is a trick that people use whenever dealing with systems that do not have the required smoothness and integrability conditions and is in general only a mathematical technique to ne … answered Jun 27 '15 by gented To calculate \langle\psi | \psi\rangle we have, using the expansion of the identity onto the momentum basis:$$ \langle\psi | \psi\rangle = \int\textrm{d}p\,\langle\psi|p\rangle\langle p|\psi\rangle …
answered Oct 2 '15 by gented
Given  \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial q}\,\delta q + \frac{\partial L}{\partial v}\,\delta v \right)= \int_{t_1}^{t_2}\textrm{d}t\,\left(\frac{\partial L}{\partial q} …
answered Sep 22 '15 by gented

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